CHEMICAL CALCULATIONS
Courtesy of Charles F. Roth
Antoine Laurent Lavoisier (1743-1794). "It was left to Lavoisier to transform a
philosophical tenet (Law of Conservation of Matter) into a fruitful scientific prin-
ciple, and to apply it to the interpretation of chemical phenomena." BERTHELOT,
La Revolution Chimique.
NEW-WORLD SCIENCE SERIES
Edited by John W '. Ritchie
CHEMICAL
CALCULATIONS
A SYSTEMATIC PRESENTATION OF THE SOLUTION OF
TYPE PROBLEMS, WITH IOOO CHEMICAL PROBLEMS
ARRANGED PROGRESSIVELY ACCORDING TO
LESSON ASSIGNMENTS
By BERNARD JAFFE, M.A.
Instructor in Chemistry, Jamaica High School
New York City
Tanker s-on-Huds on y New Tork
WORLD BOOK COMPANY
2126 Prairie Avenue, Chicago
1927
WORLD BOOK COMPANY
THE HOUSE OF APPLIED KNOWLEDGE
Established 1905 by Caspar W. Hodgson
YONKERS-ON T -HUDSON, NEW YORK -
2126 PRAIRIE AVENUE, CHICAGO
Chemistry textbooks intended to be used in
high school and beginning college classes
have been found by many teachers to contain
rather too few problems in the mathematics
of the subject. Yet in chemistry examina-
tions given under state supervision, as in
New York, and in college entrance examina-
tions perhaps three questions out of ten
require mathematical calculations. It was
primarily to strengthen the hands of teachers
by overcoming a deficiency in some of the
basal texts in chemistry that this book of
Chemical Calculations was prepared
NWSS : JCC-2
Copyright 1926 by World Book Company
Copyright in Great Britain
All rights reserved
WONTED IN tr.S.A.
PREFACE
IN an investigation of the results secured in the teaching of
high school chemistry, conducted in 1922 by Professor Samuel
R. Powers 'of Teachers College, Columbia University, one of
the outstanding weaknesses found was the inability of the
pupils to use chemical arithmetic. Professor Powers writes
concerning the first-year chemistry students in the schools
examined that "fewer than 60 per cent are able to compute
the molecular weight of gases when the weight of a liter of the
gas is given, less than 50 per cent are able to calculate a for-
mula from the molecular weight and percentage composition,
and only 40 per cent are able to make the calculations involv-
ing a knowledge of IJoyle's Law. About one out of three is
able to calculate from the percentage .composition the ratio
between the amounts of oxygen in the two oxides of sulfur.
A little more than 10 per cent are able to calculate from the
percentage composition the ratio between the amounts of
oxygen in the two oxides of nitrogen. Certainly these percent-
ages would be smaller if all the cooperating schools had used
this test. The test was used by distinctly superior schools."
The reasons for this weakness are not difficult to find. A
study of the various textbooks on elementary chemistry
reveals in most of them only a passing reference to the sub-
ject; or, when a more comprehensive treatment is given,
both teacher and pupil are handicapped by a dearth of prop-
erly graded problems to go hand in hand with the regular
daily lesson assignment. Very few teachers give their stu-
dents a systematic presentation of chemical calculations.
The ^airrr of this book is to fill the gap which teachers of
chemistry in secondary 'schools have felt exists in most of the
textbooks used. To vitalize chemistry and make it a more
VI
Preface
workable science has stimulated teachers to introduce the
solving of chemical problems, and the trend of the last few
years seems to be toward more mathematical chemistry. By
this method chemistry becomes an exact science to the stu-
dent, since he soon realizes that quantitative data are funda-
mental and that even the minutest particles of matter in their
reactions obey mathematical laws.
The problems in this book are progressively arranged ac-
cording to a number of types which the student soon learns to
recognize. They are also graded according to degree of diffi-
culty and order of lesson assignment. A minimum of algebra
is required. Since the purpose is to teach not mathematics
but chemistry, arithmetical calculations have been reduced to
a minimum. Whole numbers are used for the atomic weights
of the elements. The constant changing of atomic weights
as issued by the International Committee, and the recent
researches on isotopes, justify this practice. The student
is spared the labor of wading through a maze of cold figures.
The time saved in working out problems that involve a mul-
titude of figures can fie used in doing other problems.
In analytical work, however, it is often necessary to deal
with a large number of figures with a fine degree of accuracy.
Rapid solutions must be obtained to save the time of the
chemist. For such work the Chemists' Slide Rule (sold by
Keuffel & Esser Company) is recommended. Besides' the
CD and CI scales, this slide rule has two sets of scales with
the symbols and formulas of commonly used elements and
compounds marked at the points representing their respective
atomic and molecular weights, thus making unnecessary the
use of atomic and molecular weight tables.
This book is designed primarily to accompany any of the
more recent textbooks on general chemistry. Part II has
Preface ' vii
been added for use by first-year college students, and in those
high schools where courses in qualitative and quantitative
analysis are offered.
The author wishes to express his gratitude," for helpful
advice and suggestions, to Mr. S. Jaffe of the Boys' High
School, Brooklyn, New York ; Mr. M. Mendel of the Thomas
Jefferson High School, Brooklyn, New York; and Mr. M.
Dunay, his colleague at the Jamaica High School.
BERNARD JAFFE
SUGGESTIONS FOR THE USE OF THIS BOOK
1. Aim. This book aims to cultivate a method of attack
on mathematical problems in chemistry. With a knowledge
of the ten types of problems presented in Part I, the student
will mid little trouble in solving practically all the common
chemical problems.
2. Scope. It is the -opinion of the author that all of Part I
should be mastered by every high school student. Part II is,
on the whole, intended for more mature students, especially
first-year college students.
3. Time to begin. The introduction of quantitative laws
should be reached as early as possible. This book should
then be used in conjunction with the regular textbook. The
general procedure is to begin the study of the first-type prob-
lem about the fourth, week after the beginning of the course
and after elements, compounds, chemical changes, law of
definite proportions, oxygen, hydrogen, law of multiple pro-
portions, and the Atomic Hypothesis have been studied.
Valency and the writing of balanced equations should be
introduced early.
4. Rate of progress. No rigid rules can be set down. It
depends upon the position of chemistry in the school cur-
riculum, the age and mental attainments of the pupils, the size
of classes, and other varying factors. The work should not
be given in too large doses. In schools operated on the Dalton
or the Winnetka Plan this book is especially valuable, as it
meets the needs of the individual student.
5. Lesson assignments. The book follows the order of
topics which have become more or less standard in the teach-
ing of chemistry. The student should master each type prob-
lem before passing on to the next one. Problems should be
Suggestions for the Use of This Book ix
assigned daily from Part III to vitalize the lesson. The prob-
lems under each topic in Part III cover various type problems.
They have been selected to teach chemical facts. Problems
from the Regents Examinations and the. College Entrance
Board Examinations should be assigned from time to time
to give the student confidence in his ability to solve such
problems.
6. Problem book. All problems should be done by ,the
student in a special problem book. Problems should be
discussed in class the following day and necessary corrections
made. This develops clear thinking. The proper handling
of these problems can serve as a review.
7. Review work. The topical arrangement of Part III
facilitates reviews. The important equations clarify the
student's knowledge of laboratory and commercial prepa-
rations and fix in his mind the chemical properties and uses
of the substances studied. Part III should prove of great
value in the final review.
8. Answer book. A pamphlet containing the answers to all
the problems may be obtained by teachers from the publishers.
SUGGESTIONS FOR THE USE OF
INTRODUCTION
V1U
xv
PART ONE
CHAPTER
I. STRUCTURE OF MATTER I
II. PROBLEMS OF TYPE I : To FIND THE MOLECULAR WEIGHT OF A
COMPOUND FROM ITS FORMULA 5
III. PROBLEMS OF TYPE II : To FIND THE PERCENTAGE COMPOSI-
TION OF A COMPOUND FROM ITS FORMULA .... 8
IV. VALENCY AND THE WRITING OF FORMULAS . . . .11
V. PROBLEMS OF TYPE III : To FIND THE SIMPLEST FORMULA or A
COMPOUND FROM ITS PERCENTAGE COMPOSITION . .14
VI. PROBLEMS OF TYPE IV, BASED ON GASES AND THEIR MEASURE-
MENT I 7
Vn. PROBLEMS BASED ON THE MOLECULAR STRUCTURE OF GASES
AND VAPORS . . . . 2 5
VHI. PROBLEMS OF TYPE V, BASED ON VAPOR DENSITY, SPECIFIC
GRAVITY, WEIGHT OF A LITER, AND MOLECULAR WEIGHT OF
GASES AND VAPORS 2 8
IX. PROBLEMS OF TYPE VI : To FIND THE HYDROGEN EQUIVALENT
OF A METAL " . 3 2
X. PROBLEMS OF TYPE VII : To FIND THE TRUE FORMULA OF A
COMPOUND -35
XI. PROBLEMS BASED ON EQUATIONS 38
Problems of Type VHI : Straight-Weight Problem . . . 39
XII. PROBLEMS OF TYPE IX 4 2
I. Weight-Volume Problem 4 2
II. Volume-Weight Problem 44
XIII. PROBLEMS OF TYPE X: STRAIGHT-VOLUME PROBLEM . . 46
XIV. MISCELLANEOUS PROBLEMS 48
I. Problems Involving an Excess of One or More Reagents 48
II. Problems Involving the Most Economical Chemicals
to Use 49
III. Calculations of Mixtures Having a Common Constit- .
uent 5
XV. CHEMICAL PROBLEMS FROM NEW YORK STATE REGENTS EX-
AMINATION PAPERS (1915-1925) 5 2
xi
Xll
Contents
PART TWO
CHAPTER PAGE
XVI. To FIND THE MOLECULAR WEIGHTS OF GASES AND VOLATILE
COMPOUNDS 60
I. Dumas Method 60
II. Victor Meyer Method 61
XVII. FINDING THE MOLECULAR WEIGHTS OF NON- VOLATILE COM-
POUNDS 64
I. Elevation of Boiling Point 64
II. Depression of Freezing Point 65
XVHL FINDING THE ATOMIC WEIGHTS OF ELEMENTS EXISTING m
VOLATILE COMPOUNDS 67
XIX. FENDING THE ATOMIC WEIGHTS OF ELEMENTS WITH THE AID OF
SPECIFIC HEATS .69
XX. PROBLEMS BASED ON FARADAY'S LAWS 7*
XXI. PROBLEMS BASED ON SPECIFIC GRAVITY . ' . . . -73
XXII. VOLUMETRIC ANALYSIS (MOLAR AND NORMAL SOLUTIONS) 76
I. Standard Solutions of Oxidizing and Reducing Agents . 79
IT. Adjustment of the Strength of Solutions 81
XXIII. GRAVIMETRIC ANALYSIS 83
I. Calculations of Percentage to the Dry Basis ... 83
It. Calculations Based on the Use of Factors ... 84
XXTV. CHEMICAL PROBLEMS FROM COLLEGE ENTRANCE EXAMINATION
BOARD PAPERS 86
PART THREE
PROBLEMS BASED ON LESSON ASSIGNMENTS, WITH THE PRINCIPAL EQUA-
TIONS UNDER EACH TOPIC 95
1. Oxygen and Ozone 95
2. Hydrogen .96
3. Water and Hydrogen Peroxide 97
4. Chlorine 98
5. Hydrogen Chloride and Hydrochloric Acid ..... 99
6. Alkali Metals, Bases, and Neutralization 100
7. Sodium Salts 101
8. Sulfur and Sulfides . 102
* 9. Hydrogen Sulfide 103
10. Sulfur Dioxide 104
11. Sulfur Trioxide and Sulfuric Acid 105
12. Nitrogen and the Air 106
13. Ammonia and Ammonium Hydroxide 107
Contents xiii
CHAPTER PAGE
14. Oxides of Nitrogen 108
15. Nitric Acid and Aqua Regia 109
16. Nitrates and Nitrogen Fixation . . . . . . .no
17. Phosphorus in
18. Arsenic, Antimony, and Bismuth 112
19. Bromine and Hydrobromic Acid 113
20. Iodine and Hydrogen Iodide . .114
21. Fluorine and Etching of Glass by Hydrogen Fluoride . . .115
22. Allo tropic Forms and Varieties of Carbon n6
23. Carbon Dioxide and Carbonic Acid 117
24. Carbon Monoxide 118
25. Illuminating Gas, Water Gas, and Producer Gas .... 119
26. Reactions with Carbon in the Electric Furnace . . . .120
27. Silicon Dioxide, Glass, and Borax 121
28. Calcium, Calcium Carbonate, and Hard Water 122
29. Calcium Oxide and Calcium Hydroxide 123
30. Calcium Salts 124
31. Magnesium I2 $
32. Mercury I2 6
33- zinc 127
34. Manufacture of Iron and Steel 128
35. Iron Salts I2 g
36. Copper . I3 o
37. Silver, Gold, and Platinum 131
38. Aluminum 3:32
39. Aluminum Hydroxide and Alums . . !33
40. Aluminum Silicates and Cement !34
41. Tin and the Electrochemical Series 135
42. Lead ^
43. Nickel and Cobalt 137
44. Manganese and Chromium 138
45. Distillation Products of Coal, Wood, and Petroleum . . . - 139
46. Hydrocarbons 140
47. Alcohols !4j
48. Aldehydes, Ketones, and Ethers . ' 142
49. Organic Acids 143
50. Esters I44
51. Soaps 'i45
52. Carbohydrates . !46
APPENDIX
INTERNATIONAL TABLE OF ATOMIC WEIGHTS AND NUMBERS . . . 149
TABLE OF APPROXIMATE ATOMIC WEIGHTS OP THE COMMON ELEMENTS 150
METRIC SYSTEM AND APPROXIMATE UNIT EQUIVALENTS . . 151
xiv Contents
PAGE
THERMOMETER SCALES AND CONVERSION RULES 152
APPROXIMATE SPECIFIC HEATS or THE COMMON ELEMENTS . . .153
TABLES OF SPECIFIC GRAVITIES OF I^SO*, HCL, AND HNOs SOLUTIONS 154
SPECIFIC GRAVITY OF AMMONIA WATER SOLUTIONS AT 15 C. . . 155
CONCENTRATION AND SPECIFIC GRAVITY OF COMMON ACIDS AND BASES 155
TABLE OF SOME COMMON FACTORS FOR QUANTITATIVE ANALYSIS . 156
PRICES OF COMMON CHEMICALS 157
PRACTICE TABLE OF COMPOUNDS FROM VALENCIES . . . .158
PROBLEMS BASED ON THE THEORY OF IONIZATION . . . .160
BALANCING EQUATIONS 166
INTRODUCTION
MANY years' teaching of secondary school chemistry has
proved to me the need of such a book on Chemical Calcula-
tions as Mr. Jaffe presents. A well-thought-out problem
illustrating a chemical principle is just as important in vital-
izing the teaching of chemistry as a well-presented demonstra-
tion. The progressive arrangement of Mr. Jaffe's problems
and the clearness with which the type problems are explained
should prove a boon not only to the high school student
but to the first-year college student as well.
CHARLES H. VOSBURGH
Principal, Jamaica High School
CHEMICAL CALCULATIONS
PART ONE
CHAPTER ONE
STRUCTURE OF MATTER
As far as we know today, all matter is made up of one or
more of 92 different elements, 90 of which have already been
isolated. An element is a distinct species of matter which,
with the exception of the radioactive substances, does not
under ordinary conditions break down into simpler substances.
According to the Atomic Hypothesis, advanced by Dalton
in 1802 and since proved to be a law, all elements are made
up of small particles called atoms.
Until recently it was supposed that atoms could not be
divided," but now it is believed that they are made of particles
of negative and positive electricity. These particles are
called electrons and protons respectively. An atom is com-
posed of a positively charged nucleus, which contains an
excess of protons, ' and of one or more electrons which are
.placed at a distance from the nucleus. The nucleus is small
when compared with the total diameter of the space occupied
by the atom.
The atoms of the same element are alike, but differ from the
atoms of other elements in the number of positive and negative
charges composing them. The simplest atom, that of hydro-
gen, is considered to be made up of one positive charge, the
proton, and one electron, which is in vibratory motion and
functions when chemical union takes place.
The weight of the individual atom of even the heaviest
element known is so exceedingly small (estimated at a million
2 Chemical Calculations
millionth of a million millionth of a gram) that it would be
extremely cumbersome to use such a number. Chemists
have therefore arranged the elements according to the relative
weights of the atoms, and we call this arrangement the Table
of Atomic Weights of the Elements (page 149). Since oxygen
unites with nearly all the elements, forming oxides which can
easily be analyzed, chemists have given to oxygen the arbi-
trary weight of 16, which makes the relative weight of hydro-
gen 1.008, or approximately 1. Hydrogen is the lightest of all
the elements, hence the atomic weights of all the elements
are greater than unity, and nearly all of them are approxi-
mately whole numbers.
Atoms have the power of combining with atoms of the same
element or of other elements to form molecules, which are
aggregations of atoms chemically combined and possessing
properties quite different from those of the elements which
compose them. When atoms of different elements combine
chemically, we call the products compounds.
It is a great convenience to have a shorthand method of
representing elements and compounds. We call the abbre-
viated form of an element its symbol. Thus H is the symbol
of hydrogen. We call the abbreviated form of a compound
its formula, and we write it by joining the symbols of the
elements which compose it. Thus NaCl represents the formula
of sodium chloride.
When we bring together various substances and form mix-
tures without combining them chemically, we may form these
mixtures in any proportions whatever. In chemical union
the experimenter has no choice in the matter. The pro-
portions are determined by the substances themselves. This
is the most striking fact of chemical combinations. This
peculiar behavior is stated as the Law of Definite Proportions.
Structure of Matter 3
We often find a series of compounds which show varying
proportions of the same elements. In such cases we find that
the elements follow the Law of Multiple Proportions, which
may be stated as follows : Whenever two elements, A and B,
combine to form more than one compound, if we take a fixed
weight of A, then the weights of B wfiich unite with this fixed
weight of A are in the ratio of small whole numbers.
PROBLEMS BASED ON THE LAW OF MULTIPLE PROPORTIONS
EXAMPLE. Nitrogen and oxygen unite to form five different compounds,
whose compositions are given in the following table. The figures in the second
and third columns are determined by experiment and from these the ratio of
small whole numbers in the fourth column is found.
COMPOUNDS
FIXED WEIGHTS OF A
(NlTKOGEN)
WEIGHTS OF B
(OXYGEN)
RATIO OF.
WEIGHTS or B
IN DIFFERENT
COMPOUNDS
Nitrous oxide
28 g. of N
16 g. of O
16 to 32 to
Nitric oxide
28 g. of N
32 g. of O
48 to 64 to
Nitrogen trioxide
Nitrogen peroxide
Nitrogen .pentoxide
28 g. of N
28 g. of N
28 g. of N
48 g. of O
64 g. of O
80 g. of O
80 . or
1 to 2 to 3
to 4 to 5
1. Problem. Manganese and oxygen form five oxides : manganous oxide
(MnO), manganic oxide (MnaOa), manganese dioxide (Mn02), man-
ganese trioxide (Mn0 3 ), and manganese heptoxide (Mna0 7 ). Do
manganese and oxygen obey the Law of Multiple Proportions ?
COMPOUNDS
FIXED WEIGHTS
OF A
WEIGHTS OF B
RATIO OF WEIGHTS OF B
IN DIFFERENT COM-
POUNDS
Manganous oxide
110 g. of Mn
32 g. of O
Manganic oxide
110 g. of Mn
48 g. of O
Manganese dioxide
,110 g. of Mn
64 g. of
Manganese trioxide
110 g. of Mn
96 g. of O
Manganese heptoxide
110 g. of Mn
112 g. of O
Chemical Calculations
2. Problem. Iron and oxygen form three compounds: ferrous oxide
(FeO), ferric oxide (FeaOs), and magnetic oxide of iron (Fe 3 O 4 ). Do
iron and oxygen obey the Law of Multiple Proportions if the composi-
tions of their compounds are as follows ?
COMPOUNDS
FIXED WEIGHTS
OF .4g
WEIGHTS OF B
RATIO OF WEIGHTS OF B
IN DIFFERENT COMPOUNDS
Ferrous oxide
Ferric oxide
Magnetic oxide
168 g. of Fe
168 g. of Fe
168 g. of Fe
48 g. of O
72 g. of O
64 g. of O
CHAPTER TWO
PROBLEMS OF TYPE I: TO FIND THE MOLECULAR
WEIGHT OF A COMPOUND FROM ITS FORMULA
THE symbol of an element stands for more than the name
of the element; It designates a definite weight. Ag means
one weight of silver, one atomic weight of silver, or 108 grams
of silver.
The molecular weight of a compound is equal to the sum of
the weights of all the atoms it contains. The atomic weights "
used are the relative atomic weights; hence the molecular
weights obtained are really the relative molecular weights.
A. EXAMPLE. To find the molecular weight of potassium bromide, KBr.
Place the approximate atomic weigfits (page 150) under the symbols ana add
them.
K Br
39 + 80 = 119 = molecular weight of KBr
Very often the molecule of a compound or even of an element
contains more than one atom of a particular element. We
represent the number of these atoms by a subscript or small
number placed in the lower right corner of the symbol of that
element. Thus H 2 represents a molecule of water made up
of 2 atoms of hydrogen and 1 atom of oxygen. The subscript
1 is never written ; it is always understood.
B. EXAMPLE. Find the molecular weight of magnesium sulfate, MgSO.j.
Be sure to multiply the atomic weight of every element followed by a subscript
by the subscript.
Mg S 4 . '
24 + 32 + 16 X'4 = 24 + 32 + 64 = 120 = molecular weight
To represent more than one molecule of an element or 'com-
pound, we place a coefficient in front of its formula. This
coefficient multiplies the number of every atom in the sub-
stance. Thus 5CuS0 4 stands for 5 molecules of copper sul-
5
6 Chemical Calculations
fate, each molecule consisting of one atom of copper, one of
sulfur, and four of oxygen. In the 5 molecules, then, there
are 5 atoms of copper, 5 of sulfur, and 20 atoms of oxygen.
C. EXAMPLE. Find the molecular weight of 2CaCOa (calcium carbonate).
Be sure to multiply everything after the coefficient by the coefficient.
2Ca C O 3
2(40 + 12 + 1C X 3) or
2(40 + 12 + 48)
2(100) = 200 = weight of 2 molecules of
calcium carbonate
A radical is a group of elements acting chemically like a
single element. For example, NBU is called the ammonium
radical, and NOs is the nitrate radical. When a compound
contains more than one of the same radicals in its molecule,
a subscript is written in such a way as to multiply every atom
in that radical. This is done by enclosing the radical in
parentheses and placing the subscript outside of these paren-
theses. Thus (NH^S represents one molecule of ammonium
sulfide containing 2 ammonium radicals and 1 atom of sulfur ;
or all together, 2 atoms of nitrogen, 8 atoms of hydrogen, and
1 atom of sulfur.
D. EXAMPLE. Find the molecular weight of lead nitrate, Pb(N0 3 )2.
Remember that the subscript multiplies only the element or radical immediately
preceding it.
Pb (N O 3 ) 2
207+(14 + 16X3)'2
207 +(14 + 48)2
'207 +(62)2
207 + 124 = 331 = molecular weight of Pb(N0 3 ) 2
Compounds may contain water chemically combined in a
definite proportion by weight. This water is called water of
crystallization, and is written after the anhydrous compound
followed by a dot. Thus, crystallized sodium carbonate
Problems of Type I 7
(washing soda) contains 10 molecules of water of crystalliza-
tion and its formula is Na 2 C0 3 10H 2 0. This molecule
contains 2 atoms of sodium, 1 of carbon, 20 of hydrogen, and
13 of oxygen. A dot or period in a formula stands for a plus
sign.
E. EXAMPLE. Find the molecular weight of crystallized ferrous sulfate.
Fe S 4 7 H 2 O
56 + 32 + 16 X 4 + 7(1 X 2 + 16)
56 + 32 + 64 + 7(18)
56 + 32 + 64 + 126
278 = molecular weight of FeSO4 7H 2 O
Problems
3. Find the molecular weights of sodium bromide, NaiBr, and potassium
iodide, KI.
4. Determine the molecular weights of lithium chloride, LiCl, and zinc
sulfide, ZnS.
5. Find the molecular weights of zinc sulfate, ZnS04, copper carbonate,
CuCOs, and calcium sulfate, CaS04.
6. What are the molecular weights of manganese dioxide, Mn0 2 , and
potassium sulfate, K 2 S04?
7. What are the molecular weights of copper hydroxide, Cu(OH)2,
nickel nitrate, Ni(NO3)s, and calcium bicarbonate, Ca(HCOs) 2 ?
8. Determine the molecular weights of crystallized barium chloride,
BaCl 2 2H 2 0, and hypo, Na 2 S 2 3 5H 2 (page 102).
9. What are the molecular weights of calcium phosphate, Cas^O^a,
and gypsum, CaSC>4 2H20 ?
10. Find the molecular weights ,of 2Na2B40 7 10H 2 0, and plaster of
Paris, (CaSO 4 ) 2 H 2 O.
11. The formula for Glauber's salt is Na 2 S0 4 10H 2 0. Find the molec-
ular weight of 5 molecules of this substance.
CHAPTER THREE
PROBLEMS OF TYPE H : TO FIND THE PERCENTAGE COM-
POSITION OF A COMPOUND FROM ITS FORMULA
To find the percentage composition of a compound is to find
the percentage of each different element present in the com-
pound. Hence we divide the weight of each element by the
molecular weight of the compound and multiply the fraction thus
obtained by 100.
A. EXAMPLE. Find the percentage composition of potassium chlorate,
K Cl 3
39 + 35.5 + 16 X 3 = 122.5 = molecular weight of KC10 3
Percentage of potassium = Mo , ^ o ^ ao< X 100 - JL x 1 00 = 32%
Percentage of ch,orine - Mo , " X 100 = X 100 = 29%
Percentage of ^n - X IflO - X !00=39 %
As a check see that the total is approximately 100%. '
B. EXAMPLE. Find the percentage composition of Na 2 SO4 10H 2 O.
Na 2 S O 4 . 10 H 2
23X2 + 32 + 16X4 + 10(1 X 2 + 16) = 322 = Mol. Wt. of the compound
Now find the total atomic weight of each separate element in the compound
Thus:
Sodium = 2 atoms = 23 X 2 =46 = total atomic weight
Sulfur = 1 atom = 32 = total atomic weight
Oxygen = 14 atoms = 16 X 14 = 224 = total atomic weight
Hydrogen = 20 atoms - 1 X 20 =20 = total atomic weight
Sum of the atomic weights = 322 = Mol. Wt. of the compound
Now the percentage of Na = & X 100 = 14.3%
the percentage of S = ^ X 100 = 10.0%
the percentage of = ff$ X 100 = 69.5%
the percentage of H = ^ X 100 = 6.2%
Total = 100%
Problems of Type II 9 n
LI n- r% f^ V ; *-,'
I D K ' K Y ;,?
Problems ' % ,_ ,/Q
12. Calculate the percentage composition of th&'i
(a) water, H 2 0, (&) mercuric oxide, HgO, (c)~ ii
13. Find the percentage composition of (a)
(6) phosphorus pentoxide, P 2 0s, (c) sulfuric acid, H 2 S0 4j (d} zinc
sulfate, ZnS0 4 .
14. Calculate the percentage composition of chrome alum, KCr(SO4) 2
12H 2 O, crystallized potassium sulfate, K 2 S0 4 10H 2 0, and NiS0 4
(NH 4 ) 2 S0 4 6H 2 0.
15. Determine the percentage composition of borax, Na 2 B 4 07 10H 2 0,
and of crystallized potassium ferrocyanide, K 4 Fe(CN)e 3-H 2 0.
16. What percentage of oxygen is contained in crystallized potassium
magnesium sulfate, K 2 S0 4 MgS0 4 6H 2 0?
C. EXAMPLE. Find the weight of iron in 50 Ib. of an ore containing 80% ferric
oxide,
Since only 80% of the ore is Fe 2 3 , the weight of Fe 2 O 3 is 80% of 50 Ib., or
40 Ib.
Fe 2 Oa
56 X 2 + 16 X 3
56 X 2 + 16 X 3 = 160 = Mol. Wt.
Percentage of Fe = $ X 100 = 70%
Now the weight of all the Fe 2 3 present in the ore is 40 Ib.
Therefore 70% of 40 lb., or 28 Ib., is the weight of the Fe.
Problems
17. Potassium chlorate (KC10 3 ) contains 39% oxygen. How much
oxygen will be found in 100 grams of this compound?
18. How much arsenic (As) is available in 91 Ib. of an impure realgar ore
containing 75% of As 2 2 ?
19. How much phosphorus (P) can be obtained from 2000 Ib. of
Ca 3 (P0 4 ) 2 ?
20. If a_ skeleton weighs 25 Ib. and contains 60% Ca 3 (P0 4 ) 2 , how much
calcium (Ca) can be obtained from it?
21. How much aluminum (Al)' can be obtained from 100 Ib. of its cryo-
lite ore which, on analysis, showed the presence of 85% Na 3 AlF 6 ?
22. An ore of zinc contained 70% of ZnC0 3 . Calculate the amount of
zinc in 100 Ib. of this ore.
io Chemical Calculations
23. Which ore will give more copper, one containing 95% GuCO 3 or one
containing 80% of Cu 2 S?
24. How much phosphorus can be obtained from 170 tons of bones con-
taining 52.6% of Ca 3 (P0 4 ) 2 ?
25. The skeleton of a man weighs 23.7 Ib. and contains 59% of calcium
phosphate, CasfPO^a- What weight of P can be prepared from this
skeleton?
CHAPTER FOUR
VALENCY AND THE WHITING OF FORMULAS
VALENCY is the capacity of an element to unite with other
elements. This capacity is measured in terms of the power
of its atom to replace one or more atoms of hydrogen. Hy-
drogen is always univalent. Nitrogen, one atom of which
combines with three atoms of hydrogen, has a valency of
three. Valency is fixed after we have experimentally deter-
mined atomic weights and found the composition of com-
pounds. A knowledge of the valency of different elements
and radicals is almost indispensable in the writing of formulas.
Valencies must be memorized.
VALENCY RULES IN THE WRITING OF FORMULAS
1. Place the valency of the element as + or signs over
the symbol. The subscript after each element or radical
should be the same as the number of + or signs over
the other element or radical. EXAMPLE. The formula for
j I i __ .
antimony sulfide is c , ^4 The formula for copper
2. Whenever the subscripts are the same, they are omitted.
EXAMPLE. Ferric phosphate is written FeP0 4 and not
++-K-V--
n <~jtfo\ This rule is followed unless the subscripts rep-
"xSV^^tfe
resent the actual molecular structure of the compound.
Thus the formula of nitrogen peroxide is Na0 4 and not
NO 2 , since the molecule of nitrogen peroxide contains
2 atoms of N and 4 atoms of 0.
12
Chemical Calculations
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Valency and the Writing of Formulas 13
3. A radical acts like an element, has its characteristic va-
lency, and usually passes through a chemical reaction un-
changed. It should be placed in parentheses if followed
by a subscript greater than unity.
4. The common practice in the writing of formulas is to place
the metal, metallic radical, or more electropositive element
first, followed by the non-metal, non-metallic radical or less
+ - - +
electropositive element. Thus we write KC1 and not C1K.
Problems
26. Write the formulas of the following compounds: calcium nitrate,
aluminum phosphate, mercurous chloride, nickel sulfide.
27. Make a list of all the salts of calcium in the Practice Table of 'Valen-
cies on page 158.
28. Write the formulas for calcium sulfide, barium chlorate, ferric
ferrocyanide, sodium ferricyanide, copper acetate.
29. Make a list of all the salts of aluminum which are theoretically
possible (see Valency Practice Table, page 158).
30. Make a list of as many of the sulfites of the metals as are theoreti-
cally possible.
31. Prepare a table of as many of the salts of iron (ferric) as are theoreti-
cally possible.
32. Make a list of all the sulfides of the common metals.
33. Write the formulas of sodium potassium tartrate, silver tartrate,
sodium hydrogen carbonate, and ammonium hydrogen tartrate.
CHAPTER FIVE
PROBLEMS OF TYPE IH : TO FIND THE SIMPLEST FORMULA
OF A COMPOUND FROM ITS PERCENTAGE COMPOSITION
WHEN we already know the elements present in the com-
pound, all we need to find is the number of atoms of each of
these elements present in the molecule.
The percentage of each element in the compound is divided
by the atomic weight of that element. The quotients are
then examined and divided by their highest common factor
(H.C.F.) ; that is, by the largest number which will divide
each of the quotients and give a whole number. These
numbers are not always exactly whole numbers, but approxi-
mate, and are regarded as integers. They represent the
number of atoms of the respective elements in the compound.
The simplest formula is thus obtained. In writing the
formula we inspect the elements and arrange them so that the
electropositive element or radical comes first, followed by the
negative element or radical. We must learn to pick out those
elements which may form radicals. Thus a compound whose
composition showed 1 atom of copper, 1 atom of sulfur, and
4 atoms of oxygen is written as CuS0 4 , since 864 is the negative
sulfate radical.
A. EXAMPLE. Find the simplest formula of a compound which contains
63.6% nitrogen and 36.4% oxygen.
For N P ercenta g e composition of N _ 63.6 _ 4 51 ] 4.51 _ ^
' '---- -' i - '" --. ' 2.28
H.C.F = 2.28
2.28
2.28
Atomic 'weight of N 14
For O P ercenta ge composition of O _ 36.4 _ o os
Atomic weight of O 16
The numbers 2(N) and 1(O) represent the smallest number of the respective
elements in the compound.
The simplest formula is, therefore, (NaO) x .
14
Problems of Type III 15
This may not be the true formula, since any multiple of this formula may
be the actual formula. N-iOa, or N 6 3 will just as well satisfy the percent-
age composition as given in the problem. Temporarily, therefore, we will
designate this formula in parentheses, followed by x as the simplest
formula. Later on, under Problems of Type VII, we shall be in a position
to determine the true formula when we shaE be in command of additional
data.
Problems
34.' Calculate the simplest formula of a compound containing 7.69%
hydrogen and 92.31% carbon.
35. A compound contains 1% hydrogen, 11.99% carbon, 47,.95% oxygen,
and 39.06% potassium. Calculate its simplest formula.
36. Acetic acid consists of 40% carbon, 6.67% hydrogen, and 53.33%
oxygen. Determine its simplest formula.
37. A gas on analysis gave 96.75% iodine, 3% carbon, and 0.25% hydro-
gen. What is the simplest formula of this compound?
38. An oxide of barium gave on analysis 81% barium, and 19% oxygen.
What is its simplest formula ?
39. 63.2 parts of manganese and 36.8 parts of oxygen are found in 100
parts of an oxide of manganese. What is the simplest formula of
this oxide?
40. What is the simplest formula of a substance whose composition is
calcium, 40%; carbon, 12%; and oxygen, 48%?
41. What is the simplest formula of a compound which on analysis
showed the following composition: carbon, 54.55%; hydrogen,
9.09%; and oxygen, 36.36%?
Sometimes the percentage composition of a compound is
not given directly but from the facts of the problem we can
find this percentage composition and then proceed as above.
B. EXAMPLE. 168 Ib. of iron when oxidized produced 240 Ib. of iron oxide.
Find the simplest formula of this oxide.
The difference between the weights of iron oxide and iron will be the weight
of the oxygen. Hence the weight of the oxygen is -240 168, or 72 Ib.
For iron % com P- of Fe - H X 10Q - 1.12
At. Wt. of Fe 56 56
FM - " *W^ - 5 - 1JB
j& tam o
06"
H.C.F. = 0.6
_ = 3
1 6 Chemical Calculations
Hence 2 and 3 are the smallest numbers of atoms of Fe and O in the mole-
cule. Therefore the simplest formula is (FeOa) x .
Problems
42. 50 grams of mercury were completely converted into mercuric oxide,
yielding 54 grams of HgO. Prove that this is its simplest formula.
43. 219 grams of hydrogen chloride were formed by the union of 6 grams
of hydrogen and 213 grams of chlorine. What is the simplest
formula of this compound?
44. Bromine was allowed to act upon hot copper until it changed '32
grams of the metal into 112.5 grams of copper bromide. What is
the simplest formula of this compound ?
45. 70 grams of nitrogen completely united with hydrogen gas and
furnished 85 grams of ammonia. Find the simplest formula of
ammonia.
NOTE. Problems of this type can be checked by finding the
percentage composition of the formulas as found. This percentage
composition should, of course, agree with the data given. Thus,
in Example A, page 14, the percentage composition of N20 is
found to be :
9 v 14- 98
which is the same as given in the problem.
CHAPTER SIX
GASES AND THEIR MEASUREMENT
GASES are composed of molecules in constant and rapid
motion. These particles, being at considerable distances from
one another, may be brought closer
together by exerting pressure upon
them. The more pressure exerted,
the nearer the particles and hence
the smaller the volume the gas will
occupy. Double the pressure and
you decrease the volume to half.
This behavior of gases is expressed
by Boyle's Law (1660) which states
that the volume of a gas varies in-
versely as the pressure, when the
temperature remains the same, or,
pv = p'v' ... (1)
where p and v stand for the original
pressure and volume and p' and v'
for the new pressure and volume.
To measure the pressure of the
atmosphere, scientists make use of
the barometer (Fig. 1), con-
.-^sEEr sisting of a tube of mercury
I completely filled and then
inverted in a reservoir
filled with the same heavy
liquid. The average height
of the mercury in this tube,
at sea level, is 760 miUimeters (mm.), or about 30 inches.
This is known as standard pressure or the pressure of
FIG. 1.
i8 Chemical Calculations
1 atmosphere, and represents the weight of the gases in the
atmosphere which are pressing down upon every square cen-
timeter of the earth (equivalent to almost 15 Ib. per square
inch). This column of air is capable of supporting an equal
weight of liquid mercury in the barometer. "
A rise in temperature increases the speed of the particles
composing a gas and hence the gas tends to expand and occupy
more space. It has been found that with every lowering of
1 C. from the zero point a gas contracts ^fg- of its volume.
Theoretically, then, at 273 the gas should cease to exist.
Since all gases change to liquids before this extremely low
temperature is reached, we have not been able to verify this
conclusion. However, we call this temperature of 273 C.
the zero of the absolute temperature or T. scale.
