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Full text of "Chemical reactions and their equations;"




AND 

THEIR EQUATIONS 





MEMCAL 




COLLEGE OF PHARMACY 



O&Nfemla Oo!!e of Pharmacy 



CHEMICAL REACTIONS 
AND THEIR EQUATIONS 

HACKH 



CHEMICAL REACTIONS 
AND THEIR EQUATIONS 



A Guide and Reference Book 
for Students of Chemistry 



BY 

INGO W. D. HACKH, PH.C., A.B. 

PROFESSOR OF BIOCHEMISTRY, COLLEGE OF PHYSICIANS 



AND SURGEONS OF SAN FRANCISCO 



Q&Ilfernta, College of Pharmaos 




PHILADELPHIA 
P. BLAKISTON'S SON & CO. 

1012 WALNUT STREET 



COPYRIGHT, 1921, BY P. BLAKISTON'S SON & Co. 



THE MAPLE PRESS YORK PA 



PREFACE 

The inability to balance a chemical equation is a most 
common difficulty of students of chemistry. The writer 
when teaching at a large university actually encountered 
graduate students of chemistry who were unable to 
balance an incomplete ionic equation that involved 
oxidation and reduction, not to mention the large 
number of first and second year students who had only 
a very hazy idea of the principles. 

In order to supply students with necessary material 
and to expound the general principles of balancing equa- 
tions, this concise volume was written. It does not enter 
into a detailed discussion of physico-chemical equations, 
but confines itself mainly to a consideration of purely 
chemical equations from a technical and arithmetical 
standpoint. The writer hopes that a study of this 
volume, (which, to the writer's knowledge, is the only 
one of its kind on the subject) in connection with any 
good text, will aid toward a clearer understanding of 
chemistry on the part of the student. A detailed study 
of the following pages should enable anyone to balance 
any chemical equation rapidly and correctly. 

With pleasure the writer acknowledges his gratitude 
to Dr. Henry D : Arcy Power for many valuable sugges- 
tions and criticisms in reading the manuscript, and to 
his students, Messrs. Jordan, Konrad, and Magee for 
reading the manuscript and proofs. He also feels 
much indebted to his teachers, Profs. Edmond O'Neill, 
G. N. Lewis, and Joel H. Hildebrand. 

I.W.D.H. 

SAN FRANCISCO, CAL. 

v 

4O d "> * } (I 
6060 



TABLE OF CONTENTS 

CHAPTER I 

PAGE 

SYMBOLS 1 

Atoms, molecules, ions, ionization. 

CHAPTER II 

FORMULAS 6 

Empirical, rational, constitutional, and structural formulas. 
Valency and valence numbers. Oxidation and reduction. 
Nomenclature and terminology of compounds. 
Summary of information contained in a formula. 

CHAPTER III 

EQUATIONS (INVOLVING NO OXIDATION AND REDUCTION) ... 23 
Molecular and ionic equations. Finishing and balancing equa- 
tions. Calculations and problems. 

CHAPTER IV 

EQUATIONS (INVOLVING OXIDATION AND REDUCTION) 37 

Molecular and ionic equations. The three steps in balancing. 
Examples and problems. 

CHAPTER V 

REACTIONS AND THEIR CONTROL 48 

The speed of chemical reactions and the chemical equilibrium. 
Mechanical control by surface, catalysts, concentration, and 

pressure. Thermal control and the influence of heat. 
Electrical control, electromotive force, galvanic cells, electrolysis. 

CHAPTER VI 

TYPES OF CHEMICAL REACTIONS AND EQUATIONS 78 

Analysis, synthesis, and metathesis. Neutralization and 
hydrolysis. Combination, dissociation, displacement, and 
substitution. 



viii TABLE OF CONTENTS 

APPENDIX 

PAGE 
I. KEY TO NOMENCLATURE WITH A LIST OF RADICALS, IONS, AND 

VALENCE NUMBERS 103 

II. DISPLACEMENT SERIES OF THE ELEMENTS 107 

III. PERIODIC SYSTEM AND CLASSIFICATION OF THE ELEMENTS . . . 109 

IV. SOLUBILITY TABLE OF COMPOUNDS 115 

V. PREPARATION OF SALTS 118 

KEY TO THE EQUATIONS 119 

INDEX AND GLOSSARY OF CHEMICAL TERMS 133 






ERRATA 



Page 3: line 13. Place comma after "therefore" and take out comma 

after "atom." 

Page 101, equation No. 429. Read Zn+ + instead of Zn . 
Page 112, in table of atomic weight read 175.0 instead of 17. 50 for Lutecium. 
Page 112, line 8, read an instead of n. 

English prices : 

Page 47, question 27, last sentence would read: Fe costs 3d, Al = sh. 1.3d, 

and Mg = sh. 1.6d. 
Page 45, question 2. Price of Zn = 2 sh. 



HACKH, CHEMICAL REACTIONS AND THEIR EQUATIONS. 



CHEMICAL REACTIONS AND THEIR 
EQUATIONS 

CHAPTER I 
SYMBOLS 

Symbols. The alphabet of chemistry consists of. 
symbols. Every chemical element is represented by 
a symbol which is usually the first letter of its English 
or Latin name. Where different elements have the 
same initial letter a small additional letter is added, 
e.g. B = boron, Be = beryllium, Br = bromine, Bv = 
brevium. A symbol not only represents the kind of 
element but also stands for a certain quantity of it; 
namely the smallest amount of matter that can enter 
into combination a single atom. This amount or 
quantity is expressed in a relative number the atomic 
weight, which, as such, is a number indicating the 
proportional weight of an element in combination. 

Atoms and Molecules. We assume that atoms can 
only exist when combined in the form of a molecule; 
thus atoms of like nature combine to form an elemental 
molecule and aggregate to an element, while unlike 
atoms combine to form a compound molecule and 
aggregate to a compound. Molecules are capable of 
independent existence, that is, they may exist separately 
and not in aggregation with similar molecules. Thus 
a molecule is the smallest imaginable particle of a 
compound, an atom the smallest imaginable particle 
of an element. While atoms are chemically inde- 

l 



2 CHEMICAL REACTIONS AND THEIR EQUATIONS 

structible, 1 the molecules are susceptible to endless 
chemical change. Such changes and interchanges are 
expressed in the form of chemical equations. 

A chemical symbol has three distinct uses as it may 
represent an atom, -molecule or ion. To distinguish 
whether a symbol represents an atom, molecule, or ion 
a numeral or + or sign is used. 

Atom. The atom of an element is shown by the 
symbol alone. Thus H represents an atom of hydrogen 
with an atomic weight of 1.008; Cl stands for an atom 
of chlorine with an atomic weight of 35.45, and Fe is 
an atom of iron (ferrum) with atomic weight 56.5. 

Molecule. The molecule of an element is shown by 
the symbol with a small numeral attached indicating the 
number of atoms: H 2 or H 2 is a molecule of hydrogen 
containing two atoms of hydrogen, the molecular weight 
being the sum of the atomic weight = 2 X 1.008 = 
21016; S 4 or S 4 is a molecule of sulfur containing four 
atoms of sulfur, the molecular weight being 4 X 32.00 
= 128.00; S 6 or S 6 is a molecule of sulfur with six atoms 
of sulfur, the molecular weight of this molecule is 
6 X 32.00 = 192.00; S x or S* is a molecule of sulfur 
containing an unknown number of atoms of sulfur. 
All gaseous elements have ordinarily two atoms in their 
molecule, e.g. O 2 , N 2 , C1 2 , which indicates respectively 
a molecule of oxygen, nitrogen, and chlorine as gases. 

1 The fact that atoms are chemically indestructible does not exclude 
the possibility that they may be transformed by physical means. Mod- 
ern research, both theoretical and practical, point to still smaller cor- 
puscles or electrons as the building stones of atoms. Nevertheless 
the conception of a "structure of the atoms" will not affect the older 
conception of an atom as the "chemical unit," or an element as a type of 
atoms. Chemists will thus continue to deal with atoms, and future 
progress in the subatomic realm will merely increase the knowledge of 
the relation among elements and their probable evolution, but will not 
materially change the application of the atomic theory. 



SYMBOLS 3 

Only in a few cases have the number of atoms in a 
molecule been determined for solid elements, e.g. P 4 , 
P 8 , As 2 , As 4 . If the number of atoms in a molecule 
of a solid element is unknown, the molecule should be 
written with x, e.g. Na x , Fe x , AU, C x . However, in 
metals, this number is assumed to be one, so that in 
this case Na would represent not only an atom, but also 
a molecule of sodium (natrium). 

Ion. The ion of an element is indicated by the 
symbol with one or more positive(+)or negative ( ) 
signs attached, which represent the number of positive 
or negative charges attached to the atom of an element. 
An ion, therefore is an electrified atom, or an atom with 
a certain number of electrical corpuscles attached. 
These charges are also written as dots and dashes, thus : 
H + or H* = hydrogen ion, H-atom with one positive 
charge, Cl~ or Cl' = chloride ion, Cl-atom with one 
negative charge, Ca ++ or Ca" = calcium ion, Ca-atom 
with two positive charges, S~'~ or S" = sulfide ion, 
S-atom with two negative charges, A1+++ or Al'" = 
aluminum ion, Al-atom with three positive charges. 
These ions ordinarily exist only in solutions. The 
majority of salts dissolved in water become ionized 
because their component atoms become electrified. 
Thus NaCl (sodium chloride) when dissolved in water 
will form Na + (sodium ion) and Gl~ (chloride ion), 
while a small amount of the original sodium chloride 
remains as NaCl (molecular or non-ionized sodium 
chloride). Some salts will ionize slightly, that is, they 
will break apart (dissociate) only to a small degree. 
Ammonium chloride (NH 4 C1) when dissolved in water 
forms a few ammonium ions (NH 4 +) and a few 
chloride ions (Cl.~) the larger percentage of the sub- 
stance remaining in molecular combination (NH 4 C1). 



4 CHEMICAL REACTIONS AND THEIR EQUATIONS 

lonization or electric dissociation, therefore, is the 
breaking apart of the molecule in solution whereby 
the atoms or groups of atoms (radicals) acquire a 
positive or negative charge. It must be borne in 
mind that the properties of the ions are distinctly 
different from the properties of the atoms. Sodium is 
different from sodium ion, iron metal (Fe) differs from 
ferrous ion (Fe++) and ferric ion (Fe+++), for the ions 
are always electrified or charged atoms. 

Summary. The three ways of using a chemical symbol 
are summarized as follows: 

H = one hydrogen atom, which exists in compounds, 
2H = two hydrogen atoms, 

H 2 = one hydrogen molecule, which exists in hydro- 
gen gas, 

2H 2 = two hydrogen molecules, 
H + = one hydrogen ion, which exists in the solution 

of acids, 
2H+ = two hydrogen ions. 

Accordingly the symbols Cl, S, 28, 38, Fe, Al, stand 
respectively for one atom of chlorine, one atom of sulfur, 
two atoms of sulfur, three atoms of sulfur, one atom 
of iron, one atom of aluminum, while their respective 
atomic weights; C1 2 , SB, 2S 6 , 3S 6 , Fe, Al x , means one 
molecule of chlorine (two atoms), one molecule of sulfur 
(six atoms), two molecules of sulfur (twelve atoms), 
three molecules of sulfur (eighteen atoms), one molecule 
of iron (unknown number of atoms), one molecule of 
aluminum (unknown number of atoms); Cl~, 8 , 
28, 38, Fe++, Fe+++, A1+++, represent respectively 
one chloride ion (chlorine atom with one negative 
charge), one sulfide ion (sulfur atom with two negative 
charges), two sulfide ions (two sulfur atoms with four 



SYMBOLS O 

negative charges), three sulfide ions (three sulfur atoms 
with six negative charges), one ferrous ion (one iron atom 
with two positive charges), one ferric ion (one iron atom 
with three positive charges), one aluminum ion (one 
aluminum atom with three positive charges). 

QUESTIONS 

1. Define symbol, atom, molecule, ion, ionization. 

2. What is the symbol for an atom of bromine, selenium, potassium, 
nitrogen, copper, magnesium, phosphorus? 

3. What is the symbol for a molecule of the elements given in question 
2? 

4. What is the symbol for an ion of the elements given in question 2? 
6. What is the difference between N 2 and 2N, O 2 and O 3 , 3O 2 , 2O 3 and 

6O, F 2 , 2F-and2F? 

6. State all the information contained in the symbols: Na, Na + , 
Se, Se 4 , Se , C, As, As 2 , As 4 , Hg, Hg+, Hg + +. 



CHAPTER II 
FORMULAS 

Formula. The vocabulary of chemistry consists of 
formulas. A chemical formula is a combination of 
symbols which shows the number and kind of atoms 
in a molecule, hence, the formula tells the chemical 
composition of a substance and represents the molecule 
of a compound. However, as the symbols not only 
indicate the kind of atom, but also a certain quantity, 
called the atomic weight, it is evident that a formula 
must also include the quantitative composition of a 
compound, and the molecular weight. 

For instance, NaCl means a molecule of sodium 
chloride containing one atom of sodium or 23 parts by 
weight, and one atom of chlorine or 35.5 parts by weight. 
The total or molecular weight is naturally the sum of 
the atomic weights, that is, 23 + 35.5 = 58.5. From 
this proportion the percentage can be calculated, for 
as 58.5 parts of sodium chloride contain 23 parts of 
sodium, then 58.5 : 23 : : 100 : x, and x = 23 X 100 /58.5 
= 39.3 that is 39.3 per cent sodium; while the per- 
centage of chlorine is found by the proportion 58.5 : 
35.5 : : 100 : x, and x = 35.5 X 100 / 58.5 = 60.7 which 
means 60.7 per cent of chlorine. Two molecules of 
sodium chloride may be written either 2NaCl or (NaCl) 2- 
In the first case the two in front refefs to the whole 
following formula, in the second case the parenthesis 
is necessary as, otherwise, the formula NaCl 2 would 
indicate a compound molecule containing one atom of 
sodium and two atoms of chlorine which is wrong. 

6 



FORMULAS 



Na 2 SO 4 means one molecule of sodium sulfate contain- 
ing two atoms of sodium or 2 X 23 = 46 parts of sodium, 
one atom of sulfur or 32 parts of sulfur, 

four atoms of oxygen or 4 X 16 =64 parts of oxygen, 



which added together make 142 parts of sodium 

sulfate and the molecular weight therefore is 142.0. 

Na 2 SO 4 .7H 2 means a molecule of crystallized sodium 
sulfate in which there are in addition to the sodium, 
sulfur, and oxygen, seven molecules of water. This 
water which forms an essential part of the crystal is 
called the water of crystallization. The water of crystal- 
lization can be driven off by heat and the substance 
becomes anhydrous. 

The formula for ammonium aluminum sulfate, 
A1 2 (SO4)3.(NH4)2SO4.24H 2 O, indicates a more complex 
compound, a molecule of which contains one molecule 
of aluminum sulfate (A1 2 (804)3), one molecule of am- 
monium sulfate ((NH 4 ) 2 SO 4 ), and twenty-four molecules 
of water. Adding these molecules together, the number 
of individual atoms in this compound are: 

2 atoms of aluminum or 2 X 27.1 = 54.2 parts of Al 

3 + 1=4 atoms of sulfur or 4 X 32 = 128.0 parts of S 
2 atoms of nitrogen or 2 X 14 = 28 . parts of N 
(3 X 4) + 4 + 24 = 40 atoms of 

oxygen or 40 X 16 =640.0 parts of O 
(2X4) + (24 X 2) = 56 atoms 

of hydrogen or 56 X 1 = 56 . parts of H 



making altogether 906 . 2 parts of 

crystallized ammonium aluminum sulfate, having the 
molecular weight of 906.2. 



8 CHEMICAL REACTIONS AND THEIR EQUATIONS 

Types of Formulas. The formula for this compound 
may also be written Al 2 S4N 2 04oH5 6 . Such a formula is 
called an empirical formula for only the number and kind 
of atoms are given. From such a formula the quantita- 
tive composition of the substance can be calculated, but 
it does not indicate the way in which the atoms are 
arranged in the molecule. The arrangement of the 
atoms, that is, the way in which they group themselves, 
is shown by the rational formula. The rational formula 
given above for ammonium alum shows that sulfur is 
present in the sulfate radical (S0 4 ), and nitrogen is 
present in the ammonium radical (NH 4 ), while some of 
the oxygen and hydrogen is present as water of crystalliza- 
tion. Thus the rational formula here shows (a) two 
atoms of aluminum are linked to three sulfate radicals, 
(6) two ammonium radicals are linked to one sulfate 
radical, and (c) the crystals contain twenty-four mole- 
cules of water. 

There are four types of formulas: The empirical, the 
rational, the constitutional and the structural or graph- 
ical. The characteristics of each type are as follows: 

Empirical Formula. The empirical formula shows 
only the number and kind of atoms from which (to 
repeat) the molecular weight and the percentage com- 
position can be calculated. Thus F 2 S 3 Oi 2 simply means 
a molecule containing two atoms of iron, three atoms 
of sulfur, and twelve atoms of oxygen, the molecular 
weight being (2 X 55.9) + (3 X 32) + ( 12 X 16) = 399.8. 
Such a type of formula gives nothing further regarding 
the chemical nature and is practically never used in 
inorganic chemistry, they are used, however, in organic 
chemistry for compounds of which only the percentage 
composition is known, and which have not Been fully 
investigated. 



FORMULAS 9 

Rational Formula. The rational formula shows not 
only the number and kind of atoms, but also indicates 
the way in which the atoms are linked together. Thus 
Fe 2 (SO 4 ) 3 means that two atoms of iron are linked to 
three sulfate radicals, each sulfate radical containing 
one atom of sulfur and four atoms of oxygen. Such a 
formula gives some information regarding the chemical 
characteristics of the compound, e.g. this compound 
would give the characteristic sulfate reaction (a precipi- 
tate with barium chloride solution), etc. 

Constitutional Formula. The constitutional formula 
is a notation which throws further light upon the in- 



tramolecular arrangement of the atoms. Thus 



illustrates the valency bonds and shows that two triva- 
lent iron atoms are linked to three bivalent SO 4 groups, 
in other words it shows that the iron in this compound has 
three bonds, and that the sulfate radical has two bonds. 
Structural Formula. The structural formula is the most 
highly developed form of notation and illustrates the link- 
age of each individual atom, every bond being represented 




by a line. Thus n/^^O indicates that the iron 
Fe^0\ c /0 



is not directly linked to the sulfur atom, as may be as- 
sumed from the constitutional formula, but is connected 
by an oxygen atom to the sulfur. It also shows that 
there are six single bonds between two iron and six oxy- 
gen atoms, likewise six single bonds between three sulfur 



10 CHEMICAL REACTIONS AND THEIR EQUATIONS 

and six oxygen atoms, and finally six double bonds 
between three sulfur and six oxygen atoms. In a 
structural formula the radicals are resolved into their 
simplest components single atoms, and the connection 
between each atom is shown. Thus the above structure 
formula indicates that oxygen is always bivalent, as 
it has two bonds, while the sulfur is hexavalent having 
six bonds. 

In inorganic chemistry the rational formula is gen- 
erally used, while the constitutional and structural 
formulas together with the empirical formula are mainly 
employed in organic chemistry. In organic chemistry 
the structure theory has yielded the most magnificent 
results, for without the recognition of a structure, none 
of the synthetic substances could be manufactured such 
as indigo, camphor, dyes tuffs, and pharmaceuticals. 
A modern simplification of structure formulas for organic 
compounds has been devised in the form of structure 
symbols (see Canadian Chemical Journal, vol. 2, p. 135, 
1918; Science, vol. 48, p. 333, 1918; Chem. News, vol. 
118, p. 289, 1919). 

Summary. Summarized, the four types of formula are: 

Empirical : A1 2 C 3 9 C 3 H 4 O 4 

Rational : A1 2 (CO 3 ) 3 CH 2 (CO 2 H) 2 

Constitutional: Ab=CO 3 /COOH 

CH/ 
X 



Structural or graphic : /O\rx 



= 



H 

O 



FORMULAS 11 

Valency. What is a bond or valency? How is it 
possible to depict graphically the intramolecular arrange- 
ment of the atoms? Logically we reason that the 
number of atoms in a molecule must depend upon the 
force which holds the atoms together; furthermore we 
find that certain groups of atoms, such as the sulfate 
radical, are very stable and pass through a number of 
chemical changes as units, and in such cases the force 
holding the atoms together must be especially strong. 
At present we have no means of measuring this force 
nor do we know its mechanism. However, one of its 
manifestations is that property of the element which 
is termed valency. Valency is sometimes defined as the 
combining power of an atom in regard to hydrogen, but 
it is more exact to define valency as the combining 
capacity of an atom or group of atoms (radicals) using 
an atom of hydrogen as a unit of comparison. Hence 
if an atom of element X combines with one atom of 
H and forms a compound HX, then X is said to have 
a valency of one. If the atom of element Y can combine 
with two atoms of ' hydrogen to form a compound of 
the type H 2 Y, then the valency of Y is two, and like- 
wise if an atom of element Z combines with three atoms 
of hydrogen to give a molecule of the type H 3 Z, then 
Z has a valency of three. The structural formulas 

H \ H \ 

for these three molecules are H-X, YY, and H-)Z 

W 

in which each valency is shown by a line. In a com- 
pound the atoms which have a valency of one are 
called monovalent (e.g. H, Na, K, Cl, etc.), of two = 
bivalent (e.g. O, Ca, Ba, etc.), of three = trivalent 
(e.g. Al), of four = tetravalent (e.g. C, Si, etc.), of five = 



12 CHEMICAL REACTIONS AND THEIR EQUATIONS 

pentavalent, of six = hexavalent, of seven = heptavalent, 
and of eight = octovalent. 

Valence Number. It has become necessary to divide 
the elements into negative and positive groups (see 
Chapter 5) and accordingly in each compound there is 
a negative and positive constituent. This assumption 
leads to the consideration of positive and negative valence 
numbers (also called polar numbers). The basis for the 
assignment of valence numbers is hydrogen with a positive 
valence number of one or +1. 1 In all stable compounds 
the sum of the valence numbers must equal zero and 
by considering H always as +1 the positive or nega- 
tive character of the other elements can be established. 
Thus in HX the atom X must have a valence number of 

1, in H 2 Y the atom Y has a valence number of 2, in 
H 3 Z the atom Z has a valence number of 3. Therefore 
it follows that Cl, in HC1, must have a valence number of 

1, that O, in H 2 0, has a valence number of 2, and N, 
in NH 3 , has a valence number of 3. 

In some cases elements do not combine with hydrogen, 
but replace hydrogen (see displacement series), because 
they are more positive than hydrogen, and accordingly 
have a positive valence number. Thus Na, which 
replaces one atom of H, has a valence number of +1; 
calcium replaces two atoms of hydrogen and has a 
valency of two and valence number of +2. The 
positive valence numbers range from +1 to +8, while 
the negative valence numbers range from 1 to 4. 
The elements in their free state (uncombined) are 
assumed to have a valence number of 0. 

Univalent and Polyvalent. However, it is a common 
misapprehension to assume that a given element can 

1 Only in a very few cases has H a valence number of 1, such com- 
pounds are the hydrides, e.g. NaH, LiH. 



FORMULAS 13 

form only one type of compound, that is can have only 
one definite valency or valence number for in many 
cases an element can form two or more series of com- 
pounds, that is, have two or more definite valencies or 
valence numbers. The first class of elements are spoken 
of as univalent elements (which must not be confused 
with monovalent elements), while the second class of 
elements are called polyvalent elements. In general 
the alkali metals (Na, K, etc.), earthalkali metals 
(Ca, Ba, etc.), and earth metals (Al, etc.) are univalent, 
while the non metals and heavy metals are polyvalent. 

A univalent element may consist of monovalent atoms 
(e.g. Na), bivalent atoms (e.g. Ca), trivalent atoms 
(e.g. Al), and so on. A polyvalent element may consist 
of mono- and bivalent atoms (Cu), mono- and trivalent 
atoms (Au), bi- and trivalent atoms (Fe), bi-, tri- 
and hexavalent atoms (Cr), mono-, tri-, penta-, and 
hexavalent atoms (Cl), and so on. 

Series of Compounds. A polyvalent element, that 
is, an element having two or more valencies, will form 
two or more series of compounds and these series of 
compounds will have distinct properties and different 
names. Thus iron has three series of compounds, 
namely ferrous compounds such as FeCl 2 , FeO, etc. in 
which the valency is two and the valence number 
+2; ferric compounds such as FeCl 3 , Fe 2 O 3 , etc. in 
which the valency of iron is three and the valence number 
+3; and finally the ferrates such as Na 2 FeO 4 in which 
the valency of iron is six and the valence number +6. 
Chlorine has four series of compounds : the chlorides 
(NaCl, BaCl 2 , etc.) in which chlorine has a valency of 
one and a valence number of 1 ; the chlorites (NaOCl, 
Ca(OCl) 2 ) in which chlorine has a valency of three, 
and a valence number of +3; the chlorates (KC10 3 ) 



14 CHEMICAL REACTIONS AND THEIR EQUATIONS 

with a valency of five and valence number +5; and 
finally the perchlorates (KC1O 4 ) with a valency of seven 
and valence number of +7. Though chlorine is a nega- 
tive element, like all non metals, it has in only one series 
of compounds, the chlorides, a negative valence number, 
and in the other three series a positive valence number. 
This positive character of the valence number for a 
negative element is caused by assuming the valence 
number of oxygen in a compound to be always 2. 
This conception aids in understanding oxidation and 
reduction, as explained further on. 

Finding the Valence Number. From the formula of 
a compound the valence number of the atoms and their 
positive or negative character can readily be found by 
remembering that hydrogen in a compound is always 
+ 1, and oxygen in a compound is always 2, and that 
the sum of the valence numbers in a saturated and 
stable compound must always be zero. Thus, in water 
(H 2 O) there is H 2 = 2 X ( + 1) = +2, and O = -2, 
therefore (+2) + (-2) = 0. In lime (CaO) oxygen 
is 2, so calcium must be +2. In ferric oxide (Fe 2 O 3 ) 
there are three oxygen atoms, that is 3 ( 2) = ( 6), so 
the two iron atoms must be +6, and one iron atom +3, 
and the valence number of iron in that compound is +3. 
In a similar way the valence numbers for all other 
atoms are determined and can readily be found by the 
difference necessary to make the sum zero, hence^it is 
simply an arithmetical problem. A few examples will 
illustrate this. 

Potassium permanganate has the formula KMnO 4 . 
To find the valence number for the Mn atom add the 
valence number of K ( + 1) to the sum of the valence 
numbers of oxygen 4 (2) = 8 and get 7, thus Mn 
must have +7, To make this clearer write the known 



FORMULAS 15 

valence numbers under the symbols and form an 
arithmetical equation thus: 

K Mn O 4 

( + 1) + x + 4(-2) = and hence x = +7. 

The valence number of nitrogen in sodium nitrate is 

Na N 3 

( + 1) + x + 3(-2) = 0. In this case x = +5, thus 
the valence number of nitrogen is +5. Likewise the 
valence number of chromium in potassium bichromate 

is found: K 2 Cr 2 OT 

2( + l) + 2x + 7(-2) = 0, from this follows 
that 2x = 12 and the valence number of one chromium 
atom is +6. 

Oxidation and Reduction. If an element has several 
valence numbers, that is, if it forms several series of 
compounds, we speak of different stages of oxidation. 
As a rule oxidation can be defined as the increase or 
augmentation of the valence number, that is, a change 
from a lower to a higher valence number; while reduction 
is the decrease or diminution of the valence number, 
that is, the change from a higher to a lower valence 
number. The transformation of ferrous compounds 
(+2) into ferric compounds (+3) is thus oxidation, 
while the transformation of a ferrous compound (+2) 
into metallic iron (0) is reduction. In the former the 
ferrous compound has been oxidized, in the latter it has 
been reduced. Again if a chloride ( 1) is transformed 
into chlorine gas (0), the valence number has been 
increased and the chlorine atom in the chloride is oxi- 
dized to free chlorine. If chlorine gas (0) is further 
oxidized to a chlorite (+3), chlorate (+5), or perchlorate 



16 CHEMICAL REACTIONS AND THEIR EQUATIONS 

(+7), the valence number is further increased and 
therefore the chlorine atom further oxidized. Reduction 
is a change in the opposite direction, e.g. the chlorine 
atom in a perchlorate (+7) can be reduced to chlorate, 
chlorite, chlorine, or chloride. Schematically the case 
is illustrated by the following: 

Valence numbers: -1 +3 +5 +7 

Type of compounds : chlorides chlorine chlorites chlorates perchloratea 
Formula: XC1 C1 2 XC1O 2 XC1O 3 XC1O 4 

OXIDATION > 

REDUCTION 

in which X is a monovalent element. 

Any change from left to right is oxidation. 

Any change from right to left is reduction. 

It is evident from the diagram that the chlorides can 
not be further reduced, that the perchlorates can not 
be further oxidized, and that free chlorine gas can be 
either reduced to chlorides, or oxidized to chlorites, 
chlorates, or perchlorates. 

Similarly there are several series of sulfur compounds : 

Valence numbers: 2 -f4 4-6 

Types of compounds: sulfides sulfur sulfites sulfates 

Formula: X 2 S S X 2 SO 3 X 2 SO 4 

OXIDATION > 

< REDUCTION 

The conclusions drawn from this diagram are: 

Sulfides can be oxidized to free sulfur, sulfites, or 
sulfates. 

Sulfates can be reduced to sulfites, free sulfur, or 
sulfides. 

Free sulfur can be oxidized <to sulfites and sulfates, or 
reduced to sulfides. 

Sulfites can be oxidized to sulfates, and reduced to free 
sulfur and sulfides. 



FOKMULAS 17 

Sulfides cannot be reduced, and sulfates cannot be 
oxidized. 

It is advantageous for the student to construct such 
tables and diagrams for other elements which have 
several series of compounds, such as N, Br, As, P, Fe, Cr, 
Mn, etc., or to devise a single table in which the more 
common elements are arranged in the order of their 
compounds. 

Nomenclature. The different types or series of com- 
pounds of an element have different names. These 
names are formed by the addition of the suffixes -ides, 
-ites, -ates to a negative element, and -ous or -ic to a 
positive element. 

Endings of Names. Thus there are for negative ele- 
ments or radicals the suffixes: 

-ides for the lowest form of oxidation, characterized by 
the absence of oxygen and a negative valence number 
(nitrides, hydrides, phosphides, carbides, selenides, tel- 
lurides, arsenides, etc.). 

-ites for the lowest form of oxidation containing few 
atoms of oxygen, characterized by the lowest positive va- 
lence number (nitrites, phosphites, sulfites, arsenites, etc.). 

-ates for the normal form of oxidation containing the 
normal amount of oxygen atoms and characterized by 
the normal positive valence number (nitrates, sul- 
fates, phosphates, carbonates, arsenates, manganates, 
chromates, ferrates, etc.). 

per- -ates for the highest form of oxidation containing 
most oxygen atom's and having the highest positive 
valence number (persulfates, permanganates, perchlor- 
ates, etc.). 

For positive elements the suffixes are: 

-ous indicating the lower form of oxidation, that is, the 



18 CHEMICAL REACTIONS AND THEIR EQUATIONS 

lower positive valence number (ferrous, arsenous, mercur- 
ous, cuprous, nickelous, plumbous, etc.). 

-ic indicating the higher form of oxidation, the higher 
positive valence number (ferric, arsenic, mercuric, cupric, 
nickelic, plumbic, etc.). 

Key to Nomenclature. A list of the more common 
elements, ions, and radicals with their valence numbers 
can be found in the appendix, and this key to the nomen- 
clature of chemical compounds should enable the student 
to write the correct formula for any compound, or to 
give the correct name for any formula of inorganic 
chemistry. 

Summary of Information Contained in a Formula. 
From any chemical formula there can be deduced: 

(a) The kind of elements and the number of atoms 
which constitute the molecule. 

(6) The weight relation of these elements and the 
molecular weight of the compound. 

(c) The percentage of weight of the elements compos- 
ing the molecule. 

(d) The valency of the elements and, therefore, their 
stage of oxidation. 

Whenever the formula refers to a gas or gaseous com- 
pound there is, furthermore, expressed by a formula: 

(e) The volume relation of the gaseous constituents. 
According to Avogadro's law there are, under the same 
pressure and at the same temperature, in the same volume 
of any gas the same number of molecules, so it follows 
that in 1 liter of oxygen there are the same number of 
molecules as in 1 liter of hydrogen, and as in the forma- 
tion of water, 2 molecules of H 2 react with 1 molecule of 
O 2 , so 2 liters of H 2 are required to combine with 1 liter 
of O 2 . 

(/) The specific gravity or density of the gas. Density 



FORMULAS 19 

can be expressed in three ways according to the unit 
chosen. (1) Density with regard to oxygen = D (02) 
refers to oxygen gas for comparison and its molecular 
weight O2 = 32 is taken as a unit, hence the molecular 
weight of any other gas is also its density or D (02) = 
Mol. Wt. For H 2 the D (02) is 2, for N 2 the D (02) is 28, 
etc. (2) Density with regard to hydrogen = D (H2) 
refers to the lightest known gas as unity. In this case 
the molecular weight of the gas must be divided by the 
molecular weight of hydrogen, that is D (H2) = Mol.Wt./2. 
Hence for O 2 the D (H2) is 16, for N 2 the D (H2) is 
14. (3) Density in regard to air as unity = D (air) . 
A certain volume filled with 32 grams oxygen will weigh, 
when filled with air, 28.95 grams, hence if the density of 
air is taken as unity, the density of oxygen = D (air) is 
32/28.95 - 1.1053 and accordingly the D (air) of all 
other gases is found by dividing their molecular weight 
by 28.95, that is D (air) = Mol. Wt./28.95. 

(g) The weight of 1 liter of gas. The volume occupied 
by a mole (grammolecule) of any gas is 22.4 liters. Thus 
a mole of H 2 , pr 2 grams of hydrogen, will fill 22.4 liters, 
a mole of oxygen, or 32 grams of oxygen gas, will also 
occupy 22.4 liters under standard conditions of pressure 
and temperature, hence the simple proportion 22.4 : M 
: : 1 : x, and x = M/22.4 will give the weight of 1 liter 
of any gas. In the above equation M represents the 
molecular weight in grams ( = mole), and x the weight 
of 1 liter of gas in grams. 

(h) The volume occupied by 1 gram of any gas. In 
this case the above proportion becomes M : 22.4 : : 1 : y, 
and y = 22.4/M, for if M grams will fill 22.4 liters, then 
1 gram will occupy 22.4/M liters. 

Example. The formula NH 3 for ammonia means: 

(a) The molecule consists of one atom of nitrogen and 



20 CHEMICAL REACTIONS AND THEIR EQUATIONS 

three atoms of hydrogen making altogether four atoms 
in the molecule. 

(fe) The molecular weight is the sum of the atomic 
weights, thus 14 parts of nitrogen (by weight not vol- 
ume) and 3X1 = 3 parts of hydrogen give 17 parts of 
ammonia hence the molecular weight is 17. 

(c) The percentage of nitrogen and hydrogen is calcu- 
lated from the data given^in (&), thus in 17 parts am- 
monia there are 14 parts of nitrogen, then 17 : 14 :: 
100 : x, hence 14 X 100/17 = x, and x = 82.25 per 
cent nitrogen, while 17 : 3 :: 100 : y and y = 3 X 100/17 
gives for y the percentage of hydrogen as 17.75 per 
cent. 

(d) The valence number of H being +1, three atoms 
of H are thus equal to +3, therefore the valence number 
of nitrogen must be 3. From the table in the ap- 
pendix we see that 3 is the lowest stage of oxidation 
for nitrogen and it is evident that ammonia can not be 
further reduced, but can be oxidized to free nitrogen, 
nitrites, or nitrates. 

(e) Nitrogen and hydrogen are both gases, therefore 
one volume of nitrogen gas and three volumes of hydro- 
gen gas will combine and form ammonia. 

(/) The specific gravity or density of ammonia (1) 
compared with C>2 = 32 equals the molecular weight 
which is D( 2) = 17 for ammonia; (2) compared with 
H 2 = 1, it equals 17/2, that is D (H2 ) = 8.5; (3) com- 
pared with air = 1, it equals 17/28.95, that is D (air) = 
0.59. Ammonia is thus a little over half as light as air 
and oxygen, but 8.5 times as heavy as hydrogen. 

(g) If 22.4 liters of ammonia weigh 17 grams, then 1 
liter will weigh 17/22.4 = 0.76 gram. 

(h) If 17 grams of ammonia occupy 22.4 liters, then 1 
gram will occupy 22.4/17 = 1.31 liters. 



FORMULAS 21 

QUESTIONS 

1. Define (a) formula, (b) valency, (c) valence number, (d) oxidation, 
(e) reduction, (/) radicals, (g) monovalent and univalent elements. 

2. Write out all the information contained in the formulas (a) KI, 
(6) AgN0 3 , (c) H 3 P0 4 , (d) K 2 HP0 4 , (e) MgSO 4 .5H 2 O. 

3. Write out all the information contained in the formulas for the 
following gaseous compounds (a) HC1, (b) N 2 O 3 , (c) N 2 Os, (d) CH 4 . 

4. Calculate the molecular weights of the following compounds: 
(a) ferrous sulfate FeSO 4 , (b) ferric sulfate Fe 2 (SO 4 )3, (c) glucose C 6 Hi 2 O6, 
(d) sugar Ci 2 H 22 On, (e) bismuth nitrate Bi(NO 3 )3, (/) potassium alum 
KA1(SO 4 ) 3 .12H 2 O, (g) nitroglycerin C 3 H 5 (NO 3 ) 3 , (h) hemoglobin C 72 6- 
Hn7iNi9 4 O 2 i 3 S 3 . 

5. Calculate the percentage composition of (a) mercuric chloride 
HgCl 2 , (b) mercurous chloride HgCl, (c) manganese dioxide MnO 2 , 
(d) copper sulfate CuSO 4 .5H 2 O, (e) sodium carbonate Na 2 CO 3 , (/) 
crystallized sodium carbonate Na 2 CO 3 .7H 2 O, (g) sugar Ci 2 H 22 On, (h) 
ammonium nitrate NH 4 NO 3 , (i) ammonium sulfate (NH 4 ) 2 SO 4 , (j) silver 
chloride AgCl, (k) silver acetate AgC 2 H 3 O 2 .NB. In (d) and (/) calculate 
the percentage of water, not hydrogen and oxygen. 

6. What is the rational formula for such a compound as 



(a) NH 4 \ rn (6) Fe % ( c ) Na x 

NH,7 C 3 



\ TT^^1U 4 f 

Fe./ P 4 H/ 

7. What is the structural formula for the compounds given in 6? 

8. Determine the valence numbers of the following elements: 

(a) sulfur in H 2 S, Na 2 SO 3 , Na 2 SO 4 , FeS, Na 2 S, FeSO 4 , CaS, CaSO 4 , SO 2 , 
SO 3 , H 2 SO 4 ; (6) nitrogen in NH 3 , NH 4 OH, N 2 O 2 , N 2 O, N 2 O 3 , NO, NO 2 , 
HNO 2 , HNO 3 , NH 4 C1, KNO 2 , NH 4 NO 3 ; (c) iron in FeO, FeCl 3 , Fe(OH) 2 , 
Fe(OH) 3 , FeCNS, Fe(CN) 2 , K 4 Fe(CN) 6 , K 3 Fe(CN) 6 . NOTE. The 
cyan radical (CN~) has a valence number of 1 like Cl~; in it we 
assume C with 4, and N with -f3. 

9. Give an example of the lowest and highest stage of oxidation of 
(a) sulfur, (6) chlorine, (c) nitrogen, (d) manganese, (e) iron, (/) chromium, 
(g) mercury, (h) copper, (i) lead. 

10. Write four or more formulas of different compounds for each series 
of compounds or stages of oxidation of (a) sulfur, (b) chlorine, (c) iron, 
(d) nitrogen, and give the proper names to these compounds. 

11. Write out the rational formulas for the following compounds: 
(a) arsenous chloride, (c) arsenic chloride, (b) cerous nitrate, (d) eerie 
sulfate, (e) potassium manganate, (/) sodium permanganate, (g) calcium 
chloride, (h) calcium chlorite, (i) calcium chlorate, (j) potassium per- 



22 CHEMICAL REACTIONS AND THEIR EQUATIONS 

chlorate, (k) mercuric bromate, (I) mercurous arsenate, (m) mercuric 
arsenite, (n) ferrous bromide, (o) ferric bromide, (p) ferrous bromate, 
(q) ferric bromate, (r) potassium stannate, (s) sodium stannite, (t) lithium 
plumbite, (u) potassium plumbate, (/>) niccolous carbonate, (w) niccolic 
carbonate, (x) plumbic bichromate, (y) plumbous chromate, (z) plumbic 
chromate. NOTE. Consult the key to nomenclature in the appendix. 

12. Which of the following ores contains the highest percentage of iron : 
(a) hematite Fe 2 O 3 , (b) magnetite Fe 3 O4, (c) limonite (Fe 3 O4) 2 .3H 2 O, 
(d) siderite FeCO 3 , (e) pyrite FeS 2 ? 

13. Calculate the percentages of carbon and hydrogen in the first 
five members of the methane series of hydrocarbons which have the 
general formula C n H 2w+2 . (In this case the value of n is 1, 2, 3, 4, 5.) 

14. How many molecules of water of crystallization were present in 
the crystallized salts when (a) 10 grams of cryst. Na 2 CO 3 lose 
6.29 grams of water by heating, (6) 10 grams of cryst. BaCl 2 leave after 
heating a residue of 4.26 grams anhydrous salt, (c) crystallized ZnSCX 
loses after heating 43.8 per cent of its weight? 

15. Give the name and formula of a compound composed of (a) 
21.60 per cent sodium, 33.33 per cent chlorine, 45.07 per cent oxygen; 
(6) 70.13 per cent silver, 20.77 per cent oxygen, 9.10 per cent nitrogen; 
(c) 77.52 per cent lead, 4.49 per cent carbon, 17.88 per cent oxygen. 

16. Ten grams of the chloride of a univalent element x contain 60.6 
per cent of chlorine. If the atomic weight of chlorine is 35.5 what is the 
atomic weight of the element x, and what is its name? 

17. Find the anhydrides of the following substances by subtraction of 
H 2 (a) HN0 2 , (b) H 2 SO 4 , (c) HC1O 3 , (d) H 3 PO 4 , (e) Cu(OH) 2 , (/) A1(OH) 3 , 
(g) LiOH, ( H 4 SiO 4 , (t) Fe(OH) 3 , (j) HPO 3 . Classify these anhydrides 
as basic and acid anhydrides. 

18. 1.251 grams of the oxide of a bivalent element when reduced 
gave 1.00 gram of a fine metallic powder. What is the atomic weight of 
the metal, and what is its name? 

19. Calculate the simplest formula of the following compounds: 
(a) 44.07 per cent mercury, 55.93 per cent iodine; (b) 40 per cent calcium, 
12 per cent carbon, 48 per cent oxygen; (c) 1.59 per cent hydrogen, 
22.22 per cent nitrogen, 76.16 per cent oxygen. 



CHAPTER III 

EQUATIONS INVOLVING NO OXIDATION OR 
REDUCTION 

Equations are the sentences of chemistry as they 
express chemical changes which take place. A chemical 
change or chemical reaction is a molecular phenomena 
in which the composition of the molecule is altered. 
Usually a chemical reaction is brought about by 
mixing solutions of two different substances. The 
evidence that , a reaction has taken place is (a) the 
formation of a precipitate (insoluble substance), (6) the 
formation of gas bubbles (gaseous substance), or (c) the 
change or formation of a color (differently colored 
substance). For example, (a) by mixing a solution of 
potassium iodide and lead nitrate a yellow precipitate 
is produced, for the following reaction has taken place: 

lead nitrate and potassium iodide 

will give \ 

potassium nitrate 



and lead iodide 

(insoluble and yellow) 

If (6) hydrochloric acid (hydrogen chloride) is added 
to a solution of sodium sulfide gas bubbles will be 
observed, for the following exchange has taken place: 

23 






24 CHEMICAL REACTIONS AND THEIR EQUATIONS 

hydrogen chloride and sodium sulfide 



will give 



and 




hydrogen sulfide 
(gaseous) 



If (c) a few drops of ferric chloride are added to a 
solution of potassium sulfocyanate an intensely red color 
will be produced, due to the formation of ferric 
sulfocyanate : 

ferric chloride and potassium sulfocyanate 



will give 



and 




ferric sulfocyanate 
(dark red colored) 



In all three cases, as may be seen, one part of each 
molecule has changed places with the part of a different 
molecule. In order to write chemical equations for 
these reactions the formulas for the substances which 
act upon each other are placed to the left and connected 
by the addition sign ( + ), while the reaction products 
are written to the right separated by the sign of equality 
( = ), which means that the number and kind of atoms 
on both sides of an equation must be equal. Thus an 
equation demonstrates the principle that nothing is lost 
or gained in the course of transformation, and implies 
that the molecules to the left are converted into those 
to the right. In reading an equation the plus sign is 
rendered by and, and the equality sign by give. Balanc- 



EQUATIONS 25 

ing of an equation means sim'ply a balancing of the 
number of atoms, as both sides must be equal. 

The formulas of the compounds for the above three 
reactions are: 

(a) Pb(N0 3 ) 2 + KI = KN0 3 + PbI 2 

(6) HC1 + Na 2 S = NaCl + H 2 S 

(c) FeCl 3 + KCNS = KC1 + Fe(CNS) 3 

None of these equations are balanced. In (a) there 
are two (NO 3 ) radicals to the left, while only one appears 
at the right, and only one I atom at the left, to two I on 
the right. To balance this double the number of KI on 
the left, and KNO 3 on the right, and the correct equation 
will be 

(1) Pb(NO 3 ) 2 + 2KI = 2KNO 3 + PbI 2 . 

In (6) there are two sodium atoms at the left but 
only one at the right, and if we write 2NaCl at the 
right, this will necessitate 2 chlorine atoms at the left, 
that is 2HC1, the correct equation is then : . 

(2) 2HC1 + Na 2 S = 2NaCl + H 2 S. 

In (c) there are three chlorine atoms in FeCl 3 to the 
left, while to the right there is only one chlorine atom 
in KC1, hence writing "3KC1" to the right balances 
the chlorine atoms but gives three potassium atoms to 
the right, and as only one potassium atom is at the 
left, write 3KCNS at the left side and the balanced 
equation is : 

(3) FeCl 3 + 3KCNS = 3KC1 + Fe(CNS) 3 . 