The law expressing the change of volume of a gas due to a
temperature change is known as Charles' Law (1787) and
states that the volume of a gas varies directly as the absolute
temperature, provided the pressure remains the same. 1 By
formula we express it as
where and T stand for the original volume and absolute
temperature, and v r and T' stand for the new volume and
absolute temperature.
Since gases are measured under varying conditions of tem-
perature and pressure, and since the gas laws operate only
under certain conditions, it is necessary to correct our gas
measurements to correspond to those obtaining under standard
conditions of C. and 760 mm. pressure. We can correct
1 There are certain deviations from these gas laws whose discussion is beyond the
scope of this book. Only ideal gases follow the laws as stated above.
Gases and Their Measurement 19
for both temperature and pressure simultaneously by com-
bining equations (1) and (2) above, as follows :
PROBLEMS OF TYPE IV: TO REDUCE GAS MEASURE-
MENTS TO STANDARD CONDITIONS
A. Change in volume due to change in pressure (temperature constant)
EXAMPLE. A quantity of chlorine gas occupies a volume of 50 cc. when
the barometer reads 740 mm. What would its volume be under standard
conditions?
The temperature is not mentioned and we assume that it does not change.
Substituting in equation (1) we have :
pii = p'vf
740(50) = 760(3:)
37000 = 760 x
x = 48.7 cc. of chlorine
Alternate method. Since standard pressure is greater than the pressure
given, the volume of the gas will be smaller; i.e., ff X 50, or 48.7 cc.
Problems
46. 400 cc. of acetylene gas were measured under an atmospheric pres-
sure of 765 mm. - The gas was then subjected to a pressure of
790 mm. What new volume did the gas occupy ?
47. Ammonia gas (NHs) when measured under standard conditions
(760 mm.) occupied 4.5 liters. The pressure then changed to 745
mm. What was the volume occupied by the gas after this change
in pressure?
48. The pressure on 35 cc. of carbon monoxide gas (CO) was suddenly
changed from 765 mm. to 770 mm. What new volume did the gas
occupy ?
49. The pressure on a balloon containing 10,000 cubic meters of helium
gas was changed from 760 mm. to 730 mm. What was the new
volume of the balloon ?
20 Chemical Calculations
B. Change in volume due to change in temperature (pressure constant)
EXAMPLE. A quantity of carbon dioxide gas measured 450 cc. when the
room temperature was 22 C. What would its volume be under standard
conditions?
The pressure is not mentioned, so we assume we are working under standard
or 760 mm. pressure. Substitute in equation (2)
_ 7/_
T r
450 x
273 + 22 273
450 _ x
295 273
295 x = 450(273)
x = 416.4 cc. of carbon dioxide
Alternate method. Since standard temperature is lower than the tempera-
ture at which we are working, the volume of the gas will be lower at this
lower temperature. Hence the volume will be
Problems
50. While the pressure on 500 cc. of hydrogen gas remained steady, the
temperature was changed from 20 C. to C. What change in
volume resulted ?
51. Hydrogen sulfide was measured at 18 C. and found to occupy 15.5
cc. Next day the temperature of the gas was found to be 14.5 C.
WTmt volume did the gas occupy at this lower temperature ?
52. Methane gas, CH*, was measured at C.. and at 40 C. At the
lower temperature the volume was 22.5 cc. What volume did the
gas occupy at the higher temperature ?
C. Change in volume due to simultaneous change in temperature and pressure
EXAMPLE. At 27 C. and 800 mm. pressure a quantity of nitrogen gas
measured 20 cc. What would the volume be under standard conditions?
Gases and Their Measurement
21
Substituting in equation (3),
po _pW
T T
800(20) _ 760(a)
273 + 27 273
800(20) (273) =760(300) a;
800 (20) (273)
760(300)
x = 19.2 cc. of nitrogen
Problems
53. Reduce to standard conditions 48 cc. of nitrogen, measured at 18 C.
and 765 mm.
54. Under standard conditions a volume of oxygen gas occupied 1 liter.
Find the volume of this gas when measured at 17 C. and 770 mm.
55. Hydrogen chloride gas was subjected to a pressure of 1000 mm. and
occupied a volume of 53 cc. at 30 C. Determine the volume of the
gas under standard conditions.
56. Bromine vapor occupied 2.5 liters when measured at 57 C. and
780 mm. What theoretical volume would the vapor occupy if it
were measured at 17 C. and 750 mrn. ?
57. A gram of gunpowder produced, on exploding, 300 cc. of gases when
measured under standard conditions. What volume would these
gases occupy at the temperature pro- .
duced by the explosion (2200 C.) ? I/. 4 mm. -\
CORRECTION OP GAS VOLUMES FOR
AQUEOUS VAPOR PRESSURE '
I v
If we are measuring the volume of a
gas collected over water at 20 C. and
760 mm., the gas is saturated with
.water vapor, which, according to the
table of aqueous vapor pressure, exerts
its own pressure of 17.4 mm. Since
760 mm. represents the sum of the
aailPOU^ nrewilTP flnH the cm<; nrpsenrp FlG ' 2 ' The cor * ected atmos-
aqueous pressure ana tne gas pressure, pheric pressure ^^ the tube
the pressure of the dry gas is only 760 is 76 mm - minus 17 -* mm.
\<i A >7*o a f-n- ON (water-vapor pressure at 20 C.),
minus 17.4, or 742.6 mm. (Fig. 2) . or 742.6 mm.
22
Chemical Calculations
PRESSURE OF WATER VAPOR, OR AQUEOUS TENSION
(In millimeters of mercury)
TEMPERATURE PRESSURE
t
TEMPERATURE
PRESSURE
0.0 C. 4.6 mm.
21.5 C.
19.1 mm.
5
6.5
22.
19.7
10
9.2
22.5
20.3
10.5
9.5
23.
20.9
11
9.8
23.5
21.5
11.5
10.1
24.
22.1
12
10.5
24.5
22.8
12.5
10.S
25.
23.5
13
11.2
25.5
24.2
13.5
11.5
26.
25.0
14
11.9
26.5
25.7
14.5
12.3
27.
26.5
15
12.7
27.5
27.3
15.5
13.1
28.
28.1
16
13.5
28.5
28.9
16.5
14.0
29.
29.8
17
14.4
29.5
30.7
17.5
14.9
30
31.6
IS
15.4
40
54.9
18.5
15.9
50
92.1
19
16.4
60
149.2
19.5
16.9
70
233.8
20
17.4
80
355.4
20.5
17.9
90
526.0
21
18.5
100
760.0
Problems
58. 500 cc. of carbon monoxide at 27 C. stand in a measuring tube over
water while the barometric pressure is 770 mm. What volume
would the gas occupy under standard conditions ?
59. 200 cc. of helium is collected in a tube inverted over water. The
temperature changes from 17 C. to 27 C. while standard atmos-
pheric pressure remains unchanged. What change in volume does
the gas undergo ?
60. What change in atmospheric pressure will be necessary to reduce
20 cc. of a gas measured in a eudiometer over water at 17 C. and
760 mm. to a volume of 10 cc. at 27 C.?
Gases and Their Measurement
CORRECTION OF GAS VOLUMES FOR DIFFERENCE IN LEVEL
If a gas is collected over mercury, before the volume of that
gas is read off, the levels of the mercury inside the measuring
tube and outside must be the same (Fig. 3, C). If, as in A
(Fig. 3), the level is higher inside the measuring tube than
outside, then the gas is under a pressure which is lower than
barometric pressure by the height A. On the other hand,
if the levels of the mercury are as represented by B (Fig. 3),
then the gas is under a greater pressure than barometric read-
ing, and the pressure illustrated by height B should be added
to the barometric reading.
EXAMPLE. What is the corrected pressure under which a gas is measured if
the standard barometric reading is 760, but the level of the mercury inside
the measuring tube is 1.5 cm.
(15 mm.) above the level of the A C B
mercury in the reservoir?
The answer is 760 minus 15 mm.,
or 745 mm.
If the gas is collected over
water, which is the common
laboratory method, the
height of the column of
water which represents the
difference in levels weighs
only - that of mercury,
lo.o
since mercury is 13.6 times as heavy as an equal volume of
water.
EXAMPLE. Calculate the corrected pressure under which a gas, collected
over water, is measured if the level of the water inside the measuring tube
is 12 mm. below the level of the water in the reservoir.
"I f? mm
The answer is 760 plus 1Q ', or approximately 760.9 mm.
lo.o
FIG. 3.
24 Chemical Calculations
Problems
61. A certain quantity of hydrogen gas measured over water gave a
volume of 50 cc. at 17 C. and 770 mm. when the level inside r the
measuring tube was 1 cm. above that on the outside. Calculate
the* corrected volume of gas under standard conditions.
62. Some nitrous oxide is measured over mercury, and when the level of
the mercury is 2 cm. lower in the eudiometer than on the outside,
the volume of gas is 20 cc. If the measurements were made at
7 C. and 750 mm., what would be the corrected volume under
standard conditions ?
63. 100 cc. of air stand in a tube over mercury at 10 C. and 760 mm.
The level of the mercury within the tube is 8 cm. above that without.
What is the volume of the air under standard conditions ?
64. 200 cc. of a gas insoluble rfTwater are measured over water at 35 C.
and 700 mm. when the level of the water is 30 cm. higher within the
tube than without. What is the volume of the gas when it is again
measured at 17 C. and 760 mm., and the levels of the water within
and without the tube are the same?
65. What change in atmospheric pressure will be necessary to bring a
volume of helium gas measured at 27 C. and 770 mm. in a tube
over water from 50 cc. to 60 cc., if the temperature remains the
same?
CHAPTER SEVEN
PROBLEMS BASED ON THE MOLECULAR STRUCTURE
OF GASES AND VAPORS
IN synthesizing water, one volume of oxygen unites with two
volumes of hydrogen to form two volumes of water vapor.
Gay-Lussac in 1808 studied similar reactions, and advanced
the law which states that the relative combining volumes of
gases, and the volumes of their products, if gaseous, can be ex-
pressed in a ratio of small whole numbers.
Gay-Lussac's Law, together with Boyle's and Charles'
Laws, led Avogadro in 1811 to advance a theory in explanation
of the uniform behavior of all gases. He ascribed the cause
of this uniform behavior to the probable fact that equal
volumes of all gases and vapors under the same conditions of
temperature and pressure, contain the same number of molecules.
The actual number of these molecules, called Avogadro's
number, is staggering. It is about 6 X 10 23 molecules per
22.2 liters. (10 23 is equal to 10 followed by 23 zeros.)
PROOF THAT THE MOLECULE OF HYDROGEN CONTAINS
TWO ATOMS
From experiments we know that 1 volume . of hydrogen
unites with 1 volume of chlorine, yielding 2 volumes of
hydrogen chloride gas, or
1 volume H + 1 volume Cl >- 2 volumes HC1
According to Avogadro's Law, a cubic foot of hydrogen,
helium, hydrogen chloride, water vapor, ammonia, or any
other gas or vapor contains the same number of molecules.
Conversely, equal numbers of molecules would occupy equal
volumes. For example, 100 molecules of hydrogen, oxygen,
25
26 Chemical Calculations
and hydrogen chloride would all occupy the same volume.
Therefore, one molecule of hydrogen and one molecule of
chlorine would both occupy equal volumes, whereas two
volumes of hydrogen chloride gas would occupy twice this
volume. We may graphically represent this as follows:
1 volume + 1 volume ---- >2voIs.(By experiment)
Imolecule + lmol.- ^2/moleculels
ofH (^
must contain >2 atoms of
One of the HC1 molecules must contain at least 1 atom
of hydrogen and at least 1 atom of chlorine, since fractions
of atoms do not exist. But we have 2 of these HC1 molecules ;
therefore there must be present at least 2 atoms of hydrogen
in the hydrogen chloride formed.
Now we started with only 1 molecule of hydrogen. Since
matter can neither be created nor destroyed, we must have
started with 2 atoms of hydrogen which could have been
present only in the one molecule of hydrogen with which we
started. In other words, 1 molecule of hydrogen must con-
tain at least 2 atoms of hydrogen. Actual experimentation
has recently proved to us that the hydrogen molecule contains
2 and only 2 atoms of hydrogen. Irving Langmuir dis-
sociated the hydrogen molecule at temperatures above 3000 C.
and found the above theoretical conclusion to be experimen-
tally true.
Problems
66. Experimentally, we can determine that 1 volume of nitrogen unites
with 1 volume of oxygen to form 2 volumes of nitric oxide (NO).
Prove that the molecule of nitrogen contains at least 2 atoms of N.
Molecular Structure of Gases and Vapors 27
67. 1 volume of hydrogen unites with 1 volume of bromine vapor to
form 2 volumes of hydrogen bromide gas (HBr). Prove that the
molecule of bromine vapor contains 2 atoms of bromine.
68. 1 volume of nitrogen unites with 3 volumes of hydrogen to form
2 volumes of ammonia gas (NHs). Prove the structure of the
hydrogen molecule.
69. 3 volumes of oxygen change into 2 volumes of ozone gas when a
silent electric discharge is passed through moist oxygen. Prove that
the molecule of ozone contains 3 atoms of oxygen.
CHAPTER EIGHT
VAPOR DENSITY, SPECIFIC GRAVITY, WEIGHT OF A LITER.
AND MOLECULAR WEIGHT OF GASES AND VAPORS
IT is often necessary to compare the relative weights of
gases. Hydrogen, the lightest of gases, is used as a standard.
The vapor density (V.D.) of a gas is the number of times that
gas is as heavy as an equal volume of
hydrogen, both gases being weighed
m under standard conditions. It is really
^ the theoretical weight of 1 cc. of the gas
lcc - reduced to standard conditions.
FIG. 4. Actual volume of tir i .. r i r
i cubic centimeter. y D Weight of any volume of a gas
1000 cc.=i liter Weight of an equal volume of H
If we compare the weights of a liter of each gas, then
V T> Weight of 1 liter of a gas
Weight of 1 liter of H
Since 1 liter of hydrogen weighs 0.09 gram, therefore,
V D Weight of 1 liter of a gas
0.09 gram
From the above we obtain
Weight of a liter of a gas = V.D. X .09 gram . . (1)
Sometimes it is more convenient to compare the weight of
a gas with the weight of an equal volume of air (as standard).
By specific gravity (Sp. Gr.) of a gas is meant the weight of that
gas as compared with the weight of an equal volume of air, or
S T = Weight of any volume of a gas
Weight of an equal volume of air
Again, let us compare the weights of 1 liter of each gas.
28
Vapor Density, Specific Gravity^ etc.- , r , i 9
i.. ? n s s h
S P Weight of 1 liter of a gas
Weight of 1 liter of air .--.- ^
;-.. '" ''"-cs /! r ->"
Under standard conditions 1 liter of air weignslr293-granisr;"
therefore,
S r Weight of 1 liter of a gas
P ' r " ~ 1.293 grams
and from this we obtain
Weight of 1 liter of a gas = Sp. Gr. X 1.293 grams . (2)
Now let us substitute in equation (1) the weight of a liter
as found in equation (2).
Sp. Gr. X 1.293 = V.D. X 0.09
1.293 ...
009
PROOF THAT THE MOLECULAR WEIGHT OF A GAS IS EQUAL
TO TWICE ITS VAPOR DENSITY
The atomic weight of oxygen is 16, so that its molecular
weight is 32. Hence the vapor density of oxygen is
Weight of 1 liter of = 1.43
Weight of 1 liter of H ~ 0.09 ~~
or half its molecular weight.
The weight of a liter of ammonia gas reduced to standard
conditions has been found to be equal to 0.765 gram. Its
vapor density is, therefore, 8.5, or half its molecular weight of
17. By actual experimentation it has been found that the
molecular weight of any gas or vapor is double its vapor den-
sity, or
MoL Wt. =V.D. X 2 (4)
B ' lore
30 Chemical Calculations
By the use of the four equations derived above we can now
solve the following new type problem.
PROBLEMS OF TYPE V: TO FIND THE WEIGHT OF A LITER
OF A GAS, ITS VAPOR DENSITY, SPECIFIC GRAVITY, AND
MOLECULAR WEIGHT
A. EXAMPLE. 185 cc. of alcohol vapor weigh 0.24 gram. Find the weight
of 1 liter of the gas, and its vapor density.
1 liter = 1000 cc.
Since 185 cc. weigh 0.24 gram, 1000 cc. will weigh more, or j^ X 0.24
grains, or 1.3 grams.
The weight of a liter of this vapor is therefore 1.3 grams. (Place WOO
over the volume (cc.} and multiply by the weight (G).)
Substituting in equation (1) for the weight of a liter,
1.3 = V.D. X 0.09 or
B. EXAMPLE. The vapor density of carbon monoxide is 14. Find its
molecular weight, specific gravity, and the weight of 1 liter.
From equation (4) Mol. Wt. = V.D. X 2
Mol. Wt. = 14 X 2 = 28 = Mol. Wt. of CO
V.D.
From equation (3) Sp. Gr. =
14.4
Sp. Gr. = -M- = 0.97 = Sp. Gr. of CO
14.4
From equation (1)
Wt. of 1 liter of a gas = V.D. X 0.09
Wt. of a liter of CO = 14 X 0.09 = 1.26 grams
C. EXAMPLE. The specific gravity of ammonia gas is 0.59. Find its molec-
ular weight, vapor density, and the weight of 325 cc. of the gas.
V.D.
From equation (3) Sp. Gr. =
14.4
0.59 = - or V.D. = 0.59(14.4)
14.4
V.D. of ammonia = 8.5
From equation (4) Mol. Wt. = V.D. X 2
Mol. Wt. of ammonia = 8.5 X 2 = 17
Vapor Density, Specific Gravity, etc. 31
From equation (1) Wt. of a liter = V.D. X 0.09
Wt. of 1 liter of ammonia = 8.5 X 0.09 = 0.765 gram
Hence the weight of 325 cc. of this gas is
^- X 0.765, or 0.249 gram
(SS5
Note that here we are finding the weight of a fraction of a liter, or -rpr, and
1000
1000
Problems
70. 250 cc. of sulfur dioxide gas weigh 0.72 gram. Calculate the weight
of 1 liter, the molecular weight, vapor density, and specific gravity of
this gas.
71. If 0.74 gram of carbon dioxide occupy 370 cc., find the weight of
1 liter, the vapor density, molecular weight, and specific gravity of
this gas.
72. The molecular weight of alcohol vapor is 46. Calculate its vapor
density, specific gravity, and the weight of 1 liter.
73. The vapor density of hydrogen chloride gas is 18.25. What is its
(a) molecular weight, (6) specific gravity, and (c) the weight of
1 liter?
74. Ammonia gas has a vapor density of 8.5. Calculate its specific
gravity, molecular weight, and the weight of 167 cc. of this gas.
75. 500 cc. of phosphorus vapor weigh 2.79 grams. Calculate (a) weight
of 1 liter, (5) vapor density, (c) molecular weight, and (d) specific
gravity of P vapor.
76. Sulfur trioxide vapor shows a molecular weight of 80. Find its
vapor density, specific gravity, and the weight of 9.5 liters.
D. EXAMPLE. Find the vapor density and weight of 1 liter of chlorine gas.
Remember that the formula of chlorine gas is Clz and not Cl.
d
35.5 X 2 = 71 = Mol. Wt.
From equation (4) ~ = 35.5 = V.D.
J~i
From equation (1) 35.5 X -09 = 3.195 g. = weight of 1 liter.
CHAPTER NINE
PROBLEMS OF TYPE VI: TO FIND THE HYDROGEN
EQUIVALENT OF A METAL
THE hydrogen equivalent of a metal is that weight of the
metal which will replace one gram
of hydrogen, or combine with 8
grams of oxygen. It is equivalent
to the atomic weight of the element
divided by its valence.
The hydrogen equivalent of a
metal, capable of liberating the hy-
drogen of an acid as the free gas,
FIG. s. A weighed amount of mav be determined by reacting a
metal is placed in a jar of water we i g hed amount of the pure metal
and over it is placed a graduated "
cylinder filled with water. Acid is with an acid and calculating the
then added to the water in the jar. i_ j. r xt. VL XJT.J
The metal reacts with the add, lib- weight of the liberated hydrogen
crating H 2 , which is collected and when measured under standard
measured in the graduated cylinder. .
conditions. (Fig. 5.)
Determine the equivalent of magnesium from the. following
EXAMPLE.
data:
Weight of magnesium taken .
Measured volume of Ha gas .
. . 0.0167 gram
. . 16.60 cc. (level of water inside and
outside of measuring
tube was the same)
Temperature 17 C.
Barometer reading 755 mm.
Aqueous tension at 17 C 14.4 mm. (gas collected over water)
Deduct the aqueous tension of 14.4 from the total barometric pressure of
755 to get the corrected barometric pressure.
755 mm. - 14.4 mm. = 740.6 mm.
Now reduce the volume of hydrogen liberated to standard conditions.
32
Problems of Type VI 33
pv_ p'v'
T T'
740.6(16.60) _ 760()
273 + 17 273
= 740.6(16.60)273
290(760)
and x = 15.23 cc. of hydrogen
Under standard conditions, the weight of 1 liter of hydrogen gas is equal
to 0.09 gram, or 1 gram of hydrogen will occupy 11.1 liters.
Now since 0.0167 gram of magnesium liberated 15.23 cc. of hydrogen,
0.0167 X ^oo grams of magnesium will liberate 1 gram of hydro-
15.23
gen. Hence a 167(1 ^y iL1 , or 12.16, is the hydrogen equivalent of
15.23
magnesium.
The formula to be used for this type of problem may be
written
, . j , (1000} (11. l}(wt.in grams of metal used}
Hydrogen equvualent = , . 7-^ - ~ . > J frr ... - '-
vol. in cc. (at stand, cond.) of Hz hoerated
B. EXAMPLE. 0.127 gram of a metal react with hydrochloric acid and liberate
58 cc. of hydrogen. What is the equivalent of this metal?
0.127 gram of the metal yields 58 cc. of hydrogen.
Since a liter of hydrogen weighs 0.09 gram, therefore 0.127 gram of the metal
yields ^-1^ X 0.09, or 0.00522 gram of hydrogen. But the equivalent of a
127
metal will liberate 1 gram of hydrogen ; hence - gram of the metal
will yield 1 gram of hydrogen, or 24.3 is the equivalent of this metal.
Problems
77. Find the equivalent of aluminum from the following experimental
data : 4.25 grams of the metal when heated with potash gave 5.23
liters of hydrogen measured over water at 17 C. and 760 mm., the
level of the water in the eudiometer being the same as on the outside.
78. 1 gram of pure iron wire reacts with a dilute acid and the hydrogen
gas evolved is collected in a eudiometer tube, over water. At 27 C.
and 760 mm. the gas occupied 440 cc. What is the equivalent of
iron?
34 Chemical Calculations
79. 10 grams of tin were boiled with hydrochloric acid. When 1120 cc.
of hydrogen gas were collected the action was stopped, and the tin
was found to weigh 4.1 grams. What is the equivalent of tin?
SO. In one of Morley's experiments on the determination of the percent-
age composition of water, he found that 3.265 grams of hydrogen gas
united with 25.853 grams of oxygen. What is the equivalent of
oxygen?
81. A gram of copper oxide was heated and hydrogen passed over it until
it was completely reduced. 0.8 gram of copper was left. What is
the equivalent of copper ?
82. Phosphorus unites with hydrogen to form phosphine, which contains
91.2% phosphorus and 8.8% hydrogen. What is the equivalent of
phosphorus?
83. Bismuth was oxidized in the presence ot oxygen, yielding an oxide
which on analysis showed the presence of 89.7% bismuth and 10.3%
oxygen. Find the equivalent of this element.
CHAPTER TEN
PROBLEMS OF TYPE VII: TO FIND THE TRUE
FORMULA OF A COMPOUND
FROM the simplest formula of a compound we obtain the
simplest molecular weight by adding the atomic weights and
placing x after this number. The simplest molecular weight
must be equal to the true molecular weight, which gives us a
means of determining the value of x. By substituting this
value of x in the simplest formula, we have the true formula
we are seeking.
A. EXAMPLE. Find the true formula of a compound which contains 63.6%
nitrogen, and 36.4% oxygen, and whose vapor density is 21.9.
Under Type III (page 14) we found the simplest formula of this compound
to be (N 2 O) Z .
The simplest molecular weight of (N 2 O) X is (14 X 2 + 16)*, or 44 x.
By means of the vapor density we can find the true molecular weight.
Since V.D. = Mol. Wt.
therefore the true Mol. Wt. = 21.9 X 2 = 43.8
Now the true Mol. Wt. is always equal to the simplest Mol. Wt.
or 43.8 = 44 x
from which we obtain x = 1 (approximately)
Substituting in the simplest formula this value of x, we find the true formula
to be (N 2 O)i, or N 2 O.
Problems
84. A gas on analysis showed the following composition : 73.8% carbon,
8.7% hydrogen, and 17.5% nitrogen. Its vapor density was found
to be 80.2. Find the true formula of this compound.
85. A compound, whose vapor density was found to be 38.8, contained
carbon and hydrogen in the proportion of 92.3% and 7.7%, respec-
tively. What is the true formula of this compound?
86. Carbon,'hydrogen, an d oxygen united in the ratio by weight of 39.9%,
6.7%, and 53.4%, respectively. The vapor density of the resulting
compound was found to be 30.5. Determine the true formula of this
compound.
35
36 Chemical Calculations
87. What is the true formula of a compound containing 31.2% nickel,
3.2% hydrogen, 14.8% nitrogen, 16.9% sulfur, 33.9% oxygen, and
whose molecular weight is 1S9 ?
88. The vapor density of a substance is 21.9. It contains 27.4% carbon
and 72.6% oxygen. What is its true formula?
89. A compound contains twice as many hydrogen atoms as carbon
atoms. Its vapor density is 14.0. Find its true formula.
90. A compound contains 24.24% carbon, 4.04% hydrogen, and 71.72%
chlorine. Its vapor density is 49.6. Find its true formula.
91. The vapor density of a compound is 13.9. It consists of 85.6%
carbon, and the rest of hydrogen. What is the true formula of this
compound?
92. A compound contains 20.2% phosphorus, 10.4% oxygen, and 69.4%
chlorine. Its vapor density is 78. Find its true formula.
93. A compound is composed of an equal weight of sulfur and oxygen.
Its vapor density is 31.9. Find the true formula of this gas.
B. "What is the true formula of a compound which contains 92.3% carbon and
7.7% hydrogen, and 1.1 grams of its vapor occupy 314 cc. at standard
conditions?
For carbon,
% comp. of carbon ^.92.3 _*, PQ
At. Wt. of carbon 12~~
For hydrogen,
% comp. of hydrogen _ 7.7 _ ~ ~
At. Wt. of hydrogen 1
7 69
_^_ = 1 = smallest number of carbon
7.69
atoms in the molecule.
H.C.F. = 7.69
77
z-^r = 1 = smallest number of hydro-
7.69
gen atoms in the molecule.
The simplest formula is (CH)
The simplest Mol. Wt. is (12 + l)x = 13 x.
Since 314 cc. weigh 1.1 grams, one liter will weigh
^ X 1.1, or 3.5 grams
But the weight of 1 liter = V.D. X 0.09
therefore 3.5 = V.D. X 0.09
from which we obtain V.D. = , or 38 9
.09
Hence the true molecular weight = V.D. X 2 = 38.9 X 2 = 77.8
77.8 = 13 *
x = 6 (approximately)
Therefore, the true formula is (CH) 6 or CsHt (benzene).
We do not leave the formula as (CH) 6 , because the molecule of this com-
pound actually contains 6 atoms each of carbon and hydrogen.
Problems of Type VII 37
Problems
94. One liter of a gas weighed 1.98 grams and consisted of 27.27%
carbon and 72.73% oxygen. What is the true formula of this
compound ?
95. An oxide of sulfur was composed of 50% each of sulfur and oxygen.
2.87 grams of this gas occupied a volume of 1 liter under standard
conditions. Calculate the true formula of this compound.
96. A compound showed a composition of 94.12% sulfur and 5.88%
hydrogen. 100 cc. of the substance weighed 0.154 gram. What is
its true formula?
97. 57 cc. of a certain gas weighed 0.134 gram. Its composition was
found to be 46.15% carbon and 53.85% nitrogen. Find its true
formula.
98. A compound contains 25.93% nitrogen and 74.07% Axygen. 0.097
gram of this gas occupied, under standard conditions, 20 cc. Find
the true formula of this compound.
99. A compound has the following composition: carbon 37.6%,
hydrogen 12.5%, and oxygen 49.9%. The weight of a liter of this
vapor is 1.434 grams. What is its true formula?
100. 250 cc. of a gas weigh 0.515 gram. It contains 30.4% nitrogen and
69.6% oxygen. What is the true formula of the gas?
101. An aldehyde has the following composition: 54.7% carbon, 9.1%
hydrogen, and 36.2% oxygen. If the specific gravity of its vapor is
1.526, find the true formula of the compound.
102. 500 cc. of the vapor of a compound weighed 3.1 grams. It con-
tained 22.6% phosphorus and 77.4% chlorine. What is the true
formula of this compound?
103. 0.94 gram of a compound occupies 250 cc. It contains 42% chlorine,
1.2% hydrogen, and 56.8% oxygen. Determine its true formula.
104. 500 cc. of a gas weigh 1.03 grams. It contains 30.6% nitrogen and
69.4% oxygen. What is the true formula of this compound ?
CHAPTER ELEVEN
PROBLEMS BASED ON EQUATIONS
SINCE the symbol of an element and the formula of a com-
pound represent definite weights, an equation also represents
definite weights of the substances taking part in the reaction.
Thus
2Na + S ^Na 2 S
may be read, " 2 atoms of sodium and 1 atom of sulfur yield
1 molecule of sodium sulfide," and it may also be read as,
" 46 grams of sodium and 32 grams of sulfur yield 78 grams
of sodium sulfide."
An equation is neither chemically nor mathematically cor-
rect until it is properly balanced, and problems based on equa-
tions cannot be solved unless those equations are balanced.
Arranging the coefficients of the various substances taking
part in a chemical change so that the number of atoms of each
element is the same on both sides of the equation is called
balancing the equation (see page 166). The sum of the
weights on both sides of the equation must, of course, be the
same. We must bear in mind that even if we had no system
of writing equations, the mathematical relationship of reac-
tions would still hold. See Appendix, pages 166-170, for
exercises in balancing equations.
Problems based on chemical equations may be broadly
divided into three types :
1. Straight-Weight. Given one weight to find another
weight.
9 I a. Weight-Volume. Given a weight to find a volume.
[ J. Volume-Weight. Given a volume to find a weight.
3. Straight-Volume. Given one volume to find another
volume.
38
Problems Based on Equations 39
PROBLEMS OF TYPE VIII: STRAIGHT-WEIGHT PROBLEMS
1. Read the problem until you understand it thoroughly.
2. Obtain a balanced equation for the reaction involved.
When a compound is formed from another substance by
successive steps, it is not necessary to calculate the, inter-
mediate products.
3. The weight given is placed over the substance involved.
The weight required, designated by x grams (or any other
unit mentioned), is placed over the substance whose
weight is to be found.
4. Since the same relationship exists between the actual
weights expressed in grams, pounds, etc., as exists between
the molecular weights represented by the equation, .place
the molecular weights of only those substances involved
in the problem under those substances. Do not ignore
any coefficient.
5. Check the problem again and solve for x.
A. EXAMPLE. How many grams of sodium sulfate, Na 2 S04, are formed when
49 grams of sulfuric acid react with sodium chloride?
49 grams x grams
H 2 S0 4 + 2NaCl ^2HC1 + Na 2 S0 4
1X2 + 32 + 16 X4 23 X 2 + 32 + 16 X 4
2 + 32 + 64 46 + 32 + 64
98 142
Weight of substance given _ Weight of substance required
Mol. Wt. of substance given Mol. Wt. of substance required
49 _ x grams
98 142
Solving for *, 98 x = 49(142)
x = 71 grams Na 2 SOi
Alternate method. We can avoid the use of an equation involving * by
solving the above as follows :
Weight of substance given ^ m rf substance ^
Mol. Wt. of substance given
4Q
or X 142 = 71 grams Na 2 SO<.
40 Chemical Calculations
Problems
105. How much zinc is required to react with sufficient hydrochloric
acid to produce 4 grams of hydrogen ?
106. 432 grams of mercuric oxide, HgO, were completely decomposed by
heat. How much mercury was formed?
107. 64 grams of oxygen were obtained by heating potassium chlorate.
How much of the KC10 3 was used?
108. How much hydrogen would be required to reduce completely
20 grams of cupric oxide, CuO ?
109. 17 grams of silver nitrate, AgN0 3) were required to precipitate
completely all of the sodium chloride in a solution. How much
NaCl was present in the solution? (Equation 6, page 99.)
110. A bottle of chlorine water was exposed to the sunlight until all the
free chlorine disappeared. 20 grams of HC1 gas were found.
What weight of chlorine was present in the chlorine water? The
equation for this reaction is C1 2 + H 2 O > 2HC1 + (nascent
oxygen).
111. What weight of magnesium will be needed to react with sulfuric
acid to produce 30 grams of MgSO* ? What weight of hydrogen
would be evolved?
112. 11.5 grams of sodium completely react with water. What weight
of sodium hydroxide was produced ?
113. 12 Ib. of hydrogen were liberated by the electrolysis of water.
What weight of water was decomposed?
B. EXAMPIX. 10 grams of crystallized copper sulfate gave, after heating,
6.4 grams of the anhydrous salt. Calculate the number of molecules of
water of crystallization in the original compound.
Let x represent tlie number of molecules of water of crystallization.
Then 10 g. 6.4 g.
CuS0 4 xHzQ >- CuSO 4 + *H 2 O
The weight of the water of crystallization is the difference between the
weight of the original salt and the anhydrous salt, or, xH.zO weighs 10
6.4, or 3.6 grams.
6.4 g. 3.6 g.
Therefore, CuSO 4 *H 2 O *- CuSO4 + *H 2 O
64 + 32 + 16 X 4 (1X2 + 16)a?
64 + 32 + 64- (2 + 16)
160 18 x
Problems Based on Equations 41
Solving for x, we obtain
Weight of substance given _ Weight of substance required
Mol. Wt. of substance given Mol. Wt. of substance required
6.4 _ 3.6
160 18 *
6.4(18 *) = 160(3.6)
_ 160(3.6) _ 576 =5
6.4(18) 115.2
x = 5 molecules of EUO
( Problems
114. 10 grams of crystallized sodium sulfate lost 5.6 grams of water after
being heated. How many molecules of water of crystallization
did the original salt contain?
115. 8 grams of crystallized sodium sulfite lost 4 grams of water on
being heated. How many molecules of water of crystallization did
the crystalline salt contain?
116. 10 grams of crystallized washing soda after being heated gave
3.71 grams of the anhydrous Na 2 COs. Calculate the number of
molecules of water of crystallization in this compound.
117. If 2 grams of crystallized barium chloride lost 0.295 gram upon
being heated to constant weight, find the formula of the crystalline
salt.
118. 7 grams of crystalline calcium sulfate gave 5.536 grams of the
anhydrous compound. What is the formula of the crystalline salt ?
119. 11.5 grams of crystalline sodium carbonate produced 4.26 grams of
the anhydrous salt. Calculate the number of molecules of water of
crystallization in the original compound.
120. 10 grams of crystallized zinc sulfate gave after heating 5.62 grams
of dry ZnS0 4 . Calculate the number of molecules of water of
crystallization present in the original crystalline salt.
121. 4.88 grams of anhydrous MgSCX gave 10 grams of the crystalline
salt. How many molecules of water of crystallization does the
crystal of magnesium sulfate contain?
CHAPTER TWELVE
PROBLEMS OF TYPE IX
I. WEIGHT-VOLUME PROBLEM
(Given a weight, to find a volume)
SINCE 0.09 gram of hydrogen occupies one liter, 2 grams of
this gas will occupy 2 -j- .09, or 22.2 liters. Therefore the
gram-molecular volume of hydrogen that is, the volume
occupied by its molecular weight expressed in grams is equal
to 22.2 liters (Fig. 6). A study of a large number of gases and
vapors leads to the conclusion that the gram-molecular volume
of all gases and vapors, when measured under standard condi-
tions, is equal, approximately, to 22.2 liters.
Since the gram-molecular weight of a volatile substance
1.O6 quarts =
1 liter
\
X
RAM-MOLECULAR
VOLUME
f A cube of 11.1 inches
-I will contain 22.2 liters
j or 23.5 (juarfcs.
*
-<
2 grams hydrogen or
- 32 grams oxygen will
occupy 22.2 liters.
\
-3| in.-*
-11.1 in-
FIG. 6. Diagram showing the relative sizes of a quart bottle of milk and 22.2
liters (G.M.V.). The weight of any gas filling 23% of these milk bottles will be
its molecular weight in grams.
occupies 22.2 liters, we can use the gram-molecular volume to
represent one molecule or one volume of a gas or vapor.
The coefficient in front of the formula of a gas or vapor
represents both the number of molecules and the number of
4 2
Problems of Type IX 43
volumes of that gas or vapor. Thus 2CO stands for 2 mole-
cules of carbon monoxide, 2(12 -f 16) or 56 grams of carbon
monoxide, and also for 2(22.2) or 44.4 liters of this gas.
In solving weight-volume problems, the same relationships
hold as in the straight-weight problems, except that we may
now substitute' for the gram-molecular weight of the substance
whose volume is to be determined, its gram-molecular volume,
or 22.2 liters for each volume of that substance.
EXAMPLE. How many liters of carbon dioxide gas can be prepared from
600 grams of calcium carbonate?
500 g. x liters
CaCO 3 > CaO + CO 2
40 + 12 + 48 1(22.2)
100 22.2
Weight of substance given Volume of substance required
Mol. Wt. of substance given Volume of substance required (Gram-Mol. Vol.)
500 _ ^liters
100 22.2
* = 111 liters of C0 2
Alternate method. (Avoiding the use of #.) .
Vol. of C0 2 = Weight of substance given x Gram . Mol VoL
Mol. Wt. of substance given
Vol. of C0 a = ^ X 22.2 = 111 liters
JLUU
NOTE. By a coincidence, the same relation exists between the
gram-molecular weight and the gram-molecular volume that is,
22.2 liters as exists between the ounce-molecular weight and the
ounce-molecular volume; namely 22.2 cu. ft. This coincidence
occurs because
1 ounce = 28.3 grams and
1 cu. ft. = 28.3 liters.