These are now balanced molecular or non-ionic 
equations, in which no consideration has been given 



26 



CHEMICAL REACTIONS AND THEIR EQUATIONS 



to ionization. The number and kind of atoms are 
identical on each side of the equality sign. In balancing 
equations the correct formula for each compound 
involved is of fundamental importance, for the first step 
is always to write the formulas of all the compounds, 
as in (a), (&) and (c). Knowledge of the correct formula 
naturally requires due consideration of the valency of 
the elements and radicals concerned. The valency of 
the common elements must be memorized, and the 
valency of the radicals can be deduced from the familiar 
formulas of a few acids and bases. Thus a knowledge 
of the common acids and bases is essential for the 
derivation of the formulas of the salts. This derivation 
is demonstrated in the following equations: 



Base 

(4) NaOH 

(5) KOH 

(6) Ca(OH) 2 

(7) Ba(OH) 2 

(8) Fe(OH) 3 

(9) A1(OH) 3 

(10) 2NaOH 

(11) 3KOH 

(12) Ca(OH) 2 

(13) 3Ba(OH) 2 

(14) 2Fe(OH) 3 

(15) Al(OH), 



All these equations illustrate that a base and an acid 
will give a salt and water. This type of reaction is called 
neutralization. The equations above are given in the 
molecular or non-ionic form, in the ionic form they 
become : 



Acid 


= Salt 


+ Water 


HC1 


= NaCl 


+ H 2 O 


HN0 3 


= KNO 3 


+ H 2 O 


2HC1 


CaCl 2 


+ 2H 2 O ' 


2HNO 3 


= Ba(NO 3 ) 2 


+ 2H 2 


3HC1 


= FeCl 3 


+ 3H 2 O 


3HNO 3 


= A1(N0 3 ) 3 


+ 3H 2 O 


H 2 SO 4 


Na 2 SO 4 


+ 2H 2 O 


H 3 PO 4 


= K 3 PO 4 


+ 3H 2 O 


H 2 S0 4 


= CaSO 4 


+ 2H 2 


2H 3 PO 4 


= Ba 3 (PO 4 ) 2 


+ 6H 2 O 


3H 2 SO 4 


= Fe 2 (S0 4 ) 3 


+ 6H 2 


H 3 PO 4 


= A1PO 4 


+ 3H 2 



EQUATIONS 27 

(4a) Na+ + OH- + H+ + Cl~ = Na+ + Cl- + H 2 O 
(5a) K+ + OH- + H+ + NO 3 - = K+ + NO 3 - + H 2 O 

(lOa) 2Na+ + 2OH~ + 2H+ + SO 4 ~ = 2Na+ + 

SO 4 - + 2H 2 O 

(lla) 3K+ + 3OH- + 3H+ + PO 4 - = 3K+ + 

PO 4 + 3H 2 O and so on. 

An examination of the ionic equations (4a), (5a), (lOa), 
(lla) shows that some ions have not changed. E.g. in 
(4a) the Na + and Cl~ on the left appear also on the right, 
and therefore have undergone no change. Similarly 
the K+ and NO 3 ~ of equation (5a) remain unchanged. 
The purpose of an equation is to show a chemical reac- 
tion, that is any chemical change, and if all the ions which 
have not changed are cancelled, there remains from the 
above equations 

(16) OH- + H+ = H 2 

This is the general equation for neutralization, and 
defines neutralization as the combination of one hydroxyl 
ion with one hydrogen ion to give one molecule of water. 
By using this universal statement all the particular 
cases can readily be found by adding to the negative 
hydroxyl ion (OH~) the positive metal of any base, and 
to the positive hydrogen ion (H+) the negative non- 
metal or radical of any acid. From the proportion 1 : 1 
it is evident that any monoacid base requires one, a di- 
acid base = two, a triacid base = three hydrogen ions; 
while a monobasic acid requires one, a dibasic acid = two, 
a tribasic acid = three hydroxyl ions. Thus in equa- 
tion (10) two molecules of a monoacid base are needed 
for neutralization of one molecule of a dibasic acid. 

If only one of the reaction products is known in a 
reaction the other reaction product is found by sub- 



28 CHEMICAL REACTIONS AND THEIR EQUATIONS 

tracting the atoms of the right from those on the left. 
Thus, if the unfinished equation is BaCl 2 + Na 2 SO 4 = 
BaSO 4 + ?, the following scheme will give the unknown 
reaction product: 

On the left Ba C1 2 Na 2 S O 4 
On the right Ba - S Q 4 

Remainder C1 2 Na 2 

These remaining two chlorine and two sodium atoms 
naturally give two molecules of sodium chloride, hence 
the completed equation is 

(17) BaCl 2 + Na 2 SO 4 = BaSO 4 + 2NaCl 
Writing this equation in the ionic form: 

Ba++ + 2C1- + 2Na+ + SO 4 ~ = BaSO 4 + 2Na+ + 

2C1- 

and cancelling those ions which remain unchanged (Gl- 
and Na+) the equation becomes: 

(18) Ba++ + SO 4 = BaS0 4 

This is again a general statement, namely that every 
soluble barium salt (furnishing the barium ion) with any 
soluble sulfate (which furnished the sulfate ion) will give 
a precipitate of barium sulfate (which is insoluble). It 
is evident that an ionic equation covers many cases, 
for one may take as the soluble barium salt either BaCl 2 , 
Ba(NO 3 ) 2 , etc., and for the soluble sulfate either Na 2 SO 4 , 
K 2 SO 4 , etc., but it refers only to one specific reaction, 
e.g., the formation of barium sulfate. A non-ionic 
or molecular equation covers only one particular case, 
for from such an equation as (17) we can not make the 
deductions which were possible from (18). 

Balancing Molecular Equations. An unfinished equa- 
tion may be given in which there are more atoms in the 



EQUATIONS 29 

reaction product than in the reacting substances. A 
simple case is CuO + KC1 = CuCl 2 + ?, and as 2C1 
can not be subtracted from 1 Cl, the number of HC1 
must be doubled: 

On the left Cu O 2H 2C1 
On the right Cu C1 2 



Remainder - O 2H - 
which will form one molecule of water, thus: 
(19) CuO + 2HC1 = CuCl 2 + H 2 O 

The problem becomes more difficult when two or more 
compounds are formed, and in this case greater knowledge 
of chemistry is required. For example an experiment 
shows that, on adding a solution of sodium carbonate to a 
solution of aluminum sulfate, a white precipitate as well 
as gas bubbles are formed. Hence the reaction products 
are (a) an insoluble compound, and (6) a gaseous com- 
pound. As neither the Al, the Na, nor the SO 4 radical 
form gaseous compounds, but as the carbonates yield 
readily CO 2 , the unknown gas must be CO 2 and the un- 
finished equation is 

A1 2 (SO 4 ) 3 + Na 2 CO 3 = CO 2 + ? 

The white precipitate, which is the insoluble compound 
formed in this reaction, must be a compound of aluminum 
for all sodium compounds are soluble. At first it appears 
to be the carbonate, but as carbonic acid is given off, this 
can not be the case. The other possible compounds to 
consider would be the oxide and the hydroxide of alumi- 
num, for in this reaction the other insoluble aluminum 
compounds like the phosphates, silicates, etc. can not 
be regarded as possible reaction products, as neither of 
these acid radicals are present. The reaction occurred in 



30 CHEMICAL REACTIONS AND THEIR EQUATIONS 

watery solution hence the hydroxide is probably formed; 
if so, then water must have taken part in the reaction, 
and the unfinished equation becomes: 

A1 2 (S0 4 ) 3 + Na 2 C0 3 + H 2 = CO 2 + A1(OH) 3 + ? 
The next metal to account for in this equation is sodium, 
which loses the carbonate radical, and must combine 
with the sulfate radical to form sodium sulfate: 

A1 2 (S0 4 ) 8 + Na 2 C0 3 + H 2 = CO 2 + A1(OH) 3 + 

Na 2 SO 4 

An examination of this unbalanced equation shows that 
there are two aluminum to the left and hence 2A1(OH) 3 
must be formed. There are 3SO 4 radicals to the left, 
thus 3Na 2 SO 4 are produced this in turn requires 
3Na 2 CO 3 , and these three molecules will yield 3CO 2 . 
Finally as 2A1 (OH) 3 are formed, there are 2 X 3 = 6 
hydrogen atoms necessary and these are present in 
3H 2 O. Making these corrections, the finished and 
balanced equation is: 

(20) A1 2 (SO 4 ) 3 + 3Na 2 CO 3 + 3H 2 O = 

3CO 2 +2A1- (OH), + 3Na 2 SO 4 

As a test of correctness the following checking scheme is 
used: 

Al (SO 4 ) Na C O H 
To the left 2363 9+3 6 

To the right 2 3 6 36+66 

From this we see that one atom of oxygen from each of 
the three carbonate radicals reappears in a molecule of 
the hydroxide. 

To write equation (20) in the ionic form the soluble 
salts are written as ions, while those ions which appear 
on both sides are cancelled, and there remains: 

(21) 2A1+++ + 3CO 3 + 3H 2 O = 2A1(OH) 3 + 3CO 2 



EQUATIONS 31 

The general rule drawn from this equation is that every 
soluble aluminum salt will react with soluble carbonates 
and water to give insoluble aluminum hydroxide and 
gaseous carbon dioxide. It is again evident that 
equation (20) referred to a particular case, while equation 

(21) refers to a specific reaction which covers a number 
of particular cases, for the aluminum sulfate may be 
substituted by aluminum chloride, or aluminum nitrate, 
etc., and the sodium carbonate by potassium carbonate, 
or lithium carbonate, etc. 

Balancing Ionic Equations. In balancing an unfin- 
ished ionic equation the charges on the ions must also be 
balanced. As a rule the balancing is simple if the 
charges are balanced first, before the atoms are balanced. 
Thus in Fe ++ + + CO 3 = ?; there are three positive 
charges on the iron, and two negative charges on the 
carbonate radical, and to make their sum zero converse 
and add 2(+3) and 3 (2), that is two ferric ions 
(together six positive charges) and three carbonate ions 
(together six negative charges), therefore: 

(22) 2Fe+++ + 3CO 3 = Fe 2 (CO 3 ) 3 

It is not necessary that the sum of the charges on each side 
of an equation be zero, as in (22), it can be any integer, 
provided this sum is the same on both sides of the equation. 

Thus in (23) Pb++ + H 2 S = PbS + 2H+ 

there are two positive charges on the left which equal 
the two positive charges on the right. Similarly in 
the equation 

(24) Zn(OH) 2 + 2OH- = ZnO 2 + 2H 2 O 

there is on both sides of the equations an excess of two 
negative charges, while in 

(25) Cr 2 O 7 + 2OH- = 2CrO 4 + 2H 2 O 



32 CHEMICAL REACTIONS AND THEIR EQUATIONS 

the sum is four negative charges on both sides. The sum 
of the ionic charges may thus be different for different 
reactions, .but must be equal on both sides of the 
equation. 

The general meaning of the last four ionic equations 
is that: 

(22) any soluble ferric salt with any soluble carbonate 
will form an insoluble precipitate of ferric carbonate; 

(23) any soluble lead salt and hydrogen sulfide will give 
insoluble lead sulfide and liberate hydrogen ions ; 

(24) zinc hydroxide dissolves in bases (OH) and 
forms zincates and water; 

(25) soluble bichromates with bases give soluble chromates. 
Rules for Finishing and Balancing Equations. The 

rules in balancing and finishing equations are summarized 
as follows: 

(A) Unfinished equations in which only the reacting 
substances are given, and none or one of the reaction 
products (compare equations (17), (18), (19), (20)). In 
such cases there is usually an exchange of the positive 
atoms or radicals, according to the type MN + M'N' = 
M'N + MN', where M and M' are positive elements, and 
N and N' negative elements or radicals. The finishing 
of such problems requires: 

First, the correct formulas for the reacting substances, 
(to the left) and the correct formulas of the reaction 
products (to the right). The reaction products are 
usually gotten by exchanging the positive or the negative 
constituents of the molecules, giving due consideration 
to the valency. 

Second, balance the atoms, as given under B. 

(B) Balancing equations means equalizing the number 
and kind of atoms on both sides of the equation: Begin 
with any symbol on the left and check the same number 



EQUATIONS 33 

of symbols on the right if the number is unequal, select 
the highest number on either side of the equation, and write 
the same number before the formula containing the 
symbol on the opposite side of the equation. Hydrogen 
and oxygen atoms are often balanced by writing H 2 O on 
the opposite side of the equation. 

(C) Balancing ionic equations requires: 

First, equalizing the charges on both sides of the equa- 
tion: the sum of the charges on the left must equal the 
sum of the charges on the right. If the sums are unequal 
and they cannot be made equal in the way shown for 
equation (22), the addition of H+ or OH~ is necessary, 
depending upon whether the reaction occurs in acid or 
alkaline solution. 

Second, balance the atoms, as given under B. 

An ionic equation is always a general statement of the 
chemical properties of an ion, while the non-ionic or 
molecular equation is a particular statement of a definite 
case, or of the weight relations in a certain reaction, for 
by introducing the respective mass of the atoms and 
molecules, the proportions of weight of the reacting 
substances are found. The following questions involve 
some of these weight relations and with the aid of the 
tables in the appendix should offer no difficulty. 

QUESTIONS 

1. Define (a) equation, (b) ionic equation, (c) neutralizatio 

2. What is the cost of materials for preparing 100 kg. of p< ut 
solution of HC1 from NaCl and H 2 SO 4 , if the kg. of NaCl c<,^s p, and 
the kg H 2 SO 4 , 6^? 

3. How many grams of KBr and AgNO 3 are needed to make half a 
kg. of AgBr? 

4. Eighty grams of a solution of sulphuric acid are sufficient to dissolve 
22 grams of cupric oxide. What was the percentage by weight of sul- 
phuric acid in the original solution? 

5. Three hundred and ten grams of borax crystals are dissolved in a 

3 



34 CHEMICAL REACTIONS AND THEIR EQUATIONS 

little water and concentrated sulphuric acid is added. How many grams 
of H 2 SO 4 should be used, and how many grains of boric acid are obtained ? 
Na 2 B 4 7 .10H 2 + H 2 SO 4 = Na 2 SO 4 + 4H 3 BO 3 . + 5H 2 O. 

6. How many kilograms of KNO 3 can be made from a metric ton of 
NaNOs, and how many kilograms of KC1 must be used? (A metric ton 
equals 1,000 kg.) 

7. Calculate the loss in weight if 15 grams of marble are heated until 
the change is complete. 

8. To 32 grams of a solution of sulphuric acid an excess of BaCl 2 was 
added. The BaSO 4 obtained weighed 11.43 grams. Calculate the 
percentage of sulphuric acid in the sulphuric acid solution. 

9. How many grams of HF1 can be made from 20 grams of fluorspar 
(CaFl 2 ), and how many grams of sulphuric acid must be added? What 
will be the volume of HF1 gas, if one mole occupies 22.4 liters? 

10. How many grams of NH 3 can be obtained from 50 grams of 
ammonium chloride, and how many grams of calcium hydroxide are 
necessary? What will be the volume of the NH 3 gas? 

11. How many grams of nitric acid will be obtained by: (a) heating 
250 kg. of sodium nitrate with sulphuric acid; (6) heating 250 kg. of 
potassium nitrate with sulphuric acid? 

12. How many grams of crystallized cupric nitrate (containing 3 mol. 
H 2 O) must be heated to redness to make 50 grams of cupric oxide? 

13. How many grams of nitric acid are needed to convert: (a) 500 
grams sodium hydroxide into sodium nitrate; (6) 500 grams potassium 
hydroxide into potassium nitrate; (c) 500 grams calcium hydroxide into 
calcium nitrate; (d) 500 grams calcium oxide into calcium nitrate; (e) 500 
grams barium hydroxide into barium nitrate? 

14. A solution contains exactly 40.00 grams of sodium hydroxide in 
1 liter. Calculate the quantities by weight of (a) HC1 (b) HNOs, 
(c) H 2 SO 4 , (d) H 5 PO 4 which will be required to neutralize 10 c.c. of this 
solution. 

15. In ascertaining the strength of a dilute solution of HC1, 50 c.c. were 
measured out and neutralized with a solution of sodium hydroxide, 
containing 0.003 grams NaOH in 1 c.c. Thirty-five cubic centimeters 
of this solution was required. What is the strength of the HC1? How 
many grams HC1 are in 1 c.c.? 

16. Forty cubic centimeters of a solution of potassium hyroxide 
containing 0.01 grams of KOH in 1 c.c. were required to neutralize 
40 c.c. of a solution of H 2 SO 4 . How many grams of sulphuric acid did 
15 c.c. of the sulphuric acid solution contain? 

17. Three hundred and eighty grams of mercuric chloride are dissolved 
in water. How many grams of Kl must be added to precipitate all the 
mercury? How many grams of HgI 2 will be obtained? 

18. If 5 grams of crystallized copper sulf ate, CuSO 4 .5H 2 O, are dissolved 



EQUATIONS 35 

in water, how many grams of H 2 S are necessary to precipitate all the 
copper as copper sulfide? How many grams of FeS and 25 per cent 
HC1 are required to generate enough H 2 S for this purpose? 

19. How would 3^011 make a normal solution of the following sub- 
stances: (a) HC1, (h) HNO 3 , (c) H 2 SO 4 , (d) NaOH, (e) NH 4 OH? How 
would you make a N/10 solution of (/) Ca(OH) 2 , (g) NaCl, (h) SO 2 ; 
a 2N solution of (i) HF1, (j) H 3 PO 4 , (it) FeSO 4 ? 

20. Answer the following problems by inspection : 

(a) How much KOH will be required to neutralize 1 liter N solution of 
HNO 3 , 1 liter N solution of H 2 SO 4 , 1 liter N solution of H 3 PO 4 ? (6) 
How many grams of Ca(OH) 2 ? 

21. Write the equations for preparation of the following: 
(a) Cupric oxide from cupric carbonate; 

(6) Cupric chloride from cupric oxide ; 

(c) Ferric hydroxide from ferric sulphate; 

(d) Ferrous sulfide from ferrous sulphate; 

(c) Aluminum carbonate from aluminum nitrate; 
(/) Aluminum hydroxide from aluminum chloride. 
State in each case which substances are necessary for the reaction and 
which substances are precipitated. 

22. How would you make the following compounds: 
(a) Cupric chloride from cupric sulfate? 

(6) Cupric sulphate from cupric chloride? 

(c) Ferrous sulphate from ferrous sulfide? 

(d) Ferric chloride from ferric sulphate? 

(e) Aluminum sulfate from aluminum chloride? 
(/) Aluminum chloride from aluminum sulfate? 

As none of these compounds are insoluble, in which way should they 
be prepared? Base your answers on equations which show possible 
reactions. 

23. You are given a quantity of lead acetate solution. How would 
you prepare: (a) lead chloride, (6) lead nitrate, (c) lead sulfate, (d) lead 
chromate, (e, lead oxide, (f) lead carbonate from this solution? 

24. Starting with lime CaO how would you make (a) CaCl 2 , (fe) 
CaSO 4 , (c) Ca(NO 3 ) 2 , (d) CaCO 3 , (e) solid Ca(OH) 2 ? 

25. Describe at least three ways of making ferric chloride from ferric 
sulphate. 

26. Write the equations for the formation of: (a) Prussian blue 
Fe 4 [Fe(CN) 6 ] 3 from ferric ions and ferrocyanide ions; and (6) Thurnbull's 
blue Fe 3 [Fe(CN) 8 ] 2 from ferrous ions and ferricyanide ions. 

27. Finish the following equations: 
(a) Zn+ + -i- H 2 S = 

(6) NiCO 3 + H 2 SO 4 = 
(c) Ni ++ -f (NH 4 ) 2 S = 



36 CHEMICAL REACTIONS AND THEIR EQUATIONS 

(d) HgN0 3 + Cl- = 

(e) A1+++ + OH' = 

(/) A1(OH) 8 = A1 2 3 + ? 
(g) FeS -f H+ = H 2 S 
(h) CaF 2 + ? = CaS0 4 + ? 
(i) Ba++ + ? = BaC0 3 + ? 
0') A1+++ + ? = A1(OH) 8 + H+ 
(fc) Ca(OH) 2 + ? = Ca(SH) 2 + H 2 O 
(0 NaCl + NH 4 HC0 3 = NaHCO 3 + ? 
(m) CuO + ? + H 2 = Ba(OH) 2 -f CuS 
(n) CoCl 2 .6H 2 O = Co ++ + ? + ? 
(o) CoCl 2 .6H 2 O = CoCl 2 -f ? 
(p) MnCl 2 + (NH 4 ) 2 S = Mn(SH) 2 + ? 
(q) 6KCN + FeS0 4 = K 4 Fe(CN) 6 + ? 
28. Balance the following equations : 
(a) CaC0 3 + HN0 3 = Ca(NO 3 ) 2 + H 2 O + CO 2 
(6) SiCl 4 + NH 3 = Si(NH 2 ) 4 + NH 4 C1 

(c) SiS 2 + H 2 = 2H 2 S + Si0 2 

(d) KaSiF. + KOH = KF + K 4 SiO 4 

(e) SiF 4 + H 2 = H 4 Si0 4 + H 2 SiF 6 
(/) PC1 6 + AsF 3 = PF 5 + AsCl 3 



CHAPTER IV 

EQUATIONS INVOLVING OXIDATION AND 
REDUCTION 

Oxidation and Reduction. Oxidation and reduction 
are defined in a previous chapter as the increase or de- 
crease in the valence numbers. In an equation this 
increase or decrease of the valence number must also be 
balanced. Thus an increase of two in the valence 
number of one element must be balanced by a decrease 
of two in the valence number of some other element. 
In other words, oxidation cannot take place unless there 
is a reduction in some other element. In its simplest 
form this principle is illustrated in the displacement 
reaction : 

(26) Cu++ + Fe = Cu + Fe++ 

In this case the metallic iron (Fe = valence number O) 
acquired two positive charges from the cupric ion (Cu++ 
= valence number +2) and was oxidized (increased its 
valence number) to ferrous ion (Fe ++ = valence number 
+2), while at the same time the cupric ion was reduced 
(decreased its valence number) to metallic copper (Cu = 
valence number O). The experiment is readily per- 
formed by placing a piece of blank iron in a solution of 
any copper salt, e.g., blue vitriol, and the iron will be 
covered by a film of finely divided copper in the form of 
a brownish precipitate. If sufficient time is given for 
the reaction to take place all of the copper may be pre- 
cipitated, and the blue color of the solution will change 
to the greenish color of the ferrous ion. In this case 

37 



35 CHEMICAL REACTIONS AND THEIR EQUATIONS 

the metallic iron has replaced the copper (see also 
Chapter 5). 

Not only with positively charged elements, but also 
with negatively charged elements such a displacement 
takes place: 

(27) C1 2 + 2Br- = 2C1~ + Br 2 

In this case chlorine (O) was reduced to chloride ion ( 1), 
and the bromide ion ( 1) was oxidized to bromine (O). 
It should be noted that in equation (26) the transforma- 
tion of an element to its ion was oxidaton, while in 
equation (27) this transformation was a reduction, in the 
first case the valence number increased, in the second 
case the valence number decreased. 

Metals and Non-metals. As a general rule metals 
are oxidized, and non-metals are reduced to binary 1 
compounds, and oxidized to tertiary 2 compounds. An 
exception to this rule is the direct oxidation of non- 
metals by oxygen, e.g., burning sulfur, phosphorus, or 
carbon in air, as in these reactions binary compounds 
are formed, e.g. S0 2 , P 2 5 , CO, or CO 2 . The reason for 
this is found in the assumption that oxygen is always 
considered negative, or 2, and therefore, in non-metal 
oxides, the non-metal is positive with regard to oxygen; 
while in non-metal hydrides (e.g., H 2 S, NH 3 , CH 4 ) and 
the binary salts (e.g., KC1, CaS, A1N, FeC 2 ) the non-metal 
is always negative, as hydrogen is positive (+1) and all 
metals are positive. 