Remember that grams correspond to liters, and ounces cor-
respond to cubic feet. For other units of weight and volume, con-
version factors must be used.
44 Chemical Calculations
Problems
122. What volume of hydrogen can be obtained by the electrolysis of
90 grams of water?
123. A lump of carbon weighing 24 grams was burned in oxygen. What
volume of carbon dioxide was formed ?
124. How many liters of hydrogen sulfide gas can be obtained by the ac-
tion of an acid on 11 grams of ferrous sulfide, FeS? (Eq. 1, p. 103.)
125. 216 ounces of magnesium will be needed to react with sufficient
hydrochloric acid to liberate what volume of hydrogen ?
126. How many liters of ammonia gas can be formed by the action of 66
grams of ammonium sulfate on slaked lime, Ca(OH) 2 ? (Eq. 1 , p. 107.)
127. What volume of sulfur dioxide gas will be liberated when 32 grams
of copper are completely dissolved by concentrated sulfuric acid ?
(Eq. 6, p. 105.)
12S. How many liters of carbon dioxide gas, measured under standard
conditions, can be obtained by the action of 49 grams of sulfuric
acid on marble? CaC0 3 + H 2 S0 4 *- CaS0 4 4- H 2 + C0 2 .
129. What volume of hydrogen is needed to reduce completely 24 grams
of copper oxide, CuO, to pure copper?
H. VOLUME-WEIGHT PROBLEM
(Given a volume, to find a weight}
The same procedure is followed as in the previous example
until the equation involving x is to be written.
EXAMPLE. What weight of potassium chlorate will be needed to obtain
11.1 liters of oxygen?
# grams 11.1 liters
2KC1O 3 ^ 3O 3 + 2KC1
2(39 + 35.54-48) 3(22.2)
245 66.6
Volume of the substance given _ Weight of substance required
Gram-Mol. Vol. of substance given Mol. Wt. of substance required
11.1 _ x grams
66.6 245
(11.1)245 = 66.6 x
x = 40.8 grams KC1O S
Alternate method. (Avoiding the use of x.)
Wt. KC1O 3 = Volume of substance given ., Mol. Wt. of substance
Gram-Mol. Vol. of substance given required
= ^| X 245 = 40.8 grams '
Problems of Type IX 45
Problems
130. What weight of zinc will be required to produce 44.4 liters of
hydrogen gas?
131. How many grams of calcium carbonate will have to be treated with
sulfuric acid to produce 5.55 liters of carbon dioxide? (See Prob-
lem 12S.)
132. How many grams of FeS will be needed to prepare 88.8 liters of
hydrogen sulfide gas? (Equation 1, page 103.)
133. What weight of pure carbon will have to be burned to produce
44.4 cc. of carbon dioxide?
134. What weight of the carbon would be needed to prepare the same
volume of carbon monoxide?
135. What weight of phosphorus will have to be burned completely to
remove all the oxygen in 10 liters of air, assuming that air contains
20 % of oxygen by volume ? 4P + 3O 2 > 2P 2 O 3 .
136. What weight of copper must be dissolved by concentrated sulfuric
acid to produce 99.9 cc. of sulfur dioxide gas? (Equation 6,
page 105.)
137. What weight of sodium must be used to produce 1 liter of hydrogen
gas measured at standard conditions ?
138. How much sodium chloride will be needed to produce 111 liters of
hydrogen chloride gas? NaCl + H 2 SO 4 - NaHSO 4 + HC1.
139. What weight of sulfur will have to be burned to produce 50 cc. of
sulfur dioxide gas?
140. What weight of water must be decomposed to furnish enough
oxygen to form, with pure carbon, 44.4 liters of carbon dioxide?
CHAPTER THIRTEEN
PROBLEMS OF TYPE X: STRAIGHT-VOLUME PROBLEM
(Given one volume, to find another volume)
FROM Avogadro's Law we may conclude that the coefficients
of formulas representing gases or volatile compounds also
represent their volumes. Therefore the same relationship
exists between the gram-molecular weights of gases, their
gram-molecular volumes, and their coefficients.
In the solution of problems involving only the volumes of
gases or volatile substances, we need consider only their
volumes or coefficients. We form our equations accordingly.
A. EXAMPLE. How many liters of carbon dioxide will be formed in the
complete combustion of 12 liters of benzene?
12 liters x liters
2CsH s + 15O 2 >- 6H 2 + 12CO 2
2 vol. 12 vol.
Volume of substance given ' _ Volume of substance required
Coefficient of substance given Coefficient of substance required
12 = _
2 12
144 -=2 x
x = 72 liters of CO 2
Alternate method (avoiding the use of x).
-IT i,,_- rr\ Volume of substance given ^ - . , ,. , .
Volume LO 2 = . s_ x coefficient of substance
Coefficient of substance given
required = 1? x 12 = 72 liters of CO 2
2i *
B. EXAMPLE. What volume of air (containing 20% oxygen by volume)
would be required to bum completely 352 cc. of carbon monoxide?
352 cc. * cc.
2CO + 2 > 2CO 2
2 1
^? = | or a; = 176 cc. of oxygen
2 1
But, since tie air contains only 20% oxygen by volume, the volume of air
required in the above reaction will be 5 times that of oxygen, or 5 X 176
= 880 cc. air.
46
Problems of Type X 47
Problems
141. What volume of hydrogen must unite with 2 liters of oxygen to form
water and leave no oxygen over ?
142. If 20 liters of hydrogen react with sufficient chlorine, what volume
of hydrogen chloride gas will be formed?
143. How many liters of its component gases can be obtained by the
complete decomposition of 6 liters of ammonia gas ? 2NH 3 >
N 2 + 3H 2 .
144. Carbon monoxide passed over warm calcium hydroxide reacts
according to the equation CO + Ca(OH) 2 > CaCO 3 + H 2 .
How does the volume of carbon monoxide compare with that of
hydrogen ?
145. How many cubic centimeters of oxygen will be needed to produce
. 55 cc. of sulfur dioxide gas ?
146. What volume of air will convert 100 cc. of nitric oxide, NO, into
nitrogen peroxide, N0 2 ?
147. How many liters of air are necessary for the complete combustion of
500 cc. of acetylene gas? 2C 2 H 2 + 50 2 > 4C0 2 + 2H 2 0.
148. How many cubic feet of air will be used in the complete com-
bustion of 750 cu. ft of methane, CHi? CH* + 20 2 > CO 2
+ 2H 2 0.
CHAPTER FOURTEEN
MISCELLANEOUS PROBLEMS
I. PROBLEMS INVOLVING AN EXCESS OP ONE OR
MORE REAGENTS
THIS group of problems is really not a new type but is one
which involves the previous type problems in one form or
another. We know that substances unite chemically accord-
ing to definite weights. Therefore if we mix chemicals regard-
less of these definite weights, and a chemical reaction takes
place, a part of one or more of the reagents will be left over
after the reaction is completed. We cannot, by inspection,
tell which substance is in excess.
EXAMPLE. A mixture of 35 cc. of oxygen and 50 cc. of hydrogen is exploded.
How much water vapor is formed, and what is the nature and
volume of the gas which has not taken part in the action?
50
50 cc. of hydrogen would require XI, or 25 cc. of oxygen.
&
35
35 cc. of oxygen would require X 2, or 70 cc. of hydrogen.
But only 50 cc. of hydrogen are available, therefore some of the
oxygen will be left over. Since only 25 cc. of oxygen would be
required to react with all of the hydrogen available, 35 25, or
10 cc. of oxygen would remain.
Problems
149. 100 cc. of chlorine and 200 cc. of hydrogen are mixed and made to
unite chemically. What volume of hydrogen chloride gas was
formed, and how much gas remained unacted upon?
150. When 14 cc. of sulfur vapor and 10 cc. of oxygen combine as far as
possible to form sulfur dioxide, what volume of SOa is formed, and
how much of either factor is left?
48
Miscellaneous Problems 49
151. 4 grams of carbon and 9 liters of oxygen are combined chemically
as far as possible. How much oxygen remains uncombined ?
152. 4.8 grams of magnesium ribbon were made to react with 2 liters of
chlorine gas until no further reaction could take place. What
weight of magnesium chloride was formed, and how much of either
factor remained?
153. 100 grams of mercury and 11.1 liters of oxygen reacted chemically
until no further union could take place. How much mercuric oxide
was formed, and what weight of either substance remained uncom-
bined?
154. 234 grams of sodium chloride and 294 grams of sulfuric acid were
used to make hydrochloric acid. What weight of hydrogen chloride
gas was liberated ?
155. 17 grams of silver nitrate in solution were added to a solution
containing 2.34 grams of sodium chloride. What weight of silver
chloride precipitated out?
156. What weight of barium sulfate was precipitated when 104 grams of
reacted with a solution containing 35.5 grams of
H. PROBLEMS INVOLVING THE MOST ECONOMICAL
CHEMICALS TO USE
A. EXAMPLE. Which would be cheaper, neutralizing 50 Ib. of sulfuric acid
with commercial caustic potasli (containing 10% water) and costing 7f^
per pound, or neutralizing it with caustic soda (containing 76% NaOH) and
costing $3.10 per 100 Ib. ? How much cheaper would it be ?
The equation for the reaction using caustic soda is
xlb. 50 Ib.
2NaOH + H 2 SO 4 - ^Na 2 SO 4 + 2H 2
2(23 + 16 + 1) 2 + 32 + 64
80 98
The weight of pure NaOH = X 80 = 40.8 Ib.
9o
Since the NaOH is only 76% pure, we will need more of the commercial
NaOH. The cost of this commercial NaOH will be
X 40.8 X ^ = $1.66
.
10U 76
The equation for the reaction using caustic potash is
xlb. 50 Ib.
2KOH + H 2 SO 4 - =>- K 2 S0 4 + 2H 2 O
2(39 + 16 + 1) 2 + 32 + 64
112 98
50 Chemical Calculations
The weight of pure KOH needed would be
^ X 112 = 57.1 Ib.
98
The cost of this KOH would be (since commercial KOH is 90%, pure)
7# X 57.1 X ^, or $4.52
yu
Hence it would cost 4.52 minus $1.66, or $2.86 more if the commercial caus-
tic potash were used.
Problems
157. Which is the more economical to use in the preparation of ammonia,
ammonium sulfate at $8.75 per 100 Ib. or ammonium chloride at
12 cents per pound?
158. We can buy potassium chlorate at 7$ per pound or sodium chlorate
at 6|fp per pound. Which is the more economical oxidizing agent?
159. Which would be cheaper for preparing carbon dioxide, calcium
carbonate at l a pound, or sodium bicarbonate at $2 per 100 Ib.?
160. Would it be more economical to buy sodium peroxide (costing 9^
per pound), or to make it from metallic sodium costing 27 per
pound?
161. How much money could be saved by using either sulfuric acid at
$1.35 per 100 Ib., or 50% nitric acid costing $5 per 100 Ib., to
neutralize a solution containing 2000 Ib. of NaOH?
162. How much of a saving would be. involved in using either bromine
(47^ per pound), or chlorine (G per pound) in an oxidation process
requiring 7.1 tons of chlorine, assuming that either one could just
as well be used?
III. CALCULATIONS OF MIXTURES HAVING A
COMMON CONSTITUENT
EXAMPLE. A mixture of AgCl and Agl weighs 2.5 grams. The weight of silver
is 1.5 grams. Find the weight of AgCl and of Agl.
Let x = weight of AgCl
Let y = weight of Agl
Then x + y = 2.5 grams ........ (1)
The total weight of the silver in the mixture is 1.5 grams, therefore,
108 Ii y = 1-5 grams ....... (2)
143.5 235
Miscellaneous Problems 51
Substituting in (2) the value of y found in (1), we obtain
Ju8 x i 108 (9 5 V) = 1 5 erams
143.5 235 " &
0.753 x + 0.46(2.5 - x) = 1.5
0.753 * + 1.15 - 0.46 x = 1.5
0.293 * = 1.5 - 1.15
x = 1.2 grams of AgCl
y = 1.3 grams of Agl
Problems
163. Given a mixture of KC1 and NaCl weighing 5 grams and containing
60% chlorine. Calculate the weights of ELC1 and NaCl in the
mixture.
164. The weight of a mixture of silver chloride and silver bromide is
6 grams. On analysis 4.5 grams of pure silver were found in the
mixture. What weights of AgCl and AgBr were present in the
mixture?
165. 0.5 gram of a mixture of potassium iodide and potassium bromide
yielded 0.3 gram of KoS04- How much KBr and KE were present
in the mixture?
166. A mixture of K 2 S04 and NaaS04 weighed 1.43 grams. From this
mixture 2.136 grams of BaSO 4 were obtained. Find the weights of
potassium and sodium sulfate present in the original mixture.
167. 2.144 grams of a mixture of calcium and strontium carbonates gave
2.844 grams of a mixture of calcium and strontium sulfates. What
weights of CaCOs and SrCOs were present in the mixture?
168. If 8 grams of dolomite (CaCO 3 MgC0 3 ) liberated 3.8 grams of
carbon dioxide gas, what was the percentage of calcium and magne-
sium in this sample of dolomite?
169. A mixture of KaCOa and KHCOa weighs 2 grams and contains 0.8
gram of CO 2. What is the weight of each constituent in the
mixture?
170. A mixture of PbS04 and BaSO4 weighed 1 gram, and an analysis
showed the presence of 0.405 gram of the S0 4 radical. What
weights of the two sulfates were present in the mixture ?
CHAPTER FIFTEEN
CHEMICAL PROBLEMS FROM NEW YORK STATE
REGENTS EXAMINATION PAPERS
January, 1915
171. Find the number of liters of carbon dioxide that will result
from the addition of sufficient hydrochloric acid to 100 grams
of calcium carbonate. [10] (Atomic weights : C = 12,
= 16, Cl = 35.5, Ca = 40)
172. Find the number of volumes of hydrogen and of nitrogen that
enter into combination to form 200 liters of ammonia. [10]
June, 1915
173. How many liters of hydrogen sulfide are obtained by adding
sufficient hydrochloric acid to 88 grams of ferrous sulfide?
[10] (Atomic weights : Fe = 56, S = 32)
174. Find the cost of enough potassium chlorate at 80 cents per
kilogram to make 800 liters of oxygen. [10] (Atomic
weights : K = 39, Cl = 35.5, = 16)
January, 1916
175. Find the number of liters of hydrogen sulfide that will result
from the addition of sufficient hydrochloric acid to 100 grams
of ferrous sulfide. [10] (Atomic weights : S = 32, Fe
= 55.8)
176. Show what will be the relative volumes of the reacting sub-
stances in each of the following cases : (a) one volume of
carbon monoxide plus sufficient oxygen burns to carbon
dioxide [4], (b) one volume of marsh gas plus sufficient oxygen
burns to carbon dioxide and water [6].
June, 1916
177. Find the number of liters of ammonia gas that may be
prepared from 107 grams of ammonium chloride and sufficient
slaked lime. [10] (Atomic weights : N = 14, Cl = 35.5)
52
Problems from Regents Examination Papers 53
178. If a sample of gasoline is half hexane, CeHu, and half heptane,
C 7 Hi6, how many cubic feet of air will be necessary for the
complete combustion of 100 cubic feet of gasoline vapor?
[10] (Assume air as one-fifth oxygen by volume.)
January, 1917
179. What will a liter of oxygen weigh at 740 mm. pressure and
20 C., if it weighs 1 .43 grams under standard conditions ? [10]
180. Find the cost of making 100 liters of hydrogen sulfide, given
the following data (wholesale prices) : sulfur 3j a kilogram,
scrap iron 15j a kilogram, concentrated (assume 100%)
sulfuric acid 2 a kilogram. [10] (Atomic weights : S = 32,
Fe = 55.5)
181. How many liters of carbon dioxide will result from the com-
plete combustion of 10 liters of carbon monoxide [5] ? Find
the weight of copper precipitated from its sulfate solution by
10 grams of iron [5]. (Atomic weights : Fe=55.5, Cu=63.6)
June, 1917
182. There are five molecules of water in each molecule of crys-
tallized copper sulfate ; find the number of grams of water
required to crystallize 100 grams of anhydrous copper sul-
fate. [10] (Atomic weights : = 16, S = 32, Cu = 63.6)
183. How many liters of carbon dioxide gas will be formed by the
fermentation of 180 grams of grape sugar according to the
following equation : C 6 H 12 O 6 - >- 2C 2 H 5 OH +2CO 2 ? [10]
(Atomic weights : = 16, C = 12)
January, 1918
184. Find the number of liters of oxygen gas measured under stand-
ard conditions produced by the decomposition of 245 grams
of potassium chlorate. [10] (Atomic weights : K = 39,
Cl = 35.5, O = 16)
185. What will a liter of carbon dioxide weigh at 750 mm. pressure
and 20 C. temperature, if it weighs 1.98 grams at 776 mm.
pressure and C. temperature? [10]
54 Chemical Calculations
June, 1918
186. Name a gas composed of two elements, which, when burned
in air or oxygen, forms water and sulfur dioxide [3]. How
many liters of oxygen are required for the complete com-
bustion of five liters of this gas [7] ?
187. Find the weights and the volumes of oxygen and hydrogen
forming 180 grams of water. [10] (Atomic weight : O= 16)
188. Find the weight of ammonium sulfate necessary, when heated
with the required amount of slaked lime, to make two liters
of ammonia water, allowing 700 volumes of ammonia gas to
one volume of water for full-strength solution. [10] (Atomic
weights : S = 32, N = 14)
January, 1919
189. Find the number of grams of copper needed to prepare 10
liters of nitric oxide by interaction with moderately dilute
nitric acid. [10] (Atomic weights: O = 16, Cu = 63.6,
N= 14)
190. Find the percentage of copper in bluestone, CuSCX 5HaO.
[10] (Atomic weights : O = 16, Cu = 63.6, S = 32)
191. If 5 grams of hydrated barium chloride, heated to constant
weight, weigh 4.265 grams, find (a) the percentage of water in
the crystals [5]; (b) the number of molecules of water of
crystallization per molecule of barium chloride [5]. (Atomic
weights : O = 16, Ba = 137, Cl = 35.5)
June, 1919
192. What weight of ammonia could be obtained (theoretically)
by heating 100 pounds of ammonium sulfate with an excess
of slaked lime [7]? What volume would it occupy [3]?
(Atomic weights : N = 14, S = 32, = 16.' 1 cu. ft. of
ammonia weighs .048 Ib.)
193. What volume of oxygen would be required for the complete
combustion of 1000 cubic feet of acetylene? [10]
Problems from Regents Examination Papers 55
194. A certain gas consists of 14.29% of hydrogen in combination
with 85.71% of carbon. Measured at standard conditions, 50
cubic centimeters of the gas weighs .063 gram. Calculate the
chemical formula of the gas. [10] (Atomic weight : C = 12)
January, 1920
195. A gas is found by analysis to contain 46.15% of carbon and
53.85% of nitrogen ; 57 cc. of this gas, reduced to standard
conditions, weighs .1334 gram. From the above data derive
the chemical formula of the gas. [10] (Atomic weights:
C = 12, N = 14)
196. What weight of iron could be obtained theoretically from 200
tons of ore containing 85% ferric oxide? [10] (Atomic
weights : Fe = 56, O = 16)
June, 1920
197. What weight of sodium chloride will precipitate all the
silver from a solution made by dissolving 1 gram of silver in
nitric acid? [10] (Atomic weights: Na = 23, Cl = 35.5,
Ag = 108, N = 14, = 16)
198. How many liters of hydrogen chloride at standard conditions
will be necessary to neutralize a water solution of 10 grams
of sodium hydroxide? [10] (Atomic weights: Cl = 35.5,
Na = 23, O = 16)
199. Assuming air to contain 20 per cent of oxygen, state how many
cubic feet of air would be required for the complete combus-
tion of 1000 cubic feet of acetylene. [10]
January, 1921
200. Calculate the weight of barium chloride which, reacting with
sufficient sulfuric acid, yields 1.46 grams of barium sulfate.
[10] (Atomic weights : Ba = 137, Cl = 35.6, S = 32> = 16)
201. How many liters of chlorine (standard conditions) are re-
quired to liberate 40 grams of bromine from a solution
of potassium bromide? [10] (Atomic weights: Br = 80,
K = 39, Cl = 35.6)
56 Chemical Calculations
202. Find the volume of methane whose complete combustion
forms 10 cubic feet of carbon dioxide. [10] (Atomic
weights: C = 12, O = 16)
June, 1921
203. In the following equation, if 20 grams of Chile saltpeter are
used, state what weight of potassium nitrate will be found :
NaN0 3 + KC1 = KNOa + NaCl [10] (Atomic weights :
K = 39, Cl = 35.5, Na = 23, =-16, N = 14)
204. Magnesium acting with an acid liberates 100 cc. of hydrogen
measured at 22 C. and 780 mm. pressure. Calculate (a) the
volume of this hydrogen measured at standard conditions
[5], (6) the weight of the magnesium needed [5]. (Use .09 as
the weight of one liter of hydrogen at standard conditions and
24 as the atomic weight of magnesium.)
205. From 1 gram of -pure iron 1.43 grams of an oxide of iron are
formed ; find (a) the percentage composition of the oxide [5],
(b) the simplest formula of the oxide [5]. (Atomic weights :
Fe = 56, = 16)
January, 1922
206. Baking soda and cream of tartar in baking powder react as
follows : NaHCOg + KHCJBUOe = NaKC 4 H 4 O 6 + H 2 +
C02- How much baking soda is required for 47 grams of
cream of tartar [5] ? What volume of C02 will this free [5] ?
(Atomic weights : Na = 23, C = 12, 0=16, K = 39)
207. Name the gases formed and compute their volumes when
10 liters of methane completely unite with oxygen. [10]
(Measurements at standard conditions. Atomic weights:
C = 12, = 16)
208. If 4.265 grams of anhydrous barium chloride weigh 5 grams
when hydrated, find (a) the percentage of water of crystalli-
zation [5], (b) the number of molecules of water of crystalliza-
tion to one molecule of barium chloride [5]. (Atomic weights :
Ba = 137, Cl = 35.5, O = 16)
Problems from Regents Examination Papers 57
June, 1922
209. If 126 grams of copper react with an excess of nitric acid
according to the equation 3Cu + 8HN0 3 = 3Cu(N0 3 ) 2 +
4H 2 O + 2NO, calculate (a) the weight of copper nitrate
formed [5], (6) the volume of nitric oxide formed [5]. (Atomic
weights : Cu = 63.6, = 16, N = 14)
210. The weight of 500 cc. of a gas is 0.76 grams ; calculate (a) the
molecular weight [5], () the vapor density [5].
211. Given the equation 2CO + 2 = 2C0 2 ; state (a) the
chemical change indicated [2], (6) the number of atoms in each
molecule [2], (c) the number of molecules [2], (d) the relative
volumes [2], (e) the parts by weight [2]. (Atomic weights :
C = 12, = 16)
January, 1923
212. If 106 grams of ammonium chloride react with an excess of
sodium hydroxide, calculate (a) the number of grams of
ammonia formed [5], (&) the number of liters of ammonia
formed, measured under standard conditions [5]. - (Atomic
weights: N= 14, Cl = 35.5)
213. What volume of methane measured under standard condi-
tions can be burned to carbon dioxide and water in 50 liters of
oxygen similarly measured? [10]
214. Calculate the volume of 915 cc. of hydrogen at 27 C. and 760
mm. when changed to standard conditions [5]. What is the
equivalent of magnesium if of a gram of this metal liberates
the hydrogen mentioned in the first part of this question [5] ?
June, 1923
215. Calculate the weight of KClOs required to produce 20 liters
of oxygen. [10] (Atomic weights : K = 39, Cl = 35.5,
O= 16)
216. Find the volume >of oxygen necessary for the complete
oxidation of 100 cubic feet of water gas containing by volume
40% carbon monoxide and 60% hydrogen. [10]
58 Chemical Calculations
217. The molecular formula of a certain gas is CjNo; calculate
(a) the molecular weight [2], (6) the vapor density [2], (c) the
specific gravity (air standard) [2], (d) the weight of one liter
[2], (e) the percentage composition by weight [2], (Atomic
weights: C = 12, N= 14)
218. The percentage composition of a substance is C 39.95%,
H 6.69%, 53.36%. Deduce its formula. [10] (100 cu. cm.
of its vapor at standard conditions weighs .134 gram.
Atomic weights : C = 12, = 16)
January, 1924
219. Magnesium dissolved in acid gave 100 cc. of hydrogen at a
temperature of 22 C. and a pressure of 780 mm. ; how many
grams of metal were used? (Atomic weight: Mg = 24.3)
[10]
220. Calculate the percentage of nitrogen in crystallized copper
nitrate, Cu(N0 3 ) 2 6H 2 [5]. There are five molecules of
water in each molecule of crystallized copper sulfate ; how
many grams of water unite with 100 grams of anhydrous
copper sulfate to produce crystallized copper sulfate [5]?
(Atomic weights: O = 16, N = 14, S = 32, Cu = 63.5)
221. Find the weight of a liter of (a) chlorine [3], (6) methane [3].
Calculate the number of liters of carbon dioxide formed in
the complete combustion of 100 liters of acetylene [4].
(Atomic weights : Cl = 35.5, = 16, C = 12)
June, 1924
222. If 54 grams of silver react with an excess of nitric acid accord-
ing to the equation 3Ag + 4HNO 3 = 3AgN0 3 + 2H 2 +
NO, what weight of silver nitrate and what volume of nitric
oxide measured under standard conditions of temperature and
pressure will be formed? [10] (Atomic weights : Ag = 108,
O = 16, N = 14)
223. Calculate the molecular weight of a gas 50 cc. of which weigh
.0,67 gram. Assuming that 46.6% of it by weight is nitrogen
and the rest oxygen, calculate the chemical formula. [10]
(Atomic weights : N = 14, = 16)
Problems from Regents Examination Papers , 59
224. How many liters of hydrogen are needed to convert 30 liters
of nitrogen into ammonia? [10]
January, 1925
225. A certain solution contains .049 gram of sulfuric acid in
each cubic centimeter. How many cubic centimeters of this
solution would be needed to neutralize 100 cc. of a solution of
sodium hydroxide which contains .04 gram per cubic centi-
meter? [10]
226. What weight of calcium oxide could be obtained from 200
grams of pure calcium carbonate [6]? What volume of
carbon dioxide measured at standard conditions would be
liberated in the reaction [4] ?
227. Name the important experimental operations that are carried
out in determining the chemical formula of a newly discovered
substance, assuming that it can readily be converted into the
gaseous state [4]. Illustrate the process by calculating data
which should be approximated in actual operations carried out
to verify the formula of ethane, CaHs [6].
June, 1925
228. Calculate the weight of ammonia that can be obtained from
200 grams of ammonium chloride. What volume will this
occupy at standard conditions ? [10]
229. Find the weight of a liter of chlorine [4]. Calculate the
number of liters of carbon dioxide formed by the complete
combustion of 200 liters of acetylene [6].
230. A student on analyzing two different oxides of iron obtained
the following results : in the first he found 8 parts by weight of
oxygen combined with 28 of iron, in the second 5.5 parts of
oxygen with 14 parts of iron. Show that these results are
probably wrong because they do not agree with one of the
important laws of chemistry. [10]
PART TWO
CHAPTER SIXTEEN
TO FIND THE MOLECULAR WEIGHTS OF GASES
AND VOLATILE COMPOUNDS
THE molecular weights of gases or volatile compounds can
be determined by means of the equations derived in the study
of the vapor density of gases.
If we combine equations (1) and (4) (pages 28 and 29) ;
namely,
Wt. of 1 liter of a gas = V.D. X 0.09
and Mol. Wt. = V.D. X 2,
u* TV/T i ITT-* Wt. of a liter of a gas X 2
we obtain Mol. Wt. =
or Mol. Wt. = Wt. of 1 liter X 22.2 (since -^- = 22.2\
\ \j-uy /
In other words, the molecular weight of a gas is equal to the
weight of 22.2 liters of that- gas. Since it is inconvenient to
work with 22.2 liters, a smaller volume is weighed and the
weight of 22.2 liters is determined from the experimental data.
The most common methods in use are the Dumas, and Victor
Meyer methods.
Dumas method. A globe of measured volume (Fig. 7) is
filled with the gas or vapor, sealed, and then weighed. It is
then completely exhausted by means of an air pump, again
sealed, and the weight of the evacuated bulb determined.
The difference in weight is, of course, the weight of the gas or
vapor. From these data the weight of 22.2 liters is determined.
EXAMPLE. A globe of 500 cc. capacity was filled with carbon monoxide
gas under standard conditions. When filled it weighed 325.73 grams.
When evacuated the globe weighed 325.10 grams. Calculate the molecular
weight of this gas.
60
Molecular Weights of Gases
61
FIG. 7. A Dumas bulb of measured capacity.
The weight of the 500 cc. of the gas = 325.73
1000
Therefore, the weight of 1 liter
500
325.10 = 0.63 gram
X 0.63 = 1.26 grams
The weight of 22.2 liters, then, = 1.26 X 22.2 = 27.97 grams
Hence the molecular weight of the gas is 27.97.
Victor Meyer method. A weighed amount of liquid is
vaporized, and the air which this vapor displaces is measured
(Fig. 8). This volume is corrected to standard conditions,
and from these data the weight of the liquid, which when vapor-
ized occupies 22.2 liters, is determined. This, of course, corre-
sponds to the molecular weight of the gas. This method is
just the reverse of the Dumas method. It is applicable, not
to gases, but to liquids which volatilize without decomposition.
EXAMPLE. 0.755 gram of carbon disulfide displaced, when vaporized,
221cc. of air under standard conditions. Determine its molecular weight.
221 cc. were displaced by 0.755 grams
1000 cc. would be displaced by 0.755 X 100 grams
222 liters would be displaced by 0.755 X X 22.2 grams
62 Chemical Calculations
The molecular weight of carbon bisulfide is, therefore,
0.755(1000)22.2 = 76
221
s_r
FIG. 8- In this Victor Meyer apparatus a weighed amount of
liquid in the small vial is vaporized. The vapor displaces its own
volume of air found in the apparatus. This air, in turn, displaces
its volume of water in the graduated cylinder. The volume of
water displaced corresponds to the volume of the vapor of the liquid
whose molecular weight is to be determined.
The formula to be used for this type of problem may be
written
_ (1000) (22.2} (weight in grams of substance vaporized}
volume in cc. (at stand, cond.) of displaced air
Problems
1. Calculate the molecular weight of carbon tetrachloride, CCLi, from
the following data :
, , ,
Molecular Weights of Gases 63
Weight of CC1 4 taken 0.62 gram
Volume of air displaced 88.8 cc.
Temperature of air measured .... C.
Barometric pressure 760 mm.
2. Calculate the vapor density of carbon bisulfide from the following
data :
Weight of CS 2 taken 0.152 gram
Volume of air displaced 44.4 cc.
Temperature of air measured .... C.
Barometer reading 760 mm.
3. A Dumas bulb of 250 cc. capacity was filled with acetylene gas.
Calculate the molecular weight of the gas from the following data:
Weight of bulb plus acetylene gas . . . 253.480 grams
Weight of evacuated bulb 253.187 grams
Temperature of experiment . .' . . . 27 C.
Barometer reading 760 mm.
4. Calculate the molecular weight of chloroform from the following :
Weight of bulb plus chloroform vapor . 36.02 grams
Weight of evacuated bulb 34.34 gra,ms
Temperature of experiment 67 C.
Barometer reading 760 mm.
The Dumas bulb had a capacity of . . 400 cc.
5. Find the molecular weight of iodine from the following data obtained
by the Dumas method :
Weight of iodine taken . . . . . . 0.312 gram
Volume of air displaced 50 cc.
Temperature of experiment 227 C.
Barometric pressure 760 mm,
6. Calculate the molecular weight of a substance from the following
data obtained by the Victor Meyer method :
Weight of substance taken 0.184 gram
Volume of vapor obtained 40 cc.
Temperature of experiment ... . . . 47 C.
Barometer reading 760 mm.
CHAPTER SEVENTEEN
FINDING THE MOLECULAR WEIGHTS OF NON-VOLATILE
COMPOUNDS
WHEN a compound whose molecular weight is to be deter-
mined cannot be changed into a vapor, the preceding methods
of finding its molecular weight cannot be used. Other methods
based on certain characteristic behaviors of their solutions
are employed.
I. ELEVATION OF THE BOILING POINT
A solid dissolved in a liquid changes the boiling point
of that liquid. A definite mathematical relationship exists.
The gram-molecular weight of cane sugar (CiaHasOn)
namely, *342 grams raises the boiling point of a liter of
water from 100 C. to 100.52 C. The gram-molecular
weight of phenol (94 grams) will produce exactly the same
result. Different amounts of these substances, within cer-
tain limits, will produce a proportionate elevation of the
boiling point, and the rise of the boiling point depends upon
the number of particles of solute dissolved in the solution.
We can express this behavior mathematically, as follows :
M I Wt = we ^S^ f solute in 1000 g. of solvent ,
elevation of B, P. as found
For each solvent there is a constant k which represents the
theoretical elevation produced by a gram-molecular weight
of the solute dissolved in 1000 grams of solvent. The value
of k for water is 0.52 C. For other solvents this value of k
is different. This method is applicable to non-electrolytes
of comparatively high boiling points.
64
Weights of Non-Volatile Compounds
MOLECULAR ELEVATION OF THE BOILING POINT
SOLVENT
CONSTANT k
Water
0.52 C.
Ethyl alcohol
1.15
Ether
2.11
Benzene
2.67
Chloroform
3.66
EXAMPLE. 36 grams of glucose when dissolved in a liter of water raised the
boiling point 0.10 C. Determine the molecular weight of glucose.
From equation (5) we obtain, by substituting the values,
Mol. Wt. = 36 X *~~ =
Mol. Wt. of glucose = 187.2
(The actual molecular weight of this compound is 180.)
Problems
7. 17.1 grams of sucrose raised the boiling point of 500 grams of water
0.052 C. What is the molecular weight of sucrose?
8. 60 grams of fructose when dissolved in a liter of water raised the
boiling point of the water to 100.173 q C. What is the molecular
weight of this sugar?
9. 6.4 grams of naphthalene dissolved in 100 grams of benzene raised
its boiling point 1.33 C. What is its molecular weight?
10. 15.2 grams of camphor raised the boiling point of ether, when
dissolved in 100 grams of this solvent, 2.11 C. What is the molec-
ular weight of camphor?
11. 78 grams of menthol, when dissolved in 1000 grams of ether, raised
its boiling point 1.05 C. What is the molecular weight of menthol?
II. DEPRESSION OF THE FREEZING POINT
(RAOULT'S METHOD)
A solid dissolved in a liquid lowers the freezing point of the
liquid. Here, again, a definite mathematical relationship
exists. Each solvent has a constant of its own, representing
the fictitious lowering of the freezing point caused by dissolv-
ing one gram-molecular weight of the substance in 1000 grams
66
Chemical Calculations
of solvent. 342 grams of cane sugar and 94 grams of phenol
each lower the freezing point of 1 liter of water from C. to
- 1.87 C. The lowering of. the freezing point depends again
on the number of particles dissolved in the solution. The
mathematical formula for this behavior may be expressed as
weight, of solute in 1000 g. of solvent ,
~~ lowering of P. P. as found ^ '
MoL
MOLECULAR DEPRESSION OF THE FREEZING POINT
SOLVENT
CONSTANT k
Water
1.87 C.
Formic acid
2.77
Acetic acid
3.90
Benzene
4.90
Phenol
7.40
EXAMPLE. 9.4 grams of phenol were dissolved in a liter of water whose freez-
ing point was lowered 0.187 C. What is its molecular weight?
From equation (6) we obtain, by substituting the values,
MoLWt. =
Mol. Wt. of phenol = 94
' 0.187 0.187
Problems
12. 12.6 grams of dextrose dissolved in 100 grams of water lowered its
freezing point to 1.45 C. What is the molecular weight of
dextrose?
13. 29.5 grams of acetamide when dissolved in a liter of water lowered
its freezing point to 0.93 C. What is its molecular weight?
14. 7.5 grams of tartaric acid dissolved in 100 grams of water lowered its
freezing point 0.93 C. What is its molecular weight?
15. 0.164 gram of ethyl alcohol lowered the freezing point of 100 grams
of benzene 0. 175 C. What is the molecular weight of ethyl alcohol ?
16. 12.8 grams of naphthalene dissolved in 100 grams of benzene lowered
its freezing point 4.9 C. What is its molecular weight?
17. 42 grams of milk sugar lowered the freezing point of 1000 cc. of water
0.21 C. What is its molecular weight?
CHAPTER EIGHTEEN
FINDING THE ATOMIC WEIGHTS OF ELEMENTS
EXISTING IN VOLATILE COMPOUNDS
IN finding the atomic weight of elements existing in volatile
compounds, the simplest method to use is the following :
1. Select about six volatile compounds containing the element.
2. Determine their molecular weights from their vapor den-
sities.
3. Analyze the compounds for their percentage composition.
4. Prepare a table as given below. Select a number of
which all the weights of the element under investigation
are multiples. This H.C.F. (highest common factor) is the
atomic weight of the element.
DETERMINATION OF THE ATOMIC WEIGHT OF CHLORINE
COMPOUND
FORMULA
MOLECULAR
WEIGHT
% CL
WEIGHT OF
CHLORINE
H.C.F.
Hydrogen chloride
HC1
36.5
97.3
35.5
35.5
Phosphorus trichloride
PC1 3
137.0
77.7
106.4
or the
Bichloride of mercury
HgCl 2
270.0
26.3
70.9
atomic
Chlorine dioxide
C10 2
67.0
53.0
35.5
weight
Phosgene
COC1 2
63.5
55.9
35.5
of
Chloroform
CHC1 3
119.5
89.1
106.5
chlorine
5. The larger the number of its volatile compounds analyzed,
the surer can we be that the H.C.F. is the actual atomic
weight, and not a multiple of the atomic weight.
6. If an element gives no volatile compounds, we can deter-
mine its atomic weight with the aid of Dulong andPetit's
Law. The atomic weights found by the above method
may be used in determining molecular weights of non-
volatile compounds.
67
68
Chemical Calculations
EXAMPLE. Determine the atomic weight of sulfur by the above method.
COMPOUND
FORMULA
MOLECULAR
WEIGHT
%s
WEIGHT OF
SULFUR
H.C.F.