In the equation 

(28) 2HgO = 2Hg + 2 

the mercury of mercuric oxide was reduced, the valence 

1 Binary compounds have only two kinds of atoms, e.g., NaCl, FeCl 2 . 

2 Tertiary compounds have three different kinds atoms, e.g., KCN, 
BaSO 4 , Na 3 PO 4 . 



EQUATIONS 39 

number being decreased from 2 to 0, while the oxygen of 
mercuric oxide was oxidized, the valence number being 
increased from 2 to 0. However, in the equation 

(29) 4Na + 2 = 2Na 2 O 

metallic sodium (O) is oxidized to sodium oxide (Na = 
+ 1), while the oxygen gas (02) was reduced to an oxide 
(-2). 

Balancing Oxidation and Reduction. As seen in these 
equations, oxidation is exactly counter-balanced by re- 
duction, for the increase in valence number is the re- 
ciprocal of the decrease in valence number, in other 
words, if the increase is added to the decrease the sum is 
zero. Thus in equation: 

(26) oxidation of iron to ferric ion = increase of two, 
reduction of cupric to copper ion = decrease of two; 

(27) oxidation of two bromide ions to bromine = increase 

of 2 X 1 = 2 

reduction of chlorine to chloride ions = decrease of 
2 X -1 = -2 

(28) oxidation of two oxides to oxygen = increase of 

2X2 = 4 
reduction of two mercuric to mercury = decrease 

of 2 X 2 = -4 

(29) oxydation of four sodium = increase of 4 X 1 =4 
reduction of two oxygen to oxide = decrease of 

2 X 2 = -4 

Sum Is Zero. The sum of this increase and decrease 
is always zero in a correct equation. A more complex 
example is: 

(30) Sn + 2HN0 3 + H 2 O = H 4 Sn0 4 + N 2 O 3 . 



40 CHEMICAL REACTIONS AND THEIR EQUATIONS 

In this equation metallic tin (O) has been oxidized to a 
stannic compound (4) stannic acid, while the nitrogen 
of nitric acid (5) has been reduced to nitrous oxide (3), 
thus the tin has increased by 4, and two nitrogen atoms 
have each decreased by 2, and 2 (2) = 4. 

This increase or decrease of the valance number is 
perhaps best illustrated by writing the valence numbers 
which change under the respective symbols: 

(31) 2HBr + H 2 S0 4 = SO 2 + Br 2 + 2H 2 O 

(-1) (6) (4) (0) 

In this reaction two bromine atoms ( 1) of hydrobromic 
acid have been oxidized to free bromine (0), and the 
sulfur (6) of sulfuric acid has been reduced to sulfur (4) 
of sulfur dioxide. However in 

(32) 2HBr + MnO 2 + H 2 SO 4 = MnSO 4 + Br 2 + 2H 2 

(-1) (4) (2) (0) 

not the sulfur of sulfuric acid, but the manganese (4) of 
manganese dioxide has been reduced to manganese (2) 
of manganous sulfate. Writing the later equation in 
the ionic form : 

(33) 2Br~ + MnO 2 + 4H+ = Mn++ + Br 2 + 2H 2 O 

illustrates the rule that any soluble bromide in the 
presence of hydrogen ion (any acid) will give, with 
manganese dioxide, a manganous compound and free 
bromine. 

In balancing an unfinished equation it is always 
advantageous, and in many cases absolutly necessary, 
to strictly follow the rules. 

The Three Rules. First, balance the valence numbers 
of the elements. Write under each symbol its valence 
number, then observe which elements are oxidized or 



EQUATIONS 41 

reduced. The oxidation must be the reciprocal of the 
reduction, for if the element A has been oxidized from 
2 to 4, (increased two steps) and the element B has been 
reduced from 3 to 0, (decreased three steps) then 3 atoms 
of A will counterbalance 2 atoms of B, for 3X2 = 6, 
and 2 X (-3) = -6, and 6 + (-6) = 0. 

Second, balance the ionic charges on both sides. 
Write on each side the sum of the ionic charges of all 
the ions. This sum may be either + , , or 0, but must 
be the same on each side. If the sum is not equal and 
the reaction occurs in an acid solution, add the required 
number of hydrogen ions (H+) to the proper side, and, 
if it occurs in alkaline solution, add hydroxyl ions 
(OH~), until the sum becomes equal. 

Third, balance the atoms. Their kind and number 
must be equal on both sides. 

Strict adherence to the sequence of these three steps is 
essential for a speedy solution of the most complex 
problems; for any attempt to balance the number of 
atoms before the valence numbers and ionic charges are 
balanced may result in failure or an incorrect solution. 
A few examples will illustrate the method. 

Examples. Problem 1. Write an equation for the 
oxidation of a ferrous salt to a ferric salt by potassium 
permanganate in acid solution. Supposing the per- 
manganate is reduced to a manganous salt, we write the 
four symbols for the ions and place under each symbol 
the respective valence numbers: 

MnO 4 - + Fe++ = Mn++ + Fe+++ 
7 22 3 



Connecting the corresponding valence numbers by the 



42 CHEMICAL REACTIONS AND THEIR EQUATIONS 

dotted lines shows that manganese has been reduced 
from 7 to 2 (5 steps), while the iron is oxidized from 2 
to 3 (1 step). In order that oxidation and reduction are 
balanced, the factors are reversed : 1 X 5 steps = 5X1 
step, which means that 1 molecule of permanganate is 
sufficient to oxidize 5 molecules of a ferrous compound 
and the unfinished equation becomes: 

MnO 4 ~ + 5Fe++ = Mn++ + 5Fe+++ 

The proportion of IMn to 5Fe must be kept throughout 
the succeeding operations it may be doubled (2:10) or 
tripled (3:10) but the proportion of 1 to 5 must be 
preserved. 

The second step is balancing the ionic charges. They 
are counted and added together separately for each side 
of the equation, thus: 
on the left : 

one negative charge on the permangante = 1 
5X2 positive charges on the iron . 10 

therefore excess of positive charges 9 

on the right: 

two positive charges on the manganese = 2 
5X3 positive charges on the iron = 15 

a total of positive charges 17 

As the number of charges on both sides must be equal, 
for nothing can be gained or lost during a chemical 
reaction, some other substance must have reacted which 
furnished either 8 positive charges to the left, or yielded 
8 negative charges to the right side of the equation. 
The positive charges are usually furnished by H+ 
(acids), and the negative charges by OH~ (bases), and 



EQUATIONS 43 

as it was specified that the reaction occurred in acid 
solution, 8H+ must be added to the left in order to 
balance the charges: 



MnO 4 - + 5Fe++ + 8H+ = Mn++ + 5Fe+++. 

The sum of the ionic charges is thus equal on both sides 
for (-1) + 5(2) + 8(1) = (2) + 5(3) = 17; and the 
first and second step of balancing is accomplished. It 
remains to balance the atoms. Checking off the atoms 
leaves 4 oxygen atoms and 8 hydrogen atoms on the 
left which will naturally give 4 molecules of water. 
Therefore the completed equation is 
(34) MnO 4 ~ + 5Fe++ + 8H+ = Mn++ + 5Fe+++ + 4H 2 O. 

This ionic equation is the general statement of the 
problem and enables the construction of many equations 
for particular cases by simply supplying the ions with 
the corresponding radicals or elements. In other words, 
the permanganate (Mn0 4 ~) can be any soluble per- 
manganate, either of sodium, potassium, etc.; the ferrous 
salt can be any soluble salt, either sulfate, chloride, 
nitrate, etc.; and the acid can be either HC1, H 2 SO 4 , 
HNO 3 , etc. 

Problem 2. A nitrite solution is oxidized to a nitrate 
solution by an acidified bichromate solution : 

First, the valence numbers are written under the ions 



NO 2 - - 


+- Cr,O 7 - 


= N0 3 - - 


f 2Cr+++ 


3 


2(6) 


5 


2(3) 




= 12 


j 


= 6 











The nitrogen of the nitrite therefore has been oxidized 
from 3 to 5 increased 2 steps. The chromium of the 
chromate has been reduced from 12 to 6 decreased 6 



44 CHEMICAL REACTIONS AND THEIR EQUATIONS 

steps. To balance the factors are reversed, 6(2) = 
2(6), therefore, 

6NO 2 ~ + 2Cr 2 O 7 = 6NO 3 - + 4Cr+++ 

The proportion 6 : 2 can in this case be simplified to 3 : 1 
and the unfinished equation becomes: 

3NO 2 - + O 2 O 7 = 3NO 3 - + 2Cr+++ 
Second, the charges are counted on both sides, thus 

on the left there are 3 nitrites = 3 negative charges 

1 bichromate = 2 negative charges 

a total of 5 negative charges 

on the right there are 3 nitrates = 3 negative charges 

2 chromic ions = 6 positive charges 

a total of 3 positive charges 

To balance +3 and 5 requires either 8 positive 
charges to the left, or 8 negative charges to the right. 
As the reaction occurs in acid solution, eight hydrogen 
ions must be added which give the necessary eight 
positive charges to the left side, and the equation be- 
comes : 

3NO 2 - + Cr 2 7 - + 8H+ = 3NO 3 - + 2Cr+++ 
and the sum on both sides equals 3 positive charges: 
3(-l) + (-2) + 8(1) = 3(-l) + 2(3) = 3 

Third, the atoms are balanced. Nitrogen and chromium 
balance. On the left are (3 X 2) 7 = 13 oxygen 
atoms, while on the right there are only (3 X 3) = 9. 
This excess of 4 oxygen atoms at the left will give with 
the 8 hydrogen atoms four molecules of water and the 
finished equation becomes: 

(35) 3N0 2 - + Cr a Or- + 8H+ = 3N0 3 - + 2Cr+++4H 2 



EQUATIONS 45 

In a few cases it may happen that a substance is 
oxidized and reduced in the same reaction, that is, one 
part of the substance is being oxidized and the other is 
being reduced. It is then advisable to write the formula 
of this substance twice : 

KC1O 3 + KC10 3 = KC1 + KC1O 4 

(+5) (+5) (-1) (+7) 



This diagram shows that the chlorine (+5) of the 
chlorate is reduced to chloride ( 1), while another part 
of the chlorate is oxidized to perchlorate (7). In the 
first case the valence number has decreased 6 steps and 
in the second it has increased 2 steps. It follows, that 
for every 2 molecules of potassium chlorate giving 2 
molecules of potassium chloride, 6 molecules of potas- 
sium chlorate are transformed to 6 molecules of potas- 
sium perchlorate, hence 8 molecules of potassium chlorate 
give 2 molecules of potassium chloride and 6 mole- 
cules of potassium perchlorate, which can be simplified to 

4KC1O 3 = KC1 + 3KC1O 4 

QUESTIONS 

1. Define (a) oxidation and reduction, (b) displacement, (c) stages of 
oxidation, (d) ionic equation, and (e) non-ionic equation. 

2. What is the cost of preparing 5 kg. of hydrogen gas from zinc and 
sulfuric acid? (Assume the price of Zn to be 50^ per kilogram, and that 
of H 2 SC>4 i per kilogram.) 

3. What weight of phosphorus will completely remove the oxygen 
from 2 kg. of air? (Assume that air contains 23 per cent of oxygen 
by weight.) 

4. How many grams of oxygen can be obtained from 75 grams of a 
3 per cent solution of hydrogen peroxide? 

5. How many grams of a 6 per cent solution of hydrogen peroxide will 
convert 5 grams of lead sulfide into -lead sulf ate ? 

6. How many grams of potassium chlorate are necessary to furnish 



46 CHEMICAL REACTIONS AND THEIR EQUATIONS 

the oxygen for conversion of 25 grams metallic copper into cupric oxide? 
(6) How many grams of KC1 will be formed? (c) Can this reaction be 
written in the ionic form? (Give reasons.) 

7. How many grams of ammonium nitrate are necessary to -make 100 
grams of nitrous oxide, which is formed by heating the salt? 

8. What are the substances formed when a bromine solution is added 
to a potassium hydroxide solution? 

9. Write an equation for the reaction of (a) bromide, manganese 
dioxide, sulfuric acid to give free bromine; (6) sodium bromide, manga- 
nese dioxide, sulfuric acid to give free bromine; (c) potassium bromide, 
manganese dioxide, phosphoric acid to give bromine. 

10. How many grams of 50 per cent sulfuric acid can be made from 
10 grams of sulfur by burning the sulfur in air, and passing the sulfur 
dioxide thus formed over finely divided platinum which acts as catalyser, 
thus producing sulfur trioxide? (Give complete set of equations.) 

11. Twenty kilograms of phosphate rock, Ca 3 (PO 4 )2, are heated with 
sand, SiOo, and coke, C, in an electric furnace. The coke burns to CO, 
the phosphate is reduced to phosphorus, and CaSiO 4 is formed. How 
many grams of phosphorus are obtained ? 

12. Two grams of an alloy of Cu and Ag are dissolved in nitric acid. 
To this solution HC1 is added. The precipitated AgCl weighs 1.8 grams, 
(a) How many grams of Ag and Cu were present? (b) What is the per- 
centage of Ag and Cu in the alloy ? 

13. Thirty kilograms of an ore containing 60 per cent Sb 2 S 3 are heated 
with iron. How many kilograms of antimony are formed? 

14. A metric ton (= 1000 kg.) of pyrites containing 8 per cent of 
foreign substances is burned in a sulfuric acid plant, (a) How many 
kilograms of ferric oxide remain? (6) How many kilograms of sulfur 
dioxide are formed ? 

15. How many grams of (a) bromine, (b) iodine are produced by 
passing a current of chlorine gas through a solution of (a) 25 grams of 
magnesium bromide, (b) 10 grams of potassium iodide? 

16. How many grams of (a) water, (b) carbon dioxide are formed by 
burning 1 kg. of methane (CH 4 ) ? 

17. The hydrogen obtained from 24.2 grams of zinc and hydrochloric 
acid is passed over heated mercuric oxide. How many grams of water 
and how many grams of mercury are produced? 

18. How many grams of nitric oxide, NO, are formed when 75 grams 
of copper are dissolved in nitric acid? 

19. How many grams of chlorine can be obtained by heating 43.5 
grams of manganese dioxide with hydrochloric acid? 

20. How many grains of cryst. copper nitrate must be heated to 
redness in order to obtain 46.2 grams of NO2? The other products 
formed in this reaction are O2, H 2 O, and CuO. 



EQUATIONS 47 

21. How many grams of (a) iron, (6) zinc, (c) aluminum are required 
to precipitate 10.8 grams of silver from a silver nitrate solution? 

22. How many grams of iodine will be precipitated from a potassium 
iodide solution by the chlorine, which is formed from 21.75 grams manga- 

nese dioxide and hydrochloric acid? 

23. How many kilograms of iron and sulfuric acid will be needed to 
make 100 kg. of hydrogen gas? 

24. Balance the following equations : 

(a) Zn + Ag+ = Zn ++ + Ag 

(b) S + C1 2 = S + Cl- 

(c) PC1 3 4- KC10 3 = POC1 3 + KC1 

(d) H 3 P0 2 = PH 3 + H 3 P0 4 

(e) Sb + Cl, = Sb 2 Cl 3 
(/) Sb 2 5 = Sb0 2 + 2 

(g) KOC1 + HOC1 = KC1O 3 -f HC1 

(h) NO + C1 2 = NOC1 

(t) CdSO 4 + CdS = CdO + SO 2 

(j) CrCl 3 + H 2 = CrCl 2 + HC1 

26. Finish and balance the following equations: 

(a) HgO + C1 2 + ? = HgCl 2 -f HOC1 

(b) Bi + HN0 3 = Bi(N0 3 ) 3 + ? 

(c) Na + H 2 O = ?+ ? 

(d) As + C1 2 = ? 

(e) (NH 4 ) 2 Pt 2 Cl 6 = Pt + 2C1 2 + ? 
(/) FeS + 2 = Fe 2 3 

(g) KOH + Cl, = KOC1 + KC1 + ? 
(W KC1 + F 2 = 

(0- Al + HC1 = 

(j) CuO -f H 2 = 
(A) HgCl 2 + Hg = ? 

26. Ordinary gun powder is a mixture of saltpeter, carbon, and sulfur. 
The reaction products are K 2 CO 3 , K 2 SO 4 , K 2 S 4 , SO 2 , and N 2 . 

Complete the following reactions : 

(a) KNO 3 -f C = K 2 CO 3 + CO 2 + N 2 

(b) KN0 3 -f C + S = K 2 S0 4 + C0 2 -h N 2 

(c) KNO 3 + C + S = K 2 S 4 + CO 2 + N 2 

Assuming that these are all the reactions concerned in an explosion, 
calculate the correct proportion of saltpeter, carbon and sulfur. 

27. Construct equations for the reduction of potassium from potassium 
sulfide by means of (a) iron, (b) aluminum, (c) manganese. The reaction 
products are (a) FeS, (b) A1 2 S 3 , and (c) MgS. Which metal per pound 
will yield the largest quantity of potassium? Which metal is most 
economical? (A pound of Fe costs 5& Al = 25{, and Mg = 30^.) 
Which reaction will take place most rapidly? 



CHAPTER V 
CONTROL OF REACTIONS 

A chemical reaction is a molecular phenomenon by 
which the composition of the molecule is changed. The 
piost frequent type of reactions are those in which one 
part of the molecule is exchanged by a part of a different 
molecule thus in the reaction AB + CD = AD + CB 
we may say that B has changed places with D, or A 
has been exchanged by B for C, and we draw the infer- 
ence, that either A has a greater attractive force to D 
than to B, or that B has a greater attractive force to 
C than to A. Postponing a discussion of this force of 
attraction holding the atoms in the molecules together, 
we find that a reaction is practically never 100 per cent 
efficient. This means that not all of the AB, and not 
all of the CD has been transformed to AD and CB. In 
every reaction there is a certain and definite percentage 
of transformation, and this percentage forms the basis 
of the measurements in physical chemistry. However, 
the percentage of reacting substances and reaction prod- 
ucts is not the only factor which can be experimentally 
determined for a given reaction. Every reaction requires 
a certain space of time; some may be instantane- 
ous ; some may be so extremely slow as to be hardly recog- 
nizable. Ionic reactions or reactions which occur in 
solutions are usually instantaneous, while molecular 
reactions usually require a measurable period of time. 
For both kinds of reactions, ionic and molecular, it is 
essential that the ions or molecules come in close contact, 
for if the two ions or molecules are far apart, they cannot 

48 



CONTROL OP REACTIONS 49 

react upon each other. The contact of ions or molecules 
is usually termed ionic or molecular collision. Any 
physical means which facilitates ionic or molecular 
collisions will thus increase the possibility of a reaction 
taking place. The physical means which increase or de- 
crease the possibility of collisions and so influence a 
chemical reaction may be of a mechanical, thermal, or 
electrical nature. 

Speed of Reaction. The speed of a reaction is shown 
in the proportional amount of reaction-products (AD 
and CB) which are formed from the reacting substances 
( AB and CD) in a given period of time. Thus it becomes 
evident that the control of reactions depends entirely 
upon the regulation of the speed of the reaction. By 
application of mechanical, thermal, or electrical means, 
either alone or in combination, the speed of a reaction 
is either (a) accelerated (made to proceed faster), (6) 
retarded (made to proceed slower), (c) stopped (made to 
stand still), or (d) reversed (made to proceed in the op- 
posite direction). Hence for every reaction there are 
favorable as well as unfavorable conditions. 

THE MECHANICAL CONTROL: Surface, Catalyzers, Con- 
centration, and Pressure 

Frequency of Collisions is Influenced by Number of 
Molecules Coming in Contact. The mechanical means 
which influence the frequency of molecular collisions are 
those which regulate the number or the amount of 
molecules. The more molecules collide with each other, 
the greater the possibility of a reaction, and vice versa. 
This can be accomplished in four different ways: (1) by 
increasing the surface of contact between two substances, 
(2) by mechanical catalyzers, (3) by higher concentra- 
tion, (4) by greater pressure. 



50 CHEMICAL REACTIONS AND THEIR EQUATIONS 

Increase of Surface by Subdivision. An increase of 
the contact surface between two substances will naturally 
increase the number of different molecules coming into 
close contact, and hence increase the possibility of the 
molecules colliding. By placing a cube of iron contain- 
ing 1 c.c. in hydrochloric acid, it is evident that the 
contact surface between the solid and liquid is 6 sq. cm., 
for the edges of the cube are each 1 cm. long, and each 
side has a surface of 1 sq. cm., and there are six sides on 
the cube. If this cube is cut into eight smaller cubes, as 
shown in Fig. ] , the volume and mass of the iron, as well 






FIG. 1. Increasing the surface by subdivision. 

as the total number of iron molecules has not been 
changed, but the surface has been doubled, making it 
12 sq. cm., and consequently twice as many molecules 
of iron are in contact with hydrochloric acid. If we 
divide these cubes again into a total of 64 smaller cubes, 
the total surface of the same amount of iron will be 23 
sq. cm. It is evident that this division can be theoretically 
continued until we have single iron molecules. It is also 
evident that the smaller the cubes, the greater the sur- 
face, and the greater the contact surface between the 
solid and the liquid, therefore, the greater the possibility 
of molecular collisions. It is thus possible to predict that 
the single large iron cube will dissolve more slowly than 
the four small ones, and that the 64 smaller ones will dis- 



CONTROL OF REACTIONS 



51 



solve faster than the four small ones. The following 
table will be of interest: 

INCREASING THE SURFACE OF A 1-c.c. CUBE BY SUBDIVISION 



Edge length 


Number 
of cubes 


Total surface 


Cubes are 


Magnitude 


(a) 


(6) 


(c) 




(d) 


1 cm. 


1 


6 cm. 2 


visible 


coarse granules 


0. 1 cm. = 1 mm. 


103 


60 cm. 2 


visible 


fine granules 


0.01 cm. 


10 


600 cm. 2 


visible 


coarse powder 


0.001 cm. 


10 


6,000 cm. 2 


microscopic 


fine powder 


0.0001 cm. = In 


10 12 


6 m.2 


microscopic 


very fine powder 


0.00001 cm. 


10" 


60m.2 


ultramicro- 


colloids 


0.000001 cm. 


10>8 


600 m.2 


scopic 


colloids 


0.0000001 cm. = IMM 


1021 


6,000 m.2 


amicroscopic 


colloids 


0.00000001 cm. 


1Q2 


60,000 m.2 


or invisible 


molecules 



(a) 1 m. (meter) = 100 cm. (centimeter) = 1,000 mm. (millimeter) = 1,000,000/i 
(microns) = 1,000, 000, GOO^M (millimicrons) = 39.37 inches. 

(b) 10 3 = 1,000; and 102* = 1,000,000,000,000,000,000,000,000. 

(c) Cm. 2 = cm. X cm. = square centimeter; m. 2 = m. X m. = square meter. 

(rf) Blood corpuscles 7.5 to 15/i, bacilli 1 to 10/u, hydrogen molecules 0.20/tM. chlorine 
molecule 0.40/iM- 

Colloids. Finely divided substances usually react 
faster than coarse ones. For this reason finely powdered 
substances and substances in the colloidal state are 
reactive, as well as gases and substances in solutions. 
Colloids are simply finely divided particles of ultramicro- 
scopical size which form the transition from single 
molecules to. aggregations of molecules of microscopical 
size. 

Increase of Surface by Melting. A reaction may also 
be started when the surface of contact between two solid 
substances is increased, not by subdivision, but by melt- 
ing one substance. Thus in Fig. 2 it is evident that the 
points of contact between two substances is restricted 
when both are solid particles, but is greatly increased 
when one substance is liquid. An example is furnished 
by finely divided sulfur and iron, which can be kept mixed 



52 



CHEMICAL REACTIONS AND THEIR EQUATIONS 



together indefinitely without a reaction occurring. As 
soon, however, as the sulfur is molten, the reaction 
begins. In this case not only the number of sulfur 
molecules and iron molecules colliding with each other 
has increased, but another factor, heat, has been intro- 
duced, the effect of which forms the subject of a later 
paragraph. 

Catalyzers. A reaction may also be either accelerated 
or retarded by the presence of a third substance, a 
catalyzer or catalyst, which is, or appears to be, unchanged 
throughout the reaction but whose presence is essential 
for the procedure of the reaction. The action of a 





FIG. 2. Increasing the surface by melting. 

catalyzer is either of a mechanical or chemical nature. 
The mechanical catalyzers are such substances which, 
when finely divided, have the property to adsorb upon 
their surface the molecules of other substances. Thus, 
finely divided palladium will adsorb hydrogen gas and 
the surface of the metal will be covered with a layer of 
hydrogen molecules. These densely packed hydrogen 
molecules are much more reactive than the widely 
scattered hydrogen molecules in hydrogen gas in other 
words the hydrogen molecules have been concentrated 
and the gas has been condensed to a solid substance. 
Many important processes in industrial manufacture 
depend upon the use of catalyzers (fixation of atmos- 
pheric nitrogen, manufacturing sulfuric acid, etc.). 



CONTROL OF REACTIONS 



53 



Mass Action. The dependence of a chemical reaction 
on the concentration of the molecules is expressed in 
the law of mass action by Guldberg and Waage which 
holds that the chemical effect of molecules participating 
in a chemical reaction is proportional to their mass, that 
is, the amount or number of molecules present in a 
certain volume. In the case of solutions this mass is 
expressed as concentration which is the amount of 
substance (gram-molecules or moles) dissolved in a 
given volume of solvent (1 liter). In the case of gases 
this mass is expressed as pressure, for the pressure is an 
indication of the number of molecules present in a 






FIG. 3. Increasing the surface by concentration or pressure. 

certain volume of a gas. This is best illustrated by 
Fig. 3 in which there is in one case 10 molecules each 
in a given volume, and in the second case 20 molecules 
each in a given volume. The kinetic theory assumes 
that the molecules of a gas are in free movement and 
will therefore collide and bombard the sides of the 
vessels. The sum of these impacts upon the walls of 
the container is shown as pressure, hence by doubling 
the number of molecules in the same volume, the pres- 
sure is doubled, for the number of collisions and impacts 
has been doubled. It has been found that under normal 
conditions (atmospheric pressure of 760 mm. at 0C., 
and at sea-level) one mole of any gas will occupy a 



54 CHEMICAL REACTIONS AND THEIR EQUATIONS 

volume of 22.4 liters. (A mole is the molecular weight 
of a substance in grams, that is, a mole of H 2 is 2 grams, 
a mole of O 2 is 32 grams, etc.) The kinetic theory 
of gases has been extended to molecules in solutions 
which means that the molecules and ions in a solution 
are also free moving and exert pressure the osmotic 
pressure. 

Pressure. The more moles there are in a liter of gas 
or a liter of solution, the more molecules there are in 
this unit volume, and the higher the concentration or 
pressure, hence the greater the possibility for the mole- 
cules to collide and react. The tendency to react is 
generally increased with an increase in the number of 
molecules or ions present in a certain volume; but as, 
during the reaction, some of these molecules or ions 
have been used up or exhausted by being transformed, 
their concentration becomes less as the reaction pro- 
ceeds, yet, at the same time the concentration of the 
reaction products is increased and the tendency to 
again form the original molecules becomes greater. 
A chemical reaction therefore is never complete if the 
reaction products can again form the original sub- 
stances. If two molecules (AB and CD} react to form 
two other molecules (AD and CE) as reaction products, 
then after a certain time there will be a chemical system 
in which all four types of molecules (AB, CD, AD, and 
and CB) are present. The percentage of the different 
molecules will depend upon the speed of the reactions 
AB + CD ->AD + CB and AD + CB -+AB + CD. It 
is evident that the speed of the first reaction is greatest 
at the beginning, for the reaction is started with a high 
concentration of AB and CD, and the absence (zero 
concentration) of AD and CB, but this speed will dimin- 
ish more and more as the reaction products AD and CB 



CONTROL OF REACTIONS 55 

are formed. The speed of the second reaction is at 
first null, as no AD and CB is present, but this speed 
increases slowly with an increase in the amount or 
concentration of AD and CB. As the speed of the 
first reaction diminishes and the speed of the second 
reaction increases, there will come a time when the speed 
of both reactions are equal, and a chemical equilibrium 
is established. 

Equilibrium. In a chemical equilibrium the visible 
effects of a reaction are at stand-still and the reaction 
appears to have stopped, this does not imply, however, 
that the molecules have ceased to collide with each 
other. The molecular collisions still take place, 
but the formation of AB and CD has become equal 
to the formation of AD and CB, in other words the 
exchange of the constituents of the molecules is balanced. 
The same condition of equilibrium is reached when the 
original substances are AD and CB instead of AB and 
CD. Under definite conditions (temperature, concen- 
tration or pressure) there is for any system of substances 
only one equilibrium which is established whether the 
reaction is started with AB and CD or with AD and CB. 

Example of Equilibrium. A concrete example is 
given by the following experiment. In one closed 
vessel hydrogen gas and ferrous-ferric oxide is heated, 
and in another closed vessel iron granules and steam 
(water). After heating a certain time and keeping at a 
certain temperature, both vessels will contain the same 
proportion of hydrogen gas to steam. In the first 
container the reaction was 

(40) 4H 2 + Fe 3 O 4 = 3Fe + 4H 2 O 
while in the second vessel the reaction was 

(41) 3Fe + 4H 2 O = Fe 3 O 4 + 4H 2 . 



56 CHEMICAL REACTIONS AND THEIR EQUATIONS 

Accordingly one reaction is the reverse of the other, a 
fact which can be expressed by using the sign of a 
reversible reaction, namely two arrows: 

(42) 3Fe + 4H 2 CMFe 3 O 4 + 4H 2 

In both cases the same equilibrium was obtained, though 
the original substances were different. The percentage 
of steam and hydrogen will depend upon temperature, 
but at the same temperature this percentage is the same 
in both cases. 

Shifting the Equilibrium. The experimental condi- 
tions can be so arranged that during the reaction one 
or the other of the gases is removed. In this case the 
removal of one gas will constantly shift or destroy the 
equilibrium, and the reaction may proceed to completion. 
Thus by heating the iron granules in a tube open at one 
end, and passing a strong current of steam into this 
tube, hydrogen gas will emerge from the open end until all 
the iron is oxidized. In this case the hydrogen molecules, 
as soon as formed, are carried away and thus removed 
by the excess of steam flowing into the tube, and the 
reaction will be complete in one direction: 

(43) 3Fe + 4H 2 0->Fe 3 4 + 4H 2 

The concentration of hydrogen molecules under these 
conditions was never high enough to initiate the re verse 
reaction. On the other hand if the tube is filled with 
ferrous-ferric oxide and a stream of hydrogen gas is 
passed into it, steam will be formed until all the iron 
oxide is reduced to metallic iron: 

(44) Fe 3 O 4 + 4H 2 ->3Fe + 4H 2 O 

In this case likewise the reaction product, steam, as 
soon as formed, is carried away by the current pf hydro- 



CONTROL OF REACTIONS 57 

gen gas and the reduction of the iron oxide will proceed 
completely. 

Speed of Reaction. Theoretically, the speed of a 
chemical reaction is greatest in the beginning but di- 
minishes more and more as the reaction proceeds. As a 
rule ionic reactions are instantaneous, that is, their 
speed is too rapid to measure. Molecular reactions are 
slower and their speed can be measured and controlled 
as in the preceding example. Some reactions are too 
slow to measure, i.e. the oxidation of iron in air at ordi- 
nary temperature. By increase of concentration (or 
temperature) these reactions will proceed faster, i.e. 
burning of iron wire in oxygen. 

Ionic Equilibrium. The control of an ionic reaction 
depends upon the same principle, for the reaction will 
take place whenever the ionic equilibrium is disturbed 
or shifted by changing the concentration or the amount 
of ions present in a unit volume. Assuming for example 
that two compounds, AB and CD, are each separately 
dissolved in certain amounts of water and measure- 
ments show that the first solution contains 90 per cent 
ions while the second solution contains only 5 per cent 
ions, then an ionic equilibrium has been established 
which can be expressed by the equations: 

(a) AB ^ A+ + B- and (6) CD ^ C+ + D- 
10% 90% 90% 95% 5% 5% 

It is evident that in solution a there are 18 times as many 
ions as molecules, hence the substance AB is said -to be 
highly or strongly ionized; while in solution b there are 
nearly ten times as many molecules as ions, hence the 
substance CD is little or weakly ionized. "In both cases 
the speed of the reaction from left to right and from 
right to left has become equal. Such an ionic equili- 



58 CHEMICAL REACTIONS AND THEIR EQUATIONS 

brium, which is formed whenever a substance is dis- 
solved in water, is disturbed by either adding or 
removing AB, A+, or B~ to or from solution (a), or by add- 
ing or removing CD, C+, or D~ to or from solution (6). 
The addition or removal of molecules or ions is accom- 
plished by: (a) adding more molecules by dissolving 
more of the substance, (6) diluting the solution, (c) 
removing some ions in the form of molecular or non- 
ionized soluble compounds (like water, weak acids, weak 
bases, or weak salts), (d) removing some ions in an 
insoluble compound (precipitates), (e) removing some 
ions in a gaseous compound (gas bubbles), (/) the forma- 
tion of complex ions. In each case the ionic equilibrium 
is shifted. 

Formation of Water. Neutralization means the add- 
ing of H + to an ionic equilibrium containing OH~ 
whereby water is formed, and as water is practically not 
ionized, the removal of H + and OH" from the ionic 
equilibrium will be practically complete, that is a reac- 
tion will take place until all H+ or OH" present have 
been used up. Therefore whenever there is the possi- 
bility that water can be formed from two solutions a 
reaction is likely to take place. 

Formation of Precipitate. In precipitation a mole- 
cular and insoluble compound is formed from ionic 
constituents and by this precipitate certain ions are 
taken from the solution and thereby the ionic equilibrium 
is shifted. The ionic equilibrium BaCl 2 = Ba++ + 
2C1" can be made to proceed from left to right by either 
(a) adding more barium chloride, (6) removing Ba ++ as 
a precipitate (BaSO 4 ) by the addition of SO 4 , or (c) 
removing Cl~ as a precipitate (AgCl) by adding Ag + . 
In the last two cases the ions have been removed by a 
precipitate hence more of the molecular barium chloride 



CONTROL OF REACTIONS 59 

will ionize, and if sufficient SO 4 "~ or Ag+ is added, all 
of the ions may be precipitated, and thereby the reaction 
controlled. 

Formation of Gas. An ionic equilibrium is likewise 
shifted by the formation of a gas from certain ions, and 
the liberation of the gas will remove certain ions from 
the solution. Thus, by adding an acid (H+) to a solution 
of sodium sulfide Na 2 S = 2Na + + S~ , the formation of 
gaseous, non-ionized hydrogen sulfide will remove the 
S from the equilibrium, hence more Na 2 S will ionize 
and, if sufficient acid is added, all of the sulfide ions may 
be removed as H 2 S. Such reactions can be accelerated 
by mechanical means (e.g. agitation or passing a current 
of air through the solution) or thermal means (e.g. 
heating) or both together, for by these means the escape 
or liberation of the gas from the solution is facilitated. 

Formation of Weak Acids. The formation of weak 
acids (acids which are little ionized) will remove hydrogen 
ions. A solution of sulfuric acid contains much H+, 
that is, 

H 2 SO 4 2H+ + S0 4 - 

(small %) (large %) (large %) 

and when this is added to a soluble acetate, citrate, oxalate, 
or other salts of a weak acid, the acidity or hydrogen ion 
concentration will be reduced, as the respective slightly 
ionized acetic acid, citric acid, or oxalic acid is formed: 

(45) H+ + Ac- = HAc 

To illustrate this by experiment, fill a test tube with 
very dilute sulfuric acid and color it with an indicator 
(litmus, methyl orange, methyl violet, congo red, etc.), 
if the neutral salt (e.g. sodium acetate, sodium citrate) 
is added, the color will change and indicate a neutral 
reaction. 



60 CHEMICAL REACTIONS AND THEIR EQUATIONS 

Formation of Weak Bases. The formation of weak 
bases (bases which are little ionized) will remove hydroxyl 
ions. Thus when a solution of sodium hydroxide (strong 
base) containing a large percentage of OH~, is added to 
a solid ammonium salt, the reaction product will be am- 
monium hydroxide (weak base) which is but little ionized : 

(46) OH- + NH 4 + = NH 4 OH 

Try the following experiment: Color a diluted sodium 
hydroxide solution with a few drops of indicator and 
add some ammonium sulfate or ammonium chloride- 
note the change in color. 

Formation of Weak Salt and Complex Ion. A weak 
salt is a compound which, while soluble, is little ionized, 
MN*=?M+ + N~, with a large percentage of MN, and a 
very small percentage of M+ and N~. If such a com- 
pound is formed during a reaction its formation will 
remove certain anions and cations and thereby destroy 
the ionic equilibrium. Likewise the formation of com- 
plex ions destroys the ionic equilibrium by the removal 
of ions. A complex ion is a charged group of atoms 
or an ion combined with a neutral molecule, such 
as the ammonia complexes Cu(NH 3 )4 ++ , Ni(NH 3 ) 4 + +, 
Ag(NH 3 ) 2 + , etc.; the cyanide complexes Cu(CN) 2 + ' f , 
Ni(CN) 4 + + , etc.; certain halogen compounds, oxalates, 
ferro- and ferricyanides, etc. 

Prediction of Reaction. To summarize the facts of 
ionic equilibrium and ionic concentration: A reaction is 
likely to occur whenever there is a possibility that the 
ions of two different molecules in solution (ionic equilib- 
rium) may form (a) water, (6) precipitate, (c) gas, (d) 
weak acids, (e) weak base, (/) weak salt, or (g) complex 
ions, for by the formation of these substances the ionic 
concentration is changed (decreased). By applying 



CONTROL OF REACTIONS 61 

this rule it can be predicted whether or not a reaction 
will occur if two substances in solution are brought 
together. 1 

THE THERMAL CONTROL 

Increase or Decrease of Temperature. The thermal 
means which influence the frequency of molecular collisions 
are those by which not the number but the velocity of the 
molecules is regulated. The faster the molecules move 
the more numerous are the collisions; the slower their 
movement, the less becomes the frequency of collisions. 
Heat or a rise in temperature increases the molecular 
movement, hence increases the nnmber of molecular collis- 
ions and therefore accelerates a reaction. A rise of 10C. 
will generally double the speed of a reaction. 

States of Aggregation. It is assumed that the three 
states of aggregations, solids, liquids, and gases, are due 
to the rate of vibrations of the molecules. In solid and 
liquid substances the vibrations are comparatively slow, 
while in dissolved and gaseous substances the vibrations 
are fast and the molecules can move freely. The 
change from the solid to the liquid state, and from the 
liquid to the gaseous state is thus always a change from 
slower to faster vibrating molecules. Any change in 
the velocity or vibrating speed of the molecules manifests 
itself either in an absorption or liberation of heat. We 
speak therefore of heat of fusion or melting as well as 
heat of vaporization and mean thereby the amount of 
heat necessary to transform a solid at its melting point 
into a liquid, or a liquid at its boiling point into a gas. 

Exothermic and Endothermic. During a chemical 
reaction heat may be produced or consumed. If the 

1 For a full discussion of the control of chemical reactions and equilib- 
rium see Hildebrand's Principles of Chemistry, Chaps, xi-xiv. 



62 CHEMICAL REACTIONS AND THEIR EQUATIONS 

reaction liberates heat (exothermic reaction) then a 
compound (exothermic compound) has been formed ^as 
reaction product which contains less heat than the previ- 
ous molecules, for the molecules of one or the other 
reaction product will vibrate at a lower speed than the 
original reacting substances. Such molecules of exo- 
thermic compounds are, as a rule, stable and not readily 
decomposed. If, on the other hand, during a reaction a 
certain amount of heat is used up (endothermic reaction) 
then a compound (endothermic compound) has been 
formed whose molecules vibrate more rapidly. Endo- 
thermic compounds are, as a rule, unstable and may 
decompose so rapidly as to cause explosions. The 
relation between these two types of reactions and com- 
pounds is shown in the sketch: 

EXOTHERMIC REACTION 

(proceeds rapidly from left to right and liberates heat) 
AB + CD <= AD + CB 

ENDOTHERMIC COMPOUNDS EXOTHERMIC COMPOUNDS 
(unstable) (stable) 

ENDOTHERMIC REACTION 
(proceeds slowly from right to left and absorbs heat). 

An exothermic reaction is the change from endothermic 
to exothermic compounds from faster to slower vibrat- 
ing molecules. This change requires no heat, only an 
external start and the reaction will proceed rapidly 
and completely with the liberation of heat, it may be 
even explosive. An endothermic reaction is the change 
from an exothermic to endothermic compound, from 
slower to faster vibrating molecules. This change 



CONTROL OF REACTIONS 63 

requires a constant influx of heat and the reaction 
will proceed slowly and be limited by the tendency of 
the reaction products to again decompose (dissociation). 

The general tendency of chemical reactions is to 
produce exothermic compounds, that is, reaction products 
which develop the largest amount of heat (Thomson's rule, 
Berthelot's principle). In other words, rapidly vibrating 
molecules have the tendency to rearrange themselves and 
form slower vibrating molecules, thereby liberating a defi- 
nite amount of heat. However if external heat is applied 
to two exothermic compounds, their molecular vibrations 
may be accelerated sufficiently to form an endothermic 
compound. 

Control of Velocity. The rise or fall of temperature 
therefore is a means to increase or decrease the velocity 
of molecular vibrations which in turn controls the speed 
of molecular reactions. The application of heat 
generally has no effect upon ionic reactions. 

It is also evident that the increasing velocity of 
molecular vibrations can aid combination as well as 
decomposition, or dissociation. Dissociation or the 
breaking apart of molecules may be caused by the two 
external means lieat and electricity. The thermic 
dissociation (commonly called dissociation) is produced 
by higher temperatures and continues only as long as 
the external cause (heat) is active; the electric dis- 
sociation (properly called ionization) is produced by 
dissolving electrolytes. 

Dissociation. Thermic dissociation proceeds slowly 
with the rise of temperature and diminishes accordingly 
with a fall of temperature. I.e. hydrogen and oxygen 
begin to combine at 200C. and form steam; at 1,200C. 
dissociation begins, for some molecules of H 2 O will 
decompose into hydrogen and oxygen; at 2,500C. 



64 CHEMICAL REACTIONS AND THEIR EQUATIONS 

there is about half of the steam dissociated into H 2 and 
O 2 , and with a further rise of temperature all the mole- 
cules of steam are finally dissociated. By lowering the 
temperature combination takes place again and at 1,200 
C. only steam will be present. 

Dissociation does not depend solely upon temperature, 
but also upon pressure. Increase in pressure diminishes, 
while a decrease in pressure increases dissociation. For 
each definite pressure at a definite temperature the 
amount of dissociation, that is, the percentage of dis- 
sociated molecules of the same substance will be con- 
stant. The dissociation may be diminished by adding 
more of one of the gaseous components, for then the 
equilibrium is shifted by the introduction of a larger 
amount of one gas. Dissociation is not restricted to 
gaseous compounds and another example that heat can 
combine as well as decompose is the classical experiment 
of Lavoisier of making oxygen from mercuric oxide. 
At ordinary temperature mercury is not attacked by the 
oxygen of the air, but heating slowly transforms it into 
mercuric oxide, and if heated to a higher temperature, 
it decomposes again into mercury vapor and oxygen. 

The explanation of dissociation is found in the assump- 
tion that not only the molecules but also the atoms are 
in movement or rather vibrating. Therefore a rise of 
temperature will not only increase the speed of the 
molecules, but will also increase the speed of the vibra- 
tions of the atoms, and finally the atomic vibrations will 
become so rapid that the atoms will leave their attraction 
spheres and travel as independent units thus splitting 
the molecules. 

Action of Light. The action of light is probably of a 
similar nature. Light rays either combine or decompose. 
Thus a mixture of C1 2 and H 2 when kept in the dark 



CONTROL OF REACTIONS 65 

remains unchanged, but if brought into bright sunlight, 
the gases may combine so suddenly that the mixture 
explodes. The action of light upon the photographic 
plate and the bleaching of aniline dyes are examples of 
decomposition by light. 

THE ELECTRICAL CONTROL: Electro-motive Force 

The electrical means which influence a chemical reac- 
tion depend upon the control of the electro-motive force 
or potential energy. The relationship between chemical 
and electrical energy is close and there are various ways 
to transform one into the other and vice versa. 

Electrolysis. The transformation of electrical into 
chemical energy is illustrated by the phenomena of 
electrolysis and displacement. When an electric current 
is passed through a molten or dissolved electrolyte, that 
is, any substance which conducts the current, separation 
or electrolysis takes place. The terms employed in 
electrolysis are illustrated in the sketch: 

ELECTRODES 
or poles 



Positive electrode Negative electrode 

ANODE CATHODE 

IONS 
or charged atoms 



negative ion positive ion 
ANION CATION 

< e > 

Electrolysis can take place only when the electrolyte is 
either molten or dissolved, that is, already more or less 
dissociated or split into ions; hence it is not a decomposi- 



66 CHEMICAL REACTIONS AND THEIR EQUATIONS 

tion of the electrolyte but merely a separation of the 
ions already present. The cause of this separation or 
migration of the ions to the poles or electrodes is found 
in the charge of the atoms. The anions are negatively 
charged atoms and will migrate or move to the anode or 
positive pole, while the cations or negatively charged 
atoms move toward the cathode or negatively charged 
electrode and -there deliver their charges. The charges 
of the ions are taken up by the electrodes or poles and 
the ions which lose their charge are then deposited as 
atoms or molecules of the free elements. 

In the electrolysis of any acid the hydrogen is found 
at the cathode, and the acid radical at the anode. In 
the electrolysis of any base the metal is liberated at the 
cathode, while the hydroxyl group moves to the anode. 
In the electrolysis of any salt its metallic constituent is 
found at the cathode, while the non-metal or acid radical 
is deposited at the anode. Thus hydrogen and metals 
are considered positive elements, while non-metals and 
acid radicals are negative elements or radicals. The 
atoms or group of atoms deposited at the electrodes and 
deprived of their charges can not exist in free state, 
hence will combine and form molecules. 