Sulfur dioxide
S0 2
64
50.0
32
32, or the
Sulfur trioxide
S0 3
80
40.0
32
atomic weight
Sulfur chloride
SOClo
119
26.9
32
of
Hydrogen sulfide
Carbon bisulfide
H 2 S
CS 2
34
76
94.1
84.2
32
64
sulfur
Sulfuryl chloride
S0 2 C1 2
135
23.7
32
Problems
18. Determine the atomic weight of carbon from the following of its
volatile compounds: acetylene, C 2 H 2 ; benzene, C 6 H 6 ; formal-
dehyde, ECHO; ethyl alcohol, C 2 H 5 OH; methane CH 4 ; and
acetic acid, CH 3 COOH.
19. Determine the atomic weight of oxygen from at least five of its
volatile compounds.
20. Determine the atomic weight of phosphorus from at least four of
its volatile compounds.
21. Determine the atomic weight of hydrogen by the above method.
CHAPTER NINETEEN
FINDING THE ATOMIC WEIGHTS OF ELEMENTS
WITH THE AID OF SPECIFIC HEATS
IT is not possible to obtain, by the vapor-density method,
the true atomic weights of those elements which cannot be
vaporized. But by means of Dulong and Petit's Law (1818)
the atomic weights of such elements can be calculated. It was
found that the product of the atomic weight of a solid element
and the specific heat of that element was approximately 6.4.
Specific heat X atomic weight = 6.4 (atomic heat)
The specific heat of a solid is the amount of heat necessary
to raise the temperature of that substance 1 C. as compared
with the amount of heat necessary to raise an equal weight of
water 1 C.
Since specific heat is a physical constant which can be deter-
mined experimentally, the atomic weight of any solid element
can be found by means of this specific heat. From the per-
centage composition of compounds containing the solid ele-
ment under investigation, we can obtain a number which is
either the true atomic weight of that element or a multiple of
its atomic weight. With the aid of Dulong and Petit's Law
we can by inspection decide which is the true atomic weight.
A. EXAMPLE. 25 grams of tin are converted into stannic oxide, SnOa. The
stannic oxide is found to weigh 31.8 grams. The specific heat of tin is
0.055. Determine its atomic weight.
Since the SnOa weighed 31.8 grams and the tin originally weighed 25 grams,
the difference of 6.8 grams must be the weight of the oyxgen which united
with the tin. Hence
25 g. 6.8 g.
Sn + O 2 >- SnO 2
* 32
and x represents the atomic weight (also the Mol. Wt.) of tin.
6g
Chemical Calculations
Wt. of substance given ^ Wt. of other substance
Smce Mol. Wt. of substance given Mol. Wt. of other substance
25 _ 6.8
Therefore 7 ~ 32
and x = - = 118 = apparent atomic
b ' 8 weight
118 may be a multiple of the real atomic weight, hence to check it, we de-
termine the atomic weight by means of Dulong and Petit's Law.
Specific heat X atomic weight = 6.4
0.055(118)= 6.5 or approximately 6.4
Therefore, 118 is the true atomic weight of tin.
Problems
22. Berzelius found that 4.2 grams of MnCla yielded 9.57^ grams AgCl.
What is the atomic weight of silver if its specific heat is .055 ?
23. The same experimenter obtained 17.55 grams PbO from 16.3 grams of
lead.. The specific heat of lead is .030. What is its atomic weight?
24. A compound of hydrogen and sulfur has a molecular weight of 34.
The percentage of sulfur is 94.11%. If the specific heat of sulfur is
0.163, what is its atomic weight?
25. Weatherill, in 1924, prepared and analyzed antimony trichloride
and found that 2.75 grams SbCl 3 were equivalent to 3.9 grams of
silver. The specific heat of antimony is 0.05. Determine the atomic
weight of antimony. (SbCl 3 - > 3Ag)
26. Honigschmid and Zintl prepared and analyzed the tetrabromide of
hafnium discovered in 1923. They found that 1.92 grams of HfBr 4
were equivalent to 2.93 grams of AgBr. Determine the atomic
weight of hafnium if its specific heat is .035. (HfBr 4 - *- 4AgBr)
27. According to Berzelius, 10 grams of platinum formed 24.74 grams
of K 2 PtCl 6 . The specific heat of platinum is 0.03 ; what is its atomic
weight?
28. 50 grams of lead were changed to 73.2 grams of lead sulfate. The
specific heat of lead is 0.03 ; what is its atomic weight?
CHAPTER TWENTY
PROBLEMS BASED ON FARADAY'S LAWS
THE unit quantity of electricity, the coulomb, is that quan-
tity of electricity which will deposit 1.118 milligrams of
metallic silver from a silver salt solution. The electro chemi-
t
cal unit, the Faraday, is equal to 96,540 coulombs and cor-
responds to the electrical charge carried by one gram equiva-
lent of an ion (I gram H + , 108 grams Ag + , etc.). Univalent
ions aE carry the same quantity of electricity, while other
ions carry amounts greater than this in proportion to their
valencies.
The weights of the different substances which separate at
each electrode throughout the circuit are proportional to the
equivalent weights but independent of concentration, temper-
ature, electrode surface, and voltage.
For example, 96,540 coulombs of electricity will liberate
1 gram of hydrogen, -^ grams of oxygen, and -^ grams of
copper, and will decompose -^ grams of sulfuric acid.
A. EXAMPLE. A certain current of electricity deposited 32 grams of copper.
What weights of silver and of aluminum would this same current deposit?
Since equal quantities of electricity liberate chemically equivalent quantities
of the ions, therefore,
equivalent of copper equivalent of silver
weight of Cu deposited weight of silver deposited
e<t 108
nr H-L ^^-
32 ~ x
x 108 grams of silver
Similarly, f = f
x = 9 grams of aluminum
B. EXAMPLE. A certain current liberated 5 grams of chlorine. What weight
of calcium was deposited from the fused calcium chloride?
72 Chemical Calculations
equivalent of CI _ equivalent of calcium
grains of Cl liberated grams of calcium deposited
3Sr5 4O
| = (since Ca is divalent)
o x
35.5 x = 100
x = 2.8 grams of Ca
1 Problems
29. A 'current deposited 162 grams of silver. What weight of copper
could this same current deposit?
30. 1 gram of potassium is deposited electrically. What weight of
copper will the same current liberate from a copper sulfate solution?
31. A certain current liberated 50 grams of hydrogen in an hour. What
weight of nickel would the same current deposit on spoons in a
nickel sulfate bath?
32. Two electrolytic cells are arranged in series. One cell contains
acidulated water and the other a solution of lead nitrate. When
35 cc. of hydrogen (standard conditions) have been liberated, what
weight of lead nitrate has been decomposed ?
33. What weights of copper can be replaced by (a) 50 grams of ferrous
iron, (b) 65 grams of zinc, (c) 54 grams of aluminum, and (d) 12 grams
of magnesium?
34. How many coulombs are carried by, and will deposit (a) 56 grams of
silver, (b) 17.75 grams of chlorine, and' (c) 100 grams of mercury (ic) ?
CHAPTER TWENTY-ONE
SPECIFIC GRAVITY OF SOLUTIONS
THE specific gravity of a liquid is the number of times that
liquid is as heavy as an equal volume of water. The specific
gravity of a solution multiplied by its volume equals the
weight of the solution, or,
Sp. Gr. X Vol. in cc. = Wt. in grams
The specific gravity of a solution multiplied by its volume and
again multiplied by the percentage of material in solution is
equal to the weight of solute in the solution.
Thus 1 cc. H 2 S04, of specific gravity 1.4 and containing
50 per cent H 2 S0 4 by weight in solution, weighs 1.4 grams
and contains 1.4 X 0.50 or 0.70 gram of actual sulfuric acid
dissolved in the water.
EXAMPLE. How many cubic centimeters of HaSO^ of Sp. Gr. 1.64, containing
72% sulfuric acid by weight, are required to react completely with 12 grams
of magnesium?
This is a straight-weight problem.
x grams 12 grams
H 2 S0 4 + Mg =- MgS0 4 + H 2
2 + 32 + 64 24
98
12 = x_ 2 ^ 9g
24 98' '
# = 49 grams of 100% H 2 SO 4
Now since 1 cc. of the acid used contains 1.64 X 0-72 grams of pure H 2 S04,
we will require
49
1.64(0.72)
or 41.5 cc. of the above acid.
Problems
35. What volume of HC1 of Sp. Gr. 1.1, containing 20% HC1 by weight,
will be required to unite with 112 grams of iron to produce ferrous
chloride?
73
74 Chemical Calculations
36. Nitric acid of Sp. Gr. 1.37 contains 59.4% HN0 3 by weight. Cal-
culate the number of cc. of this acid which will be needed to neutralize
180 grams of NaOH.
37. How many cubic centimeters of 30% HC1, Sp. Gr. 1.15, must be
added to an ammonium hydroxide solution in order to form 10.7
grams of NH 4 C1?
38. How many cubic centimeters of 0.91% ammonia water, Sp. Gr. 0.996,
are required to precipitate all the aluminum in 17.1 grams of
A1 2 (S0 4 ) 3 ?
39. How many cubic centimeters of 75% HN0 3 , Sp. Gr. 1.44, will be
required to dissolve completely 3.2 grams of pure copper?
40. Ammonium hydroxide of Sp. Gr. O.S9S contains 29% NH 3 by weight.
What weight of ammonium sulfate wiU be formed by the neutraliza-
tion of 500 cc. of sulfuric acid of Sp. Gr. 1.825 containing 91%
H 2 S04?
DILUTION OF SOLUTIONS TO A GIVEN DEGREE
OF SPECIFIC GRAVITY
When two mutually soluble substances are mixed, the re-
sulting solution is not always equal to the sum of the original
volumes. If we assume, however, that no change in volume
takes place, the calculation is as follows :
If V is the volume of one solute of Sp. Gr. S, and V is
the volume of another solute of Sp. Gr. S', and if R is the vol-
ume of the resulting solution whose Sp. Gr. is S", then,
VS + V'S' = RS"
But we are assuming that V + V is equal to R, therefore,
VS + V'S' = (V + V')S" .... (1)
A. EXAMPLE. Two lots of sulfuric acid, one of Sp. Gr. 1.72, and the other
of Sp. Gr. 1.32, are available. What volumes of these acids must be mixed
to make 500 cc. of H 2 SO 4 of Sp. Gr. 1.6?
Let x - volume of H 2 SO 4 of Sp. Gr. 1.72
Let y = volume of H 2 S0 4 of Sp. Gr. 1.32
Since the sum of the weights must be equal, by substituting in equation (1)
we obtain,
(1.72)* +(1.32)y = 1.6(500) =800 grams . . . . (2)
Specific Gravity of Solutions 75
We have assumed that no change in volume takes place, therefore
* + y = 500 cc.
or x = 500 y
Substituting in (2) this value of x, we obtain,
(1.72) (500 - y)+(1.32)y = 800
860 - 1.72 y + 1.32 y = 800
- 0.4 y = - 60
y = 160 cc. of H 2 S0 4 (Sp. Gr. 1.32)
and therefore * = 350 cc. of H 2 S0 4 (Sp. Gr. 1.72)
Problems
41. Two lots of hydrochloric acid are available. One lot has a specific
gravity of 1.2 and the other of 1.05. What volumes of these acids
must be mixed to make a liter of HC1 of Sp. Gr. 1.16?
42. A manufacturer requires 100 liters of nitric acid of Sp. Gr. 1.45. He
has on hand an acid of Sp. Gr. 1.5, and another of Sp. Gr. 1.31.
What volumes of these acids must he mix?
43. A chemist requires 100 cc. of an ammonia solution of Sp. Gr. 0.89.
He has in his laboratory some ammonia water of Sp. Gr. 0.882 and
some other solution of Sp. Gr. 0.97. How much of each must he
use to obtain the required ammonia water?
44. It is required to make 1500 cc. of sulfuric acid of Sp. Gr. 1.6. Two
acids are available, one having a Sp. Gr. of 1.54 and another of
Sp. Gr. 1.84. What volumes of these acids must be mixed?
45. 500 cc. of nitric acid of Sp. Gr. 1.19 must be prepared from two nitric
acid solutions of Sp. Gr. 1.50 and 1.11. What volumes of the two
acids must be used?
46. A chemist mixes a liter each of ammonia water of Sp. Gr. 0.99 and
0.91. What is the specific gravity of the resulting solution?
CHAPTER TWENTY-TWO
VOLUMETRIC ANALYSIS
VOLUMETRIC analysis involves the analysis of substances
by means of solutions of which the chemical value is accurately
known. Substances which can be obtained in a very pure con-
dition, like Na 2 CO 3 , AgN0 3 , KMn0 4 , Nad, and oxalic acid,
can be accurately weighed and the desired volume of solution
prepared. These solutions are called standard solutions and
can be used to standardize other solutions whose strength
cannot be so well adjusted. Thus we can determine the
strength of hydrochloric acid by means of a standard sodium
carbonate solution. The analytical operation is called a
titration. The standard solution is placed in a graduated
vessel, like a burette, and is then dropped into a solution of the
substance to be estimated. By means of a suitable indicator,
the completion of the chemical reaction taking place between
the standard solution and the substance tested is shown by
a change in color, formation of a precipitate, or some other
visible sign. This point is called the end-point.
In determining the strength of a HC1 solution the indicator
used is methyl orange, which changes color when the reaction
Na 2 C0 3 + 2HC1 >- 2NaCl + H 2 + C0 2
is completed.
Standard solutions are usually prepared as normal solutions
or molar solutions.
A mole of a substance is its molecular weight in grams. For
example, a mole of crystallized barium chloride (BaCl 2 2H 2 0)
is 244.4 grams. A molar solution is one containing the gram-
molecular weight of the solute dissolved in 1000 cc. of so-
lution.
76
Volumetric Analysis 77
Problems
47. Find the weights of a mole of each of the following : hydrochloric
acid, nitric acid, sulfuric acid, and acetic acid.
48. Calculate the weights of a mole of each of the following bases:
sodium hydroxide, calcium hydroxide, ammonium hydroxide, and
aluminum hydroxide.
49. Determine the molar weights of each of the following salts : sodium
chloride, crystallized sodium carbonate, anhydrous copper sulfate,
and sodium potassium tartrate.
50. What is the molar strength of a solution containing 87 grams of
potassium sulfate in 250 cc. of water?
A normal solution, is one containing one equivalent of the
substance in 1000 cc. of the solution (not the solvent). In
calculating the weights to be used in preparing normal solu-
tions, therefore, the reaction in which the solution is to be
used must be known.
By the equivalent of an element we mean that weight of the
element which will replace or react with one gram of hydrogen.
It corresponds to the atomic weight of the element divided
by its valence. Thus the equivalents of chlorine, oxygen,
and copper (cupric) are 35.5, 8, and 31.78, respectively.
The equivalent of an acid is that weight of the acid con-
taining one gram of hydrogen. The equivalent of HC1 is
36.5, the equivalent of H 2 S0 4 is , and of H 3 P0 4 -r-.
2 o
The equivalent of a base is that weight of the base con-
taining 17 grams of the hydroxyl (OH) group. The equiva-
lent of NaOH is 40, and of Fe(OH) 3j iM.
o
The equivalent of a simple normal salt is that weight of the
salt which contains one equivalent of the 'metal or metallic
radical. The equivalents of NaCl, CuS0 4 , A1C1 3 , and Fe 2 (S04) 3
co e 159.6 133.5 , 399.68 .. ,
are 58.5, , , and - , respectively.
78 " Chemical Calculations
The equivalent of an acid salt, a basic salt, and a complex
salt will depend upon the hydrogen, hydroxyl, metal, or radical
with reference to which the equivalent is to be determined.
009 909 909
Thus , , and - are the equivalents of calcium, hydro-
a 4: D
gen, and the phosphate radical in CaHXPO^.
A. EXAMPLE. How would you prepare 15 cc. of a N solution of potassium
permanganate?
A normal solution of KMnO< contains one equivalent of KMnO 4 in 1000 cc.
of the solution. One equivalent of KMnO 4 is equal to 39.1 + 54.9 + 64,
or 158 grams. 1000 cc. of a normal solution would contain 158 grams of
KMnO4. But we need a solution, therefore, 1000 cc. =^ solution would
6 o
1 'iS
contain i-5, or 26.33 grams. The problem calls for only 15 cc. Hence we
6
should use 26.33 X , or 0.395 gram, to which enough water is added
to make the final volume 15 cc.
B. EXAMPLE. What is the normality of a solution containing 35 mg. of silver
(as AgNOa) per cubic centimeter of solution?
A normal solution of AgNOs should contain the atomic weight of silver
(a univalent element) expressed in grams per liter of solution. Therefore,
it should contain only 107.88 grams of silver. Since it contains 50 grams
per 1000 cc., its normality is
NOTE. If 10 grams of salt are dissolved in 90 grams of water,
such a solution is called a 10% salt solution and refers to the per-
centage of solute by weight in the solution. If 100 cc. of salt
solution contains 10 grams of salt, we have a solution which may be
described as containing 100 grams of salt per liter.
Problems
51. Calculate the amount of KNOa necessary to make 5 liters of
solution of potassium nitrate. 5
52. How many grams of salt does half a liter of a 0.3 N sodium iodide
solution contain?
Volumetric Analysis 79
53. How would you prepare 25 cc. of a tenth-normal solution of sodium
carbonate?
54. What -is the normality of a solution containing 75 mg. of copper, as
CuSCh, per cubic centimeter of solution? ^
55. How many grams of potassium permanganate must 695 cc. of
solution contain?
56. What volume of sulfuric acid solution would be required to
5
neutralize a solution of sodium carbonate containing 1 gram NaaCOa ?
57. How much of a solution of hydrochloric acid will be needed to
neutralize a limewater solution containing 0.02 gram of Ca(OH) 2
in solution? jq-
58. What weight of silver nitrate will be required to make 10 liters of
solution?
59. 1 cc. of a silver nitrate solution contains 0.0095 gram of silver.
Calculate the normality of this solution.
60. How much pure potassium hydroxide must be dissolved in 1900 cc.
of water to make a half-normal solution?
61. What weight of Na 2 C0 3 10H 2 must be dissolved in 250 cc. of
water to make a tenth-normal solution?
62. What weight of a 20% solution of NaOH will be required to neutralize
8 grams of a 50% solution of nitric acid?
63. How much of a 5% solution of hydrochloric acid will just neutralize
100 grams of a 50% solution of potassium hydroxide?
64. What is the normality of a solution of copper sulfate containing
0.016 gram of copper per cubic centimeter of solution?
I. STANDARD SOLUTIONS OF OXIDIZING AND REDUCING AGENTS
To prepare a standard solution of an oxidizing or reducing
agent we must calculate what fractional part of the mole of
the oxidizing or reducing agent contains one equivalent. This
value will depend upon the reaction involved.
Some of the common oxidizing agents, and their reactions
showing the number of atoms of oxygen acting as oxidizers,
follow.
(1) Chlorine
C1 2 + H 2 >- 2HC1 + O
So Chemical Calculations
(2) Manganese dioxide
Mn0 2 - > MnO +
(3) Potassium nitrate
2KN0 3 - ^ K 2 + 2ND + 30
(4) Potassium permanganate
2KMn0 4 - ^ K 2 + 2MnO + 50
(acid solution)
(5) Hydrogen peroxide
H 2 2 - ^-H 2 +
(acid solution)
(6) Potassium permanganate
2KMn0 4 - >- K 2 + 2Mn0 2 + 30
(neutral and alkaline solution)
A. EXAMPLE. How much KMn0 4 must be weighed to make 100 cc. of N
solution of potassium permanganate to be used in an acid solution? 5
KMnC>4 in acid solution yields 5 atoms of oxygen.
Now since 5 oxygen atoms represent 10 equivalents, the quantity of
KMnO* required to make a liter of normal solution will, according to
definition, be equal to the weight of 2KMn0 4 divided by 10, or, 31.6 grams.
Hence to make 100 cc. of solution will require
5
31.6 100
Problems
65. How many grams of potassium permanganate must 235 cc. of
N
- solution contain, for use as an oxidizing agent in the presence of
sulfuric acid ?
66. A liter of solution to be used in an oxidation reaction contains
50.5 grams of KNO 3 . What is the normality of this solution?
67. What weight of KMn0 4 must 500 cc. of a - solution contain, for use
2i
as an oxidizing agent in an alkaline solution?
Volumetric Analysis 81
68. Potassium bichromate is to be used as an oxidizing agent in an
acid solution. How would you prepare 250 cc. of a solution of
5
this reagent ? K 2 Cr 2 7 > Cr 2 3 + K 2 + 30.
69. What weight of K4Fe(CN) 6 .' 3H 2 O should a normal solution contain
for use as a reducing agent? The equation for this reaction is
10K4Fe(CN) 6 3H 2 + 2KMn0 4 + 8H 2 S0 4 >
10K3Fe(CN) 6 + 6K 2 S0 4 + 2MnSO 4 + 38H 2
II. ADJUSTMENT OP THE STRENGTH OF SOLUTIONS
The normalities of two -solutions vary inversely as the
N V
volumes they occupy. That is, , Obviously, then,
if TV equals N', V must be equal to V, or solutions of the same
normality occupy equal volumes.
This means that 1 cc. N HC1 is equivalent to 1 cc.
N NaOH, and 1 cc. N H 2 S0 4 is equivalent to 1 cc.
NCa(OH) 2 .
A. EXAMPLE. A solution of HN0 3 is of such, strength that 1 cc. is equivalent
to 0.045 gram of HNO 3 . How much water must be added per liter to make
the solution?
1
1 cc. N HN0 3 contains 0.083 gram HN0 3 per liter, therefore the given
solution is normal. But the required solution is to be
63 &
Since normalities vary inversely to volumes, therefore
HlL= * "r x where x = volume of- H 2 to be added.
*N 1
Solving for x, we obtain,
x = - liters or 428.6 cc. H 2 must be added.
7
B. EXAMPLE. If you had on hand a silver nitrate solution, and wanted
5
to make a solution containing 15 mg. of silver per cubic centimeter, to what
extent would you have to dilute the original solution?
Let x = volume of water to be added to 1 cc. of the solution.
5
82 Chemical Calculations
Then x + 1 = total number of cubic centimeters of final solution.
The weight of the silver does not change in the dilution, therefore,
(i)
Substituting for ^ AgN0 3 the value 107 - 88 + 14 + 48 Qr g4 mg> ^ ^
5 o
tain for equation (1) 15(a; -{- 1) = 34
15 * + 15 = 34
x = 1.267 cc. water
Problems
70. 1 cc. of sulfuric acid is equivalent to 0.025 gram of sodium hydroxide.
How much water must be added per liter to make it --?
5
71. How much water must be added to 1900 cc. of a sodium hydroxide
solution, 1 cc. of which is equivalent to 0.045 gram of HaSO^, to
make it half normal?
72. We have on hand 175 cc. of a solution of 0.35 normality. We wish to
change this to a tenth-normal solution. How much water must we
add?
73. We have on hand a normal solution of NaOH and some hydrochloric
acid of unknown strength. 5 cc. of the NaOH solution required
7.5 cc. of the acid for neutralization. Determine the strength of the
acid solution, and what volume of water would be required for
dilution to make the solutions of equal normality?
74. 1 cc. of a solution of H 2 S0 4 is equivalent to 0.03 gram of NaOH.
How much water must be added per liter to make a half normal
solution?
CHAPTER TWENTY-THREE
GRAVIMETRIC ANALYSIS
THE term gravimetric analysis means essentially the weigh-
ing of a substance of known composition obtained from the
material to be analyzed, and from this, calculating the amount
of an element or radical present in the original material.
I. CALCULATIONS 01 PERCENTAGE TO THE DRY BASIS
A. EXAMPLE. An ore contains 25% zinc and 8% water. What is the percent-
age of zinCj on the dry basis ?
100 grams of zinc ore contain 25 grams Zn and 8 grams B^O.
After all the water is removed, the ore weighs 92 grams, of which 25 grams
are zinc. Therefore, the percentage of zinc on the dry basis is
25X100
B. EXAMPLE. A load of ore contains 30% copper, and, after exposure, 28%
copper and 15% water. Find the percentage of water in the original load.
100 grams of the original sample contain 30 grams Cu and x grams H 2 O
After exposure 28 grams Cu and 15 grams H 2 O
The weight of Cu on the dry basis in the original sample was .
100 x
The weight of Cu in the exposed load was ^ X *??
100 15
But the actual weight of the copper has not changed, therefore,
30 X 100 _ 28 X 100
100 - x 100-15
28(100 - *) = 30(100 - 15)
28 x = 250
* = 8.93% of water
C. EXAMPLE. A sample of ore weighed 50 grams. When heated to 50 C.
it lost 5% water. The partially dried sample was found to contain 3%
water. Find the amount of water in the original sample of ore.
50 g. ore = 47.5 g. partially dried ore + 2.5 g. water.
47.5 g. partially dried ore = 47.5(0.97) g. dry ore + 47.5(0.03) g. H 2 O.
Hence the total weight of the water = 2.5 g. + 47.5(0.03) g. = 3.925 g.,
which amounts to X 100 = 7.85% water.
50
83
84 Chemical Calculations
Problems
75. An ore contains 8% moisture and 15% copper. What percentage of
copper does the ore contain, on the dry basis?
76. What is the percentage of manganese in a dry ore which, before
being heated, contained 12% moisture and 60% manganese ?
77. An ore contained 15% lead and 10% zinc as well as 8% of moisture.
What are the percentages of the two metals in the ore, on the dry
basis?
78. An ore contained 12% water. When dried it contained 55.5% of
metal. Find the percentage of metal in the dry ore.
79. An ore contained 1.23 grams of Fe 3 4 and 0.105 gram of water.
Calculate the percentage of magnetic oxide of iron present, on the
dry basis.
80. The dry sample of an iron ore showed the presence of 65% iron. The
natural ore contained 10% water. What was the percentage of iron
in the natural ore?
81. A load of ore is found to contain 5.5% water. It is rained upon and
found to contain 9.5% water, and to weigh 5000 Ib. Find the weight
of the original load of ore.
82. A sample of coal contained 8% of ash, on the dry basis. What is the
percentage of ash in the natural coal containing 4% of moisture?
83. On analysis, an. ore is found to consist of 47% of metal. When
partially dried it is found to contain 7% water and 51% copper.
Find the percentage of water in the original ore.
84. A sample of coal contained 1.2% sulfur and 2.4% water when partially
dried. The original sample contained 1.1% sulfur. Find the per-
centage of water in the original sample of coal,
H. CALCULATIONS BASED ON THE USE OF FACTORS
A chemical factor is that number by which the weight of a
compound must be multiplied to give the weight of the de-
sired constituent. This corresponds to the percentage of the
required substance in a given compound divided by 100.
Thus the factor of N0 2 in KN0 3 is -^2l, or 14 + 32 ,
KN0 3 39 + 14 + 48
or 0.4554. (See Table of Factors on page 156.)
Factors are often used to simplify chemical calculations.
Gravimetric Analysis 85
This is accomplished by weighing out a small multiple of the
factor for analysis. Then if x grams of a substance taken for
analysis gives 'P grams of precipitate, and if F represents the
factor of that precipitate with respect to the substance ana-
lyzed,
FP
percentage of substance sought = X 100.
x
If a reaction takes place in more than one step, consider only
the ratio between the actual weights involved.
EXAMPLE. How much of a sample of steel should an analyst take so that
when all of the phosphorus in the steel is converted into MgaPaO?, this
weight of Mg2P2Oz in grams shall equal the percentage of phosphorus?
The factor for phosphorus in MgaPsO? is equal to
2(31)
= 0.278526
Mg 2 P 2 7 24.3 X2 + 31X2 + 16X7
1 gram Mg 2 P 2 Or 0.278526 gram P ~ 1% phosphorus
(=c= stands for " is equivalent to ")
The weight of steel ~ 100% P =c= 27.8526 grams steel.
Problems
85. How many grams of zinc ore must be taken for an analysis so that
each milligram of ZnzP^Of shall be equivalent to 0.1 % zinc?
86. What weight in grams of bicarbonate of soda (NaHC0 3 ) must be
taken in order that the weight of NaoSC>4 in milligrams into which
it is converted shall be equivalent to 100% of bicarbonate?
87. How many grams of gold ore must be taken for an analysis so that
each milligram of gold shall be equivalent to 1 troy ounce per ton ?
(A ton contains 29,166 troy ounces.)
CHAPTER TWENTY-FOUR
CHEMICAL PROBLEMS FROM COLLEGE ENTRANCE
EXAMINATION BOARD PAPERS 1
1916
88. Describe in detail how you should determine experimentally
either (1) the hydrogen equivalent of some metal, or (2) the
weight of a liter of some gas.
89. If the percentages of calcium in two samples of limestone are
40% and 80% respectively, what conclusion should you draw in
regard to the specimens? State the law upon which your
conclusion is based. (Atomic weights : Ca 40, 16, C 12)
90. A body of air at constant pressure occupies a volume of 500 cc
at 20 C. At what temperature will its volume be 1000 cc. ?
91. How many cc. of ammonia gas, NH 3 , under standard con-
ditions, can be obtained from 5 grams of ammonium phos-
phate, (NH 4 ) 3 P0 4 , by treating it with sodium hydroxide in
excess? (NH 4 ) 3 P0 4 + 3NaOH = Na 3 P0 4 + 3H 2 + 3NH 3 .
(One liter of ammonia weighs 0.77 gram.)
92. Ten liters of a gas whose formula is CsHeO are completely
burned in oxygen. How many liters of oxygen gas are required
for its complete combustion? How many liters of carbon
dioxide should result? All volumes are measured under the
same conditions of temperature and pressure.
93. If the temperature of a gas enclosed at an initial pressure of one
atmosphere is decreased from 37 C. to 27 C. without change
in volume, what will be the final value of the pressure?
1917
94. Illustrate the law of multiple proportions by considering the
composition by weight of two compounds containing the same
elements.
1 From Examination Questions, Fourth Series, 1916-1920. Copyright 1916
1917, 1918, 1919, 1920, by the College Entrance Examination Board Pub-
lished by Ginn & Co., Boston and New York. Used by permission.
86
Problems from College Entrance Examinations 87
95. Excess of phosphorus is allowed to act on air at atmospheric
pressure in a sealed flask. When the reaction is complete,
(1) what is the pressure (approximately) in the flask, the
temperature remaining the same ? (2) How does the weight
of the flask and its contents after the reaction compare with
the weight before? State the law upon which the latter
conclusion is based.
96. If 50 grams of calcium carbonate, CaC0 3 , are added to
50 grams of hydrochloric acid, HC1, in water solution,
which substance remains in excess?
97. What volume of nitric oxide, NO, measured under standard
conditions, can be obtained by adding 10 grams of copper to
excess of dilute nitric acid according to the equation :
3Cu + 8HN0 3 = 3Cu(N0 3 ) 2 + 4H 2 + 2ND?
(One liter of nitric oxide at C. and 760 mm. weighs 1.34
grams.)
98. State Gay-Lussac's Law of Combining Volumes. What
volume of oxygen is required for the complete combustion of
five liters of acetylene, C^Ha? All measurements are to be
made under the same conditions of temperature and pressure.
99. One liter of gas A weighs 1.98 grams. One liter of gas B
weighs 0.77 gram, under the same conditions. The molec-
ular weight of A is 44. What is the molecular weight of B ?
1918
100. What is the percentage of ammonia in cuprammonium sulfate,
CuS0 4 4NH 3 H 2 0? (One liter of carbon dioxide at C.
and 760 mm. weighs 1.98 grams.)
101. What weight of pure sodium bicarbonate, NaHCO 3 , must be
treated with excess of acid to produce 250 liters of carbon
dioxide measured at C. and 1520 mm. pressure? (One
liter of carbon dioxide at C. and 760 mm. weighs 1.98
grams.)
102. What weight of sulfur dioxide will occupy a volume equal to
that of 42 grams of nitrogen at the same temperature and
pressure?
88 Chemical Calculations
103. What volume of a solution of hydrochloric acid containing
73 grams per liter would suffice for the exact neutralization
of the sodium hydroxide obtained by allowing 0.46 gram of
metallic sodium to act on water?
1919
104. How many grams of sodium hydroxide will be required to
neutralize 9.8 grams of sulfuric acid? Name the products
formed and determine the weight. of each product.
105. What volume will one liter of gas at 136.5 absolute have at
136.5 C., pressure remaining the same?
106 . Apply Avogadro 's Law in finding the volume of carb on dioxide
formed from the burning of 4 liters of carbon monoxide in
oxygen, provided all of the gases are measured at the same
temperature and pressure.
1980
107. Describe an experiment illustrating the quantitative char-
acter of chemical action, including apparatus, method, and
precautions. Show how the data found in the experiment
warrant the conclusion reached. The data may be repre-
sented by letters.
108. If 540 grams of silver react with an excess of concentrated
sulfuric acid according to the equation
2Ag + 2H 2 S0 4 > Ag 2 S0 4 + 2H 2 +,S0 2 ,
what weight of silver sulfate, and what volume of sulfur
dioxide measured under standard conditions of temperature
and pressure, will be formed ?
109. A dirigible balloon at sea level contains 400,000 cubic feet of
gas measured at a pressure of 774 mm. of mercury and at a
temperature of 27 C. It rises to an elevation at which the
pressure is reduced to 430 mm. and the temperature to
23 C. What is the volume of the gas under the latter
conditions, assuming that none is allowed to escape from the
containers ?
Problems from College Entrance Examinations 89
CHEMICAL PROBLEMS FROM COLLEGE ENTRANCE
EXAMINATION BOARD PAPERS
(Comprehensive Examinations)
1916
110. A certain quantity of magnesium dissolved in acid gave
exactly 100 cc. of dry hydrogen at a temperature of 22 C. and
a pressure of 780 mm. How many grams of metal were used ?
Compute the result to three significant figures. (A liter of
hydrogen at C. and 760 mm. weighs 0.09 gram.)
111. At standard temperature and pressure, one liter of gaseous
element A unites with three liters of the gaseous element B ,
to make two liters of the gas C. If each molecule of C
contains one atom of A, what is the simplest formula for a
molecule of the element A? Show clearly how you arrive at
your conclusion.
112. One gram of pure iron forms 1.43 grams of an oxide. Find
(a) the percentage composition of this oxide of iron, (6) its
simplest formula, and (c) the equivalent weight of iron in the
compound.
113. Calculate the percentage of oxygen in crystallized copper
nitrate, Cu(N0 3 ) 2 6H 2 0.
114. What weight, and volume at C. and 760 mm., of carbon
dioxide can be obtained by treating an excess of sodium acid
carbonate with 490 grams of sulfuric acid containing 20%
H 2 S0 4 ?
115. A compound has the following composition : carbon 54.67%,
hydrogen 9.11%, oxygen 36.22%. Find the simplest formula
for this substance.
116. A body of air at constant pressure occupies a volume of 500 cc.
at 20 C. At what temperature will its volume become
1000 cc.?
117. How many liters of ammonia gas, measured under standard
conditions, can be obtained when 20 grams of sodium hy-
droxide react with an excess of ammonium sulf ate ?
QO Chemical Calculations
1917
118. State the Law of Multiple Proportions and illustrate it by
the consideration of two compounds of carbon and hydrogen
containing respectively 75% and 92.3% of carbon.
119. How many grams of hydrochloric acid can be obtained by
heating 10 grams of crystallized ferric chloride, FeQ 3 6H 2 O,
with concentrated sulfuric acid?
120. What is the percentage of iron in iron alum, K 2 S0 4 Fe 2 (S0 4 ) 3
24H 2 0?
121. What volume would be occupied by 14 grams of CO under
standard conditions ?
122. If 10 liters of CO under standard conditions are heated to
27.3 C. without change in pressure, what volume will the
gas occupy?
123. What volume of oxygen is necessary to combine with 5 liters of
CO under the same conditions of temperature and pressure?
124. A certain chemical compound is found by analysis to contain
92.3% of carbon and 7.7% of hydrogen. What is the simplest
formula which can express its composition ? If the molecular
weight is 78, what is the formula?
125. Define the term "molecular weight" and give one method for
determining a molecular weight.
126. State Gay-Lussac's Law of Combining Gas Volumes, and
illustrate with an example. State the hypothesis which
serves to explain the facts generalized by this law.
127. 30 grams of chlorine are to be united with hydrogen. What
volume of hydrogen is required at C. and 760 mm., and
what volume of HC1 results under the same conditions ?
128. A balloon requires for adequate inflation 5000 gram-molecular
volumes (or an equal number of gram molecules) of hydrogen
measured at normal temperature and pressure. What is the
cost of inflating such a balloon if scrap iron at lj per kg. and
H 2 S04 containing 20% acid at 4^ per kg., are used for generat-
ing the hydrogen?
129. 1.0085 grams of pure tin foil were oxidized with nitric acid,
and the final weight of the highly ignited oxide was
Problems from College Entrance Examinations 91
1.2790 grams. From the data submitted, find the percentage
composition and the simplest formula of this oxide of tin.
1918
130. Calculate the weight of MnC>2 required to produce chlorine
enough to fill a flask of 4 liters capacity when the barometer
stands at 760 mm. and the temperature is C.
131. What weight of manganese dioxide would be required if the
flask were filled at 760 mm. and 273 C. ?
132. What is a normal solution of an acid and of an alkali?
133. 10 cc. of ordinary household ammonia are neutralized by the
addition of 40 cc. of a normal solution of HCL What weight
of ammonia gas, NH 3 , is contained in each cubic centimeter
of the original solution ?
134. 1.8 grams of magnesium displace from acid 1820 cc. of hydro-
gen, measured dry at 740 mm. and 20 C. Find the volume of
this gas under standard conditions and determine the valence
of magnesium.
135. 10 grams of crystallized ammonium carbonate, (NH 4 ) 2 C0 3
H 2 0, are heated until completely decomposed into ammonia,
carbon dioxide, and steam. What will be the total volume of
these products at 273 C. and 760 mm. pressure?
136. State Avogadro's Law and show in detail how this law guides
the chemist in determining the molecular weights of gases and
vapors.
137. 2 grams of finely divided nickel are heated with a quantity of
sulfur slightly in excess of that needed for complete com-
bination until the chemical action is completed and the excess
of sulfur is vaporized. The weight of the product formed is
3.09 grams. Find the percentage composition and simplest
formula of this compound.
1919
138. How many cc. of oxygen, measured at C. and 760 mm., can
be obtained by the complete decomposition of 1 kg. of 3%
solution of H^Oa, into water and oxygen?
92 Chemical Calculations
139. What is the percentage of silica, Si0 2 , in the mineral analcite,
NaaSiOa Al 2 (Si0 3 ) 3 2H 2 0?
140. State the Law of Multiple Proportions and give two sub-
stances to illustrate this law, with the percentage composition
in each case.
141. 100 liters of dry air at 20 C. and 760 mm. contain 0.078 gram
of carbon dioxide. What is the proportion by volume of COg
present in the air?
142. How many cc. of carbon dioxide measured at C. and
760 mm. will be absorbed by a solution containing 2 grams of
sodium hydroxide, to form sodium carbonate and water ?
143. The formula of a gaseous compound is C 2 H 2 . Calculate its
percentage composition and its vapor density.
144. Give directions for making a normal solution of sulfuric acid
and one of sodium hydroxide.
145. 10 cc. of a normal solution of sulfuric acid require 2.5 cc. of a
solution of NaOH for complete neutralization. Find the
concentration of the solution of NaOH in grams per liter.
146. Calculate the molecular weight of carbon dioxide gas from the
following data, showing every step in the calculation :
Weight of flask filled with C0 2 , dry . 102.38 grams
Weight of flask filled with air, dry . 101.56 grams
Temperature of measured gases . . C.
Pressure of measured gases .... 770 mm.
Volume of flask 1200 cc.
One liter of air weighs 1.29 grams at C. and 760 mm.
147. 5 liters of CO are mixed with just enough oxygen for complete
combustion and the mixture is ignited. If the original gases
are measured at 20 C., at what temperature will the product
occupy the same volume, the pressure remaining the same?
1920
148. How much cream of tartar (HKC 4 H 4 6 ) should be mixed with
1 kg. of baking soda (NaHC0 3 ) so that neither ingredient will
be in excess?
149. What volume of carbon dioxide at C. and 760 mm. may be
obtained from the baking powder thus made?
Problems from College Entrance Examinations 93
150. What is the relation between volume and pressure of a gas
when the temperature is constant? between volume and
temperature when the pressure is constant? between tem-
perature and pressure when volume is constant?
151. From the standpoint of the molecular hypothesis, explain the
fact that (1) 40 gallons of oxygen may be forced into a small
steel cylinder, (2) gas escapes from a glass of soda water,
(3) chlorine gas, although heavier than air, escapes from an
upright vessel.
152. Tell how you would determine experimentally the equivalent
of some element or the molecular weight of some gas.
153. A portable gas stove used in heating a room burns natural gas
(assumed to be pure methane, CH 4 ) and consumes in one
hour 10 cubic feet of gas measured at a pressure of 30 inches
of mercury and a temperature of 60 F. What volume of
oxygen is consumed and what space is occupied by the carbon
dioxide which is produced, when measured under conditions
of temperature and pressure previously specified ?
154. 2 gram molecules of the oxide RO combine with 44.8 liters of
oxygen at C. and 380 mm. The volume of the product
under the same conditions is nearly equal to that of the oxygen
taken. Calculate the formula of the product.
155. 50 cc. of a solution containing 40 grams of NaOH per liter
neutralize 40 cc. of a solution of HC1. What is the con-
centration of the acid solution in grams per liter?
156. How many grams of potash alum, K2S04 A^SO^s 24HaO,
can be made from 69 grams of bauxite (AUOs 2H2O) ?
157. A liter of a certain elementary gas weighs 7.5 grams, while a
liter of hydrogen gas under the same conditions weighs 0.09
gram. The atomic weight of the element composing the gas
is 75. How many atoms are there in a molecule of the gas ?
158. A certain weight of potassium carbonate, heated with 10
grams of sand, gives 25.6 grams of potassium silicate accord-
ing to the equation K 2 C0 3 + Si0 2 >- K 2 Si0 3 + C0 2 .
Calculate the weight of the potassium carbonate.
159. What is the percentage of sulfur trioxide in white vitriol
(ZnS0 4 -7H 2 0)?
94
Chemical Calculations
160. Assume x to be a gaseous element having a valence of 3 and
containing 2 atoms per molecule. To make 2 volumes of its
gaseous compound with hydrogen, how many volumes of each
constituent are required? State the law upon which your,
conclusion is based.
PART THREE
PROBLEMS BASED ON LESSON ASSIGNMENTS, WITH" THE
PRINCIPAL EQUATIONS UNDER EACH TOPIC
1. OXYGEN AND OZONE
Lab. Prep. <
Com. Prep. <
Chem. Prop.
f 2HgO
.2KC10 3
2H 2
3O 2 >
I air)
4P+50 2 -
oxides)
,2Ag + 20 3
->- 2Hg 4- -f 2 (Decomposition of mercuric oxide)
> 2KC1 + <f 3O 2 (Decomposition of KC10 3 )
-> 2H 2 + O 2 (Electrolysis of water)
20 3 (Passage of a silent electric discharge through
^PjOs (Unites with non-metals to form
>- Ag 2 2 + 20 2 (Very strong oxidizing agent)
1. Calculate the percentage composition of Ca(C10s)2-
2. How much mercury and oxygen can be obtained from 108 grams of
mercuric oxide?
3. What weight of oxygen can be obtained from 245 Ib. of potassium
chlorate?
4. How many grams of cupric oxide (CuO) could be obtained by heating
1 gram of copper in the air?
5. What weight of carbon dioxide is formed by burning 240 Ib. of coal
containing 11% impurities?
6. A certain weight of potassium chlorate was heated. The KC1 formed
weighed 149 grams. Wha^weight of KClOa was used?
7. How much mercury will remain after obtaining 160 grams of oxygen
from mercuric oxide?
8. What is the vapor density of ozone if 325 cc. weigh 0.702 gram?
9. 1000 liters of pure oxygen are required. What weight of water
would have to be decomposed?
10. How many liters of oxygen would be required to burn 6.2 grams of
phosphorus completely?
11. What volume of ozone can be obtained from 32 grams of oxygen?
12. 112 grams of iron gave 160 grams of Fe 2 3 . What is the atomic
weight of oxygen?
95
Lab. Prep.
2. HYDROGEN
Zn + H 2 S0 4 *- | H 2 + ZnS0 4 (Action of a metal on
an acid)
2Na + 2H 2 > 2NaOH + 1 H 2 (Action of sodium on
water)
f 2H 2 - > 2H 2 + 2 (Electrolysis of water)
Com. Prep. j 3Fe + 4H 2 - > Fe 3 O 4 + |4H 2 (Action of steam on hot
[ iron)
D, / H 2 + C1 2 - *" 2HC1 ( Stron S affinity for chlorine)
Chem. Prop. - Cu + H20 (Redudng agent)
13. Calculate the percentage of hydrogen present in microcosmic salt,
HNaNH 4 P0 4 4H 2 0.
14. How many liters of steam result from the burning of 1500 cc. of
hydrogen?
15. How many grams of zinc would be required to replace all the hydro-
gen from 49 grams of sulf uric acid ? What weight of ZnS04 would
be formed?
16. 6.5 grams of zinc react with an excess of dilute hydrochloric acid.
What weights of the products will be formed?
17. A balloon holds 200 kg. of hydrogen when filled. What weights of
zinc and sulfuric acid are necessary to furnish sufficient gas to fill
this balloon?
18. Calculate the weight of steam that could be reduced by a ton of red-
hot coke containing 95% pure carbon.
19. What weight of hydrogen must unite with 500 cc. of oxygen and
leave no oxygen free?
20. By the action of 69 grams of sodium on water, how many liters of
gas will be formed when measured over water at 20 C. and 765 mm.,
the height of the water outside the tube being 13 cm. above the
height of the water inside the measuring tube ?
21. Which will produce more hydrogen, the action on water of 46
grams of sodium or 78 grams of potassium?
22. Calculate the hydrogen equivalent of cadmium, if 0.112 gram of
cadmium released 24.4 cc. of hydrogen at 20 C. and 740 mm.
23. A certain weight of copper oxide when heated in a current of hydro-
gen lost 60 grams of oxygen and formed 67.5 grams of water. What
is the atomic weight of hydrogen ?
24. 500 cc. of a gas weigh 0.38 gram. It is composed of hydrogen
and nitrogen, the hydrogen weighing -j^- of the whole. What is the
true formula of this gas?
96
3. WATER AND HYDROGEN PEROXIDE
2Hg + 02 > 2H2O (Passage of electric spark through
. p , mixture of O 2 and H 2 )
P- ^ Ba0 2 + H 2 S0 4 ^|BaSO 4 + H 2 2 (Action of barium
peroxide and sulfuric acid)
CuS0 4 + 5H 2 >- CuSO* 5H 2 (Water of crystalliza-
tion)
Chem. Prop.
CaO + H 2 * Ca(OH) 2
CwHsiOn > 12C + HHjO
H 2 2 *- H 2 + -f O (Instability of H 2 2 )
25. Calculate the percentage composition of copperas, FeS04 7H 2 0.
26. What per cent of water does alum, KA1(S04) 2 ' 12H20, contain?
27. The weight of 1 liter of water vapor was found to be equal to 0.81
gram. Calculate its specific gravity, and molecular weight.
28. Calculate the weight of 70% sulfuric acid necessary to react with
16.9 grams of barium peroxide to produce pure hydrogen peroxide.
29. How many cubic centimeters of free oxygen are liberated when 84.5
grams of pure hydrogen peroxide decompose?
30. The following results were obtained in the gravimetric determination
of water: Loss of weight of CuO 10.83 grams, weight of water
formed 12.2 grams. Calculate the ratio in which hydrogen and
oxygen combined.
31. How many liters of steam result by burning 5 liters of hydrogen?
32. 15 cc. of oxygen collected in a eudiometer measured over mercury,
were mixed with hydrogen gas until the mixture measured 22.4 cc.
The gases were then exploded by means of an electric spark. What
gas was left, and what was the volume of the resulting gas?
33. How many grams of Sr(N0 3 ) 2 5H 2 O may be obtained from 100
grams of a mineral containing only 55% of SrC0 3 ?
34. Find the formula of crystallized copper sulfate, 7.84 grams of which
lost 2.79 grams of water after being heated.
35. 70 grams of a solution of hydrogen peroxide on decomposing gave 2.6
liters of oxygen. What was the percentage of TI^Oz in the solution?
36. In one of Dumas' experiments on the composition of water, the
following data were obtained :
Weight of CuO + tube before experiment . . 334.596 g.
Weight of CuO + tube after experiment . . 314.236 g.
Weight of drying tube before experiment . . 426.358 g.
Weight of drying tube after experiment . . . 449.268 g.
Determine the composition of water by weight.
97
Lab. Prep.
Com. Prep.
Chem. Prop.
4. CHLORINE
4HC1 + Mn0 2 - *- MnCl 2 -f 2H 2 + | C1 2 (Oxidation
of HC1)
2KC1 + Mn0 2 + 2H 2 S0 4 - >- K 2 S0 4 + MnS0 4 + 2H 2
2NaCl + 2H 2 - > 2NaOH + j H 2 + \. C1 2 (Elec-
trolysis of brine)
Cu + Cla - >- CuCl 2 (Unites with metals to form chlo-
rides)
2FeCl 2 + C1 2 - > 2FeCl 3 (Oxidizing agent)
H 2 + C1 2 - > 2HC1 4-^0 (Liberates nascent from
water)
37. Find the percentage of chlorine present in bleaching powder con-
taining 90% of CaOCl 2 .
38. How many grams of chlorine can be made by the use of 174 grams
of manganese dioxide?
39. How many grams of manganese chloride are formed in the above
reaction?
40. A load of tin cans was treated with chlorine and 26.1 Ib. of stannic
chloride, SnCU, were obtained. How much tin was present in the
tin cans?
41. The vapor density of chlorine .was found to be 35.5. Determine
its specific gravity, weight of a liter, and molecular weight.
42. What weight of sodium chloride must be decomposed to liberate
100 liters of chlorine gas?
43. How many cubic centimeters of oxygen will be liberated when a
volume of chlorine water, containing 1 gram of chlorine in solu-
tion, is completely decomposed?
44. What volume of chlorine can be obtained by the electrolysis of a
brine solution containing 5 kilograms of sodium chloride?
45. If 261 grams of pyrolusite (91% pure MnOa) are treated with an
excess of hydrochloric acid, what is the volume of the gas liberated?
46. 5 liters of chlorine are burned in a sufficient volume of hydrogen
gas. What volume of hydrogen chloride gas is produced ?
47. 91 grams of silver when heated in chlorine yield 121 grams of silver
chloride. What is the atomic weight of chlorine ?
48. A chemist determined the atomic weight of chlorine found in NHiCl
contained in some mine water by finding the ratio of Ag to AgCl.
The following figures were obtained : 0.8 gram of silver gave 1.06
grams of AgCl. What was the atomic weight of the chlorine found?
5. HYDROGEN CHLORIDE AND HYDROCHLORIC ACID
f Nad + H 2 S0 4 >- NaHSO* + ^HCl (Action of H 2 SO 4
Lab. Prep. < on a chloride)
I H 2 + C1 2 > 2HC1 (Direct union of the elements)
Com. Prep. 2NaCl + H 2 SO 4 > Na 2 S0 4 + -f 2HC1
Zn + 2HC1 >-|H 2 + ZnCl 2 (Reacts with some metals
lib era ting .H 2 )
. _. , ZnO + 2HC1 >- ZnCl 2 + H 2 O (Forms water with
Cnem. Prop. < . ,,. . , .
metallic oxides)
Nad + AgN0 3 >- NaN0 3 + f AgCl (Test for a
chloride)
49. The vapor density of hydrogen chloride gas was found to be 18.24.
Determine the weight of a liter of this compound as well as its spe-
cific gravity (air standard).
50. 164 cc. of hydrogen chloride gas under standard conditions weighed
0.269 gram. Find (a) its vapor density, and (i) the weight of 5 liters
of this gas.
51. Find the number of grams of hydrogen chloride set free by the com-
plete reaction between 32 grams of sulfuric acid and sufficient sodium
chloride.
52. How many grams of potassium chloride and sulfuric acid (65% by
weight) are necessary to produce 2.8 grams of hydrogen chloride ?
53. 28.4 grams of sodium sulfate are found after sulfuric acid reacts with
sodium chloride. How much of the latter was used?
54. A 25% hydrochloric acid solution has a specific gravity of 1.125.
Determine the weight of such an acid necessary to dissolve com-
pletely 56 grams of iron to form ferrous chloride, FeCl2.
55. 50 cc. of hydrogen are liberated when hydrochloric acid reacts with
a metal. What weight of acid was decomposed? "
56. 143.5 grams of silver chloride are precipitated out of a potassium
chloride solution by means of silver nitrate. How much KC1
reacted with the silver nitrate?
57. What volumes of hydrogen and chlorine will be necessary to prepare
5.5 liters of hydrogen chloride gas by direct union of the elements?
58. 20 cc. of chlorine gas and 25 cc. of hydrogen gas are placed in a
measuring tube and electric sparks passed through the mixture.
What volume of hydrogen chloride can be theoretically obtained?
What volume of any other gas will be found in the tube ?
59. Calculate the volume of hydrochloric acid, containing 20% HC1 by
weight (Sp. Gr., 1.1), necessary to neutralize 37.37 grams of calcium
hydroxide.
99
6. ALKALI METALS, BASES, AND NEUTRALIZATION
Com. Prep. 2NaOH > 2Na + | H 2 + | O 2 (Electrolysis of fused
NaOH)
[ 2K + 2H 2 >-2KOH -f- -f Ho (Liberates hydrogen from
water)
Chem. Prop. 2Na -f- 2 *- Na 2 O 2 (Unites with oxygen readily)
2Li -f- Cla - 2LiCl (Reacts -with chlorine vigorously)
NaOH + HC1 *- NaCl + H 2 (Neutralized by acids)
60. What percentage of sodium and nitrogen does sodium cyanide
(NaCN) contain?
61. What weight of metallic sodium can be obtained by the electrolysis
of SO Ib. of fused sodium hydroxide containing 2% impurities?
62. A gram of rock gave on analysis 0.113 gram of K 2 PtCl 6 . Find the
percentage of potassium available in this ore.
63. Which will require more water to dissolve completely: 45 grams
of sodium or 79 grams of potassium?
64. What weight of potassium chloride can be obtained by reacting
500 cc. of chlorine with sufficient potassium? How much potassium
will be required ?
65. 10 grams of a mixture of NaCl and KC1 yielded 20.5 grams of AgCl.
Determine the percentage of each salt in the mixture.
66. A piece of sodium is thrown on water. The sodium hydroxide
formed just neutralized 98 grams of sulfuric acid. What weight of
sodium was used?
67. A certain weight of sodium hydroxide was neutralized with sulfuric
acid. The solution on evaporation yielded 15.5 grams of sodium
sulf ate. What weight of caustic soda was neutralized ?
68. Calculate the volumes of hydrogen and oxygen produced during the
electrolysis of 112 grams of fused potassium hydroxide. What
weight of potassium would be formed at the same time?
69. Calculate the formula of a compound having the following compo-
sition : potassium, 16% ; platinum, 40.4% ; and chlorine, 43.6%.
70. What volume of chlorine will be required to change completely
70 grams of h'thium to its chloride? What weight of lithium chlor-
ide will be formed ?
71. Caustic potash costs 7% per pound and contains 10% of water.
Caustic soda sells at $3.25 per 100 Ib. and contains 24% of water.
Which would be cheaper to use in neutralizing a given weight of
sulfuric acid?
7. SODIUM SALTS
f NaCl + NH 4 OH + CO 2 ^ NaHCO 3 + NH 4 C1 (Solvay
Com. Prep. < process for the manufacture of Na 2 C0 3 )
l2NaHCO 3 >- Na 2 C0 3 + H 2 + 1 C0 2
Lab. Prep. 2NaOH + H 2 + CO 2 > Na 2 C0 3 + 2H 2
Na 2 CO 3 + 2HC1 >- 2NaCl + H 2 -f- 1 C0 2 (Decom-
posed by acids liberating carbon dioxide)
Chem. Prop. Na 2 C0 3 10H 2 + C0 2 >- 2NaHCO 3 + 9H 2
NaHCO. + KHC 4 H 4 O 6 ^ KNaC 4 H 4 6 + H 2 +4 C0 2
(Action of baking powder)
72. How much sodium carbonate can be obtained from a ton of sodium
bicarbonate?
73. Sodium bicarbonate sells at $2 per 100 lb., and sodium carbonate
at $1.30 per 100 lb. Which would be cheaper to use in the prepara-
tion of 5 cu. ft. of carbon dioxide ?
74. How many grams of sulfuric acid will be needed to liberate all the
carbon dioxide from 22 grams of washing soda?
75. A fire extinguisher holds 2.5 lb. of sodium bicarbonate in solution.
What weight of sulfuric acid will be required to liberate all of its
carbon dioxide?
76. 2000 lb. of- sodium bicarbonate are available. What weight of potas-
sium acid tartrate must be used to make enough baking powder
to use all of this available bicarbonate ?
77. What volume of carbon dioxide will be liberated when hydrochloric
acid reacts with 42 grams of sodium bicarbonate ?
78. 256 grams of sodium bicarbonate are heated. What weight of
sodium carbonate and what volume of carbon dioxide are formed?
79. 4.3 grams of Rochelle salts (KNaC-^Oe) were found present in a
cake after baking. What weight of baking powder was used in the
recipe?
80. Equal weights being taken, which will neutralize a greater amount
of acid, sodium carbonate or sodium bicarbonate?
81. Hydrochloric acid completely neutralized potassium acid carbonate
(KHCOs), and 50 cc. of carbon dioxide were liberated. What
weight of the bicarbonate was neutralised?.
82. 8 grams of crystallized ?
being heated. Derive
83. What weight of baking
bicarbonate and 282 lh
8. SULFUR AND SULFEDES
5 -f- Oa > S0 (Unites with non-metals to form sulfides)
Cu + S >- CuS (Unites with metals to form sulfides)
Chem. Prop. { 2S + C1 2 *- S 2 C1 2 (Preparation of sulfur chloride)
4S + 6NaOH + 2H 2 >- Na 2 S 2 3 5H 2 + 2Na 2 S
I (Prep, of " hypo ")
84. What is the percentage of sulfur and carbon present in carbon bi-
sulfide CS 2 ?
85. It is required to prepare 100 tons of pure hypo. What weight of
sulfur is needed to prepare it from sodium hydroxide ?
86. What weight of sulfur can be obtained from an ore weighing
120 Ib. and containing 87% of FeS? What weight of iron could
this ore furnish ?
87. How many pounds of sulfuric acid can be manufactured from 160
tons of pure sulfur?
88. 95 grams of copper sulfide were obtained by burning 100 grams of
copper in sulfur vapor. How much free copper remained after the
reaction was completed?
89. On analysis, IS grams of a compound of iron and sulfur yielded
8.4 grams of iron and the rest of sulfur. ' What is the simplest for-
mula of this compound?
90. 2 grams of silver sulfide were found on some silverware. How much
of the silver was tarnished?
91. With copper selling at 14 J^ per pound and sulfur at $20 per ton what
would be the cost of 4000 Ib. of copper sulfide, assuming that the
cost, excluding chemicals, is 2j per pound?
92. How many liters of sulfur dioxide can be formed by burning one
kg. of sulfur ore containing 95% of free sulfur? What would be
the weight of this oxide?
93. What volume of chlorine will be necessary to prepare 67.5 grams
of sulfur chloride?
94. What volumes of sulfur vapor and chlorine gas will be formed when
5 liters of sulfur chloride vapor are completely decomposed into its
elements?
95. Assuming that sulfur vapor contains 6 atoms to the molecule, what
volume of sulfur vapor will be formed when 116 grams of mercuric
sulfide are completely decomposed?
9. HYDROGEN SULFIDE
Lab. Prep. FeS + 2HC1 > FeCl 2 + | H 2 S (Action of an acid on
a sulfide)
' 2H 2 S + 30 2 - 2H 2 + 2SO 2 (Complete combustion
of H 2 S)
2H 2 S + O 2 >- 2H 2 O + 1 2S (Incomplete combustion
of H 2 S)
Chem. Prop. { 2Ag + H 2 S >- Ag 2 S + H 2 (Tarnishing of silver)
CuSO 4 + H 2 S J-|CuS + H 2 S0 4 (Precipitates metals
from their salts)
Pb(C 2 H 3 2 )2 + H 2 S -f PbS + 2HC 2 H 3 O 2 (Test for
a sulfide)
96. The weight of a liter of hydrogen sulfide is 1.54 grams. Find its
vapor density, specific gravity, and molecular weight.
97. Which compound contains more sulfur, arsenic sulfide (AsaSs) or
mercuric sulfide (HgS) ?
98. A compound of hydrogen and sulfur has a molecular weight of 34.
The percentage of sulfur in the compound is 94.11. What is the
true formula of this compound?
99. What weight of hydrogen sulfide gas reacted with a copper sulfate
solution and precipitated 32 grams of copper sulfide ?
100. What was the volume of this hydrogen sulfide gas when measured
over water at 17 C. and 765 mm. ?
101. A solution containing hydrogen sulfide produced 0.2 gram of lead
sulfide when mixed with a lead acetate solution. What was the
weight of HaS present in the solution?
102. How much more lead sulfide will 17 grams of hydrogen sulfide
precipitate from a lead nitrate solution, than an equal weight
of ammonium sulfide?
103. How much H 2 S will be required to combine with all of the zinc in
a solution containing 6.8 grams of zinc chloride? What weight of
ZnS will be formed?
104. How many grams of ferrous sulfide must be used to prepare 5 liters
of hydrogen sulfide gas?
105. What weight of ferrous chloride would be produced in the above
reaction?
106. What volume of HaS is necessary for the complete precipitation of
40 grams of copper sulfate in solution? What will be the weight
of thisH 2 S?
103
Lab. Prep.
10. SULFUR DIOXIDE
S + Q 2 - SO 2 (Burning of sulfur)
Na 2 SO + H 2 SO 4 > Na 2 S0 4 + H 2 + | SO 2 (Acid on
a sulfite)
Cu + 2H 2 S0 4 > CuS0 4 + 2H 2 O + ^ SO 2 (Cone.
H 2 S0 4 on copper)
Com. Prep. 4FeS 2 + 110 2 ^2Fe 2 3 + |8S0 2 (Burning of iron
sulfide)
( S0 2 +H 2 O - H 2 S0 3 (Acid anhydride of sulfurous acid)
Chem. Prop. | 2KMn0 4 + 5H 2 S0 3 >- K 2 S0 4 + 2MnS0 4 + 2H 2 S0 4
1 + 3H 2 (Reducing agent)
107. The weight of a liter of sulfur dioxide is 2.86 grams. Calculate
its vapor density, specific gravity, molecular weight, and the weight
of 550 cc. of the gas.
108. How much sodium sulfite must react with sulfuric acid to produce
128 grams of sulfur dioxide?
109. What volume of sulfur vapor containing 4 atoms to the molecule
would be used in preparing 2.5 liters of sulfur dioxide gas?
110. What weight of sulfur dioxide is formed by burning 60% of a lot of
sulfur weighing 760 kilograms?
111. How much oxygen would be required to oxidize completely 4 1 grams
of sulfurous acid?
112. What weight of copper must react with sulfuric acid to produce 11.1
liters of sulfur dioxide?
113. What volume of sulfur dioxide, measured under standard condi-
tions, can be obtained by burning 1200 kilograms of iron pyrites?
114. What weight of sulfurous acid can be obtained from 5 liters of its
anhydride?
115. An unknown weight of potassium sulfite is acted upon by sulfuric
acid and 55 cc. of SO 2 is liberated. What weight of the sulfite
was decomposed?
116. What volume of air unites with sulfur when the latter is burned
to form 100 cc. of sulfur dioxide?
117. It is required to prepare 3.1 Ib. of liquid sulfur dioxide. What
volume of air (containing 20% oxygen by volume) will be required
to synthesize this weight of sulfur dioxide?
118. 500 cu. ft. of sulfur dioxide were liberated during the burning of
a load of FeS 2 . What weight of this iron compound was roasted
during the process?
104
11. SULFUR TRIOXIDE AND SULFHRIC ACID
2SO 2 + 2 > 2SO 3 (In the presence of Pt catalyst)
Com. Prep. HzSQt + SQj ^ ^SO* S0 3 (Fuming sulfuric acid)
(Contact H 2 so 4 S0 3 + H 2 >- 2H 2 S0 4 (Concentrated sul-
process)
furic acid)
f HoO + 2SO 2 + N 2 3 +0 2 >- 2SCvOH-ONO (Nitrosyl
(Chamber ir -j\
i sulfuric acid)
process) ^ 2SQ2 _ QH _ ONO + ^ ^ 2H 2 S0 4 + N 2 3
f Cu + 2H 2 SO 4 ^ CuSO 4 + 2H 2 O + S0 2 (Action of con-
centrated sulfuric acid on copper)
Chem. Prop ' C 12 H 22 U >- 11H 2 + 12C (Dehydrating action)
MnS0 4 +BaCl 2 > ^ BaSO 4 +MnCl 2 (Test for a sulfate)
119. How much sulfur trioxide will be needed to change 490 Ib. of sulfuric
acid to fuming sulfuric acid?
120. What weight of fuming sulfuric acid can be obtained from 196
grams of sulfuric acid?
121. Calculate the weight of barium sulfate which is formed when a
solution containing 52 grams of anhydrous barium chloride is
treated with sufficient sulfuric acid.
122. How much sodium chloride must react with sulfuric acid to pro-
duce 42.6 grams of dry sodium sulfate?
123. What weight of slaked lime can be neutralized by 98 grams of sul-
furic acid?
124. How much zinc sulfate can be formed from 81 grams of zinc oxide?
125. Calculate how many grams of silver sulfate and of copper sulfate
could be made from a dime weighing 2| grams, and containing
10% of copper.
126. The manufacture of sulfuric acid from iron pyrites may be repre-
sented by the equation 4FeS 2 + 8H 2 + 150 2 >- 2Fe 2 3 +
8H 2 S04. How much iron pyrites would be required to make 1000
kg. of sulfuric acid?
127. What volume of barium chloride solution containing 61 grams of
BaCla 2H 2 per liter will be required to precipitate all of the
sulfate in 10 grams of crystallized ferrous sulfate?
128. How much 68% sulfuric acid (specific gravity = 1.6) can be ob-
tained from one ton of sulfur?
129. How many tons of pyrite containing 40% of available sulfur must
be used to make 1000 tons of sulfuric acid, of specific gravity 1.7?
130. How many cubic feet of air are necessary for the conversion of
1000 cubic feet of sulfur dioxide gas to sulfur trioxide?
105
12. NITROGEN AND THE AIR
f 4P + air (5O 2 ) - > 2P 2 5 + nitrogen (Impure)
. wrep. | NH4NOs - ^ 2H,0 + N 8 (Pure)
Chem. Prop.
3Mg + N a
nitrides)
N 2 + 3H 2 >
2NaOH + C0 2 -
on C0 2 in air)
CaCl 2 + 2 H 2 -
H 2 in air)
>- Mg 3 N 2 (Unites with metals to form
2NHs (Unites with H 2 to form ammonia)
> Na 2 C0 3 + H 2 O (Action of NaOH
> CaCl 2 2H 2 (Action of CaCl 2 on
131. What is the percentage of nitrogen in sodium nitrate, NaNOs?
132. The weight of a liter of helium is 0.1S gram. Find its vapor den-
sity, specific gravity, molecular weight, and the weight of 5 liters
of the gas.
133. Lord Rayleigh, in one of his experiments, obtained 66.5 cc. of argon
from 8 liters of air. What is the percentage of argon by volume in
the air?
134. Find the true formula of a gas containing 30.4% of nitrogen and
the rest of oxygen. A liter of this gas weighs 2.06 grams. .
135. If a man inhales 19 cu. ft. of air per hour, what weight of oxygen
does he require per day? (Assume that the air contains 20%
oxygen by volume. 1 cu. ft. of air weighs 36.5 grams.)
136. Compute the molecular weight of nitrogen from the f ollowing data :
222.5 cc. of nitrogen at 12 C. and 760 mm. were found to weigh
0.26 gram.
137. Calculate the weight of air needed to burn 64 grams of sulfur, if
the air contained 23.2% of oxygen by weight.
138. What weight of ammonium nitrite must be used to prepare 70 cc. of
nitrogen under standard conditions?
139. 500 cc. of nitrogen gas completely unites with burning magnesium.
What weight of magnesium nitride is formed?
140. 167 cc. of nitrogen weigh 0.21 gram. Determine the number of
atoms in the molecule of nitrogen.
141. What weight of ammonium nitrite would have to be decomposed to
furnish 25 cc. of nitrogen gas when measured over water at 17 C.
and 760 mm-?
1 06
13. AMMONIA AND AMMONIUM HYDROXIDE
Lab. Prep. (NH 4 ) 2 S0 4 + Ca(OH) 2 - >- CaSO 4 + 1 2NH 3 + 2H 2 O
f N a + 3H 2 - =- 2NH 3 (Haber process)
Com. Prep, i CaC 2 + N 2 - > CaCN 2 + C \ (Cyanamid
{ CaCN 2 + 3H 2 O . > CaCO 3 + | 2NH 3 / process)
' / NH 3 + H 2 - *- NH 4 OH (Its water solution is a base)
Cfcem. .Prop, + ^ - ^ 4NQ + & ^ Q ( Ox idation of ammonia)
142. The specific gravity of ammonia is 0.597. Find its vapor density,
molecular weight, and the weight of 1 liter.
143. How many grams of ammonia gas can be obtained from 107 grams
of ammonium chloride by heating it with slaked lime?
144. How much oxygen will be required for the complete oxidation of
51 grams of ammonia?
145. 10 grams each of HBr and NH 3 are mixed. What weight of NH4Br
results? What is left over, and what is its volume, measured
under standard conditions?
146. 684 grams of ammonium sulfate are heated with sodium hydroxide.
What weight of ammonia escapes? What volume?
147. 20 grams of calcium cyanamid are treated with water. How much
ammonia, by volume, is produced?
148. Under standard conditions, how many liters of ammonia can be pre-
pared by heating a mixture of 11.4 grams of ammonium sulfate
with calcium hydroxide?
149. What weight of ammonium chloride will be formed when 1 liter
of ammonia gas comes in contact with 1 liter of hydrogen chloride
gas?
150. What weight of pure calcium cyanamid will be needed to prepare
10 liters of ammonia gas ?
151. What weight of air, containing 75% of nitrogen by weight, would
have to be liquefied to furnish sufficient nitrogen to make 1000
liters of ammonia gas by the Haber process?
152. What volume of nitrogen would be absorbed by 80 grams of calcium
carbide in changing the carbide into calcium cyanamid? What
weight of carbon will be formed ?
153. 100 cc. of dry ammonia are decomposed by electricity. Find the
volume of each of the gases formed when measured under standard
conditions.
107
14. OXIDES OF NITROGEN
Lab. Prep.
Chem. Prop.
2H 2 0+<f N 2 (Laughing gas, nitrous
NH 4 N0 3 -
oxide)
3Cu + SHNO 3 > 3Cu(NO 3 ) 2 -f- 4H 2 O + 2NO (Nitric
oxide)
2NO + O 2
to air)
N 2 3 + HjO
HN0 2 )
-*- 2HNOs (N 2 0s is the acid anhydride of
- 2N0 2 ' (NO changes to N0 2 on exposure
- 2HN0 2 (N 2 3 is the acid anhydride of
N 2 5 + H 2
HN0 3 )
2N0 2 + H 2
- HN0 2 + HNO 3
154. The weight of a liter of nitric oxide is 1.32 grams. Find its vapor
density, specific gravity, and molecular weight.
155. 290 cc. of nitrous oxide weigh 0.574 gram. Calculate the vapor
density, specific gravity, and weight of 1 liter of this gas.
156. What weight of the anhydride can be theoretically obtained from
21 grams of nitric acid?
157. What weight of water will react with 27 Ib. of nitric anhydride to
form nitric acid?
158. 90 grams of nitric oxide are obtained by the action of copper on
nitric acid. How much copper nitrate was formed at the same
time?
159. How much nitrous acid (10% by weight) can be made from 19 grams
of nitrogen .trioxide?
160. What weight of ammonium nitrate must be . heated to produce
5 liters of laughing gas ?
161. 5 grams of copper are completely dissolved in nitric acid. What
volume of nitric oxide is liberated, the gas being measured at
standard conditions?
162. 53 cc. of nitric oxide are liberated in a reaction and the gas is then
exposed to the air. What volume of oxygen combined with the
nitric oxide, and what weight of nitrogen peroxide was. produced?
163. What weight of ammonium nitrate must be decomposed to fill a
balloon of 100 liters' capacity when measured at 25 C. and 760 mm.
pressure, with laughing gas?
164. How many cubic centimeters of air are necessary to change com-
pletely 65 cc. of nitric oxide to the peroxide?
165. 5 liters of laughing gas are decomposed into its elements. What
volumes of the products are formed?
108
15. NITRIC ACID AND AQUA KEGIA
Lab. Prep. NaN0 3 + H 2 S0 4 >- NaHS0 4 -f- HN0 3 (Action of
HaSO4 on a nitrate)
Com. Prep. 2NaNO s + H 2 S0 4 > Na 3 S0 4 + 2HN0 3 (Same as lab.
prep.)
Cu + 4HNO 3 > Cu(N0 3 ) 2 + 2H 2 + 2NO 2 (Cone.
HNOs used)
3Cu + 8HN0 3 >- 3Cu(NO,) a + 2NO + 4H 2 (Cold
Chem. Prop. <
dil. HNO 3 used)
3HC1 + HN0 3 ^2H 2 + NO + 3C1 (Action of aqua
regia)
. Au + 3C1 - AuCls (Action of aqua regia on gold)
166. How many grams of nitric acid can be obtained by heating
425 grams of sodium nitrate with sulfuric acid?
167. What weight of gold will be dissolved by aqua regia capable of fur-
nishing 142 grams of nascent chlorine?
168. Chile saltpeter, containing 12% impurities, is treated with 49 Ib.
of sulfuric acid. What weight of nitric acid will be produced?
169. 20 grams of sodium bisulfate (NaHSO^ were produced by the
action of sodium nitrate and sulfuric acid. What weight of nitric
acid was produced at the same time ?
170. 189 grams of concentrated nitric acid reacted with an excess of
copper. What volume of nitrogen peroxide was liberated?
171. What weight of 65% nitric acid should be taken just to dissolve
27 grams of silver?
172. What volume of nitrogen peroxide can be obtained by the action
of 32 grams of copper on concentrated nitric acid?
173. What weight of commercial potassium hydroxide, containing 10%
water, can be neutralized by 105 grams of 50% HNOs (Sp. Gr.,
1.315)?
174. Gold is dissolved in aqua regia and 5.3 grams of gold chloride
are formed. How much chlorine, by volume, was furnished by the
aqua regia?
175. 2.7 grams of silver are dissolved in nitric acid. The resulting silver
nitrate is treated with a solution of sodium chloride. What weight
of silver chloride is precipitated?
176. Nitric acid, containing 50% by weight of HNOs, is sold for $5
per 100 Ib. What would be the cost of neutralizing 5 tons of slaked
lime (Ca(OH) 2 ) with this acid?
177. What weight of aqua regia will be needed to dissolve completely
5.3 grams of gold? Assume the use of 100% acids.
109
16. NITRATES AND NITROGEN FIXATION
- 2NO ]
2NO 2 \ (Arc process)
n- x- KT J 2 S - > HN 2 + HN0 3 J
Com. Fixation of N : \ ~ , , T _ r p, M , ~ w
CaC 2 + N 2 - > CaCN 2 + C \(Cyanamid
CaCN 2 +3H 2 - ^2NH 3 + CaCOs/ process)
N 2 + 3H 2 - >- 2NH 3 (Haber process)
f NaN0 3 + KC1 - >- Nad + KN0 3 (Prep, of KNO 3 )
Chem. Prop. I 2KNO 3 + 3C + S - >- K 2 S + 3C0 2 + N 2 (Explosion of
[ gunpowder)
178. Calculate the percentage of nitrogen present in nitroglycerine, whose
formula is Cs
179. What weight of saltpeter (KNO 3 ) can be formed from 170 Ib. of
impure Chile saltpeter containing 90% NaNOs ?