Secondary Reaction Products. In many cases these 
newly formed molecules react either with the material 
of the electrode or with the solvent producing new com- 
pounds as secondary reaction products. In such cases, 
not the elements of the electrolyte, but secondary reac- 
tion products are found. If an electric current is 
passed through a solution of sodium chloride (electro- 
lyte), the sodium (cation) migrates to the negative 
electrode (cathode), and the chlorine (anion) moves to 
the positive electrode (anode) and forms respectively 
sodium metal and chlorine gas, but these free elements 



CONTROL OF REACTIONS 67 

both react with water and thus give sodium hydroxide, 
at the cathode and at the anode hydrochloric acid 
and hypochloric acid. These important secondary re- 
action products account for many phenomena and 
their formation is utilized in many industrial processes 
for the manufacture of chemicals. Formerly it was 
believed that the electric current electrolyzed or de- 
composed water into hydrogen and oxygen gas. Now 
we know that absolutely pure water is a very poor 
electrolyte and will not conduct the electric current, 
hence it cannot be decomposed. However if an electro- 
lyte is added, i.e., if the water is acidulated with sulfuric 
acid, then these two gases will be produced, not as 
primary, but as the secondary reaction products. What 
really happens is the decomposition of the sulfuric acid 
which is dissociated or ionized into the cations H+ and 
anions SO 4 ~~. The electric current simply separates 
these ions locally and the H + of the sulfuric acid delivers 
its charge at the cathode and forms H 2 molecules; 
while the anions SO 4 ~ ~ migrate to the anode, deliver 
their charges, and react with water to form sulfuric acid 
and oxygen: 

4H+ = 2H 2 + 4(+) at cathode, 
2SO 4 - + 2H 2 O = 2H 2 SO 4 + O 2 + 4(-) at anode 

Combining these two reactions and cancelling the sul- 
furic acid there remains: 

(47) 2H 2 = 2H 2 + 2 

which is the visible part of this electrolysis, for when the 
gases are collected the volume of hydrogen will be 
exactly twice the volume of oxygen. The sulfuric acid 
formed at the anode dissociates again : 

(48) 2H 2 SO 4 = 4H+ + 2SO 4 



68 CHEMICAL REACTIONS AND THEIR EQUATIONS 

and its respective cations and anions migrate once more 
to the cathode and anode and repeat the process. In 
the electrolysis of acidulated water there is thus a 
continuous dissociation and combination of the acid in 
which the constituents of water take part, and by which 
hydrogen and oxygen are formed as secondary reaction 
products. 

Electrochemical Equivalent. The amount of substance 
deposited or liberated at the electrodes depends upon 
the amount of electricity which passes through the 
solution. The amount of electricity is measured in 
coulombs or ampere-seconds. The same quantity of 
electricity in the same length of time will separate the 
ions at both electrodes in proportion to their equivalent 
weight (Faraday's rule). The equivalent weight is the 
atomic weight divided by the valency. To deposit a 
gram equivalent (atomic weight /valence = x grammes) 
of any ion requires 96,540 coulombs. Hence of copper 
(Cu++) 63,6/2 = 31.8 grams, of silver (Ag+) 107.0 
grams, of ferrous ion (Fe++) 56/2 = 28 grams, of ferric 
ion (Fe+++) 56/3 = 18.7 grams will be deposited by 
96,540 coulombs. The total electrical charge on a gram 
equivalent of any ion is therefore 96,540 coulombs as 
this amount of electricity is required to neutralize the 
total charge of the gram equivalent; and the total 
charge on a mole or gram atom of an ion is 96,540 times 
its valency. 

Electro-motive Force. The force with which the 
ions hold this charge is called electro-affinity or electro- 
motive force. This force varies for different ions. To 
the strong ions belong the cations K + , Na + , Li+, and 
the anions NO 3 ~, F~, Cl~, SO 4 ; while to the weak ions 
belong the cations Hg++, AU+++, Pt+++ + , and the anions 
O , S , CN~. (It is well to note the distinction 



CONTROL OF REACTIONS 69 

between the ionic charge and the force with which the 
ionic charge is held.) 

A substance of strong electro-motive force is one which 
produces strong ions, that is, the atoms hold the electrical 
charge firmly. A substance of weak electro-motive 
force is one whose atoms hold the electrical charge less 
firmly, and thus gives weak ions. 

Displacement. If a substance capable of yielding 
strong ions be added to a solution of weak ions, then the 
weak ions will give up their charge and become electri- 
cally neutral, while the solution will contain the strong 
ions. When a piece of metallic zinc (strong electro- 
motive force) is placed in a solution containing lead 
ions (weak electro-motive force), zinc ions are formed, 
and metallic lead is deposited as a black spongy mass : 

(49) Zn + Pb++ = Zn++ + Pb 

Metallic lead in turn will precipitate copper (weaker 
electro-motive force) from a solution of cupric ions: 

(50) Pb + Cu++ = Pb++ + Cu 

Metallic copper will discharge mercuric ions and deposit 
mercury : 

(51) Cu + Hg++ = Cu++ + Hg 

The weak electro-motive force of mercury is stronger 
than the electro-motive force of silver, therefore it will 
take the ionic charge from the silver ions and crystalline 
silver will be deposited: 

(52) Hg + 2Ag+ = Hg++ + 2Ag 

In the same manner electro-negative elements of 
stronger electro-motive force will become ionized and 
form anions if they come in contact with anions of 



70 CHEMICAL REACTIONS AND THEIR EQUATIONS 

weaker electro-motive force, hence chlorine gas will 
react with bromine ions: 

(53) C1 2 + 2Br- = 2C1~ + Br 2 
Bromine will react with iodide ions: 

(54) Br 2 + 21- = 2Br- + I 2 
Iodine is stronger than sulftde ion : 

(55) I 2 + S = 21- + S 

Displacement Series. The sequence in which the 
elements act in this manner is according to the preceding 
equations for the positive metals Zn > Pb > Cu > Hg 
>Ag, and for the negative non-metals Cl > Br> I >S. 
Zinc is thus more positive than silver, and silver more 
negative than zinc. Chlorine is more negative than 
sulfur, or sulfur is more positive than chlorine. Positive 
and negative are thus relative terms and it is possible 
to place all elements in such a relative series which is 
termed the displacement series. In the displacement 
series, given in the appendix, the most negative elements 
are at the beginning, and these will displace as anions all 
that follow; while the elements at the end are the most 
positive and will displace as cations all above them. 

From the above it becomes evident that, if a piece of 
iron is dipped into a copper salt solution, metallic copper 
will be deposited and the iron will go into solution: 

(56) Fe + Cu++ = Fe++ + Cu 

The cupric ion has given its charges to the iron and has 
become electrically neutral copper, while the previously 
electrically neutral iron has taken this charge and become 
ferrous ion. This transmission of the electric charge 
takes place simultaneously and may occur at any point 
of contact between the iron and the copper ions. 



CONTROL OF REACTIONS 



71 



Voltaic Cells. The transformation of chemical energy 
into electrical energy is demonstrated by voltaic cells or 
batteries. Thus the experimental conditions of the above 
reaction (56) can be so arranged that the direct contact of 
iron and copper ions is prevented, but the exchange of the 
charges or electrons is made possible by a conducting 
connection between the iron and the copper ions which 
causes a stream of electron to pass, thus producing an 
electric current in the conducting connection (metal 




FIG. 4. A voltaic cell or electrical battery. 

wire). These experimental conditions are illustrated in 
a galvanic battery or voltaic cell. It is essential in 
such experimental conditions that the reacting sub- 
stances are locally separated, yet connected by a con- 
ducting media, for only then are the electrons forced to 
travel through the conductor and produce an electric 
current when the circuit is closed. The conditions are 
schematically represented in Fig. 4. 

The farther apart the two metals selected as electrodes 
stand in the displacement series, the greater will be the 
produced electro-motive force or potential, measured in 



72 CHEMICAL REACTIONS AND THEIR EQUATIONS 

volts. Hence Fe and Zn will produce a smaller voltage 
than Al and Pb or Mg and Cu; likewise Zn and Cu 
or Zn and Ag will produce a greater voltage than Zn 
and Fe. 

Example of Cell. In the figure a rod of iron and a rod 
of copper are used as electrodes. The iron stands in 
a solution of ferrous sulfate, -the copper in a solution of 
cupric sulfate, and the solutions are separated by a 
porous wall or semi-permeable membrane. If the 
concentration of the ferrous sulfate solution is low, and 
the concentration of the copper sulfate solution is high 
the metallic iron tends to go into solution, that is, acquire 
two positive charges and thus form ferrous ions, while, 
at the same moment, a cupric ion will deliver two positive 
charges to the copper rod and be deposited as metallic 
copper. The iron rod will lose weight; the copper rod 
will gain weight. The iron electrode becomes . " less 
positive" that is "more negative," and the copper 
electrode becomes more positive. The reactions which 
take place are : 

at the cathode (57) Fe + 2( + ) = Fe++ 
at the anode (58) Cu++ = Cu + 2(+) 

and by adding these two equations, and cancelling the 
two positive charges which appear on both sides, equa- 
tion (56) results. 

Both reactions (57 and 58) occur simultaneously 
and proceed until an equilibrium is established, 
and the current ceases. If the ferrous ion concentra- 
tion is very low in the beginning and the cupric ion 
concentration very high, the battery will live longer. 
If, however, at the beginning the ferrous ion concentra- 
tion is high, and the cupric ion concentration is low, 
little or no current will be produced. 



CONTROL OF REACTIONS 73 

Secondary Cells. By the proper selection of the 
materials used as electrodes and solutions, many of the 
batteries can be arranged so that the reaction is reversi- 
ble. In such a case an electric current passed through 
the cell in the reverse direction will cause the reverse 
reaction and the original state will be resumed. Such 
would be the case in Fig. 4 if the copper were connected 
with the negative pole of a dynamo, and the iron with 
the positive pole so that electrolysis would take place. 
Such a battery is called a secondary cell or accumulator. 

Accumulator. At present the lead accumulator is 
the most practical secondary cell for common use. 
In discharging this accumulator, the lead is the cathode, 
the lead oxide the anode, and the following reaction 
takes place: 

at the cathode (59) Pb + 2(+) = Pb++ 
at the anode (60) PbO 2 + 2(-) = PbO 2 ~ 

Thus at the cathode the lead will dissolve and send 
positive lead ions into solution, and the cathode becomes 
less positive hence negative. From the anode negatively 
charged PbO 2 ions go into solution, and the anode thus 
loses negative charges or electrons, it becomes less 
negative, hence more positive. As there is an excess 
of sulfuric acid in the accumulators the lead ions form 
insoluble lead sulfate: 

(61) Pb++ + S0 4 = PbS0 4 

The plumbate ions formed at the anode react with the 
hydrogen ions of the sulfuric acid and give plumbic 
acid, which in turn reacts with the sulfuric acid and 
yields ultimately lead sulfate. 

(62) PbO 2 - + 2H+ = H 2 PbO 2 

(63) H 2 PbO 2 + 2H+ + SO 4 = PbSO 4 + 2H 2 O 



74 CHEMICAL REACTIONS AND THEIR EQUATIONS 

The endproduct in both cases is thus lead sulfate and 
theoretically the accumulator is entirely exhausted or 
discharged when all the lead arid lead oxide has been 
transformed into lead sulfate; but practically this 
condition never happens as the reaction is reversible. 
When an accumulator is exhausted a chemical equilib- 
rium is established in which the reactions 59 to 63 have 
ceased and a large amount of lead sulfate is present. 
By passing an electric current from an outside source 
through the accumulator the lead sulfate is decomposed 
again into Pb and PbO 2 , as the reactions have been 
reversed. In charging the accumulator the reactions 
are at the anode 

(64) PbS0 4 = Pb++ + S0 4 - 

(65) Pb++ = Pb + 2(+) 

Adding these two reactions together : 
(A) PbS0 4 = Pb + 2(+) + S0 4 . 
The reactions at the cathode are : 

(66) PbS0 4 + 2H 2 = 2H+ + SO 4 + H 2 Pb0 2 

(67) H 2 PbO 2 = PbO 2 - + 2H+ 

(68) Pb0 2 = Pb0 2 + 2(-) 

Adding these three reactions together: 

(E) PbS0 4 + 2H 2 = Pb0 2 + SO 4 + 4H++ 2(-) ;' 

and comparing equation (A) with (5) we find the two 
negative charges liberated in (J3) counterbalanced by 
the two positive charges of (A). By the addition of 
(A) to (B) it is possible to represent the reactions occur- 
ring at the electrodes during charge and discharge of 
the accumulator: 



CONTROL OF REACTIONS 

anode 



charge 



2PbSO 4 + 2H 2 O <=* Pb 

discharge 




electrolyte 

+ 2so 4 - +!H+ 

electrolyte 

Chemical Affinity. The preceding paragraphs show 
the influence of mechanical, thermal, and electrical means 
and the corresponding phenomena have been discussed 
as mass action, kinetic or free energy, and electro-motive 
force of the atoms. However some reactions are affected 
by another factor which can not be grouped under these 
three headings. This fourth factor depends upon the 
nature of the substance and is expressed by a certain selec- 
tive tendency or chemical affinity of one type of atoms 
for another type of atoms. The chemical affinity mani- 
fests itself in certain elements which are grouped together 
in the periodic chart (see appendix). Thus the elements 
located around boron have an affinity for nitrogen and 
yield nitrides (BN, etc.), while those in the neighbor- 
hood of lithium have an affinity for hydrogen and give 
hydrides (LiH, etc.). The reason of chemical affinity 
is not due to electro-motive force and is yet unknown. 1 

Summary. By physical means (mechanical, thermal, 
electrical) the speed of a reaction can be (a) accelerated, 
(6) retarded, (c) stopped, or (d) reversed. 

Mechanical means when applied to ionic and molecular 
reactions control the number of molecules per unit 
volume: (a) subdivision of solid substances (colloids), 
(6) concentration of solutions, (c) pressure of gases, (d) 

1 For a discussion of affinity see American Journal of Science, vol. 46, 
page 490, 1918. 



76 CHEMICAL REACTIONS AND THEIR EQUATIONS 

catalysers. In molecular reactions an increase in the 
number of molecules will increase the frequency of 
molecular collisions and thereby increase the possibility 
of reaction. In ionic reactions the control of the concen- 
tration is brought about by the removal of anions or 
cations; thus the possible formation of (a) water removes 
H+ and OH~, (6) precipitate removes either cation or 
anion, (c) gas removes either cation or anion, (d) weak 
acid removes H+, (e) weak base removes OH~, (/) weak 
salt removes cation or anion, (g) complex ions remove 
cation or anion. 

Thermal means are generally applied to molecular 
reactions and consist in control of temperature or the 
velocity of molecular vibrations.. Raising the tempera- 
ture increases the velocity of the molecules, thus increas- 
ing the frequency of molecular collisions and hence 
accelerating the speed of a reaction. 

Electrical means when applied to ionic reactions 
control the local separation of ions. The density of the 
electric current will determine the speed of the reaction. 

QUESTIONS AND PROBLEMS 

1. What will be the effect of (a) little, (b) much water on the hydrolysis 
of BiCl 3 (equation 160) ? 

2. Write equations for the reaction of bromine with (a) aluminum, 

(b) iron, (c) magnesium, (d) sodium, and arrange them according to their 
relative speeds as deduced from the displacement series. 

3. By heating the hydroxides and carbonates of (a) calcium, (b) copper, 

(c) iron, (d) nickel, (e) silver, the respective metal oxides are formed. 
Write equations and state which compound requires the lowest, and 
which compound the highest temperature for its decomposition. 

4. By adding water to (a) Ca 3 N 2 , (b) Ca 3 P2, (c) Ca 3 As 2 , calcium- 
hydroxide is formed. What are the other reaction products and which 
of these three reactions will proceed most rapidly and why? 

5. By heating aluminum with (a) antimony, (b) arsenic, (c) bismuth, 

(d) phosphorus, the respective binary compounds are formed. Which 
compound will require the least amount of heat? 

6. If the metals Al, Ga, Fe, In, Mg, Zn, are heated in a current of 






CONTROL OF REACTIONS 77 

chlorine gas, the respective chlorides are formed. Write the equations 
and state which metal requires the lowest, and which metal requires the 
highest temperature to start the reaction. 

7. By heating TiCl 4 with metallic Zn or Sn, the TiCU is reduced to 
TiCl 3 and the bichlorides of Zn or Sn are formed. Write equations 
and state which reaction occurs more readily. 

8. Arrange according to velocity the reactions of water with metallic 
(a) calcium, (b) copper, (c) aluminum, (d) lithium, (e) iron, (/) magnesium, 
(g) potassium, (h) sodium. 

9. GeO 2 can be reduced to metallic Ge with either H2 or C. Which 
reaction proceeds more readily? Which compound does hydrogen 
gas reduce more readily GeO 2 or GeS2? Write equations for each. 

10. Predict what will happen if metallic magnesium is heated with 
(a) SiO 2 , (b) GeO 2 , (c) PbO 2 . Which reaction proceeds more readily? 

11. Carbon dioxide can be reduced to carbon by (a) calcium, (6) 
magnesium, (c) potassium, (d) sodium. Write the equations and 
arrange them in the order of their speed of reaction. 

12. Point out which metals are soluble in sulfuric acid under the evolu- 
tion of hydrogen gas: (a) aluminum, (b) copper, (c) iron, (d) magnesium, 
(e) silver, (/) tin, (g) zinc? 

13. Which element reduces HAuCl 4 most readily : (a) copper, (6) iron, 
(c) phosphorus, (d) tin? 

14. Arrange the following oxides in the order in which carbon will 
reduce them most readily to metals: (a) aluminum oxide, (6) iron oxide, 
(c) lead oxide, (d) magnesium oxide, (e) tin oxide, (/) zinc oxide. Write 
the respective equations. 

16. Give a reason why iodine replaces chlorine in the equation: 
2KC1O3 + I 2 = 2KIOs -f C1 2 . (This reaction occurs in the presence 
of HNOs as catalyser). 

16. By heating metallic Mg and RbOH the following reaction occurs: 
2RbOH + 2Mg = 2Kb + 2MgO + H 2 . Why is hydrogen gas formed? 
Why is it not possible for the reaction Mg(OH) 2 -f- 2Kb = 2RbOH -f 
Mg to occur? 

17. If ammonium bichromate is heated green Cr 2 Oa is formed. What 
are the possible reaction products ? How could you test by experiment 
whether the oxygen or nitrogen is oxidized ? (Write two or more equa- 
tions of possible reactions.) 



CHAPTER VI 
TYPES OF REACTIONS AND THEIR EQUATIONS 

Reactions and their equations are divided into types. 
Thus, if a complex molecule is broken apart into simpler 
or elementary constituents, the process is called analysis; 
the reverse process of constructing or building up 
complex molecules from elementary constituents is 
called synthesis; while the exchange of parts of one 
molecule with parts of another molecule is termed 
metathesis. The first two terms are explained by the 
following scheme: 



Type of 
reaction 



ANALYSIS 



marble 

(calcium carbonate) 
CaCO 3 



SYNTHESIS 



lime 

(calcium oxide) 
CaO 



y \ 

calcium oxygen 
Ca O 



calcium oxide 
CaO 



\ 



carbon- 
dioxide 
C0 2 



/ \ 

carbon oxygen 
C O 2 



X 

carbon dioxide 
C0 2 



calcium carbonate 
CaC0 3 

78 



Type of 
substance 

complex, 
compound 



simple 
compound 



element 



simple 
compound 



complex 
compound 



TYPES OP REACTIONS AND THEIR EQUATIONS 79 

Analysis in the above example involves the reactions: 

(69) CaCO 3 ->CaO + CO 2 

(70) 2CaO-2Ca + O 2 

(71) CO 2 -+C + O 2 

while synthesis is expressed by the equations: 

(72) CaO + CO 2 -CaCO 3 

(73) 2Ca + O 2 -2CaO 

(74) C + O 2 



An examination of these two sets of equations reveals 
the fact that in (69) and (72) there are no changes in the 
valence number of the elements involved, while in (70), 
(71), (73), and (74) there is a change in the valence 
number of the elements, therefore analysis and synthesis 
may or may not involve oxydation and reduction. It 
is also evident that (72) is the reverse of (69), (73) the 
reverse of (70), (74) the reverse of (71). An analytical 
reaction of type (69) is termed decomposition, while the 
reverse synthetical reaction (72) is addition in these 
two types there is no change of valency. The analytical 
reaction of type (70) or (71) is known as reduction, while 
the reverse synthetical reaction (73) or (74) is usually 
termed oxydation. But the terms oxydation and reduc- 
tion are ambigiuous for they also mean an increase or 
decrease in the valence number of an element, therefore 
it is more exact to use combination for oxydation, and 
division for reduction when speaking of reactions. 

An example of metathesis: 

(75) CaCl 2 + Na 2 CO 3 = CaCO 3 + 2NaCl 

shows that there is an exchange of parts in a molecule 
without change in the valence number. One of the 
most common types of metathesical reactions is neutrali- 
zation which has previously been defined as the reaction 



80 CHEMICAL REACTIONS AND THEIR EQUATIONS 

between an acid and a base to give water and a salt. 
An example is: 

(76) Ca(OH) 2 + 2HC1 = CaCl 2 + 2H 2 O 
while the reverse reaction is hydrolysis : 

(77) CaCl 2 + 2H 2 O = Ca(OH) 2 + 2HC1 

The interrelation of these different types of reactions 
is shown in the following diagram in which the arrow 
indicates the direction of the reaction, M and X a metal 
or positive radical, N and Y a non-metal or negative 
radical. 

TYPES OF REACTIONS 

A. Reactions involving no oxidation and reduction 
(i.e. no increase or decrease in the valence numbers of 
the elements): 

I. ADDITION 

MN + XN = MN.XN 



II. DECOMPOSITION 



= MXN 2 



III. METATHESIS MN + XY = MY + XN 



IV. NEUTRALIZATION 
V. HYDROLYSIS MOH+HN-MN+HOH 



B. Reactions involving oxidation and reduction (i.e. 
increase or_ decrease in the valence numbers of the 
elements) : 

VI. COMBINATION 



VII. DIVISION M MN 



TYPES OF REACTIONS AND THEIR EQUATIONS 81 

VIII. DISPLACEMENT | M + XN = MN + X 

?= -=* \ N + MY = MN + Y 

IX. SUBSTITUTION } |2M +.XN = MN +MX 



X. RESTITUTION 



2N + MY = MN + YN 



Examples of these ten types of reactions are given in 
the text and later the complex reactions are discussed. 
Complex reactions are those which involve two or more 
types of reactions. 