ISO. 1.49 grams of potassium chloride in solution are treated with a
solution of sodium nitrate. What weight of potassium nitrate is
theoretically possible?
181. What weight of ammonia can be produced from 1 kilogram of
cyanamid containing only 90% CaCN2?
182. 50.5 Ib. of KNOs are used in the making of gunpowder. What
weights of sulfur and carbon must be added to the potassium
nitrate?
183. The gunpowder in the above problem was exploded. The tempera-
ture attained by the reaction was 2000 C. What volume of C0 2
was liberated?
184. A volume of nitric oxide is exposed to the air and 52 cc. of nitrogen
peroxide are formed. How much NO was originally present (the
gas being measured under standard conditions) ?
185. How many liters of nitric oxide can be obtained by oxidizing 500 cc.
of ammonia gas?
186. What weight of calcium carbide will be needed to furnish enough
calcium cyanamid to give 5 liters of ammonia?
187. 5 liters of air, containing 78% nitrogen and 21% oxygen by volume,
are treated with an electric spark. What volume of nitric oxide
would be formed if the reaction were not reversible ?
17. PHOSPHORUS
_ . (In electric furnace)
Com. Prep. ] , pn , 9W 9 n
L.a3(rU 4 J 2 -f- ji.ti2oU 4 T ^llnU >-
2CaSO 4 + CaH 4 (PO 4 ) 2 2H 2 (Superphosphate)
4P + 30 2 > 2P 2 3 (Incomplete oxidation of P)
4P + 5O 2 ^2P 2 O 5 (Complete combustion of P)
T^_ i PaO 5 + 3H 2 > 2H 3 P0 4 (Acid anhydride of phos-
Cnem. Prop. < , . .,..
phone acid)
Ca 3 P 2 + 6H 2 O >- 3Ca(OH) 2 + | 2PH 3 (Preparation
of phosphine)
188. What is the percentage composition of disodium acid phosphate
(Na 2 HP04) ?
189. What is the percentage of phosphorus in Thomas slag whose com-
position might be considered to be (CaO)s PaOs SiOs?
190. A compound was vaporized and its vapor density was found to
be 76.9. It contained 22.7% phosphorus, 10.9% oxygen, and 66.4%
chlorine. Find the true formula of this compound.
191. What weight of oxygen will be consumed in combining with 0.62
gram of phosphorus?
192. Calculate the amount of phosphorus available in a skeleton weigh-
ing 20 Ib. and containing 70% calcium phosphate.
193. One pound of phosphorus serves to tip one million matches. How
many matches can be tipped from the phosphorus obtained from
a human skeleton weighing 20 Ib. and containing 50% Ca 3 (P0 4 )2?
194. The price of phosphorus in the New York market is 68 per pound.
What is the value of the phosphorus found in a human skeleton
which weighs 25 Ib. and contains 60% of calcium phosphate?
195. How much superphosphate can be prepared from 1500 Ib. of bones
containing 75% calcium phosphate? What weight of CaS0 4 is
formed at the same time?
196. What weight of water is necessary for the complete hydrolysis
of 27.5 grams of phosphorus trichloride? PCls + 3HaO >-
H 3 P0 3 + 3HC1.
197. 5.5 liters of phosphine were liberated by the action of calcium
phosphide and water. What weight of Ca(OH)2 was formed at
the same time?
Com. Prep.
Chem. Prop.
18. ARSENIC, ANTIMONY, AND BISMUTH
2As 2 S 3 + 90 2 > 2As 2 3 + 6SO 2 (Roasting of the
sulfide)
As-O 3 + 3C *- 2As + SCO (Reduction of the oxide)
Sb 2 S 3 + SFe *- 2Sb + 3FeS (Action of Fe on the
sulfide)
AsCls + 3H 2 >- ^ AsH 3 + 3HC1 (Formation of arsine)
2AsH 3 + 30 2 *- 3H 2 + As 2 3 (Marsh's test for
arsenic)
.2SbCl 3 +3H 2 S >|Sb 2 S 3 +6HCl (Precipitation of Sb 2 S 3 )
198. Mispickel (FeAsS), orpiment (As 2 S 3 ), realgar (As 2 S 2 ), and white
arsenic (As 2 3 ) are the principal ores of arsenic. Determine
the percentage of arsenic in each.
199. What .volume of carbon monoxide will be formed in the reduction
of 33 grams of As 2 3 by carbon?
200. What volume of hydrogen will be needed to prepare 5.5 liters of
arsine from AsCls?
201. A compound contains 96.2% arsenic and 3.8% hydrogen. What
is its true formula if the vapor density of the compound is found
to be 38.9?
202. Wood's metal consists of Bi 4 parts, Pb 2 parts, Sn 1 part, and Cd
1 part. How much bismuth is necessary to make 10 Ib. of this
easily fusible alloy?
203. What weight of iron will be required to extract completely all
the antimony from 42 Ib. of antimony trisulfide?
204. Wood's metal is composed of 50% bismuth, 12.5% each of tin and
cadmium, and 25% lead. What weight of lead oxide, PbO, would
have to be reduced to furnish sufficient lead to make 100 Ib. of
Wood's metal?
205. The stomach of a poison victim was analyzed and 1.5 grams of ar-
senic were found. Howmuch Paris green, Cu(C 2 H 3 02)2 Cu 3 (As0 3 ) 2 ,
was swallowed by the victim?
206. 55 grams of bismuth formed 61.4 grams of Bi 2 3 . What is the
atomic weight of this element?
19. BROMINE AND HYDROBROMIC AGED
2KBr + Mn0 2 + 2H 2 S0 4 - ^
K S SO
l-PBrs + 3H 2 - >- H 3 PO 3 + SHBr (Pure)
Com. Prep. MgBr a + C1 2 - s- MgCl a + Br 2
/ 2KBr + C1 2 - * 2KC1 + Br 2 (Replacement)
Cnern. Wop. K 2 SO 4 + 2H 2 + S0 2 + Br 2
207. A liter of bromine gas weighs 7.2 grams. What are its vapor den-
sity, specific gravity, and molecular weight ?
208. How much sodium bromide must be heated with sulfuric acid and
manganese dioxide to obtain 10 grams of bromine?
209. In the preparation of bromine 48 grams of manganese sulfate were
found among the products of the reaction. What weight of potas-
sium salt was decomposed?
210. 238 Ib. of potassium bromide are treated with sufficient sulfuric acid
to decompose it completely. What weight of hydrogen bromide
will be formed?
211. 0.35 part of magnesium bromide was found in 1000 parts of sea
water. How much of this water would be required to furnish 10 Ib.
of bromine ?
212. What weight of phosphorus tribromide will be needed to prepare
162 kilograms of pure HBr?
213. A solution of sodium bromide was treated with an excess of silver
nitrate and 0.6 gram of silver bromide was precipitated. What
weight of NaBr was present in the original solution?
214. What weight of manganese dioxide must be used to obtain 1 liter
of bromine vapor from KBr?
215. Chlorine gas is passed through a solution of potassium bromide
until 5 grams of bromine are liberated. What weight of the bro-
mide was decomposed? What volume would this bromine occupy
at57C.?
216. 45 grams of magnesium chloride were obtained from a bromide.
What volume of chlorine, measured at 20 C. and 760 .mm., took
part in the reaction?
"3
20. IODINE AND HYDROGEN IODIDE
(2KI + Mn0 2 + 2H 2 S0 4 >-
Lab. Prep. K 2 S0 4 + MnS0 4 + 2H 2 + I 2
I PI, + 3H 2 ^ H 3 P0 3 + 3HI (Pure)
Com. Prep. 2NaI0 3 + 3Na 2 S0 3 + 2NaHS0 3 >
5Na 2 S0 4 + H 2 + 1,
f2Na 2 S 2 3 + Is >-2NaI + Na 2 S 4 O 6 (Soluble in hypo)
Chem. Prop. \ 2KI + Br 2 > 2KBr + I 2 (Replacement)
[ H 2 S + I 2 > 2HI + S (It is a mild oxidizing agent)
217. 90 cc. of hydrogen iodide gas weigh 0.516 gram. Calculate its
vapor density, specific gravity, weight of a liter, and molecular
weight.
218. Find the molecular weight of hydrogen iodide if the specific gravity
of the compound is 4.42.
219. A compound contains 96.7% iodine, 3.05% carbon, and 0.25%
hydrogen. Its vapor density is 197. Find the true formula of this
compound.
220. How much potassium iodide is needed to prepare 63.5 grams of
iodine?
221. How much iodoform, CHI 3 , can be obtained from 1 kg. of
iodine?
222. How much sulfuric acid must be used to obtain all the iodine
available in 83 Ib. of potassium iodide ?
223. Which will liberate more free iodine from an ammonium iodide
solution 50 grams of chlorine or 100 grams of bromine?
224. A ton of kelp will give 2 Ib. of iodine. How much potassium
iodide can be prepared from the iodine available in 10.5 tons of
this kelp?
225. How much free iodine is necessary to oxidize 1 gram of hypo?
(The crystal contains 5 molecules of water of crystallization.)
226. What volume of chlorine gas must be used just to liberate all the
iodine in a liter of potassium iodide solution containing 33.2 grams
of the salt?
227. How much phosphorus tri-iodide must be reacted with water to
produce 500 cc. of hydrogen iodide gas ?
228. What volume of bromine vapor must be passed through a sodium
iodide solution to precipitate 3.81 grams of iodine?
114
Chem. Prop.
21. FLUORINE AND ETCHING OF GLASS BY HYDROGEN
FLUORIDE
Lab. Prep. CaF 2 + H 2 S0 4 >- CaS0 4 -f- -f 2HF (The common prepa-
ration is the same)
SiO 2 + 4HF ^ 4 SiF 4 + 2H 2 O (Etching of glass)
3F 2 + 3H 2 O >- 6HF + O 3 (Action of F 2 on water)
CaSiO 3 + 6HF >- 1 SiF 4 + CaF 2 + 3H 2 (Action of
HF on insoluble silicates)
229. What is the percentage of fluorine in cryolite, NasAJFe, and in
fluorite, CaF 2 ?
230. Determine the percentage composition of (NaF) 3 A1F 3 both by
elements and salts.
231. A liter of silicon fluoride weighs 4.68 grams. Calculate the molec-
ular weight, vapor density, and specific gravity of this vapor.
232. A quantity of fluor-spar is treated with sulfuric acid, and, as a
result, 3.4 grams of calcium sulfate is obtained. What weight
of fluor-spar was decomposed?
233. What weight of quartz will a liter of hydrofluoric acid, containing
30% HF by weight, dissolve? (Sp. Gr. of 30% HF is 1.12.)
234. Assuming the formula for glass to be CaO Na 2 6SiO2, what
weight of HF would be required to etch away 28.9 grams of this
glass ?
235. What weight of calcium fluoride must be used to prepare 5 liters
of hydrogen fluoride gas?
236. 5 liters of hydrogen fluoride were liberated when sulfuric acid
reacted with fluor-spar. What weight of calcium sulfate was pre-
pared at the same time?
237. What volume of hydrogen fluoride gas is needed to reduce the
weight of a glass plate 2 grams? The formula for glass is taken as
CaO Na 2 - 6Si0 2 .
238. What weight of hydrogen fluoride would be required to react
with quartz to furnish 5 liters of silicon fluoride ?
239. What volume of fluorine must react with water to produce 500 cc.
of hydrogen fluoride gas? What volume of ozone will be formed
at the same time?
115
22. ALLOTROPIC FORMS AND VARIETIES OF CARBON
Lab. Prep. Ci 2 H 22 Ou >- 11H 2 + 12C (Pure carbon from cane
sugar)
C + Oz > CO 2 (Complete combustion of carbon)
2C + 2 >- 2CO (Incomplete combustion of carbon)
Chem. Prop. C + CuO *- Cu + CO (Carbon a reducing agent)
C + C0 2 > 2CO (Reduction of C0 2 by C)
C + H 2 > CO + H 2 (Prep, of water gas)
240. 171 Ib. of dry cane sugar are treated with concentrated sulfuric
acid. What weight of pure carbon will result ?
241. How much steam, by weight and volume, can react with 120 grams
of pure carbon?
242. A manufacturer requires 6 Ib. of pure carbon made from cane
sugar. What weight of the latter would have to be decomposed
to supply this weight of pure carbon?
243. What weight of coke containing 90% carbon would be needed to
reduce 10 tons of hematite containing 85% FeoOs?
244. 5 grams of coke, on burning in a sufficient supply of oxygen, gave
17 grams of carbon dioxide. Find the percentage of carbon in
the coke.
245. Determine the volume of gas, measured at standard conditions,
which can be obtained by the complete combustion of 12 grams
of pure carbon.
246. What volume of air would be required for the complete .combustion
of 1 kg. of coal containing 80% of carbon?
247. Calculate the weight of air necessary to burn a ton of coal con-
taining 90% carbon. Air contains 23% of oxygen by weight.
248. What weight of hematite, containing 79% of Fe 2 3 , can be reduced
by 1000 Ib. of coke, containing 98% pure carbon?
249. Carbon monoxide was passed over hot copper oxide. 134 grams
of carbon dioxide were formed, and 48.7 grams of oxygen were lost
by the copper oxide. Determine the atomic weight of carbon from
these data.
250. An organic compound was completely oxidized and the carbon
dioxide formed was absorbed, and found to weigh 5.3 grams.
What weight of carbon did the organic compound contain?
116
23. CARBON , DIOXIDE AND CARBONIC ACID
[ CaCOs + 2HC1 *- CaCl 2 + 1 CO 2 + H 2 (Acid on a
Lab. Prep. < carbonate)
I C + Os >- COz (Complete combustion of carbon)
C COa + H 2 s- H 2 COs (Acid anhydride of carbonic acid)
I 6C0 2 + 5H 2 O >- C 6 H ]0 O5 + |6O 2 (Photosynthesis)
Chem. Prop. I NaHC0 8 + HKCilLA >- NaKC 4 H 4 6 -f H 2 + 4 C0 2
(Action of baking powder)
I H 2 C0 3 + CaC0 3 >- Ca(HCO 3 ) 2
251. What percentage of carbon, by weight, is contained in carbon diox-
ide and carbon monoxide?
252. The weight of a liter of carbon dioxide is 1.98. Find its vapor
density, specific gravity, molecular weight, and the weight of 275.75
cc. of the gas.
253. How many grams of pure calcium carbonate must be used to
prepare 22 grams of carbon dioxide by the action of hydrochloric
acid?
254. 25 grams of marble, containing 5% of inert material, are dissolved
in hydrochloric acid. Calculate the weights of the substances
formed.
255. HOW many grams of sulfuric acid reacting with an excess of sodium
carbonate will be required to form 220 grams of carbon dioxide?
256. What weight of carbon dioxide will be required to 'precipitate
the calcium in 3.7 grams of calcium hydroxide, as calcium car-
bonate ?
257. What weight of carbon dioxide will be used to change limewater
into 81 grams of calcium bicarbonate?
258. If a piece of pure graphite weighing 1 kg. is burned in oxygen,
what volume of carbon dioxide will be produced?
'259. What weight of starch can a plant manufacture from 20 liters of
carbon dioxide gas?
260. What volume of oxygen is liberated in this reaction?
261. Calculate the weights of sodium bicarbonate and cream of tartar
found in 100 grams of baking powder. What volume of carbon
dioxide can this baking powder liberate?
117
24. CARBON MONOXIDE
f HCOOH > CO -f H 2 (Decomposition of formic acid)
Lab. Prep. \ COOH COOH >- CO + C0 2 + H 2 (Oxalic acid and
[ concentrated H 2 S0 4 )
/ H 2 O + C *- H 2 + CO (Production of water gas)
Com. Prep. ^ C02 + c ^ 2 CO (Reduction of C0 2 )
f2CO + 2 > 2C0 2 (Combustion of carbon monoxide)
Chem. Prop. ^ CQ + Q,, ^ CO C1 2 (Formation of phosgene)
262. The weight of a liter of carbon monoxide is 1.25. Calculate its
vapor density, specific gravity, and molecular weight.
263. The specific gravity of carbon monoxide is '0.968. Calculate
its molecular weight.
264. A gas contains 43% carbon and 57% oxygen. 33 cc. of this gas
weigh 0.042 gram. Calculate the true formula of the compound
from these data.
265. 20 grams of carbon are heated in the presence of 88 grams of carbon
dioxide. What weight of carbon monoxide is formed? What
weight, if any, of carbon remains?
266. What volume of carbon dioxide must be passed over glowing char-
coal to form 42 grams of carbon monoxide?
267. 108 Ib. of steam are passed over glowing coal. ' What volume of
carbon monoxide will the resulting water gas contain?
268. 25 cc. of carbon monoxide were prepared from formic acid. What
weight of the acid was decomposed?
269. What weight of coke, containing 95% carbon, will be needed to
reduce completely 3.5 liters of carbon dioxide gas?
270. What weight of formic acid will be needed to liberate 100 cc. of
carbon monoxide measured under standard conditions?
271. What volume of oxygen is required for the combustion of 10 cu. ft.
of carbon monoxide?
272. 0.442 gram of phosgene gas occupies 100 cc. Determine its molec-
ular weight, specific gravity, vapor density, and the weight of
9 liters.
118
26. ILLUMINATING GAS, WATER GAS, AND PRODUCER GAS
r m Pr.n 2 2 C
uom. *rep. ^C + air (N 2 + 2 ) - >- 2CO + N 2 (Producer gas)
(2CO + 2 - -2C02 (Burning of carbon monoxide)
Chem. Prop, j 2CO + 2H 2 + 20 2 - J-2C0 2 -f 2H 2 (Burning of water
I gas)
273. What volume of steam is required for the production of 1000 liters
of water gas?
274. Water gas contains 40% of hydrogen and 60% of carbon monoxide
by volume. What volume of air is necessary for the complete
combustion of 1000 cu. ft. of such a gas? (Assume air to contain
20% oxygen by volume.)
275. What weight of coal is required to produce 1000 cu. ft. of water
gas? (Note. The molecular weight of a substance expressed in
ounces occupies 22.2 cu. ft.)
276. Analysis of a producer gas showed the presence of 30% of carbon
monoxide. How much carbon was burned in preparing 1000 cu. ft.
of this fuel?
277. 10 grams of. carbon were treated with steam, and water gas was
produced. The water gas thus obtained was completely burned
to carbon dioxide and water vapor. What volumes of these gases
were formed?
278. What weight of glowing coal, containing only 95% carbon, would
have to be treated with steam in order to make 10,000 cu. ft. of
water gas?
279. 5 liters of producer gas containing 27% carbon monoxide were
burned as fuel. What weight and what volume of carbon dioxide
were produced?
280. A kilogram of glowing carbon was treated with steam and the
resulting water gas was burned in oxygen. Determine the volume
of carbon dioxide produced.
281. A natural gas from the Pittsburgh region contained 26% hydrogen
and 65% methane (CH 4 ). What weight of water would be formed
in completely burning 500 liters of this gas?
119
Cfcem. Prop.
26. REACTIONS WITH CARBON IN THE ELECTRIC FURNACE
CaO + 3CT *- CaC 2 + CO (Preparation of calcium
, carbide)
Com. Prep. \ g.^ + 3C ^ Sic + 2 CO (Preparation of carborundum)
Q _{_ 2S > CS 2 (Preparation of carbon disulfide)
CaC 2 + 2H 2 >- I C 2 H 2 + Ca(OH) 2 (Action of CaC 2
on water)
CS 2 + 30 2 *- C0 2 + 2S0 2 (Complete combustion of CS 2 )
CS 2 + 2 > C0 2 + 2S (Incomplete combustion of CS 2 )
282. What weight of carbon bisulfide can be produced from 3000 Ib.
of sulfur?
283. How much carborundum can be made theoretically from a ton of
sand composed of 90% Si02?
284. Assuming that 75% of the carbon is converted into carborundum,
what weight of carbon will be needed in the production of 5 tons
of silicon carbide?
285. How much sand containing 5% impurities must be used to obtain
a theoretical yield of 5 tons of carborundum ?
286. What weight of sulfur will be formed by the incomplete combustion
of 38 grams of carbon bisulfide?
287. 500 cc. of the vapor of carbon disulfide weigh 1.71 grams.
Calculate its vapor density, specific gravity, molecular weight,
and the weight of 35 cc.
288. What weight of sulfur will be produced in the incomplete combus-
tion of 350 cc. of carbon bisulfide vapor?
289. What weights of the elements must be combined to produce 5 liters -
of carbon bisulfide vapor?
290. How much dry calcium carbide will be needed to produce 510 cc.
of acetylene gas?
291. What amount of calcium oxide must be used to make sufficient
calcium carbide to furnish 5000 liters of acetylene gas, measured
under standard conditions?
292. What volume of sulfur dioxide is produced in the complete com-
bustion of 50 cc. of carbon bisulfide vapor?
293. What weight of calcium carbide must be used in order to fill an
acetylene reservoir holding 25 liters of the gas at 17 C. and 760 mm.
pressure?
27. SILICON DIOXIDE, GLASS, AND BORAX
Cliem. Prop.
SiO 2 + 2KOH >- K 2 Si0 3 + H 2 (Si0 2 reacts with
bases, forming silicates)
SiO 2 + 2NaOH > Na 2 SiO 3 + H 2 (Preparation of
water glass)
SiO 2 + H2O *- H 2 SiO 3 (Si0 2 the acid anhydride of
silicic acid)
Na 2 C0 3 + CaC0 3 + 6SiO 2 ^ Na 2 O CaO 6Si0 2 +2CO 2
(Formation of glass)
SiCU + 4H 2 O >- 4HC1 + Si(OH) 4 (Action of SiCl 4
on water)
Na 2 B 4 7 + H 2 S0 4 + 5H 2 > Na 2 S0 4 + 4H 3 BO 3
(Boric acid from borax)
294. What percentage by weight does crystallized borax (Na 2 B 4 CV 10H 2 O)
lose on being heated until it has lost all of its water of crystalliza-
tion?
295. How much of the element boron is present in 5 grams of anhydrous
borax ?
296. Find the percentage of silicon in the following silicates : talc
[H 2 Mg 3 (Si0 3 ) 4 ] and clay [H 2 Al 2 (Si0 4 ) 2 H 2 0].
297. A compound contains 29.2% silicon, 66.7% oxygen, and 4.10%
hydrogen. Find the simplest formula of this compound.
298. How much commercial NaOH containing 10% water would have
to be used to prepare 244 grams of water glass?
299. When 39 grams of silicic acid are dehydrated, what weight of silicon
dioxide remains?
300. How much calcium oxide would have to be used as flux in order
to remove 60 kg. of SiOa present in an ore? How much slag would
result ?
301. The reaction representing the formation of glass may be repre-
sented by the above equation (4). What weights of materials must
be used to make 100 kg. of glass?
302. Find the weights of materials necessary to make 1000 Ib. of potas-
sium lead glass.
303. What weight of sulfuric acid would be required to convert 100
tons of sodium borate, containing 20% impurities, to boric acid,
H 3 B0 3 ?
28. CALCIUM, CALCIUM CARBONATE, AND HARD WATER
Com. Prep. CaCI 2 > Ca -f C1 2 (Electrolysis of fused CaCl 2 )
f Ca(OH) 2 + C0 2 *- f CaCOa, + H 2 O (Formation of
insoluble CaCO 3 )
CaCOj, -f H 2 O + CO 2 > Ca(HCO 3 ) 2 (Formation of
soluble Ca(HCQ s ) 2 )
CaSO 4 -f NajCOa > | CaCOs + Na 2 SO 4 (Softening of
Cnem. Jtrop. \ permanent hard water)
Ca(HCO 3 ) 2 + Ca(OH) 2 >-^2CaCO 3 + 2 H 2 O (Soften-
ing of temporary hard water)
2C 17 HuCOONa+CaS04 J-Na 2 S0 4 + 1 (Ci 7 H 35 COO) 2 Ca
(Action of hard water on soap)
304. Determine the simplest formula of a compound containing 40.1%
calcium, 12^- carbon, and 47.9^ oxygen. The molecular weight
of this compound is 100. What is its true formula?
SOS. What weight of calcium chloride must be decomposed to produce
5 grams of pure calcium?
306. What volume of chlorine, measured under standard conditions,
would be liberated in the above operation?
307. What weight of water would be required to dissolve completely
15 grams of calcium? What weight of calcium hydroxide would
be formed?
308. How much slaked lime, containing 90% Ca(OH) 2 , will be needed
to precipitate completely a solution containing 243 grams of the
bicarbonate of magnesium?
309. What volume of carbon dioxide is liberated when 48.6 grams of
calcium bicarbonate are heated?
310. What weight of potassium stearate, Ci 7 H3 5 COOK, would be re-
quired to precipitate 5 grams of magnesium sulfate dissolved in 100
liters of water?
311. If 100 grams of BaCl 2 give 112.1 grams of barium sulfate, find the
atomic weight of barium.
312. A water contains 1.2 grams of calcium bicarbonate per gallon. What
weight of calcium hydroxide would be required to soften 1000 gal.
of this water?
313. 100 grams of CaCl 2 2H 2 and 69 grams of K 2 C0 3 are mixed in
solution. Calculate the weights of calcium and potassium salts
in filtrate and precipitate.
29. CALCIUM OXIDE AND CALCIUM HYDROXIDE
CaCO 3 >- CaO + ^ CO 2 (Heating limestone)
, CaO + H 2 > Ca(OH) 2 (Slaking of lime)
Com. Prep, -j CaO + 3C ^ CaC2 + CQ ( Preparation of ca l c ium car-
bide)
Ca(OH) 2 + C0 2 ^|CaCO 3 + H 2 O (Test for carbon
, dioxide)
Chem. Prop, -j Ca(OH ) 2 + Si02 ^ CaSiO 3 + H 2 (Forms a silicate
with sand)
314. How many tons of lime can be made from 100 tons of limestone,
containing 90% CaCos?
315. How many pounds of calcium carbide can be made theoretically
from a long ton of lime?
316. What weight of water will be necessary to slake completely 28 Ib. of
quicklime?
317. How many pailfuls of water, each pail having a capacity of
50 Ib. of water, will be necessary to slake completely a ton of
quicklime?
318. 2 tons of water are just sufficient to slake a load of lime. The lime
is made into mortar, which is then used in plastering a house. What
weight of carbon dioxide may be chemically absorbed by the
mortar?
319. What weight of lime containing 80% CaO will be required to make
128 Ib. of calcium carbide ?
320. How much lime can be made from 5 tons of limestone containing
95% calcium carbonate?
321. What volume of carbon dioxide will be available in the above
reaction?
322. What volume of carbon dioxide must be absorbed by 14 grams of
quicklime before it is completely air-slaked?
323. What will be the final weight of this air-slaked lime?
324. What volume of carbon dioxide can be obtained from the heating
of 1 kg. of marble containing 95% of calcium carbonate?
325. How many gallons of water, each weighing 8.35 Ib., will be needed
to make 222 Ib. of slaked lime from quicklime?
123
30. CALCIUM SALTS
f 6H 2 + Ca 3 (P0 4 ) 2 + 2H 2 S0 4 > 2CaS0 4 2H 2 O
+ Ca(H 2 P04)a 2H 2 (Superphosphate of lime)
CaS0 4 + 2H 2 >- CaS0 4 2 H 2 (Gypsum)
2CaS0 4 2H0 ^3H 2 +(CaS0 4 ) 2 H 2 O (Plaster of
Paris)
Com. Prep. | CaCXa + C + Na 9 COj > CaC0 3 + 2NaCN (Prepara-
tion of sodium cyanide)
Ca(OH). + C1 2 ^ H0 + CaCl OC1 (Bleaching
powder)
CaCl OC1 + 2HC1 *- CaCl 2 + H 2 + |2C1 (Decom-
position of bleaching powder)
326. \Vhatisthepercentageofchlorineiiibleachingpowder?
327. Determine the percentage composition of crystallized superphos-
phate of lime.
328. What volume of chlorine will react with 33.3 grams of dry calcium
hydroxide in the preparation of bleaching powder?
329. What weight of plaster of Paris can be made from 8.6 tons of
gypsum?
330. What volume of chlorine will be liberated when hydrochloric acid
completely reacts with 76.2 grams of bleaching powder?
331. What weight of sodium cyanide can be prepared from 400 Ib. of
calcium cyanamid?
332. How much superphosphate of lime can be prepared from 100 tons
of bones containing 70% calcium phosphate?
333. A manufacturer must turn out 50 tons of bleaching powder. What
weight of slaked lime and what volume of chlorine will he need ?
334. What volume of steam will be given off when 8.6 grams of gypsum
are completely burned?
335. How much superphosphate of lime, containing 2 molecules of
water of crystallization, can be prepared from 50 tons of an ore
containing 75% pure calcium phosphate?
336. A tube containing calcium chloride weighs 200 grams. 80 liters of
air is sent through the tube, which is again weighed and found
to be 201.7 grams. Calculate from these data the percentage of
moisture in the air.
124
31. MAGNESIUM
Com. Prep. MgCI 2 >- Mg + C1 2 (Electrolysis of fused MgCls)
' 2Mg + 2 >- 2MgO (Burns in air with, a bright white
light)
3Mg + N 2 > Mg 3 N 2 (Unites to a small extent with
pt, T> i nitrogen of the air)
P> < Mg -r"H 2 S0 4 + 7H 2 ^ MgS0 4 7H 2 + -f H 2
(Formation of Epsom salt)
Mg + 2H 2 s- Mg(OH) 2 + -f H 2 (Reacts with boiling
water to liberate H 2 )
337. Find the percentage of magnesium in: talc, H 2 Mg 3 (Si03)4; ser-
pentine, (MgFe)sSi a 7 2H 2 0; and carnallite, MgCl 2 KC1 6H 2 0.
338. Determine the percentage of water of crystallization in Epsom
salt, MgS0 4 7H 2 0.
339. Find the weight of magnesium obtainable from one ton each of
dolomite (CaC0 3 MgC0 3 ) and magnesite (MgC0 3 ).
340. How much Epsom salt can be obtained from one pound of pure
magnesium?
341. What volume of nitrogen will be required to react completely with
magnesium and form 10 grams of magnesium nitride ?
342. What volume of air will be required to burn completely 2 grams of
magnesium to magnesium oxide? Assume air to contain 21%
oxygen.
343. A certain weight of magnesium was placed in boiling water and
50 cc. of hydrogen gas measured at 17 and 760 mm. were liberated.
What weight of water was decomposed ?
344. A compound contained 13% sulfur, 71.5% oxygen, 9.8% magne-
sium, and 5.7% hydrogen. What is the simplest formula of this
compound?
345. What volume of H 2 SC>4 containing 70% H 2 S0 4 (Sp. Gr., 1.61) will
be needed to react completely with 4 grams of magnesium ?
346. Magnesium ribbon reacted with 20 cc. of steam. What weight of
hydrogen was formed?
347. What volume of chlorine, measured at 20 and 760 mm., can be ob-
tained by the electrolysis of 190 Ib. of fused MgCl 2 ?
125
32. MERCURY
Com. Prep. HgS + 2 > Hg + S0 2 (Extraction of Hg from cinna-
bar)
Cu + Hg(N0 3 ) 2 >-| Eg + Cu(N0 3 ) 2 (Replacement of
Hg by copper)
HgCl 2 + Hg > 2HgCl (Reduction' of HgCI 2 with Hg)
Chem. Prop. | 2HgCl 2 + SnCl 2 > SnCl 4 + 2HgCl (Action of SnCl 2 on
bichloride of mercury)
2HgCl + SnCl 2 s- SnCl 4 + f 2Hg (Reduction of HgCl
by stannous chloride)
348. The specific gravity of mercury vapor is 6 .9 1 . Determine its molec-
ular weight, the weight of one liter of its vapor, and its vapor
density.
349. Calculate the simplest formula of a mercury compound containing
92.6% mercury and the remainder oxygen.
350. What weight of mercury must be added to 10.84 grams of bichlo-
ride of mercury to change it completely into mercurous chloride
(calomel) ?
351. How much mercury fulminate, Hg(OCN)a, can be prepared .from
100 Ib. of mercury?
352. One ton of an ore containing 85% of HgS was roasted. What weight
of mercury was obtained?
353. What volume of sulfur dioxide gas was liberated in the above re-
action?
354. A strip of copper weighing 2 grams is placed in a solution containing
an equal weight of mercuric nitrate. What weight of mercury was
deposited?
355. Stannous chloride was added to a solution containing 4.42 grams
of bichloride of mercury. One gram of calomel (HgCl) was pre-
cipitated. What weight of stannous chloride was used ?
356. Calculate the volume of oxygen obtained by heating 36 grams of
mercuric oxide.
357. What weight of mercury and what volume of sulfur dioxide can
be prepared from 100 Ib. of cinnabar, containing 92% HgS?
126
33. ZINC
f ZnCOs *- ZnO + | CO 2 (Roasting of Smithsonite)
Com. Prep. < 2ZnS + 30 2 -> 2ZnO + j 2S0 2 (Roasting of zinc blende)
[ ZnO + C >- Zn + | CO (Reduction of the oxide)
Zn + 2KOH > K 2 ZnO 2 + <f H 2 (Action with a strong
base)
Chem. Prop. ZnCl 2 + H 2 S >-| ZnS -f- 2HC1 (Formation of ZnS)
ZnS0 4 + BaS >- BaS0 4 + ZnS (Preparation of litho-
pone)
358. Calculate the percentage of zinc in the following compounds : zinc
carbonate, zinc sulf ate, zinc, oxide, and zinc sulfide.
359. A compound contains 43.9% water, 22.7% zinc, 11.1% sulfur,
and 22.3% oxygen. What is the simplest formula of this com-
pound?
360. Find the formula of a zinc sulfide which contained 67% zinc and
33% oxygen.
361. The vapor density of zinc vapor was found to be equal to 32.5.
Determine the weight of a liter of this vapor, and its molecular
weight.
362. What weight of carbon is needed to reduce 243 Ib. of zinc oxide,
the carbon being oxidized only to carbon monoxide?
363. How much zinc would be necessary to produce 10 grams of crys-
tallized zinc sulfate (ZnS0 4 7H 2 0) ?
364. What weight of carbon dioxide would be given off on heating 1 kg.
of zinc carbonate?
365. What volume of hydrogen sulfide would be needed to precipitate
19.4 grams of zinc sulfide from a sufficient amount of zinc chloride
solution?
366. What volume of air would be used in the roasting of 48.5 grams of
zinc sulfide?
367. What weights of ZnSO* and BaS will be needed to make 33 tons of
lithopone ?
127
34. MANUFACTURE OF IRON AND STEEL
' Fe 2 3 + 3C - - 2Fe + 1 SCO (Reduction of hematite
with cokej
Fe 2 s + SCO - >- 2Fe + |3C0 2 (Reduction of hema-
_ ^ . tite with CO)
om. *rep. j CaCO;j + SiQ2 - ^ CaSiO 3 + 4- C0 a (Production of slag)
2CaO + 2S -f 3O 2 - *- 2CaS0 4 (Removal of S)
6CaO + 4P + 50s - - 2Ca 3 (P0 4 ) 2 (Removal of P)
3Fe -f- C -- >- FesC (Formation of iron carbide)
36S. Calculate the percentage of iron in e?.ch of these ores: hema-
tite, FesOs; magnetite, FesCu; siderite, FeC0 3 ; and limonite,
Fe 2 3 3H 2 0.
369. What volume of air would be required to oxidize all of the carbon
present in 5 tons of cast iron, containing 3.5% carbon?
370. It is required to produce 100 tons of steel containing 0.5% of carbon
from pig iron containing 4.5% carbon. What volume of air will be
needed?
371. 100 tons of iron ore containing 5% calcium carbonate is to be
smelted. What weight of sand must be added to remove completely
all this calcium as slag?
372. How many tons of coke containing 98% carbon are needed to reduce
500 tons of hematite containing 88%
373. What volume of carbon monoxide is necessary to reduce 3.2 grams
of Fe 2 3 ?
374. An ore of iron showed on analysis 7% sand and 10% limestone.
What weight of flux would have to be added in the smelting of 100
tons of this ore?
375. It is required to make 100 tons of cast iron containing 95% iron.
What weight of an iron ore containing 90% siderite, FeC0 3 , would
be required?
376. What volume of carbon monoxide is necessary to reduce 2.976
grams of hematite containing 93 %
377. 9.5 grams of iron yielded 13.6 grams of FeaOa. Calculate the
atomic weight of iron.
128
35. IRON SALTS
f 4KCN + Fe(CN) 2 *- K 4 Fe(CN) 6 (Preparation of po-
I tassium ferrocyanide)
Com. Prep, j 2 K 4 Fe(CN) 6 + C1 2 >- 2KC1 + 2K 3 Fe(CN) 6 (Prepara-
l tion of potassium ferricyanide)
Lab. Prep. FeCl 3 + 3NH 4 OH > 3NEUC1 + f Fe(OH) 3 (Precipita-
tion of ferric hydroxide)
3Fe + 4H 2 > Fe 3 O 4 + ^ 4H 2 (Action of hot iron on
steam)
2FeCl 2 '+ CI 2 >- 2FeCl 3 (Oxidation of ferrous chloride)
2FeCI 3 + H 2 > 2HC1 + 2FeCl 2 (Reduction of ferric
Chem. Prop. { chloride)
4FeCl 3 + 3K 4 Fe(CN) 6 > 12KC1 + ^ Fe 4 (Fe(CN) 6 ) 3
(Prussian blue test for a ferric salt)
3FeCl 2 + 2K 3 Fe(CN) 6 > 6KC1 + 1 Fe 3 (Fe(CN) 6 ) 2
(TurnbuU's blue test for a ferrous salt)
378. Determine the percentage composition of Prussian blue.
379. What is the percentage composition of green vitriol (FeS04 7H20) ?
380. What is the simplest formula of a compound containing 15.2% Fe,
42.4% K, 19.6% C, and 22.8% N?
381. Let us take for the formula of limonite (Fe 2 3 ) 2 3H 2 0. What
weight of iron could 10 tons of this ore furnish?
382. What weight of ferric chloride could be theoretically reduced by
5 liters of hydrogen gas?
383. What weight of Prussian blue could be formed from 73.6 Ib. of
potassium ferrocyanide ?
384. How much ferrous chloride would be needed to prepare 2.71 grams
of TurnbuU's blue?
385. What weight of FeS04 would result by the action of 100 grams of
pure iron on sulfuric acid? What volume of hydrogen would be
evolved?