I. ADDITION 

Ordinarily two binary compounds combine and form 
a more complex compound. The general type is MX + 

NX = MX.NX = MNX 2 . The oxysalts of metals are 
addition products: 

(78) K 2 O + SO 3 = (K 2 O.SO 3 ) = K 2 SO 4 

(79) CaO + C0 2 = (CaO.CO 2 ) = CaCO 3 

Many sulfides form compounds of this type: 

(80) PbS + Sb 2 S 3 = (PbS.Sb 2 S 3 ) = PbSb 2 S 4 

(81) 3CuS + Sb 2 S 3 = (3CuS.Sb 2 S 3 ) = 2Cu 3 SbS 3 

Some acids and bases may be considered addition pro- 
ducts of water and a metallic or non-metallic oxide: 

(82) H 2 + N 2 6 = (H 2 O.N 2 O 6 ) = 2HNO 3 

(83) H 2 O + SO 3 = (H 2 O.SO 3 ) = H 2 SO 4 

(84) H 2 O + CaO = (H 2 O.CaO) = Ca(OH) 2 

(85) H 2 O + Na 2 O = (H 2 O.Na 2 O) = 2NaOH 

(86) H 2 O + NH 3 = (H 2 O.NH 3 ) = NH 4 OH 

Other compounds belonging to addition products are: 

(87) 3PC1 6 + P 2 O 5 = 5POC1 3 ' 

6 



82 CHEMICAL REACTIONS AND THEIR EQUATIONS 

(88) NH 3 + HC1 = NH 4 C1 

(89) Ni++ + 6NH 3 = Ni(NH 3 ) 6 ++ 

(90) Cu++ + 4NH 4 OH = Cu(NH 3 ) 4 ++ + 4H 2 O 

(91) AgCl + 2NH 4 OH = 2Ag(NH 3 ) 2 Cl + 2H 2 O 

(92) Zn(OH) 2 + 2NH 4 C1 + 4NH 3 = Zn(NH 3 ) 8 Cl 2 + 

2H 2 O 

(93) NaOH + CO = NaHC0 2 or HCOONa 

(94) As 2 S 3 + 3(NH 4 ) 2 S = 2(NH 4 ) 3 AsS 3 

(95) SnS + (NH 4 ) 2 S 2 = (NH 4 ) 2 SnS 3 

(96) SnS 2 + (NH 4 ) 2 S = (NH 4 ) 2 SnS 3 

(97) Fe(CN) 2 + 4KCN = K 4 Fe(CN) 6 

(98) Zn(CN) 2 + 2KCN = K 2 Zn(CN) 4 

(99) Ni(CN) 2 + 2KCN = K 2 Ni(CN) 4 

(100) Co(N0 2 ) 3 + 3KNO 2 = K 3 Co(NO 2 ) 6 

II. DECOMPOSITION 

Decomposition is a chemical reaction in which a 
complex compound is broken apart into simpler com- 
pounds without any change in the valence numbers of 
the elements involved it is the reverse reaction of 
addition : 

(101) NH 4 C1 = NH 3 + HC1 

(102) Cu(OH) 2 = CuO + H 2 O 

(103) H 4 SnO 4 = H 2 SnO 3 + H 2 O 

(104) Cu(NO 3 ) 2 = CuO + N 2 5 

(105) Te(OH) 6 = TeO 3 + 3H 2 O 

(106) 2B(NH 2 ) 3 = B 2 (NH) 3 + NH 3 = 2BN + 4NH 3 

To this type belongs dehydration, which is the reaction 
taking place when crystals containing water of crystal- 
lization are heated and lose part or all of their water. 
Dehydration usually proceeds in steps, depending upon 
temperature, thus 



TYPES OF REACTIONS AND THEIR EQUATIONS 83 

beginning at 19C.: 

MgSO 4 .7H 2 O = MgSO 4 .6H^O + H 2 O 
beginning at 38C.: 

MgSO 4 .6H 2 O = MgSO 4 .2H 2 O + 4H 2 O 
beginning at 112C.r 

MgSO 4 .2H 2 O = MgSO 4 .H 2 O -f H 2 O * 
beginning at 203C.: 

MgSO 4 .H 2 O = MgSO 4 + H 2 O 

The complete dehydration is expressed by: 
(107a) MgSO 4 .7H 2 O = MgSO 4 + 7H 2 O 

In this case the dehydrated magnesium sulfate, when 
heated to about 400C., will begin to decompose accord- 
ing to 

(107b) MgSO 4 - MgO + SO 3 

Many other sulfates behave similarly. The dehydration 
and decomposition of copper sulfate is expressed in the 
equations : 

CuSO 4 .5H 2 O = CuSO 4 .3H 2 O -f 2H 2 O, begins at 27C. \ 

CuSO 4 .3H 2 O = CuSO 4 .H 2 O + 2H 2 O, begins at 93C. | dehydration 

CuSO 4 .H 2 O = CuSO 4 + H 2 O, begins at 11 5C. J 

2CuSO 4 = CuSO 4 .CuO + SO 3 , begins at 660C. | decomposi- 

CuSO 4 .CuO = 2CuO 4- SO 3 , begins at 710C. J tion 

The complete dehydration is thus: 
(108a) CuSO 4 .7JI 2 O = CuSO 4 + 7H 2 O 

and the complete decomposition: 
(108b) CuSO 4 = CuO + SO 3 

UI. METATHESIS 

Metathesis is a common reaction. The general type 
MN + XY = MY + XN indicates a double decomposi- 



84 CHEMICAL REACTIONS AND THEIR EQUATIONS 

tion in which an exchange of the elements or radicals 
has taken place. 
Some examples are: 

(109) AgNO 3 + NaCl = AgCl + NaNO 3 

(110) CaO + 2HC1 = CaCl 2 + H 2 O 

(111) CaS + 2HC1 = CaCl 2 + H 2 S 

(112) Ca 3 P 2 + 6HC1 = CaCl 2 + 2PH 3 

(113) A1 2 O 3 + 6HC1 = 2A1C1 3 + 3H 2 O 

(114) A1 2 S 3 + 6HC1 = 2A1C1 3 + 3H 2 S 

(115) Al 2 Se 3 + 6HC1 = 2A1C1 3 + 3H 2 Se 

(116) Al 2 Te 3 + 6HC1 = 2A1C1 3 + 3H 2 Te 

(117) FeS + 2HC1 = FeCl 2 + H 2 S 

(118) 2Na 3 As + 6HC1 = 2AsH 3 + 6NaCl 

(119) SnCl 2 + H 2 S = SnS + 2HC1 

(120) CaO + H 2 S = CaS + H 2 O 

(121) CaCO 3 + Na 2 S = CaS + Na 2 CO 3 

(122) 2KC1 + H,SiF, = K 2 SiF 6 + 2HC1 

(123) 3CaO + Na 3 AlF 6 = 3CaF 2 + Na 3 AlO 3 

(= Al(ONa),) 

(124) CaSiO 3 + 3H 2 F 2 = SiF 4 + CaF 2 + 3H 2 O 

(125) WO 3 + 2KOH = K 2 WO 4 + H 2 O 

(126) K 2 WO 4 + H 2 SO 4 + H 2 O = K 2 SO 4 + WO(OH), 

(127) 2A1(OH) 3 + 3CS 2 = A1 2 S 3 + 3H 2 S + 3CO 2 

(128) H 2 SO 4 + P 2 O B = SO 3 + 2HPO 3 

(129) SnCl 2 + Na 2 CO 3 + H 2 O = Sn(OH) 2 + 

2NaCl + CO 2 

(130) CaC 2 + 2H 2 = C 2 H 2 + Ca(OH) 2 

(131) A1 4 C 3 + 12H 2 O = 3CH 4 + 4A1(OH) 3 

(132) PH 4 I + KOH = PH 3 + KI + H 2 O 

(133) 2H 3 BO 3 + 3PC1 5 = B 2 O 3 + 3POC1 3 + 6HC1 

(134) 2CuCl + Na 2 CO 3 = Cu 2 O + 2NaCl + CO 2 

(135) CuClo + Na 2 CO 3 = CuCO 3 + 2NaCl 

(136) Mo0 3 + 2KOH = K 2 MoO 4 + H 2 O 

(137) Au,O, + 2KOH = 2KAu0 2 + H 2 






TYPES OF EEACTIONS AND THEIR EQUATIONS 85 

In the preceding list only molecular equations are 
given some of which can be put into ionic form. Typi- 
cal ionic equations of metathesic reactions are : 

(138) Ag+ + Cl- = AgCl 

(139) Ba++ + SO 4 = BaS0 4 

(140) Ba++ + CrO 4 = BaCrO 4 

(141) Cu++ + 20H- = Cu(OH) 2 

(142) Co++ + 2OH- = Co(OH) 2 

(143) Zn++ + 2OH- = Zn(OH) 2 

(144) Zn(OH) 2 + 2OH- = Zn0 2 - + 2H 2 O 

(145) Fe++ + 2CN- = Fe(CN) 2 

(146) Ni++ + 2CN- = Ni(CN) 2 . 

(147) Zn++ + 2CN- = Zn(CN) 2 

(148) 2Cu++ + Fe(CN) 6 = Cu 2 Fe(CN) 6 

(149) Cu(OH) 2 + 4NH 4 OH = Cu(NH 3 ) 4 ++ + 20H- 

+ 4H 2 

(150) 2Na 3 SbS + 6H+ = Sb 2 S 6 + 6Na+ + 3H 2 S 

(151) BaO 2 + 2H+ = Ba++ + H 2 O 2 

(152) PtCl 6 - + 2K+ = K 2 PtCl 6 

(153) Cu++ + H 2 S = CuS + 2H+ 

IV. NEUTRALIZATION 

As previously stated, the general equation for neu- 
tralization is H+ + OH- = H 2 O. This equation is 
irrespective of the acid and base employed. The salt 
formed differs in accordance with the acids and bases 
used, but the formation of water is the principal earmark 
of neutralization (compare equations 4 to 15). By 
attaching a negative nonmetal or acid radical, N, to the 
positive hydrogen ion and to the negative hydroxyl ion 
a positive metal or basic radical, M, the general type of 
neutralization is M(OH) + HN = MN + HOH and 
hence it is a metathesical reaction. By substituting any 



86 CHEMICAL REACTIONS AND THEIR EQUATIONS 

metal or positive radical for M, and any nonmetal or 
negative radical for N, a great number of particular 
reactions are possible, some of which are given in equa- 
tions 4 to 15. 

V. HYDROLYSIS 

Hydrolysis is a metathesical change brought about 
by the action of water and is the reverse reaction of 
neutralization, the general type is MN + HOH = 
M(OH) + HN. Hydrolysis takes place when the salt 
of a weak acid, of a weak base, or a compound, whose 
constituents are not far apart in the displacement series, 
is dissolved or brought into contact with water. 

Hydrolysis of salts of a weak acid : 

(154) CH 3 COONa + H 2 O = NaOH + CH 3 COOH 

(155) Na 2 CO 3 + H 2 O = NaOH + NaHCO 3 

(156) NaHCO 3 + H 2 O = NaOH + H 2 CO 3 = NaOH + 

H 2 O + CO 2 

(157) 2BaS + 2H 2 O = Ba(OH) 2 + Ba(SH) 2 

(158) Ba(SH) 2 + 2H 2 O = Ba(OH) 2 + 2H 2 S 

Hydrolysis of salts of a weak base: 

(159) CuSO 4 + 2H 2 O = Cu(OH) 2 + H 2 SO 4 

(160) BiCl 3 + H 2 O = BiOCl + 2HC1 

(161) BiOCl + 2H 2 O - Bi(OH) 3 + HC1 

(162) Aid, + 3H 2 O - A1(OH) 3 + 3HC1 

Hydrolysis of compounds whose constituent elements 
are not far apart in the displacement series : 

(163) PC1 3 + 3H 2 = 3HC1 + P(OH) 3 ( = H 3 PO 3 ) 

(164) PC1 5 + 5H 2 = 5HC1 + P(OH) 5 ( = H,PO 4 + 

H 2 0). 

(165) SbCl 3 + H 2 = SbOCl + HC1 

(166) SiCl 4 + 2H 2 O = SiO 2 + 2HC1 

(167) CC1 4 + H 2 = COC1 2 + 2HC1 



TYPES OF REACTIONS AND THEIR EQUATIONS 87 

(168) BN + 2H 2 O = HBO 2 + NH 3 

(169) Ca 3 As 2 + 6H 2 O = 2AsH 3 + 3Ca(OH) 2 

(170) WC1 6 + 5H 2 O = H 4 WO 5 + 6HC1 

(171) PB 3 + 3H 2 O = 3HBr + P(OH) 3 

(172) PBr 5 + H 2 O = POBr 3 + 2HBr 

(173) POC1 3 + 3H 2 O = H 3 PO 4 + 3HC1 

Under certain conditions even distinct polar compounds 
hydrolyse: 

(174) MgCl 2 + H 2 O = MgO + 2HC1 

(175) 2NaCl + H 2 O = Na 2 O + 2HC1 

These polar compounds are called "strong" salts while 
the compounds enumerated in the list which have con- 
stituents near together in the displacement series, are 
termed "weak" salts. The former ionize strongly, 
the latter ionize but slightly if at all. 

VI. COMBINATION 

Combination is the union of two elements resulting 
in the formation of a compound: M + N = MN. 
While this equation apparently is the simplest type of 
reaction, it involves the principles of oxidation and 
reduction for M is usually oxidized and N is reduced. 
Ordinarily combination is called oxidation, and the 
reverse reaction is termed reduction. In a strict sense, 
however, combination as well as division involves 
oxidation and reduction. The common terminology, 
while practical in some respects, is thus ambigious and 
should be avoided. Some examples of combination or 
" oxidation" are: 

(176) 2H 2 + O 2 = 2H 2 O 

(177) Ca + C1 2 = CaCl 2 

(178) Fe + S = FeS 




.:-: = 



(202) 4KMnO< = 2K^IM>, + 3O* 

203 Mn(OH), = HMnO. + H,O 

204 2Pb(XO,), = 2PbO 



A related reaction is dinociaiion, that 

apart of molecules at 




pbee (see Chapter 5). ExamtpU: At 

(206) 2Fe-O, = 4Te + 3Os 

At low temperature: 

(206) 4Fe + 3O- = 2Fc 



Displacement is a A*Mg-J reaction in 

imimatlMar element in a compoiincL The 
for metals is M + YX = MX + T; 

N - YX = YX - X E:--- .1 -vil 
can replace another metal, or a nonmetal can replace 

metal, M, is more electro-positive than Ike metal, Y. 




nonmetal r X. The electro-positive or negative < 
appendix). Kramples off this type of reactinag are: 



(.207) Cu + HgCls = OiCI, + Hg 

(208) Fe + CnCls - Fed, + . 

(209) Zn + Fed, = ZnO, + Fe 

(210) 2K + ZnCl, = 2KQ + & 

211 Zn -I- 2AgCl = ZnCl. + Ag 

212 Fe + PbS - FeS + Pb 

212a + KFe(C30. - KCN + Fe 



90 CHEMICAL REACTIONS AND THEIR EQUATIONS 

Displacement by nonmetals: 

(213) Br 2 + 2AgI = 2AgBr + 21 

(214) C1 2 + 2AgBr = 2AgCl + Br 2 

(215) F 2 + 2AgCl = 2AgF + C1 2 

(216) 3F 2 + 3H 2 O = 6HF + O 3 

(217) 2F 2 + 2H 2 O = 2H 2 F 2 + 2 

(218) O 2 + 4HC1 = 2H 2 O + 2C1 2 

(219) O 2 + 2MgCl 2 = 2MgO + 2C1 2 

(220) 5O 2 + 4PBr 3 = 2P 2 O 5 + 6Br 2 

(221) Br 2 + H 2 S = 2HBr + S 

The many displacement reactions which involve hy- 
drogen can be grouped into two types. In the first type 
of displacement the hydrogen of acids or bases is being 
replaced by a more electro-positive metal. In this case 
hydrogen is reduced and hence acts as an oxidizing agent : 

(222) Zn + H 2 S0 4 = ZnSO 4 + H 2 

(223) 2A1 + 6HC1 = 2A1C1 3 + 3H 2 

(224) Mg + 2HNO 3 = Mg(NO 3 ) 2 + H 2 

(225) Zn + 2KOH = Zn(OK) 2 + H 2 (= K 2 ZnO 2 ) 

(226) Al + 2KOH = KA1O 2 + K + H 2 

(227) 2Mg + 2KOH = MgO + 2K + H 2 

In the other cases the hydrogen is oxidized, and so acts 
as a reducing agent: 

(228) 2Na 2 O + H 2 = 2NaOH + 2Na 

(229) CuO + H 2 = H 2 O + Cu 

(230) Fe 2 O 3 + 3H 2 = 3H 2 + 2Fe 

(231) 2AgCl + H 2 = 2HC1 + 2Ag 

(232) W0 3 + 3H 2 = 3H 2 + W 

The first set of reactions (222-227), in which hydrogen 
is replaced by a metal, depends upon the electromotive 
force of the metal and the following general rule is 
deduced: Whenever the metal, M, is more positive 



TYPES OF REACTIONS AND THEIR EQUATIONS 91 

than hydrogen (see displacement series) it will take 
the charge from the hydrogen ion and liberate hydrogen 
gas, hence these metals are all soluble in acids according 
to the general equation 2M + 2H+ = 2M+ + H 2 . 

IX. SUBSTITUTION 

Substitution is a chemical reaction which resembles 
displacement; however, the element which interacts 
with a compound does not liberate the replaced element 
but combines with it according to the general type 
2X + MN = XN + XM in the case of metals, and 
2Y + MN = MY + NY in the case of nonmetals : 

(233) 2NH 3 + HgCl 2 = NH 4 C1 + HgNH 2 Cl 

(234) C1 2 + CH 4 = HC1 + CH 3 C1 

(235) C1 2 + H 2 O = HC1 + HOC1 

(236) 61 + 5AgF = IF 5 + 5AgI 

(237) 30 2 + 2PbS = 2S0 2 + 2PbO 

A related reaction is: 

(238) 4Na + 3SiF 4 = Si + 2Na 2 SiF 6 

X. RESTITUTION 

The reverse reaction of substitution is re-substitution 
or restitution, the general type being MN + MY = 
M + NY. To this class belong: 

(239) 2PbO + PbS = 3Pb + SO 2 

(240) 2CuO + Cu 2 S = 4Cu + SO 2 

(241) Cu 2 S + CuSO 4 = 3Cu + 2SO 2 

(242) CO + H 2 = H 2 + C0 2 

Similar and related reactions are: 

(243) 3BaSO 4 + BaS = 4BaO + SO 2 

(244) BaCO 3 + C = BaO + 2CO 
(246) CaC 2 + N 2 = CaCN 2 + C 



92 CHEMICAL REACTIONS AND THEIR EQUATIONS 

(246) 4BaSO 4 + 4C = 4BaO + S0 2 + 4CO 

(247) BaCO 3 + 3Mg + C = BaC 2 + 3MgO 

(248) 2KNO 3 + 10K = 6K 2 O + N 2 

XI. OXIDATION AND REDUCTION 

There are many reactions not classified under the 
previous titles which are characterized by oxidation 
and reduction. These reactions must be studied and 
in each case it should be determined which elements 
are oxidized and which are reduced, for only thus is it 
possible to get a clear conception of the meaning of 
oxidation and reduction. 

A number of such reactions are : 

(249) 2SO 2 + 2 = 2SO 3 

(250) 2KNO 3 = 2KNO 2 + O 2 

(251) PC1 6 = PC1 3 + Cl, 

(252) 2CrO 3 + 2NH 3 = Cr 2 3 + N 2 + 3H 2 O 

(253) 2CrO 3 + 12HC1 = 2CrCl 3 + 3C1 2 + 6H 2 O 

(254) 2CuSO 4 + 4KCN = 2CuCN + 2K 2 SO 4 + C 2 N 2 

(255) 4Co(OH) 3 + 4H 2 SO 4 = 4CoSO 4 + O 2 + 10H 2 O 

(256) 3KC1O = KC1O 3 + 2KC1 

(257) 2KC1O 3 = KC1O 4 + KC1 + O 2 

(258) Na 2 SO 4 + 40 = Na 2 S + 4CO 

(259) 2MnO 2 + 2K 2 O + O 2 = 2K 2 MnO 4 

(260) H 2 TeCl 6 + 2SO 2 + 4H 2 O = Te + 2H 2 SO 4 + 

6HC1 

(261) 2KOH + 2NO 2 = KNO 2 + KNO 3 + H 2 O 

(262) 2KOH + 2C1O 2 = KC1O 3 + KC10 2 + H 2 O 

(263) 2P + 5Br 2 + 6H 2 O = lOHBr + 2HPO 3 

(264) Na 2 S 2 O 3 + Br 2 + H 2 O = Na 2 SO 4 + 2HBr 

(265) 10NO + 6KMnO 4 + 9H 2 SO 4 = 10HNO 3 + 

6MnSO 4 + 3K 2 SO 4 + 4H 2 O 

(266) Au + 4HC1 + HNO, = HAuCl 4 + NO + 2H 2 



TYPES OF REACTIONS AND THEIR EQUATIONS 93 

(267) PbO 2 + HNO 3 + HNO 2 = Pb(N0 3 ) 2 + H 2 O 

(268) 4Hg 2 CrO 4 = 2Cr 2 O 3 + 4Hg 2 O + 3O 2 

(269) 4Hg 2 O = 8Hg + 2O 2 

(270) 2HgCl 2 + SO 2 + 2H 2 O = 2HC1 + 2HgCl 

+ H 2 S0 4 

(271) P 4 + 3KOH + 3H 2 O = PH 3 + 3KH 2 PO 2 

(272) Ag 3 As + 3AgNO 3 + 3H 2 O = 6Ag + As(OH) 3 

+ 3HNO 3 

(273) 3Ca(PO 3 ) 2 + IOC = P 4 + Ca s (P0 4 ) 2 + 10CO 

(274) 2Ca(P0 3 ) 2 + 2Si0 2 + IOC = 2CaSiO 3 + 

10CO + P 4 

(275) 2Ca 3 (PO 4 ) 2 + 6SiO 2 + IOC = 6CaSiO 3 + 

10CO + P 4 

(276) TiO 2 + 2C + 3C1 2 = TiCl 4 + 2COC1 

(277) 4RbO 2 + 4H 2 = 4RbOH + 2H 2 O + O 2 

(278) 2RbO 2 + H 2 O = 2RbOH + O 2 

(279) 4K 2 FeO 4 + 10H 2 O = 4Fe(OH) 3 + 3O 2 + 8KOH 

(280) 2K 4 Fe(CN) 6 + C1 2 = 2K 3 Fe(CN) 6 + 2KC1. 