386. What volume of chlorine would have to be passed over hot iron
to form 100 grams of ferric chloride, provided all the chlorine
reacted with the iron?
387. What weight of ferric salt could be obtained by the oxidation of
381 grams of ferrous chloride with free chlorine ?
388. What volume of chlorine will be required to change 18.4 grams of
potassium ferrocyahide to the ferricyanide ?
129
36. COPPER
f 4CuFeS 2 + 110 2 > 4FeO + SS0 2 + 2Cu 2 O (Roasting
Com. Prep. < of copper pyrites)
(.Cu 2 + C >-2Cu + CO (Reduction of cuprous oxide)
3Cu + SHX0 3 >- 3Cu(N0 3 ) 2 + 2NO + 4H 2
(Action of HN0 3 on Cu)
2Cu + 0* + 2H 2 S04 >- 2CuS0 4 + 2H 2 O (Action of
r>t, TH- ) dilute HoSO 4 on Cu in presence of air)
Cfaem. Prop. \ Cu + 2HaSO4 ^ CuS Q 4 + ^ S02 + 2 H 2 (Action with
concentrated H 2 S04)
Ag 2 SO 4 + Cu > CuS0 4 +|2Ag (Replacement of Ag
by copper)
389. Determine the percentage composition of the following copper
ores: copper pyrites, CuFeS ; malachite, Cu2(OH)aCO3; azurite,
Cu 3 (OH) a (C0 3 ) 2 ; and cuprite, Cu 2 0.
390. A compound was found to contain 53.1% silver, 31.1% copper, and
15.8% sulfur. Determine the simplest formula of this compound.
391. How many grams of copper must be heated to form 32 grams of
copper oxide?
392. What weight of Cu 2 could be reduced by 100 Ib. of carbon, assum-
ing all the carbon is oxidized to carbon dioxide?
393. What weight of copper will a ton of ore, containing 80% malachite,
yield?
394. What weight of silver will a gram of copper replace in a solution
containing 6 grams of silver sulfate?
395. A strip of copper was placed in a silver sulfate solution. After a
while it was taken out and found to have lost 2.3 grams. What
weight of copper sulfate was formed?
396. What weight of silver will be precipitated by copper from a solution
containing 3.12 grams of silver sulfate?
397. What volume of nitric oxide is liberated when 8 grams of copper is
completely dissolved by nitric acid? What weight of copper
nitrate is formed?
398. A copper coin liberated 20 grams of S0 2 from H 2 S04. What is
the weight of the coin?
130
37. SILVER, GOLD, AND PLATINUM
3Ag + 4HN0 3 *- 3AgNO s + 2H 2 + NO (Action of
HN0 3 on silver)
2Ag + 2H 2 S0 4 > Ag 2 S0 4 + 2H 2 O + S0 2 (Action with
cone. H 2 S0 4 )
Ag 2 S0 4 + Cu *~ CuS0 4 + |2Ag (Replacement of Ag
Chem. Prop. \ by copper)
AgN0 3 + KC1 ^| AgCl + KNO 3 (Test for a chloride)
KCN + AgCN - KAg(CN) 2 (Formation of a complex
silver salt)
4AgOH + HCHO >- 3H 2 + C0 2 + ^4Ag (Reduction
of Ag salt by formaldehyde)
399. Determine the percentage of silver present in each of these two
silver ores : pyrargyrite, Ag 3 SbS 3 , and prousite, Ag 3 AsS 3 .
400. A compound contains 16% potassium, 40.4% platinum, and 43.6%
chlorine. What is the simplest formula of this compound? '
401. What weight of PtCl 4 can be obtained by dissolving 19.5 grams of
platinum in aqua regia ?
402. A silver dollar weighs 26.5 grams. It contains 90% silver. What
weight of silver nitrate could be prepared from this coin?
403. Calculate the weights of the sulfates of silver and copper which
could be made from a dime weighing 2.48 grams and containing
10% copper.
404. What weight of pure copper would be needed to replace completely
all the silver in a liter of silver sulfate containing 10 grams of the
salt in solution?
405. On dissolving a batch of platinum ore in aqua regia, 41 grams of
chlorplatinic acid, H 2 PtCl 6 , were obtained. What weight of
platinum was present in the ore?
406. How many grams of silver can be obtained from 85 grams of silver
nitrate by simple replacement ?
407. One gram of silver is dissolved in nitric acid and to it are added 0.2
gram of sodium chloride. How many grams of silver salt remain
in solution?
408. What volume, in cubic centimeters, of nitric oxide is evolved when
54 grams of silver are completely dissolved in nitric acid? What
is the weight of this gas?
38. ALUMINUM
Com. Prep. 2A1 2 3 *- 4A1 + 30 2 (Electrolysis of bauxite)
' 2A1 4- GHC1 > 2A1C1 3 + 1 3H 2
2A1 + 6H 2 SC>4 >- A1 2 (S0 4 ) 3 + 3S0 2 + 6H 2 O (Cone.
H 2 S0 4 dissolves aluminum)
,2Al + 6NaOH > 2Na 3 A10 s + 1 3H 2 (Action of alumi-
Crieni. Prop. \ m ^ m ITT . +% n K;I;T,^ ei,it;^v, O f NaOH)
Fe 2 3 + 2A1 * A1 2 3 + 2Fe (Thermit process)
Cr 2 3 +2Al >-Al a Oj+2Cr (Preparation of Cr by
means of Al)
409. How much aluminum is available in 204 tons of bauxite, containing
85% AlaOs?
410. A compound contains 13% aluminum, 54.2% fluorine, and 32.8%
sodium. What is its simplest formula?
411. Magnalium is an alloy containing 95% aluminum and 5% mag-
nesium. What weight of bauxite would be required to prepare
1 kg. of this alloy?
412. What weight of carbon electrode would have to be replaced after
51 tons of aluminum oxide have been decomposed in the Hall
electrolytic process?
413. It is necessary to produce 1 ton of molten iron for a thermit welding
operation. How much aluminum will be required?
414. What weight of chromium metal can be produced by the thermit
process from 76 kg. of a chromium ore containing 50% of chromium
oxide?
415. 10 grams of anhydrous aluminum chloride were prepared by the
action of hydrochloric acid on aluminum. What weight of alumi-
num was dissolved?
416. 4.77 grams of crystallized aluminum sulfate lost 2.3 grams of water
after being heated. Find its formula.
417. 0.37 gram of aluminum liberated 0.04 gram of hydrogen from a
strong solution of sodium hydroxide. Knowing the valency of
aluminum to be 3, find the atomic weight of aluminum.
418. What weights of materials must be used in the thermit process to
form 14 Ib. of molten iron?
132
39. ALUMINUM HYDROXIDE AND ALUMS
f A1 2 (S0 4 ) 3 + 6NH 4 OH >|2A1(OH) 3 + 3(NH4) 2 SO 4
Lab. Prep. | (Precipitation of A1(OH) 3 )
l3Ca(OH) 2 + A1 2 (S0 4 ) 3 >-|2Al(OH) + 3CaS0 4
Com. Prep. K 2 SO 4 + A1 2 (S0 4 ) 3 + 24H 2 O > 2KA1(S0 4 ) 2 12H 2
(Common alum)
f 3KOH + AI(OH) 8 ?- K 3 A10 3 + 3H 2 (With strong
bases, A1(OH) 3 acts as an acid)
I 3H 2 S0 4 + 2A1(OH) 3 < *- A1 2 (S0 4 ) 3 +'6H 2 (With strong
] acids, A1(OH) 3 acts as a base)
2A1(OH) 3 ^A1 2 3 + 3H 2 O (Al(OH). when heated
forms A1 2 3 )
419. Determine the percentage composition of chrom alum (KCrSO^z
12H 2 0, and ammonium alum, NH 4 A1(S0 4 )2 12H 2 0.
420. What weight of aluminum hydroxide can be precipitated by 17.1 Ib.
of aluminum sulfate?
421. 117 grams of aluminum hydroxide is to be prepared from aluminum
sulfate. What weight of ammonia (NHs) will be required?
422. How much aluminum oxide can be obtained from 39 grams of
aluminum hydroxide?
423. It is required to prepare 2.34 grams of dry aluminum hydroxide.
What weight of aluminum sulfate will be needed?
424. What weight of commercial NaOH (containing 10% water) will be
required to neutralize 195 grams of aluminum hydroxide?
425. What weights of potassium and chromium sulfates will be needed
to prepare 99.8 grams of chrome alum, KCr(S0 4 ) 2 12H 2 0?
426. What weight of crystallized ferric ammonium alum can be made
from 40 grams of ferric sulfate and 15 grams of ammonium sulfate?
427. How many pounds of water are found chemically combined in 237
Ib. of common alum?
428. 19 grams of aluminum hydroxide lost, on heating, 6.56 grams,
leaving A1 2 3 . What is the molecular weight of aluminum hy-
droxide?
429. 0.918 gram of Al(OH)s is obtained from an aluminum alloy
weighing 2.5 grams. What percentage, of aluminum is present in
the alloy?
40. ALUMINUM SILICATES AND CEMENT
5CaC0 3 + 2HAlSi0 4 - >- 2CaSiO 3 + Ca^AlO^ + 5CO a + H 2 O
(limestone) + (clay) - > (cement) (Preparation
of cement)
Ca 3 (A10 3 } 2 4- 6HsO - > 3Ca(OH) 2 + 2HaAlQs (Setting of cement)
Felspar, KAlSuOa | g ome common aluminum silicates
Mica, KA1S10 4 found in nature
Clay, HsAMSiOOs H 2 J
430. Determine the percentage composition of the aluminum silicates
given above.
431. Determine the percentage composition of mica by elements and
oxides.
432. What weight of water will be needed in the setting of 60 Ib. of
cement?
433. Assuming that the first equation above represents the chemical
reaction for the making of cement, what weights of limestone and
clay should be heate$ together to obtain 100 tons of cement?
434. A compound contains 24.7% potassium, 17.1% aluminum, 17.7%
silicon, and 40.5% oxygen. Determine its simplest formula.
435. What volume of carbon dioxide is liberated in the making of 5.92
kg. of cement?
436. 120 tons of a clay containing 80% HAlSiCU are available. What
weight of cement can be made from this clay?
437. A compound contains 14% potassium, 10% aluminum, 30% silicon,
and 46% oxygen. Determine the simplest formula of this sub-
stance.
438. What weight of aluminum is contained in a sample of garnet con-
taining 98% of CasAlsCSiO^s and weighing 7.5 g. ?
439. In the complete setting of a sample of cement, 3.7 grams of Ca(OH) 2
were produced. What weight of limestone was used in the making
of this cement ?
440. What volume of carbon dioxide measured at 17 and 760 mm. is
liberated in the making of 1184 tons of cement?
41. TIN AND THE ELECTROCHEMICAL SERIES
Com. Prep. Sn02 + C > Sn + ^ C0 2 (Reduction of cassiterite)
Sn + 2HC1 > SnCl 2 -f -f H 2 (Action of tin on hot con-
centrated hydrochloric acid)
Sn + 4H 2 S0 4 > Sn(S0 4 ) 2 + 2S0 2 + 4H 2 O (Tin acts
Chem. Prop. { like copper on HaSO*)
SnCl 2 +Zn ' >- ZnCl 2 + i' Sn (Replacement of tin by zinc)
2HgCl + SnCl 2 > 2Hg + SnCU (Complete reduction
of a mercury salt)
Electrochemical Series : Na Ca Mg Al Zn Fe Ni Sn Pb
H Cu As Hg Ag Pt Au
441. Determine the percentage composition of the two chlorides of tin.
442. A compound contains 52.6% tin, 31.4% chlorine, 1.8% hydrogen,
and 14.2% oxygen. What is the simplest formula of this com-
pound?
443. How much carbon would be needed for the reduction of 151 kg. of
tin oxide?
444. 50 grams of free mercury were obtained by the reduction of a
solution of mercuric chloride with a stannous chloride solution.
What weight of bichloride of mercury was reduced ?
445. What weight of carbon will be required for the reduction of 302 Ib.
of cassiterite, containing 75% Sn0 2 ?
446. What weight of ferric chloride could be reduced to the ferrous con-
dition with the aid of 9.5 grams of stannous chloride?
447. In the reduction of a sample of SnOa, 250 cc. of carbon dioxide,
measured under standard conditions, were obtained. What was
the weight of oxide reduced?
448. What weight of zinc would be needed to replace completely all the
tin in 10 liters of a solution containing 38 grams of stannous chloride
per liter ?
449. 10 grams of tin were heated in concentrated HC1, and as a result
55 cc. of hydrogen, measured at 17 C. and 760 mm., were given off.
What weight of the tin actually dissolved in the acid ?
450. What weight of tin will be needed to replace completely all the
copper present in a solution containing 25 grams of crystallized
copper sulfate ?
42. LEAD
f 2Pb + 30 2 >-2PbO + |2S0 2 (Roasting of galena)
Com. Prep, i PbS + 2PbO > SPb + | S0 2 (Reaction of the oxide and
[ sulfide of lead)
f Pb + 2HC 2 H 3 2 >- Pb(C 2 H 3 2 ) 2 + >f H 2 (Lead soluble
in acetic acid)
Pb+2HX0 3 *-Pb(N0 3 ) 2 +^ H 2 (Lead soluble in
nitric acid)
Chem. Prop. { ^Q^ 6pb + 4C Q 2 + 2H 2 - 2[Pb(OH) 2 2PbCO 3 ]
(Manufacture of white lead)
Pb(NO s )j + 2 CrO 4 > 2KN0 3 + ^ PbCrO* (Test for
a cliromate)
451. Determine the percentage composition of white lead as made by
the Dutch process, if its composition is expressed by the formula
written above.
452. A compound was found to contain 90.65% lead and 9.35% oxygen.
WTiat is its simplest formula?
453. Find the simplest formula of a compound which on analysis gave the
following composition : 62.45% lead, 8.7% nitrogen, and 28.85%
oxygen.
454. A sample of galena from Colorado showed on analysis the presence
of 95% PbS. 956 tons of this ore were roasted. What weight of
lead was obtained?
455. What weight of lead would have to be treated to obtain one ton of
white lead?
456. At the prices of 10?! per pound for lead and 6$ per pound for glacial
acetic acid, what would be the cost of the chemicals used in prepar-
ing one ton of pure lead acetate?
457. What weight of lead will be found in a volume of tetraethyl lead,
Pb(C 2 H 5 ) 4 , weighing 9.69 grams?
458. Potassium chromate is added to a solution containing lead nitrate,
and 6.46 grams of lead chromate are precipitated. What weight of
lead nitrate was present in solution?
459. In converting 111.5 Ib. of PbO into PbsO-j, what volume of oxygen
is used?
460. 8.78 grams of PbO yielded 8. 15 grams of lead. What is the atomic
weight of lead?
136
43. NICKEL AND COBALT
r 2NiS + 30 2 - >- 2NiO + | 2S0 2 (Roasting of nickel
Com. Prep. | sulfide)
1 NiO + C - >- Ni + /}> CO (Reduction of nickel oxide)
Ni+2HN0 3 - >- Ni(N0 3 ) 2 +^H 2 (Reacts readily
with HN0 3 )
Cfcem. Prop. CQ + 2HNOs - ^ Co(N0 3 ) 2 + | H 2
. Ni + 4CO - >- Ni(CO) 4 (Formation of nickel carbonyl)
461. Determine the amount of Ni present in the nickel ammonium sul-
fate crystals, NiSO^NH^SCX GH^O, used in the nickel plating
. bath.
462. Determine the percentage composition of cobaltite (CoAsS),
crystallized cobalt nitrate, Co(NOa)2 6HoO, and nickel glance
(NiAsS).
463. Cobalt is obtained chiefly from cobalt glance, whose formula may
be represented as CoAso CoSa- Determine the weight of cobalt
available in 1 ton of this ore containing 50% cobalt glance.
464. How much of a 5% solution by weight of cobalt nitrate can be
obtained by dissolving 5.9 grams of cobalt in nitric acid?
465. 50 cc. of carbon monoxide completely react with metallic nickel.
What weight of nickel carbonyl is produced?
466. 27.3 Ib. of nickel sulfide (NiS) are roasted. What weight of nickel
oxide and what volume of sulfur dioxide are produced?
467. Which will produce a greater volume of hydrogen : 6 grams of nickel
or 5.5 grams of zinc reacting with nitric acid?
468. A manufacturer requires 1 Ib. of pure nickel as a catalyst in the
hydrogenation of an oil. What weight of pure NiSCX must be
treated to obtain this weight of catalyst?
469. What volume of carbon monoxide will be liberated by the decom-
position of 5 liters of the vapor of nickel carbonyl?
470. What weight of nickel will be left in the above operation?
471. 12.5 grams of nickel oxide were reduced in a stream of hydrogen
and lost 2.67 grams in weight. The specific heat of nickel is 0.095.
Calculate the atomic weight of this element.
44. MANGANESE AND CHROMIUM
Com. Prep.
>- 2A1 2 3 + 3Mn (Reduction of the
Al 2 0s + 2Cr (Preparation of chro-
Chem. Prop. ,
2K 2 Cr0 4 + H 2 (Reduction of
> 2KC 2 H 3 2 + | PbCrO 4
f3MnO 2 + 4A1 -
oxide with Al)
Cr 2 3 + 2A1
mium)
Mn 4- 2HC1 - MnCl 2 + -f H 2 (Reacts with HC1)
5Na 2 S0 3 + 2KMn0 4 + 3H 2 S0 4 > 5Na 2 SO 4 + K 2 S0 4
+ 2MnS0 4 + 3H 2 O (KMnO 4 an oxidizing agent)
2K 2 Cr0 4 + H 2 S0 4 > K 2 Cr 2 7 + H 2 + K 2 S0 4
(Oxidation of K 2 CrO 4 )
K 2 Cr 2 7 + 2KOH >
K 2 Cr0 7 )
K 2 CrO 4 + Pb(C 2 H 3 O 2 ) 2 -
(Test for a chromate)
472. Determine the percentage composition of chromite, Fe(Cr02)2, and
hausmannite, Mns0 4 .
473. What Is the factor of K 2 in potassium permanganate?
474. In 100 parts of a compound are found 63.2 parts of manganese and
36.8 parts of oxygen. What is its simplest formula?
475. What weight of chrome yellow, PbCr04, is formed when a solution
containing 9.7 grams of potassium chromate is precipitated with
lead acetate?
476. WTiat weight of aluminum powder must be used in the thermit
process to obtain all the chromium from 100 grams of C^Os?
477. A solution containing 4.5 grams of sodium sulfite completely
decolorized a solution of potassium permanganate. What weight
of KMn0 4 was present in the solution?
478. A pyrolusite ore was found on analysis to contain 85% Mn0 2 .
What weight of manganese could be obtained by the thermit process
from 10 tons of this ore?
479. What weight of sodium sulfite can be oxidized by 79 grams of
potassium permanganate?
480. What volume of hydrogen will be evolved when 100 cc. of HC1
containing 20% HC1 (Sp. Gr., 1.1) completely reacts with
manganese?
481. What weight of chrome yellow will be precipitated by lead acetate
added, in sufficient quantities, to a solution containing 9.7 Ib. of
potassium chromate?
138
45. DISTILLATION PRODUCTS OP COAL, WOOD, AND
PETROLEUM
DESTRUCTIVE
DISTILLATION
FRACTIONAL DISTILLATION
Coal (2000 Ib.)
Wood
Petroleum
Illuminating gas 10,000 cu. ft.
Coke .... 1325 Ib.
Gas fuel
Gasoline CeHn and OiHii
Naphtha CrHw and CaHis
Hydrogen sulfide . (variable)
( benzene"]
Coal tar < phenol > . 12 gal.
I pitch J
Pyroligneous/ ac f cacid
acid acet f e , . .
I wood alcohol
Kerosene CioHza and Ci6H34
Lubricating o Is
Paraffin CasHis and CasHss
Vaseline CasEUs and CziELw
482.
determine its
Assuming the composition of vaseline to be C 2
percentage composition.
483. A ton of coal yielded 5 Ib. of ammonia gas. What is the percentage
of nitrogen in the coal?
484. Determine the percentage composition of petroleum ether, assuming
that it consists of equal parts of CsHi 2 and CeHi4.
485. The vapor density of one of the hydrocarbons found in gasoline is
49.9. It contains 83.9% carbon and 16.1% hydrogen. Determine
its true formula.
486. What weight of oxygen would be used in -completely burning
10 grams of petroleum ether, assuming it to be composed of equal
parts of C5Hi2 and CeBDu?
487. What volume of air would be required for the complete combustion
of 1 Ib. of kerosene, assuming the formula of kerosene to be CioH 2 2?
488. A coal, on analysis, showed the presence of 0.75% nitrogen. What
volume of ammonia could be obtained from 100 tons of this coal,
assuming all its nitrogen to be changed to ammonia during the
destructive distillation?
489. An analysis of coal gas showed the presence of 50% hydrogen, 35%
methane, and 5% carbon monoxide, by volume. What volume of
oxygen will be needed to burn completely 1000 cu. ft. of this gas?
490. What volume of carbon dioxide would be produced in the above
reaction?
A ton of coal gave 10,000 cu. ft. of illuminating gas containing 30.5%
methane and 5% carbon monoxide by volume. What weight of
carbon dioxide would be produced by the complete combustion of
this volume of gas ?
491.
46. HYDROCARBONS
METHAXE SERIES
ETHVLENE SERIES
ACETYLENE SERIES
CKj methane
C 2 He ethane
CaHg propane
CHCI 3 chloroform 1 halogen
CHI S iodoform > substitution
CCU carbona j products
C 2 H 4 ethylene
CsHe propylene
C 2 H 2 acetylene
BENZENE SERIES
CoH 6 benzene
CoHsCH 3 toluene
2C S H 6 + 15O 2 >- 12CO 2 4- 6H 2 O (CO 2 and H 2 always formed by the
complete combustion of a hydrocarbon)
CS- + 3C1 >- S 2 C1 2 + CCU (Preparation of carbon tetrachloride)
CH 3 aH 5 + 3HN0 3 ^3H 2 + CH 3 C 6 H 2 (N0 2 ) 3 (Formation of TNT)
492. Calculate the percentage composition of propane, C 3 H 8 .
493. The weight of a liter of methane is 0.72. Calculate its vapor
density, specific gravity, and molecular weight.
494. Any member of the paraffin (methane) series may be represented by
the formula C n H 2n+ 2- Determine the formula of that member of
this series containing IS atoms of carbon in its molecule.
495. Determine the percentage composition of all of the four chlorine
substitution products of methane (CH 3 C1, CH 2 C1 2 , CHC1 3 ,
ccy.
496. A liter of ethane weighs 1.34 grams. Find its molecular weight,
vapor density, and specific gravity.
497. Propane when vaporized gives a vapor density of 22. It contains
81.81% carbon and 18.19% hydrogen. F.rom these data calculate
its true formula.
498. 0.58 gram of acetylene gas occupies a volume, under standard con-
ditions, of 500 cc. Determine its specific gravity, vapor density,
molecular weight, and weight of 1.5 liters.
499. The vapor density of ethane was found to be 15. Determine the
true formula of this compound which contains 80% carbon.
500. How many grams of oxygen will be necessary for the complete com-
bustion of 3.2 grams of marsh gas?
501. Find the weights of each of the products formed in the complete
combustion of 17 grams of camphene, CioHie-
502. What weight of iodine would be required to prepare 19.7 Ib. of
iodoform?
140
47. ALCOHOLS
C 5 Hi 2 6 ^2C 2 H 5 OH +-f2C0 2 (Preparation of ethyl alcohol by the
fermentation of glucose)
2CH 3 OH + 2 >- 2H 2 + 2HCHO (Formaldehyde, test for wood
or methyl alcohol)
C 2 H 5 OH + GNaOH + 4I 8 >-| CHI 3 + HCOONa + 5NaI + 5H 2 O
(lodoform test for ethyl alcohol)
C C H 5 OH + 3HN0 3 =>- C 6 H 2 OH(NO 2 ) 3 + 3H 2 (Preparation of picric
acid from phenol)
503. Determine the percentage composition of glycerine, C 3 H 6 (OH) 3 ,
and wood alcohol, CH 3 OH.
504. 167 cc. of alcohol vapor weigh 0.34 gram. It is composed of 52.2%
carbon, 13% hydrogen, and 34.8% oxygen. Determine the true
formula of this alcohol.
505. What weight of phenol (C 6 H 5 OH) can be obtained from 234 grams
of benzene (C 6 H 6 ) ?
506. What weight of alcohol will be .needed to prepare 197 Ib. of
iodoform ?
507. What weight of formaldehyde could be theoretically obtained from
112 grams of denatured alcohol containing 2% wood alcohol?
508. 23 grams of ethyl alcohol are burned. What weights of CO 2 and
H20 are formed?
509. What weight of alcohol can be obtained by completely transforming
9 grams of glucose containing 10% water?
510. 100 grams of wine were treated with sodium hydroxide and iodine.
91.97 grams of iodoform were precipitated. What percentage of
alcohol was present in the wine?
511. What weight of glycerine can be obtained from 89 Ib. of beef fat
whose formula might be considered to be CsH^CnHasCOO^?
(See the fifth equation under Esters, page 144.)
512. A manufacturer must make 100 tons of picric acid. What weight of
phenol will he require, and what will he have to pay for this at the
price of 23 i per pound?
513. What volume of carbon dioxide results from the fermentation of
4.5 grams of glucose?
514. Upon oxidation, methyl alcohol was changed to 500 cc. of formal-
dehyde. What weight of methyl alcohol was oxidized ?
515. 5.8 grams of acetone were dissolved in 100 grams of water. The
boiling point was raised 0.52 C. What is the molecular weight
of acetone?
141
48. ALDEHYDES, KETONES, AND ETHERS
2CH 3 OH 4- 2 *- 2H 2 Q + 2HCHO (Prep, of formaldehyde)
3HCHO >- H 3 C 3 H 3 O S (Paraformaldehyde)
CsEUCHs 4- 2C1 2 + H 2 > 4HC1 + C 6 H 5 CHO (Benzaldeliyde or oil of
bitter almonds)
(CH s COO) 2 Ca >- CaCOs + CH 3 COCH 3 (Formation of acetone)
C 2 HsHSO 4 + CoHsOH > H 2 SO 4 + (C 2 H 5 ) 2 O (Preparation of ether)
HCHO + Ag 2 O -HCOOH + ^2Ag (Use of formaldehyde in sil-
vering mirrors)
516. Determine the percentage composition of (a) acetone, (b) ethyl
ether, and (c) formaldehyde.
517. A compound contains 40% carbon, 6.8% hydrogen, and 53.2%
oxygen. Its vapor density is 14.95. Find the true formula of this
compound.
518. 250 cc. of the vapor of ether weigh O.S3 gram under standard
conditions. Calculate its vapor density, specific gravity, and
molecular weight.
519. What weight of acetone can be prepared from 79 Ib. of calcium
acetate?
520. What weight of pure ethyl alcohol will be required to manufacture
37 Ib. of ethyl ether?
521. A manufacturer must make 424 Ib. of oil of bitter almonds. What
weights of toluene (C 6 H 5 CH 3 ) and chlorine must he use ?
522. What weight of formaldehyde will be required to precipitate 12
grams of silver from a silver solution, to cover the back of a mirror?
523. What volume of air will be required to oxidize sufficient wood
alcohol to produce 5 grams of formaldehyde?
524. What volumes of carbon dioxide and water vapor are formed when
1.48 grams of ether are completely burned?
525. 1.5 grams of silver are precipitated out of a silver solution by
formaldehyde as the reducing agent. What weight of ECHO was
required?
142
49. ORGANIC ACIDS
Oxalic acid, COOH COOH (Obtained from sawdust)
Tartaric acid, H2C4ELi06 (Obtained from fermentation of grapes)
Citric acid, HsCeBUOr (Found in lemons and limes)
Stearic acid, CnEUsCOOH (Obtained from fats)
C 2 H 6 OH + O 2 - *- CHsCOOH + H 2 (Acetic acid from ethyl alcohol)
C 3 H 6 (Ci7H3 5 COO) 3 + 3H 2 - > C 3 H5(OH) 3 + 3C 17 H 36 COOH (Hydrol-
ysis of fat)
526. Determine the percentage composition of citric acid and of tartaric
acid.
527. How much Rochelle salt (KNaC4H 4 6 ) could be prepared from
100 Ib. of tartaric acid?
528. How much cream of tartar, KHC^H^e, could be prepared from
10 kg. of tartaric acid?
529. A compound contained hydrogen, oxygen, and carbon in the propor-
tion of 4.6%, 67.2%, and 28.2%, respectively. Its molecular weight
was found to be 150 by the depression 'of the freezing point method.
Find its true formula.
530. A barrel of wine went sour and 10.3 Ib. of acetic acid were found
present. What weight of pure grain alcohol was oxidized?
531. What weight of ethyl alcohol on oxidation yielded 36.6 grams of
acetic acid?
532. What weight of stearic acid can be made from a fat consisting
entirely of glycerol stearate and weighing 4.45 Ib. ?
533. What weight of tartaric acid would be needed to make 64.6 Ib. of
tartar emetic whose formula is
534. Oxalic acid contains 26.8% carbon, 2.2% hydrogen, and 71% oxygen.
Determine its simplest formula.
535. What volume of air would be required to change completely 2.3 kg.
of ethyl alcohol into acetic acid?
536. The vapor density of acetic acid was found to be 30. Find (a) the
weight of 40 cc. of its vapor, (6) its specific gravity, and (c) its
molecular weight.
Preparation
50. ESTERS
CHsCOOH + C 2 H 5 OH - > H 2 O + CH 3 COOC 2 H 5
(Ethyl acetate)
HN0 3 + C 2 H 6 OH - >- H 2 + C 2 H 5 N0 3 (Ethyl nitrate)
3HN0 3 + C 3 H E (OH) 3 - >- 3H 2 + C 3 H5(N0 3 )3 (Nitro-
glycerine)
CHsCOOH + C 5 HuOH - > H 2 + C 6 HnCOOCH 3
(Amyl acetate or banana oil)
C 3 Hs(OOCH3 B C 17 ) 3 + 3H 2 O - > CsHtfOH), + 3C n HCOOH
(Hydrolysis of glyceryl stearate or fat)
4CjHB ( NOj j j - ^ 12C O 2 + 10H 2 O + 6N 2 + O 2
(Decomposition of nitroglycerine) .
537. Determine the percentage composition of ethyl nitrate.
538. A compound contains 26.4% carbon, 15.4% nitrogen, 5.5% hydro-
gen, and 52.7% oxygen. Its molecular weight is found to be 91.
What is its true formula?
539. It is required to prepare 110 grams of ethyl acetate. What weights
of alcohol and acid will be needed ?
540. What weight of nitric acid will be used in the making of 2270 tons
of nitroglycerine?
541. What weight of glacial acetic acid would be required to prepare
13 Ib. of amyl acetate (banana oil) ?
542. What weight of ethyl alcohol would be needed for the production of
273 grams of ethyl nitrate?
543. How much glycerine will be needed for the manufacture of 1 135 tons
of nitroglycerine?
544. How much glycerine would be available from 108 Ib. of a fat con-
taining 70% glyceryl stearate ?
545. Oil of wintergreen is made by the action of methyl alcohol (CHsOH)
and salicylic acid, Ce^OHCOOH. Determine the weights of the
substances necessary for the manufacture of 10 Ib. of this ester.
546. A manufacturer gets an order for 450 Ib. of banana oil. What
weights of amyl alcohol and acetic acid must he have to fill this
order?
144
Com. Prep.
51. SOAPS
f CnHssCOOH + NaOH - > H 2 + Ci 7 H 35 COONa
(Sodium stearate, hard soap)
, CnHgsCOOH + KOH - >- H 2 + Ci 7 H 36 COOK (Potas-
"' ' sium stearate, soft soap)
2Ci 7 H 35 COONa+CaS04 - >- Na 2 S0 4 + f (Ci7H 3 5COO) 2 Ca
(Calcium stearate, insoluble soap)
C 3 H 5 (CnH 36 COO) 3 + SNaOH - > 3NaCi 7 H 35 COO
+ C 3 Hs(OH) 3 (Saponification of glyceryl stearate)
C 3 H 5 (COOCi 7 H 35 ) 3 -f 3H 2 O - >- 3HCi 7 H 36 COO
+ C3H 5 (OH) 3 (Hydrolysis of glyceryl stearate)
547. Determine the percentage composition of sodium stearate.
548. What weight of glycerine can be obtained from 19SO Ib. of glyceryl
stearate?
549. What weight of stearic acid can be obtained from 17.8 Ib. of glyceryl
stearate ?
550. Determine the simplest formula of a fatty acid containing 76%
carbon, 12.7% hydrogen, and. 11. 3% oxygen.
551. A soap contains 12.1% potassium, 67% carbon, 10.9% hydrogen, and
10% oxygen. What is its simplest formula?
552. What weight of CaCl 2 2HaO will be needed to precipitate com-
pletely 10.2 grams of sodium stearate?
553. A soft soap is added to some hard water and results in the precipi-
tation of 5.9 grams of magnesium stearate. What weight of MgSC>4
was present in the water?
554. The saponification value of a fat is the number of milligrams of
KOH required to saponify 1 gram of the fat. What weight of KOH
would be required to saponify 89 kg. of a fat whose composition
might be represented as
555. A gallon of hard water was treated with an excess of hard soap and
0.2 gram, of calcium stearate precipitated. What weight of CaSO*
was present in the water?
556. 1 liter of Clarke's Hardness Solution 'contains 2.29 grams of CaClj
in solution. What volume of this solution will be required to pre-
cipitate 4.58 grams of-soft soap whose formula might be considered
to be Ci 7 H 35 COOK?
Starch,
Cellulose, (C 6 Hio0 5 )y
Glucose,
Sucrose,
52. CARBOHYDRATES
6C0 2 + 5 H 2
synthesis)
CeHwOs
*~460 2 + C6H 10 5 (Photo-
6C + 5 H 2 (Action of cone.
EjSOO
CeHioOs + H 2 > C 6 Hi 2 O 6 (Hydrolysis of
starch)
CiaH 22 Oii + H 2 >- CeHijOe -J- CsH^Oa
(Inversion of cane sugar)
2C s Hio0 5 +6HNO s .
cellulose)
6H 2 + C 12 Hi 4 4 (N03)6 (Production of nitro-
557. Determine the percentage composition of (a) cellulose, (&) sucrose,
and (c) glucose.
558. What is the simplest formula of a carbohydrate which gave on
analysis 42.1% carbon and 57.9% water?
559. A compound contained 40% carbon, 6.7% hydrogen, and 53.3%
oxygen. 90 grams of this substance lowered the freezing point of
1000 grams of water to - 0.93 C. What is the true formula of
this compound?
560. What weight of pure carbon can be obtained from 171 Ib. of dry
sucrose?
561. 12.6 grams of dextrose dissolved in 100 grams of water raised the
boiling point 0.364 C. What is its molecular weight ?
562. What weight of fructose (CeH^Oe) can be obtained by the complete
inversion of 6.84 grams of cane sugar?
563. A plant manufactured 4.05 grams of starch. What volume of
oxygen did the plant liberate during the process?.
564. It has been calculated that an average man exhales 450 liters of
carbon dioxide in a day. Calculate the weight of starch a plant
could manufacture from this CCV
565. In photosynthesis what volume of oxygen is liberated by the plant
during its manufacture of 1 Ib. of starch?
566. What weight of sucrose must be inverted to furnish 72 Ib. each of
glucose and fructose?
567. What weight of cotton,
required to make 1188
given above to reprei
will be
ing the equation
APPENDIX
/ '
r 4 ^/ f r>
INTERNATIONAL TABtE^GF ATOMIB
CHEMldtXEXEMENTS (1925)
THE
Q-lFlm/IT
ATOMIC
ATOlife.,]
^VAvr-1
rswwf.-
^AxoVJ
\\-
,
\''..'"'i":i
,
- - '--'-
-. ^-sL'. ;.'
1" "^'"
it ..in;..
Aluminum .
Al
13
26.97 1
Molybdenum
Mo
42
96.0
Antimony .
Sb
51
121.77
Neodymium
Nd
60
144.27
Argon . .
A
18
39.91
Neon . . .
Ne
10
20.2
Arsenic . .
As
33
74.96
Nickel . .
Ni
28
58.69
Barium . .
Ba
56
137.37
Nitrogen
N
7
14.008
Beryllium .
Be
4
9.02
Osmium . .
Os
76
190.8
Bismuth . .
Bi
83
209.00
Oxygen . .
8
16.000
Boron . -
B
5
10.82
Palladium .
Pd
46
106.7
Bromine . .
Br
35
79.916
Phosphorus
P
15
31.027
Cadmium .
Cd
48
112.41
Platinum .
Pt
78
195.23
Calcium . .
Ca
20
40.07
Potassium .
K
19
39.096
Carbon . .
C
6
12.000
Praseodym-
Cerium . .
Ce
58
140.25
ium . .
Pr
59
140.92
Cesium . .
Cs
55
132.81
Radium . .
Ra
88
225.95
Chlorine . .
Cl
17
35.457
Radon . .
Rn
86
222.
Chromium .
Cr
24
52.01
Rhodium
Rh
45
102.91
Cobalt . .
Co
27
58.94
Rubidium .
Rb
37
85.44
Columbium
Cb
41
93.1
Ruthenium
Ru
44
101.7
Copper . .
Cu
29
63.57
Samarium .
Sm
62
150.43
Dysprosium
Dy
66
162.52
Scandium .
Sc
21
45.10
Erbium . -
Er
68
167.7
Selenium
Se
34
79.2
Europium .
Eu
63
152.0
Silicon . .
Si
14
28.06
Fluorine . .
F
9
19.00
Silver . .
Ag
47
107.880
Gadolinium
Gd
64
157.26
Sodium . .
Na
11
22.997
Gallium . .
Ga
31
69.72
Strontium .
Sr
38
87.63
Germanium
Ge
32
72.60
Sulfur . .
S
16
32.064
Gold . . .
Au
79
197.2
Tantalum .
Ta
73
181.5
Hafnium
Hf
72
178.3
Tellurium .
Te
52
127.5
Helium . .
He
2
4.00
Terbium
Tb
65
159.2
Holmium -
Ho
67
163.4
Thallium .
Tl
81
204.39
Hydrogen .
H
1
1.008
Thorium
Th
90
232.15
Indium . .
In
49
114.8
Thulium .
Tm
69
169.4
Iodine . .
I
53
126.932
Tin ...
Sn
50
118.70
Indium . .
Ir
77'
193.1
Titanium .
Ti
22
48.1
Iron . . .
Fe
26
55.84
Tungsten .
W
74
184.0
Krypton
Kr
36
82.9
Uranium
U
92
238.17
Lanthanum
La
57
138.90
Vanadium .
V
23
50.96
Lead . . .
Pb
82
207.20
Xenon . .
Xe
54
130.2
Lithium . .
Li
3
6.940
Ytterbium .
Yb
70
173.6
Lutecium .
Lu
71
175.0
Yttrium . .
Y
39
88.9
Magnesium
Mg
12
24.32
Zinc . . .
Zn
30
65.38
Manganese .
Mn
25
54.93
Zirconium .
Zr
40
91.
Mercury
Hg
80
200.61
149
ELEMENT
SYMBOL
ATOMIC
WEIGHT
COMMON
VALENCE
ELEMENT
SYMBOL
ATOMIC
WEIGHT
COMMON
VALENCE
Aluminum
Al
27
HI
Iron
Fe
56
II, III
Antimony
Sb
122
HI
Lead
Pb
207
H
Arsenic
As
75
HI
Lithium
Li
7
I
Barium
Ba
137
n
Magnesium
Mg
24
n
Bismuth.
Bi
209
HI
Manganese
Mn
55
II, IV
Boron
B
11
HI
Mercury
Hg
200
I, II
Bromine
Br
80
I
Nickel
Ni
59
II
Cadmium
Cd
112
H
Nitrogen
N
14
III, V
Calcium
Ca
40
H
Oxygen
16
II
Carbon
C
12
IV
Phosphorus
P
31
III,V
Chlorine
Cl
35.5
I
Platinum
Pt
195
IV
Chromium
Cr
52
HI
Potassium
K
39
I
Cobalt
Co
59
H
Silicon
Si
28
IV
Copper
Cu
64
I,H
Silver
Ag
108
I
Fluorine
F
19
I
Sodium
Na
23
I
Gold
Au
197
I, HI
Strontium
Sr
87
II
Helium
He
4
Sulfur
S
32
II, IV
Hydrogen
H
1
I
Tin
Sn
119
11,1V
Iodine
I
127
I
Zinc
Zn
65
II
150,
THE METRIC SYSTEM
1 decimeter (dm.) = 0-1 meter (m.)
1 centimeter (cm.) = 0.01 meter (m.)
1 millimeter (mm.) = 0.001 meter (m.)
1 kilometer (km.) = 1000 meters (m.)
1 decigram (dg.) = 0.1 gram (g.)
1 centigram (eg.) = 0.01 gram (g.)
1 milligram (mg.) = 0.001 gram (g.)
1 kilogram (kg.) = 1000 grams (g.)
UNIT EQUIVALENTS (approximate)
1 inch (in.) ' = 2.54 centimeters (cm.)
1 square inch (sq. in.) = 6.45 square centimeters (sq. cm.)
1 square meter (m.) = 10.76 square feet (sq. ft.)
1 cubic inch (cu. in.) = 16.4 cubic centimeters (cc.)
1 cubic foot (cu. ft.) = 28.3 liters (1.)
1 cubic foot (cu. ft.) = 7.48 gallons (gal.)
1 liter (1.) = 61 cubic inches (cu. in.)
1 liter (1.) = 1.06 quarts (qt.)
1 gallon (gal.) = 231 cubic inches (cu. in.)
1 cubic meter (cu. m.) = 1.31 cubic yards (cu. yd.)
1 cubic meter (cu. m.) = 35.32 cubic feet (cu. ft.)
1 gram (g.) = .15.43 grains (gr.)
1 ounce (oz.) Avoirdupois = 28.35 grams (g.)
1 kilogram (kg.) = 2.2 pounds (Ib.)
1 ton (T.), short, 2000 Ib. = 907.2 kilograms (kg.)
1 ton (T.), long, 2240 Ib. = 1016 kilograms (kg.)
1 liter of water = 2.2 pounds (Ib.)
1 cubic foot of water = 62.4 pounds (Ib.)
1 gallon of water . = 8.35 pounds (Ib.)
1 atmosphere pressure = 1033 grams per square centimeter
THERMOMETER SCALES
On the Centigrade scale, the zero point is the height of the mer-
cury when the thermometer is immersed in a mixture of ice and
water when the atmospheric
pressure is 760 mm. The 100
mark is that point representing
373"A. tne nei S nt of tne mercury when
the thermometer is immersed in
steam vapor under the same
condition of pressure. On the
Fahrenheit scale, the freezing
po mt of water is marked at 32
an( i the boiling point of water
at 212, making the absolute
zero on this scale 459.4.
On the Absolute scale, the ab-
solute zero is 273 C. The
boiling point of water reads
373 on this scale (Fig. 9).
Thermometer Conversion Rules
1. " To convert Centigrade read-
ings to the corresponding
Fahrenheit readings, multi-
ply the Centigrade reading
by 1.8 and add 32.
EXAMPLE. What is the Fahren-
heit reading corresponding to
18 C.?
18 X 1.8 + 32 = 64.4 F.
2. To convert Fahrenheit readings to the corresponding Centigrade
readings, subtract 32 degrees from the Fahrenheit reading and
' divide by 1.8.
EXAMPLE. What is the Centigrade reading corresponding to 100 F ?
Absolute zero
Fahrenheit Centigrade
thermometer thermometer
scale scale
FIG. 9.
Absolute
scale
IS2
3. To convert Centigrade to Absolute scale readings, add 273 to
the Centigrade.
EXAMPI. What is the Absolute scale reading corresponding to 43 C. ?
43 + 273 = 230 Absolute.
4. To convert Fahrenheit to Absolute scale readings, first convert
to Centigrade reading and then add 273.
EXAMPLE. What is the Absolute scale reading corresponding -to 68 F. ?
68-32
1.8
+ 273 = 20 -f 273 = 293 Absolute.
APPROXIMATE SPECIFIC HEATS OF THE COMMON ELEMENTS
(Measured at room temperature)
ELEMENT
SPECIFIC HEAT
ELEMENT
SPECIFIC HEAT
Aluminum
0.215
Gold
0.030
Antimony
0.050
Iron
0.115
Arsenic
0.075
Lead
0.030
Barium
0.060
Magnesium
0.225
Bismuth
0.035
Manganese
0.095
Boron
0.305
Mercury
0.035
Cadmium
0.055
Nickel
0.095
Calcium
0.155
Platinum
0.030
Carbon
0.205
Silicon
0.125
Chromium
0.085
Silver
0.055
Cobalt
0.085
Tin
0.055
Copper
0.085
Zinc
0.090
153
SPECIFIC GRAVITY OF SULFURIC ACID SOLUTIONS AT 15 C.
(Per cent = grams of acid per 100 grams solution)
SPECIFIC
GRAVITY
PER CENT
HjSCu
SPECIFIC
GRAVITY
PER CENT
HsSOi
SPECIFIC
GRAVITY
PER CENT
HjS04
1.03
4.49
1.40
50.11
1.64
71.99
1.07
10.19
1.42
52.15
1.66
73.64
1.11
15.71
1.44
54.07
1.68
75.42
1.15
20.91
1.46
55.97
1.70
77.17
1.19
26.04
1.48
57.83
1.72
78.92
1.23
31.11
1.50
59.70
1.74
80.68
1.27
35.71
1.52
61.59
1.76
82.44
1.31
40.35
1.54
63.43
1.78
84.50
1.32
41.50
1.56
65.08
1.80
86.90
1.34
43.74
1.58
66.71
1.82
90.05
1.36
45.88
1.60
68.51
1.83
92.10
1.38
48.00
1.62
70.32
1.84
99.20
SPECIFIC GRAVITY OF HYDROCHLORIC ACID SOLUTIONS AT 15 C.
(Per cent = grams of acid per 100 grams solution)
SPECIFIC
GRAVITY
PER CENT
HCi.
SPECIFIC
GRAVITY
PER CENT
HO.
SPECIFIC
GRAVITY
PER CENT
HCL
1.01
2.14
1.08
16.15
1.15
. 29.57
1.02
4.13
1.09
18.11
1.16
31.52
1.03
6.15
1.10
20.01
1.17
33.46
1.04
8.16
1.11
21.92
1.18
35.39
10.5
10.17
1.12
23.82
1.19
37.23
1.06
12.19
1.13
25.75
1.195
38.16
1.07
14.17
1.14
27.66
1.20
39.11
SPECIFIC GRAVITY OF NITRIC ACID SOLUTIONS AT 15 C.
(Per cent = grams of acid per 100 grams solution)
SPECIFIC
GRAVITY
PER CENT
HNOj
SPECIFIC
GRAVITY
PER CENT
HNOs
SPECIFIC
GRAVITY
PER CENT
HNOs
1.01
1.90
1.21
33.82
1.41
67.50
1.03
5.50
1.23
36.78
1.43
72,17
1.05
8.99
1.25
39.82
1.45
77.28
1.07
12.33
1.27
42.87
1.46
79.98
1.09
15.53
1.29
45.95
1.47
82.90
1.11
18.67
1.31
49.07
1.48
86.05
1.13
21.77
1.33
52.37
1.49
89.60
1.15
. 24.84
1.35
55.79
1.50
94.09
1.17
27.88
1.37
59.39
1.51
98.10
1.19
30.88
1.39
63.23
1.52
99.67
154
SPECIFIC GRAVITY OF AMMONIA WATER SOLUTIONS AT 15 C.
(Per cent = grams of NH 3 per 100 grams solution)
SPECIFIC
GRAVITY
PER CENT
NH,
SPECIFIC
GRAVITY
PER CENT
NH.
SPECIFIC
GRAVITY
PER CENT
NH.
0.882
34.95
0.922
21.12
0.962
9.35
0.886
33.25
0.926
19.87
0.966
8.33
0.890
31.75
0.930
18.64
0.970
7.31
0.894
30.37
0.934
17.42
0.974
6.30
0.898
29.01
0.938
16.22
0.978
5.30
0.902
27.65
0.942
15.04
. 0.982
4.30
0.906
26.31
0.946
13.88
0.986
3.30
0.910
24.99
0.950
12.74
0.990
2.31
0.914
23.68
0.954
11.60
0.994
1.37
0.918
22.39
0.958
10.47
0.998
0.45
CONCENTRATION AND SPECIFIC GRAVITY OF
COMMON ACIDS AND BASES
REAGENT
SPECIFIC
GRAVITY
PER CENT
BY WEIGHT
Cone. HC1
Dilute HC1
1.20
1.05
39%
10%
Cone. H 2 S0 4
Dilute H 2 S0 4
1.84
1.09
98%
13%
Cone. HN0 3
Dilute HN0 3
1.42
1.11
70%
19%
Cone. NH 4 OH
Dilute NH 4 OH
0.90
0.96
28%
10%
5 N Acetic acid
1.04
30%
Commercial NaOH contains 10% EkO
Commercial KOH contains 20% H 2 O
155
TABLE OF SOME COMMON FACTORS FOR QUANTITATIVE
ANALYSIS
WEIGHED AS
REQUIRED
FACTOR
A1 2 3
Al
0.530
NH 4 C1
NH 3
0.318
MgjAsiOr
As 2 3
0.637
BaS0 4
BaCI 2
0.892
AgBr
Br
* 0.431
CaS0 4
CaO
0.412
CO 2
CaC0 3
2.274
AgCl
NaCl
0.408
PbCr0 4
Cr 2 3
0.235
CuO
Cu
0.799
Ag
CN
0.241
Agl
HI
0.545
PbS0 4
PbS
0.789
MgiPjOi
MgS0 4
1.081
MgSO 4
MgO
0.335
HgO
HgS
1.074
(NH 4 ) 2 PtCl
N0 3
0.279
P S 6
P0 4
1.338
K,Ptd,
Pt
0.402
SiF 4
Si0 2
0.578
NaCl
Na 2
0.530
NajCOj
Na 2 O
0.585
BaSO 4
S 2 O 3
0.240
S*0 2
SnCl 2
1.258
-ZnO
ZnS
1.197
156
PRICES OF COMMON CHEMICALS
(Prevailing in the New York market, March, 1926)
ACIDS
BASES
Acetic acid, 56% . . .
Boric acid
Hydrochloric 35% . . .
. $6.09, 100 Ib.
. . 9||i per Ib.
. . 90(4 100 Ib.
Ammonia water, 90% . . .
Slaked lime
Caustic soda
Caustic potash
. 6^(4 per Ib.
Ii4 per Ib.
$3.10 per 100 Ib.
Hydrofluoric, 30% . . .
. . 6(4 per Ib.
Nitric 56%
. $5.75, 100 Ib.
Sulfuric 98% ....
. $1.25, 100 Ib.
METALS
NON-METALS
Aluminum
Antimony
. . 28(4 per Ib.
. . 19(4 per Ib.
. . 45(4 per Ib.
Bromine
Chlorine
Helium
Hydrogen
Iodine
. 47(4 per Ib.
5|4 per Ib.
. 6j4 per cu. ft.
. 2(4 per cu. ft.
$4 65 per Ib
Bismuth
. . $3.35 per Ib.
. . 14|(4 per Ib.
Gold
$20.67 per troy oz.
9f (i per Ib
Nitrogen
2jfi per cu. ft.
f< per 100 cu. ft.
68j per Ib.
. . 85(4 per Ib.
Phosphorus (red)
. $89 per 75 Ib.
Sulfur, roll (mines) .
Nickel
. . 34(4 per Ib.
Flowers of Sulfur . . . ^ .
$3 per 100 Ib.
. . $120 per oz
Silver ....
. . 27f{ per Ib.
G5J(< per Ib.
. . 8|jS per Ib
SALTS
MISCELLANEOUS
Ammonium chloride .
Alcohol, ethyl
. $4 75 per gal
Ammonium sulfate . .
. $2.70, 100 Ib.
. . 5|i per Ib
Calcium carbide
6(* per Ib
. $2 00 100 Ib
Carbon tetrachloride . . .
Chloroform
. 6}j!perlb.
. 28f! per Ib.
25rf npr 1h
Calomel
. $1.37 per Ib.
. $5.25 100 Ib
. . 4|fi per Ib
Cyanamid . . . $1.90 per ammonia unit
Ether . IfWnerTh.
Epsom salts
. $2.00, 100 Ib.
Ferrous sulfide ....
. $2.60, 100 Ib.
Ethyl acetate ....
. $1.15 per gal.
Lead acetate
Mercury bichloride
. . 15fSperlb.
. .$1.31 per Ib.
Hydrogen peroxide (25 v) . .
Paris green
7(4 per Ib.
. 19(i per Ib.
5^(4 per Ib
Phenol
22(5 per Ib.
Potassium chlorate
. . 8|(i per Ib.
Toluene
. 35ji per gal.
Potassium chloride . . .
Potassium permanganate
. $34.55 per ton
. . 14jfiperlb.
White arsenic
., 3j(S perlb.
White lead
Wood alcohol
. lO&perlb.
57(4 per gal.
$2 100 Ib
Sodium bromide ....
. . 48j! per Ib.
OXIDES
Sodium carbonate ....
Sodium chlorate ....
. $1.30, 100 Ib.
. . Sfjiperlb.
T> 'J
1 14 nor ll-i
Sodium chloride ....
. . $12 per ton
Sodium nitrate
$2.72 per 100 Ib.
.75j per 100 Ib.
. . 3it<perlb.
L"
$1.05, 100 Ib.
Af (i' ''H
Zinc sulfate
$18 per ton
. . 27jS per Ib.
Sulfur dioxide
. . 8^ perlb.
157
PRACTICE TABLE OF COMPOUNDS FROM VALENCIES
ACETATE
BICARBON-
ATE
BRO-
HIDE
CAR-
BONATE
CHLORATE
CHLO-
RIDE
Aluminum
Ammonium
Barium
Ba(C 2 H 3 2 ) 2
Ba(HCO 3 )s
BaBrj
BaCOa
Ba(C10 3 ) 5
BaClj
Sismuth
Calcium
Chromium
Cupric
Gold
Hydrogen
Ferrous
Ferric
Lead
Lithium
Magnesium
Mercurous
Mercuric
Potassium
Silver
Sodium
Stannous
Zinc
158
PRACTICE TABLE OF COMPOUNDS FROM VALENCIES (Continued]
IO-
DIDE
NITRATE
NITRITE
OX-
IDE
PHOS-
PHATE
SOT-
PATE
STJL-
FITE
S0L-
PIDE
Aluminum
Ammonium
Barium
Bali
Ba(NO,),
Ba(N0 2 )a
BaO
Ba,(P0 4 )j
BaS0 4
BaSOi
BaS
Bismuth
Calcium
Chromium
Cupric
Gold
Hydrogen
Terrous
? erric
,ead
jithium
Magnesium
tfercurous
Mercuric
'otassium
Silver
Sodium
Stannous
Zinc
159
APPENDIX (Continued')
PROBLEMS BASED ON THE THEORY OF IONIZATION
WE have noted (pages 64-66) that the gram-molecular
weights of sucrose and phenol raise the boiling point and
lower the freezing point of water to 100.52 C. and 1.87 C.,
respectively. The gram-rnolecular weights of NaCl, NaOH,
and HNO 3 will each raise the boiling point and lower the
freezing point of water almost twice these amounts. The
gram-molecular weights of Na 2 S0 4 and H 2 S04 each in a liter
of water will produce results nearly three times as great.
Now, since the depression of the freezing point and the eleva-
tion of the boiling point depend upon the number of particles
of solute, dissolved in the solution, why should not the gram-
molecular weights of all substances which contain the same
number of particles have the same effect on the freezing and
the boiling points?
Another curious problem presents itself. Sugar and phenol
when dissolved in water do not conduct an electric current.
They are non-electrolytes. Solutions of NaCl, HC1, NaOH,
Na 2 S04, and H^SCX are all good electrolytes. There seems
to be some relationship between the conductivity of a solution
and the power of a substance to affect abnormally the freezing
point and the boiling point of solutions. By actual experi-
mentation it has been found that those substances which give
a normal lowering of the freezing point and rise of the boiling
point are non-electrolytes, while those that conduct the
current give abnormal results.
To explain the above problems a young Swedish chemist,
Svante Arrhenius, advanced his Theory of Electrolytic Disso-
ciation or lonization, in 1887. His theory, briefly put, stated
that when a crystalloidal solute rendered a solution a conductor
of electricity, the current was carried by particles catted ions
carrying definite electric charges proportional only to their
160
Problems Based on the Theory of lonization 161
valencies. These ions were produced by the breaking down
or dissociation of the molecules of the solute into positively
charged particles and negatively charged particles.
Thus water ionized NaCl into sodium ions, written Na + , and
chlorine ions, written Cl~. Sulfuric acid dissociated into
H + , H + , and 50 4 , potassium hydroxide into K + and OH~,
while sugar and phenol did not dissociate at all. Since the
current is carried by ions, a sugar solution could not conduct
the current. Since the number of particles (ions) actually
present in the NaCl solution is double that of the sugar
solution, NaCl lowers the freezing point of water twice as much
as sugar. Ions differ in their properties from atoms because
of the charges which ions carry.
Electrolytes do not all dissociate to the same extent. Acids
all contain hydrogen ions in solution, and acid properties are
due to the presence of these hydrogen ions. In fact, the
number of these H + determines the strength of the acid. Thus
HNOs, which is more than 90% ionized, is much more active
than acetic acid, which is only 1 % ionized. The facility with
which equi-molar solutions of different substances conduct the
current depends chiefly upon the degree of ionization. There-
fore, by means of conductivity measurements we can determine
the degree of ionization of a substance in solution.
A. Degree of ionization
Since the conductivity of a solution and its degree of dis-
sociation are intimately related, we may express this relation-
ship by the -equation
a = = degree of dissociation or ionization,
Aoo
where \v is the equivalent conductivity of a gram-equivalent
of the substance in v liters of water,
162
Appendix
and Aoo is the equivalent conductivity of a gram-equivalent
of the substance at infinite dilution that is, when the
dissociation of the electrolyte is complete. Both \v and
Xoo are expressed in reciprocal ohms. 1
By measuring, then, both \v and Xoo, or rather their recip-
rocals, we can determine the proportion of molecules com-
pletely ionized in solution. A table showing the degree of
ionization of some common chemicals follows :
DEGREE OF IONIZATION IN 0.1N SOLUTIONS AT 18 C.
Hydrochloric acid .... 92%
Nitric acid 92
Sulfuric acid 61
Phosphoric acid 27
Hydrofluoric acid .... 15
Tartaric acid 8
Acetic acid 1.3
Carbonic acid 0.17
Boric acid 0.01
Potassium hydroxide ... 91
Sodium hydroxide .... 91
Barium hydroxide .... 77
Ammonium hydroxide ... 1.3
Potassium chloride .... 86%
Ammonium chloride .... 85
Sodium chloride 84
Sodium nitrate ...'... 83
Potassium acetate .... 83
Silver nitrate 81
Sodium bicarbonate .... 78
Barium chloride 77
Potassium sulfate 72
Sodium tartrate 69
Zinc sulfate 40
Copper sulfate 39
Mercuric chloride 1
Mercuric cyanide Minute
EXAMPLE. The equivalent conductivity of KC1 at 18 C. is 130.1 at infinite
dilution, and 112 in a 0.1 N solution. What is its degree of ionization in
0.1 N solution?
a = _*! = 112 = 0.86 or 86%
Xoo 130.1
1 Every conductor offers some electrical resistance. The greater the conductance
the less the resistance. Resistance is therefore the reciprocal of conductance. The
practical unit of resistance is the ohm, which is the resistance offered to a current by a
thread of mercury at C., i sq. mm. in cross-section and 106.3-011. in length. The
unit of conductance is the reciprocal ohm or mho.
Problems Based on the Theory of lonization 163
Problems
1. The equivalent conductivity of acetic acid at 18 C. is 347 at infinite
dilution, and 4.67 in a tenth normal solution. What is the degree of
dissociation of acetic acid in 0.1 N solution?
2. The equivalent conductivity of a sodium acetate solution of 0.08 N
is 63.4. At infinite dilution it is 78.1, Calculate the degree of
ionization of this sodium acetate.
3. At 18 C. the equivalent conductivity of LiCl is 101.4 mhos at infinite
dilution. For a 0.01N solution it is 93.6. What is the degree of
ionization of this lithium chloride?
4. The equivalent conductivity at 18 C. and infinite dilution of HI
is 384 mhos. The equivalent conductivity of the 0.5 N solution is
345. What is its degree of ionization?
5. The equivalent conductivity of HC1 at 18 C. is 379 at infinite dilu-
tion, and 351 in 0.1 normal solution. Find the value of oc.
6. The equivalent conductivity of potassium, acetate at 18 C. and
infinite dilution is 100. For a half normal solution it is equal to 71.6.
What is the degree of ionization of K^HgOa ?
7. At infinite dilution the equivalent conductivity of K.OH is 249.
For a 0.2 N solution it is 206. Calculate its degree of dissociation.
8. The equivalent conductivity of acetic acid is 1.32 mhos for a normal
solution. Its equivalent conductivity at infinite dilution is 350.
What is its degree of dissociation for a normal solution?
9. The equivalent conductivity of a normal HC1 solution is 301 mhos.
At infinite dilution it is equal to 380. Calculate its degree of ioniza-
tion for a normal solution.
LAW OF MASS ACTION AS APPLIED TO REACTIONS BETWEEN
ELECTROLYTES
The direction in which a reversible reaction will go depends
upon the concentrations, expressed in gram moles per liter,
of all the substances involved in the reaction. This behavior
is expressed by the Law of Mass Action, first advanced by
Guldberg and Waage in 1864. The equation for this law may
be written as
r _U]x[g]
*~~TcT~'
where [A], [B], and [C] are the concentrations of the reacting
164
Appendix
substances, and K is the rate at which the reaction would pro-
ceed were the reacting substances originally present con-
stantly maintained at unit concentration. The value of K
for each action is the same. Whenever one of the reacting
substances is removed from the sphere of the reaction, the
reaction proceeds in one direction to completion. This occurs
when (a) a gas is liberated, (b) a precipitate is formed, or
(c) water is formed, since the ionic concentrations of these
three classes of substances become zero.
The most common reactions are those which occur between
electrolytes where concentration of ions are considered,
lonization is an excellent example of a reversible reaction.
When NaCl dissolves in water, Na + and Cl~ are formed, and
as they move about in solution they frequently meet and
re-form NaCl molecules. Soon they reach a state of equilib-
rium which is dependent almost entirely upon the amount of
solvent.
B. Solubility product
From the Law of Mass Action it follows that for a saturated
solution of a difficultly soluble salt, such as silver acetate,
AgC 2 H 3 2 * r Ag+ + C 2 H 3 2 -
the product of the ionic concentrations is a constant, called
the Solubility Product (S, P.}. The concentration is expressed
in gram moles per liter of solution. Thus
[Ag+] X [C 2 H 3 2 -] = S. P. (constant)
SOLUBILITY PRODUCTS OF SOME COMMON SALTS
BaCO 3 1.9 X 10- 9
BaC,0 4 1.2 X 10^
BaSO 4 1.0 X10- 10
CaCO 3 2.8X10- 9
CuS 8.5 X 10- 45
PbS 4 X 10~ 28
HgS 4X10" 54
AgBr 4 X 10~ 23
AgCl 2 X 10- 10
ZnS IX 10- 23
Problems Based on the Theory of lonization 165
EXAMPLE. Analysis shows that the concentration of a saturated solution of
silver acetate at 18 C. is approximately 0.06. Find its solubility product.
The concentration of both [Ag+] and [C 2 H 3 2 ~] is 0.06. In such a dilute
solution the dissociation of the silver acetate will be practically complete.
Hence we can write
S.P. = (0.06) X (0.06) =0.0036.
Problems
10. The solubility of silver chloride is 0.00016 grams per liter. What
is its solubility product?
11. The solubility of barium sulfate is 1 X 10~ 5 grams per liter. Deter-
mine its ion product concentration (S. P.).
12. The solubility of Ag 2 Cr0 4 is 2.5 X 10~ 2 . Calculate its solubility
product.
13. The solubility of Pb 3 (P0 4 ) 2 is 1.4 X KH. What is its ion product
concentration?
14. The solubility product of AgCl is 2 X lO" 10 . Calculate its solubility
in grams per liter.
15. The solubility of calcium chromate in gram molecules per liter
at room temperature is 0.03. Calculate its S. P.
BALANCING EQUATIONS
THE important thing to remember about balancing equations
is to attack the problem in a definite way. While there is no
one way of balancing all chemical equations, the student, by
practice, will soon learn to find the easiest scheme to balance
each particular equation. The following schemes will help
balance most of the common equations.
1. Pick out 'the compound with the largest number of atoms
and proceed as follows :
KBr + H 2 S0 4 + Mn0 2 >- K 2 S0 4 + MnS0 4 + H 2 O + Br 2 .
Let us start with K 2 S0 4 . This molecule has 2 atoms of K ;
therefore 2 atoms of K must be present on the left side of the
equation. Since we cannot change the number of atoms in
KBr, we place the coefficient 2 in front of the formula, 2KBr.
There are 2 sulfate radicals S0 4 on the right of the equation,
one in MnS0 4 and the other in K 2 S0 4 . Therefore we add the
coefficient 2 in front of H 2 S0 4 . The equation may now be
written as
2KBr + 2H 2 S0 4 + Mn0 2 ^K 2 S0 4 + MnS0 4 + H 2 O + Br 2 .
Now we have 4 atoms of H on the left side of the equation and
only 2 on the right. Hence we add the coefficient 2 in front
of H 2 0. This gives us 2 atoms of oxygen on the right side
(irrespective of the in S0 4 ), which is the same as the number
of oxygen atoms on the left side of the equation' (irrespective
of the in S0 4 ). The equation may now be written :
2KBr + 2K 2 S0 4 + Mn0 2 ^ K 2 S0 4 + MnS0 4 + 2H 2 + Br 2 .
Now check the equation to see whether there is the same
number of atoms of each element on each side of the equation.
1 66
Balancing Equations 167
Remember that, as a rule, radicals do not change during a
chemical transformation.
EXERCISES. Balance the following equations :
1. CaO + C >- CaC 2 + CO
2. H 2 O + N 2 5 > HN0 3
3. Mg+ HC1 * MgCl 2 + H 2
4. Si0 2 + C >- SiC + CO
5. Fe 2 3 + Al > A1 2 3 + Fe
6. NaHCO 3 * . Na 2 C0 3 + H 2 + C0 2
7. AgNOs + H 2 S - Ag 2 S -f HN0 3
8. CaC 2 + H 2 >- C 2 H 2 + Ca(OH) 2
9. Pb(N0 3 ) 2 + K 2 Cr0 4 + PbCr0 4 + KN0 3
10. A1 2 (S0 4 ) 3 + NH 4 OH ^ A1(OH) 3 + (NH 4 ) 2 S0 4
11. C 3 H 6 (OH) 3 + HN0 3 >- C 3 H 5 (N0 3 ) 3 + H 2 6
12. CeHiaOe >- C 2 H 6 OH + C0 2
13. K 3 Fe(CN) 6 + FeCl 2 >- Fe 3 [Fe(CN) 6 ] 2 + KC1
14. K 4 Fe(CN) 6 + FeCl 3 * Fe 4 [Fe(CN) 6 ] 3 + KC1
15. Ca 3 (P0 4 ) 2 + Si0 2 + C ^ CaSi0 3 + CO + P
2. Use the atom of a gas rather than its molecule in balancing,
and then make the necessary changes as follows :
C 6 H 6 + > C0 2 + H 2
On the left side of the equation there are 6 carbon atoms and
i68 Appendix
6 hydrogen atoms. Therefore we ought to have. 6CO 2 and
3H 2 O, or,
C 6 H 6 + >- 6C0 2 + 3H 2
On the right of the equation we now have 12 + 3 or 15 atoms
of oxygen. Hence on the left side we should have 15 atoms
of oxygen. The equation now becomes
C 6 H 6 + 150 >- 6C0 2 + 3H 2
But oxygen has 2 atoms to the molecule, therefore
C 6 H 6 + 150 2 >- 6C0 2 + 3H 2 0.
Since we have multiplied the number of oxygen atoms by 2,
we should do the same with all the remaining substances in
the equation. The balanced equation is therefore
2C 6 H 6 + 150 2 -- 12C0 2 + 6H 2 0.
EXERCISES. Balance the following by inserting the proper coefficients :
16.
c +
>-
CO
17.
Fe +
>
Fe 2 3
IS.
P +
>-
P 2 5
19.
Na +
H 2
-> NaOH +
20.
S0 2 +
>- S0 3
21.
Fe +
>
Fe 3 O 4
22.
Sb +
Cl >
SbCl 3
23.
Mg +
N s
> MggN 2
24.
N-f
H ^
NH 3
25.
H 2 S +
O :
>- H 2 O + S0 2
H
Balancing Equations 169
26. KC10 3 > KC1+
27. C 2 H 2 + ^ C0 2 + H 2
28. CH 4 + O > 'C0 2 + H 2
29. Fe + H 2 >- Fe 3 4 + H
30. Al + NaOH >- Na 3 A10 3 + H
3. Use the acid and base rather than their anhydrides and
then make the necessary changes as follows :
Hot concentrated sulfuric acid is a strong oxidizing agent and
when it reacts with copper is itself reduced to sulfurous acid,
H 2 S0 3 ; thus,
Cu + H 2 S0 4 *~ CuS0 4 + H 2 + H 2 S0 3 .
H 2 S0 3 , however, is unstable and at the temperature of the
reaction breaks down into H 2 0, and S0 2 , its anhydride.
Hence the equation is now
Cu + H 2 S0 4 > CuS0 4 + H 2 -f S0 2 + H 2 O
Since we now have 2H 2 6, we shall need 2H 2 S0 4 and the bal-
anced equation becomes
Cu + 2H 2 S0 4 > CuS0 4 + 2H 2 + S0 2
Carbonic acid is also unstable and breaks down into CO 2
and H 2 O.
Similarly, ammonium hydroxide breaks down into NH 3 and
H 2 0.
EXERCISES. Balance the following by using the proper anhydrides
and coefficients :
31. Na 2 S0 3 + H 2 S0 4 > Na 2 S0 4 + (H 2 S0 3 )
32. CaC0 3 + HC1 >- CaCl 2 + (H 2 C0 3 )
170 Appendix
33. Ca(OH) 2 + NH 4 C1 >- CaCl 2 + (NH 4 OH)
34. CaC0 3 + H 2 S0 4 ^ CaS0 4 + (H 2 C0 3 )
35. Ba(DH) 2 + (NH 4 ) 2 S0 4 *- BaS0 4 + (NH 4 OH)
36. A1(OH) 8 + - (NH 4 ) 2 S0 4 ^ AI 2 (S0 4 ) 3 + (NH 4 OH)
37. NaHCOs + HKC4H 4 O 6 >- NaKC 4 H 4 6 + (H 2 CO 3 )
38. Ag+ H 2 S0 4 *- Ag 2 S0 4 -f (H 2 S0 3 ) + H 2 0^
39. Sn+ H 2 S0 4 >- Sn(S0 4 ) 2 + (H 2 S0 3 ) + H 2
40. H 2 2 + (H 2 S0 3 ) >- H 2 S0 4 + * H 2
4. Sometimes no amount of ordinary manipulation will
enable the student to balance an equation within a reasonable
length of time. In such cases it is best to memorize some of
the coefficients and work out the rest of the coefficients.
The equation Cu + HN0 3 >- Cu(N0 3 ) 2 + HaQ + NO
is very difficult to balance unless we know some of the co-
efficients which are found in the balanced equation. If we
memorize 3Cu and 8HN0 3 , the problem of balancing the
equation becomes a simple one.
3Cu will give 3Cu(N0 3 ) 2 ,
8HN0 3 will give 4H 2 O 3 3Cu(N0 3 ) 2 , and 2NO.
The balanced equation is then
3Cu + 8HN0 3 > 3Cu(N0 3 ) 2 + 4H 2 + 2NO.
EXERCISES. Balance the following equations :
41. SAg + 4HN0 3 ^ AgN0 3 + NO + H 2
42. SKI + 5H 2 S0 4 *- K 2 SO 4 + H 2 + H 2 S + I 2
Balancing Equations 171
43. 10FeS0 4 + 8H 2 S0 4 + 2KMn0 4 *- K 2 S0 4 + MnS0 4
. + , Fe 2 (S0 4 ) 3 + H 2
44. 4Zn + lOHNOs > Zn(N0 3 ) 2 + NH^Os + H 2
45. 2NaI0 3 + 3Na 2 S0 3 + 2NaHS0 3 > Na 2 S0 4 + H 2
TUT,
' UIBRAR
285
imiiiimiiiimimiiimmiiimiimiu*
NEW-WORLD SCIENCE SERIES
Edited by John W. Ritchie
Laboratory Manualof
CHEMISTRY
BY GEORGE HOWARD BRUCE
Horace Mann School for Boys, Teachers College,
Columbia University
THIS Manual was tried out for a number of years
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Sixty experiments are outlined. All of them have
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Laboratory Manual of Chemistry may be used with any
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I Introductory |
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= YONKERS-ON-HUDSON, NEW YORK 1
| , 2126 PRAIRIE AVENUE, CHICAGO |
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NEW-WORLD SCIENCE SERIES |
Edited ly John W. Ritchie |
QUALITATIVE 1
ANALYSIS I
i By WILLIAM C. COOPER 1
i Professor of Chemistry, De Paul University 1
THIS is a student's working manual giving a concise |
course in analysis. It presupposes a foundation f
\ course in general inorganic chemistry. Among its distinc- |
I tive features are: f
i 1. It is written to the student just as the instructor might |
I talk to him. The extremely clear directions enable f
I the student to go right ahead with his own work. f
I 2. It explains the different steps in analysis so that the |
| student knows why he does certain things as well as |
| how to do them. |
1 3. The directions are given in sequence as needed, so f
I that it is not necessary to refer from one part of the |
1 book to another in order to complete a test. f
I 4. Equations representing various types of reactions and |
| ' by-products at each step are included. |
1 5. Precautions are explained and directions for removing |
1 interfering factors are given at the time needed. |
1 This is a compact but complete course that will give |
I students a quick and sure means of analysis. Confining s
I itself to the proper method of procedure, with reasons ^for |
I the different steps, it does not attempt to go> deeply into g
1 additional theory. The clear directions alone make this |
1 a most practical manual for a first course in the subject |
I and for use in laboratories. |
1 Cloth. ,v+142 pages. Price $1.52 f
1 WORLD BOOK COMPANY |
1 YONKERS-ON-HUDSQN, NEW YORK |
| 2130 PRAIRIE AVENUE, CHICAGO |
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NEW-WORLD SCIENCE SERIES
Edited by John W. Ritchie
EXPERIMENTAL
ORGANIC CHEMISTRY
By AUGUSTUS P. WEST, PH. D.
Professor of Chemistry, University of the Philippines
HERE is a textbook of college grade, somewhat different
from similar books now in general use, since it is a
combination of textbook and laboratory manual in which the
theoretical discussions and laboratory experience are
blended.
1 Only the more important compounds are dis-
cussed.
2 Dangerous or very difficult experiments for
beginners are omitted.
3 Review tables are given at frequent intervals.
4 Special emphasis is laid upon the exact prep-
aration of organic compounds.
5 Questions relating to the experiments are
placed at the ends of chapters.
Unlike many laboratory manuals this book gives the equa-
tions to explain the experiments. Over one hundred experi-
ments are outlined with the aliphatic and aromatic com-
ppund_s. The appendix contains (a) general laboratory
directions, (b) the list of the approximate quantity of
materials required, (c) the apparatus for one student, (d)
a table of the atomic weights of the common elements, and
(e) a list of books for reference.
The index has been most carefully prepared, and the book
is illustrated with many drawings and with an unusually
large number of diagrams. The text matter in this book
has been supplied to students in mimeographed form for a
number of years; thus, this book is the outgrowth of a care-
fully worked out method of instruction that is giving excel-
lent results. The printed directions for students save much
of *the laboratory instructor's time, because they are clear
and complete.
Cloth. xiii+^.6g pages
Price $3.20
WORLD BOOK COMPANY
YoNKERs-oN-HunsoN, NEW YORK
2126 PRAIRIE AVENUE, CHICAGO
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