(281) K 2 Cr 2 7 + 3H 2 = Cr 2 O 3 + K 2 O + 3H 2 O 

(282) NC1 3 + 3H 2 = NH 3 + 3HOC1 

(283) 2S 2 C1 2 + 2H 2 O = SO 2 + 3S + 4HC1 

(284) SC1 4 + 2H 2 O = SO 2 + 4HC1 

(285) 2Co(OH) 2 + Br 2 + 2NaOH = 2Co(OH) 3 + 

2NaBr 

(286) 2H 2 S 2 3 + 21 = 2HI + H 2 S 4 O 6 

(287) 5N 2 H 4 + 4KIO 3 + 4HC1 = 5N 2 + 4KC1 + 

41 + 12H 2 O 

(288) 4HNO 3 + 4CH 2 O = 4NO 2 + 3H 2 O + CO 2 

(289) NH 4 N0 3 = N 2 O + 2H 2 O 

(290) H 2 N 2 O 2 = N 2 O + H 2 O 

(291) HNO 2 + NH 2 OH = H 2 N 2 O 2 + H 2 O 

(292) T1C1 + Cl, = T1C1 3 

(293) 2Mn(OH) 3 + MnO 2 + 2H 2 = 3MnO + 5H 2 O 

(294) 2(NH 4 ) 2 MoO 3 + O 2 = 2MoO 3 + 4NH 3 + 2H 2 O 



94 CHEMICAL REACTIONS AND THEIR EQUATIONS 

The above reactions are all in the non-ionic or molecu- 
lar form; it is a helpful exercise to transcribe these equa- 
tions into the ionic form in cases where the substances 
are ionized. ^Examples of typical ionic equations are: 
Oxidation by H 2 O 2 : 



(296) ^Mn++ + H 2 O 2 = ^OH~ = $Mn(OH) 3 

(296) 2Fe++ + H 2 2 + 40H~ = 2Fe(OH) 3 

(297) 2Co++ + H 2 O + 4OH- = 2Co(OH) 3 

(298) 2Cr++ + H 2 O 2 + 4QH- = 2Cr(OH) 3 

(299) ^KI + H 2 2 = 2KOH + I 2 

The metals Cr and Mn can be further oxidized: 

(300) 2Mn(OH) 3 + H 2 2 = 2Mn0 2 + 4H 2 O 

(301) 2MnO 2 + H 2 O 2 + 2OH~ = 2Mn0 3 - + 2H 2 O 

(302) 2Cr(OH) 3 + 20H- = 2Cr0 2 - + 4H 2 O 

(303) 2CrO 2 - + 3H 2 O 2 + *OH- = 2CrO 4 - + 4H 2 O 

By adding the equations for manganese (295, 300) and 
chromium (298, 302, 303) together, the total equation 
of the reaction taking place becomes: 



(304) 2Mn++ + 2H 2 O 2 + 4OH- = 2Mn0 2 + fH 2 O 

(305) 2Cr++ + 4H 2 2 + 80H~ = 2CrO 4 - + 8H 2 O 

As shown by these equations, hydrogen peroxide will act 
as an oxidizing agent in alkaline solution. In acid 
solution, however, it will act as a reducing agent: 

(306) Cr 2 O 7 + 3H 2 O 2 + 8H+ = 2Cr+++ + 7H 2 O + 

80 2 

(307) 2MnO 4 - + 3H 2 2 + ?H+ = 2Mn0 2 + 4H 2 O + 30 

(308) 2MnO 2 + 2H 2 2 + 4H+ = 2Mn++ + 4H 2 + 

20 2 

By adding the last two equations (307, 308) together, 
and cancelling Mn0 2 which appears on both sides, the 
total equation becomes: 






2 



. TYPES OF REACTIONS AND THEIR EQUATIONS 95 

j 

(309) 2MnO 4 - + 5H 2 O 2 + 6H+ = 2Mn++ + 8H 2 O + 

5O 2 
Reduction by NO 2 ~: 

(310) Cr,O T + 3NO 2 - + 8H+ = 2Cr+++ + 3NO 3 - 

+ 4H 2 O 

(311) 2MnO 4 ~ + 5NO 2 ~ + 6H+ = 2Mn++ + 5NO 3 ~ 

+ te 2 o 

Reduction by HI : 

(312) Cr 2 O 7 + 6HI + 8H+ = 2Cr+++ + 3I 2 + 7H 2 O 

(313) 2MnO 4 - + 10HI + 6H+ = 2Mn++ + 5I 2 + SH 2 O 

(314) H 2 SO 4 + 2HI = H 2 SO 3 + I + H 2 

(315) H 2 SO 3 + tel = H 2 S + $1, + $H 2 O 

(316) H 2 SO 4 + SHI = H 2 S + 4I 2 + 4H 2 O 

Reduction by Fe++: 

(317) Cr 2 O 7 + 6Fe++ + 14H+ = 2Cr+++ + 6Fe+++ 

+ 7H 2 

(318) MnO 4 - + 5Fe++ + 8H+ = Mn++ + 5Fe+++ + 

4H 2 

(319) 3Hg++ + 6Fe++ = 3Hg + 6Fe+++ 

Reduction by H 2 S: 

(320) H 3 AsO 4 + H 2 S = H 3 AsO 3 + S + H 2 O 

(321) 2H 3 AsO 3 + 3H 2 S = As 2 S 3 + fH 2 O 

(322) 2H 3 AsO 4 + |H 2 S = As 2 S 3 + ?S + te 2 O 

Oxidation by HN0 3 : 

(323) 3Zn + 8H+ + 2NO,- = 3Zn++ + 2ND + 4H 2 O 

(324) 3Cu + 8H+ + 2NO 3 - = 3Cu++ + 2NO + 4H 2 O 

(325) 3Hg + 8H+ + 2NO 3 ~ = 3Hg++ + 2NO + 4H 2 O 

(326) 3Sn + 4H+ + 4NO 3 - = 3H 2 SnO 3 + 4NO 

(327) 3PbS + 8H+ + 2NO 3 ~ = 3Pb++ + 3S + 2NO 

+ 4H 2 



96 CHEMICAL REACTIONS AND THEIR EQUATIONS 

Oxidation by H 2 S0 4 : 

(328) Cu + H 2 S0 4 = CuO + SO 2 + H 2 O 

(329) CuO + 2H+ = Cu++ + H 2 

(330) Cu + 4H+ + SO 4 = Cu++ + SO* + 2H 2 O 

Oxidation by HNO 2 : 

(331) Co(N0 2 ) 2 + 2HN0 2 = Co(NO 2 ) 3 + NO + H 2 O 
Oxidation and reduction of Hg~: 

(332) 2Hg+ + H 2 S = Hg + HgS + 2H+ 

(333) 2HgCl + 2NH 3 = NH 4 C1 + HgCNH 2 )Cl + Hg 

(334) 2Hg+ + 2CN- = Hg + Hg(CN) 2 

The student should devise laboratory experiments to 
prove the correctness of these reactions, and should be 
able to predict what will happen if certain chemicals 
react upon each other. He should tell whether the 
solutions must be acidified or made alkaline; he should 
know what color-changes or precipitations to expect, and 
be able to write the specific non-ionic equations for the 
above ionic reactions. 

XII. COMPLEX REACTIONS 

There are many cases in which the reaction products 
interact upon each other and produce a second set of 
reaction products. The equations of these reactions 
seem most complex when secondary reactions are 
included in a single equation; but this complexity disap- 
pears when more than one equation is written. The 
following shows the various steps of such reactions, in 
different equations, and these, added together, form the 
equation of a complex reaction. 

(335) 2A1C1 3 + 3H 2 = A1(OH) 3 + 6HC1 (hydrolysis) 

(336) 3Na 2 S 2 3 + 6HC1 = GNaCl + 3S + 3SO 2 

(decomposition) 



TYPES OF REACTIONS AND THEIR EQUATIONS 97 

Added together and cancelling HC1 leaves: 

(337) 2A1C1 3 + 3Na 2 S 2 3 + 3H 2 O = Al(OH), + 

6NaCl + 3S + 3SO 2 

Reduced to the ionic form 337 becomes : 

(338) 2A1+++ + 3S 2 O 3 + 3H 2 O = Al(OH), + 3S + 

3SO 2 
Likewise the reaction: 

(339) 2AgNO 3 + 2KOH = 2KNO 3 + H 2 O + Ag 2 O 
is composed of a metathesis and decomposition: 

(340) 2AgN0 3 + 2KOH = 2KNO 3 + 2AgOH 

(341) 2AgOH = A g2 O + H 2 O 

Other reactions of this type may involve amphoteric 
substances, as, for example: 

(342) A1C1 3 + 3NaOH = Al(OH), + 3NaCl (metathe- 

sis) 

(343) A1(OH) 3 = H 3 A1O 3 (amphoteric substance) 

(344) H 3 A1O 3 + 3NaOH = Na 3 AlO 3 + 3H 2 O 

(neutralization) 

Adding these three equations together and cancelling the 
intermediate products gives: 

(345) A1C1 3 + 6NaOH = 3NaCl + Na 3 AlO 3 + 3H 2 O 
Similar equations are: 

(346) Pb(NO 3 ) 2 + 2KOH = Pb(OH) 2 + 2KNO 3 

(347) Pb(OH) 2 = H 2 PbO 2 

(348) H 2 PbO 2 + 2KOH = K 2 PbO 2 + 2H 2 O 

(349) Pb(NO 3 ) 2 + 4KOH = 2KNO 3 + K 2 PbO 2 + 

2H 2 
and 

(360) SnCl 2 + 2NaOH = 2NaCl + Sn(OH) 2 
(351) Sn(OH) 2 = H 2 SnO 2 



98 CHEMICAL REACTIONS AND THEIR EQUATIONS 

(362) H 2 Sn0 2 + 2NaOH = Na 2 SnO 2 + 2H 2 O 

(353) SnCl 2 + 4NaOH = 2NaCl + Na 2 Sn0 2 + 2H 2 O 

In a similar way some sulfides are formed : 

(354) 3HgCl 2 + 2H 2 S = 4HC1 + 2HgS.HgCl 2 

(355) 2HgS.HgCl 2 + H 2 S = 2HC1 + 3HgS 

(356) 3HgCl 2 + 3H 2 S = 6HC1 + 3HgS 

and 

(357) 2PbCl 2 + H 2 S = 2HC1 + PbS.PbCl 2 

(358) PbS.PbCl 2 + H 2 S - 2HC1 + 2PbS 

(359) 2PbCl 2 + 2H 2 S = 4HC1 + 2PbS 

It may be that only a part of a double salt will react, 
and the resulting equation will appear complex, e.g. 

(360) 2(NH 4 ) 3 AsS 3 + 6HC1 = 6NH 4 C1 + As 2 S 3 + 3H 2 S 

However, this equation becomes simple if the double salt 
is considered as composed of two salts, e.g. 

(361) 2(NH 4 ) 3 AsS 3 = As 2 S 3 .3(NH 4 ) 2 S 
and one of these reacts: 

(362) 3(NH 4 ) 2 S + 6HC1 = 6NH 4 C1 + 3H 2 S 
Other examples are 

(363) 2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 = K 2 SO 4 + 

2MnS0 4 + 8H 2 O + 5O 2 

which is composed of the three steps: 

(364) 2KMn0 4 + H 2 SO 4 = 2HMnO 4 + K 2 SO 4 

(365) 2HMnO 4 + 5H 2 O 2 = 2Mn(OH) 2 + 4H 2 O - 5O 2 

(366) 2Mn(OH) 2 + 2H 2 SO 4 = 2MnSO 4 + 4H 2 O 

or again: 

(367) 2AuCl 3 + 3H 2 O 2 + 6NaOH = 2Au + 6NaCl + 

3O 2 + 6H 2 



TYPES OF REACTIONS AND THEIR EQUATIONS 99 

which consists of the three reactions: 

(368) 2AuCl 3 + GNaOH = 2Au(OH) 3 + GNaCl 

(369) 2Au(OH) 3 = Au 2 O 3 + 3H 2 O 

(370) Au 2 O 3 + 3H 2 O 2 = 2Au + 3O 2 + 3H 2 

In the following complex reactions two or more steps 
can be recognized, and accordingly two or more equations 
should be written for every case: 

(371) MnO 2 + 4HC1 = MnCl 2 + 2H 2 O + C1 2 

(372) 2BaCl 2 + K 2 Cr 2 O 7 - H 2 = 2BaCr0 4 + 2KC1 

+ 2HC1 

(373) MnCl 2 + 2KOH - H 2 O 2 = 2KC1 + H 2 O + 

MnO(OH) 2 

(374) 3NH 3 + 3NaOCl = 3NaCl + NH 4 OH + H 2 O 

When problems involve a series of incomplete oxida- 
tions and reductions, it is well to construct separate 
equations for each reaction. An example of this type is 
offered by the following experiment: By adding sulfuric 
acid to potassium iodide a number of reaction products 
are formed. These reaction products are hydroiodic 
acid, iodine, sulfur, sulfur dioxide, and hydrogen sulfide 
besides potassium sulfate and water. It is impossible to 
see from a single equation the changes which have 
occurred, and each oxidation and reduction must be 
treated separately: 

A. The formation of hydroiodic acid: 

(375) 2KI + H 2 SO 4 = K 2 SO 4 + HI 

B. The formation of sulfur dioxide and iodine: 

(376) 2KI + 2H 2 SO 4 = K 2 SO 4 + 21 + SO 2 + 2H 2 O 

C. The formation of sulfur and iodine : 

(377) 6KI + 4H 2 SO 4 = 3K 2 SO 4 + 61 + S + 4H 2 O 

D. The formation of hydrogen sulfide and iodine: 

(378) SKI + 5H 2 S0 4 - 4K 2 S0 4 + 81 + H 2 S + 2H 2 O 



100 CHEMICAL REACTIONS AND THEIR EQUATIONS 

These equations show that in A there is a simple meta- 
thesis, the replacement of a weaker acid by a stronger 
acid; in B the iodide is oxidized to iodine, and the hexa- 
valent sulfur is reduced to tetravalent sulfur, in C to free 
sulfur, and finally in D to sulfide. It is quite possible 
th&t the reactions B, C, and D, are secondary reactions, 
as the HI formed in reaction A may be decomposed by 
the sulfuric acid. If such is the case the KI of the equa- 
tions should be replaced by HI, and the unfinished equa- 
tions are: 

(6) HI + H+ + S0 4 = I + S0 2 + H 2 

(c) HI + H+ + S0 4 = I + S + H 2 

(d) HI + H+ + S0 4 = I + H 2 S + H 2 

If the equations (375), (376), (377) and (378) are added 
the following "monster" equation results: 

(379) 18KI + 12H 2 S0 4 = 9K 2 SO 4 + 2HI + 161 + 

SO 2 + S + H 2 S + 10H 2 O 

QUESTIONS AND PROBLEMS 

1. Classify the following reactions according to type and note partic- 
ularly whether oxidation and reduction is involved. If so, which ele- 
ments are oxidized and which are reduced? 

(380) AgNO 3 4- NaNO 2 = AgNO 2 + NaNO 3 

(381) NaCl + AgOH = NaOH + AgCl 

(382) 2Hg+ + Cr0 4 ~ = Hg 2 CrO 4 

(383) Fe(OH), + 3HBr = FeBr 3 + 3H 2 O 

(384) PtS + H, = Pt + H 2 S 

(385) MnCl 2 + Na 2 CO 3 = MnCO 3 + 2NaCl 

(386) 2K 2 MnO 4 + H 2 O = 2KOH + K 2 Mn 2 O 6 + O 2 

(387) Mn+ + + 2C1- + 4H 2 O = MnCl 2 .4H 2 O 

(388) CdCl 2 H- 4RbCl = Rb 4 CdCl 6 

(389) Hg++ + Hg = 2Hg+ 

(390) BeO + H 2 = Be(OH) 2 

(391) T1+ -f Cl- = T1C1 

(392) T1NO 3 + NaCl = T1C1 + NaNO 3 

(393) 2MnO 2 + 2H 2 S0 4 = 2MnSO 4 + 2H 2 O + O 2 
(396) TiCl 4 + 4H 2 = H 4 TiO 4 + 4HC1 

(396) Th(N0 3 ) 4 - Th0 2 + N 2 O 6 



TYPES OF REACTIONS AND THEIR EQUATIONS 101 

(397) KMgCl 3 = KC1 + MgCl 2 

(398) KOH + H 2 S = H 2 O + KSH 

(399) SiO 2 + 2CaF 2 -f H 2 SO 4 = 2CaSO 4 + SiF 4 + 2H 2 O 

(400) CH 4 + 2O 2 = CO 2 + 2H 2 O 

(401) CS 2 + 3O 2 = CO 2 + 2SO 2 

(402) B 2 S 2 + 6H 2 O = 2B(OH) 3 + 3H 2 S 

(403) H 3 BO 3 = HBO 2 + H 2 O 

(404) 2HBO 2 = B 2 O 3 -f H 2 O 

(405) NaH 2 PO 4 = NaPO 3 + H 2 O 

(406) 2Na 2 HPO 4 = Na 2 P 2 O 7 + H 2 O 

(407) BaO 2 + 2HC1 = BaCl 2 + H 2 O 2 

(408) PbO 2 + 4HC1 = PbCl 4 + 2H 2 O 

(409) MnO 2 + 4HC1 = MnCl 4 + 2H 2 O 

(410) PbCl 4 = C1 2 + PbCl 2 

(411) MnCl 4 = d a + MnCl 2 

(412) 2HBr + H 2 SO 4 = Br 2 + SO 2 + 2H 2 O 

(413) KBr + KC10 3 = KC1 + KBrO 3 

(414) 21 + H 2 S = 2HI -h S 

(415) 2HC1 + F 2 = 2HF + C1 2 

(416) 2AsH 3 + 3Na = 2Na 3 As + 3H 2 

(417) 2H 2 S + SO 2 = 3S + 2H 2 O 

(418) 6F + 3H 2 O = 6HF + O 3 

(419) 2KI + O 3 + H 2 O = 2KOH + I 2 + O 2 

(420) As 4 O 6 + 3C = As 4 + 3COi 

(421) 4SbCl 3 -f- 6H 2 O = Sb 4 O 6 + 12HC1 

(422) 2Ga(NO 3 ) 3 = Ga 2 O 3 + 3N 2 O 5 

(423) NaNOj + Pb = NaNO 2 + PbO 

(424) A1 2 (C0 3 ) 3 = A1 2 3 + 3C0 2 

(425) A1 2 (S0 4 ) 3 + K 2 S0 4 = 2KA1(SO 4 ) 2 

(426) A1(OH) 8 = HA1O 2 + H 2 O 

(427) ZnSO 4 .7H 2 O = ZnSO 4 + 7H 2 O 

(428) ZnSO 4 .7H 2 O = Zn ++ + SO 4 ~ + 7H 2 O 

(429) ZnSO 4 = Zn~ + SO 4 ~ 

(430) Co(OOC) 2 = Co + 2C0 2 

(431) CoCO 3 = CoO + CO 2 

(432) 2Cr(OH) 3 + 3H 2 SO 4 = Cr 2 (SO 4 ) 3 + 3H 2 O . 

(433) Cr0 4 ~ - Pb ++ = PbCr0 4 

(434) Pb ++ + 21- = PbI 2 

(435) PbCl 4 + 2H 2 O = Pb0 2 + 4HC1 

(436) Te + SO 8 = TeO + SO 2 

(437) HAuCl 4 = AuCl + C1 2 -h HC1 

(438) HAuCU = And, + HC1 

(439) Ag 2 O + H 2 O = 2AgOH 

(440) Ag+ -f OH- = AgOH 



102 CHEMICAL REACTIONS AND THEIR EQUATIONS 

(441) 2AgOH = Ag 2 + H 2 O 

(442) 2A1 + 6HC1 = 2A1C1, + 3H 2 

(443) 2A1C1, = 2A1 + 3C1 2 

(444) AlCla + 3H 2 O = A1(OH) 3 + 3HC1 
(446) Al(OH), + 3HC1 = A1C1 3 + 3H 2 O 
(446) 2A1 + 3S = A1 2 S 3 

2. Write reactions (380) and (381) in the ionic form and deduce their 
general meaning and the experimental conditions. 

3. Devise a laboratory experiment to prove the correctness of reactions 
(382), (389), (433), and 434. 

4. Translate the meaning of reaction (387) into common language. 

6. What is the difference between the reactions (407), (408) and (409)? 
What equation results if *(408) is added to (410), and (409) is added to 
(411)? 

6. Can you predict from the displacement series that the reactions 
shown in equations (414), (415), (418), (420), (384), and (416) will take 
place? On what grounds do you base your predictions? 

7. What is the meaning of equations (427), (428) and (429), and what 
actual phenomena or reactions correspond to these equations? 

8. Is there an essential difference between the reactions expressed in 
(430) and (431)? 

9. What is the difference or similarity between (431) and (422), (426) 
and (427), (396) and (403) and (404)? 

10. Compare equations (387) and (428) and state if either of these 
reactions is a reversible one? 

11. What will happen if Br 2 is substituted for F 2 in reaction (415)? 
What will happen if HBr is taken instead of HC1? 

12. Write the complex reaction obtained by adding equations (437) 
and (438) together. Describe the experimental conditions. 

13. What is the meaning of equations (439), (440), and (441)? Devise 
experiments for each reaction. 

14. State in common terms the information contained in equations 
(442), (443), (444), (445). Suggest for each equation the experimental 
conditions under which the reaction takes place. 

16. How will the speed of reaction (446) be affected if S is replaced by 
(a) Se, (6) Te, (c) O? 

16. Why does the Te in (436) rob the S of a part of its O? Suggest an 
explanation. 

17. What is the difference between equations (391) and (392)? 

18. What takes place in the reactions expressed in equations (400) and 
(401)? Give a common term for the phenomena. 

19. Does S replace O, or does O replace S in reaction (402) ? Is this 
reaction in harmony with the displacement series? 



APPENDIX I 

KEY TO NOMENCLATURE OF CHEMICAL 
COMPOUNDS 

This key to the nomenclature of inorganic compounds 
enables the beginner to construct numerous formulas 
of compounds whose names are given, and likewise to 
find the correct name for a given formula. It will also 
aid in finding possible reaction products in cases where a 
precipitate or a change of color occurs during the reaction. 

To construct the formula for a compound it is only 
necessary to join the radicals or ions in such a way 
that all the valencies or charges are satisfied or neutral- 
ized. The formula for "cupric sulfate" is found by 
joining Cu++ (cupric ion) and SO 4 (sulfate ion) to 
CuSO 4 . The formula for " ferric chromate" is found by 
joining 2Fe +++ (= six positive charges) with 3Cr04 
(six negative charges) to Fe 2 (Cr0 4 ) 3 . Likewise cupric 
chlorate requires Cu+ + and 2C1O 3 ~, that is Cu(ClO 3 )2. 
Cuprous phosphide is Cu 3 P, and cupric phosphide is 
Cu 3 P 2 . 

By finding the ion or radical under the formulas the 
name of a given formula can be constructed : FeS shows 
iron to be bivalent (Fe^*), hence ferrous sulfide, while 
in Fe 2 S 3 the iron is trivalent (Fe+++), hence ferric 
sulfide. Likewise the formulas Nad, NaClO 2 , NaClO 3 , 
NaC10 4 represent respectively sodium chloride, sodium 
chlorate, sodium chlorate, and sodium perchlorate. 

The column indicating the color of the ion or radical 
is of advantage in laboratory work where the formation 

103 



104 



APPENDIX 



Name of element, ions or 
radicals 



Valence 
number 



Symbol of 



Element 



Cation 



Anion 



Color of 
ions 



Aluminum 

aluminum ion 

aluminate ion 

Ammonia 

ammonium 1 

Antimony 

antimonous 

antimonic 5 

antimoniate 5 

antimonite 3 

Arsenic 

arsenpus 3 

arsenic 5 

arsenide 

arsenite 

arsenate 5 

Barium 

barium 2 

Bismuth 

bismuth 

bismuthyl 

Boron 

boron 3 

borate 

tetraborate 3 

Bromine 

bromide 1 

bromite 

bromate 5 

Cadmium 

cadmium 2 

Calcium 

calcium 2 

Carbon 

carbide 4 

carbonate 4 

bicarbonate 4 

Cerium 

cerous 3 

eerie 4 

Chlorine 

chloride 1 

hypochlorite 1 

chlorite 3 

chlorate 5 

perchlorate 7 

Chromium 

chrompus 2 

chromic 3 

chromite 3 

chromate 5 

bichromate 5 

Cobalt 

cobaltous 2 

cobaltic 3 

Copper 

cuprous 1 

cupric 2 

copper ammonium 2 



Al 

NHa 
Sb 



As 

Ba 
Bi 
B 

Br 

Cd 
Ca 
C 

Ce 
Cl 

Cr 

Co 
Cu 



AS+ 



BiO + 

B+++ 



Cd ++ 
Ca ++ 



Ce + 



A10 2 - 



SbO?- 



Cs'" 
AsOs 
As04 



B0 3 " 



Br~ 
|Br0 2 - 
BrOr 



C" 

co 3 ~- 

HC0 3 - 



ci- 

cio- 

cio 2 - 

ClOj- 



Cr0 2 - 
CrOs- 
Cr 2 0; 



Co ++ 
Co + ++ 

Cu+ 

Cu+ + 

Cu(NH 8 )4 



blue 

purple 

green 

yellow 

orange 

red 
(unstable) 

colorless 
blue 
deep blue 



APPENDIX 



105 



Name of element, ions, or 
radicals 



Valence 
number 



Symbol of 



Element Cation 



Anion 



Color of 
ions 



Fluorine 

fluoride - 1 

Gold 

aurpus 

auric 

Hydrogen 

hydrogen 1 

hydride -1 

Iodine 

iodide ~ 1 

iodate 5 

periodate 7 

Iron 

ferrous 

ferric 

ferrate 6 

Lead 

plumbpus 

plumbic 4 

plumbate 4 

Lithium 

lithium 

Magnesium 

magnesium 2 

Manganese 

manganous 

manganic 3 

manganate 5 

permanganate 7 

Mercury 

mercurpus 1 

mercuric. ... 2 

Nickel 

niccolpus 2 

niccolic 

Nitrogen 

nitride 3 

nitrite 3 

nitrate 5 

Oxygen 

oxide -2 

Phosphorous 

Ehosphide 3 

ypophosphite 1 

phosphite 3 

phosphate 5 

metaphosphate 5 

monophosphate 5 

pyrophosphate 5 

Platinum 

platinpus 2 

platinic 4 

Potassium 

potassium 1 



F 
Au 

H 
I 

Fe 

Pb 

Li 

Mg 
Mn 

Hg 

Ni 
N 

.O 

P 



Pt 
K 



Au ++ 
H+ 



Fe ++ 
Fe +++ 



Pb+ 
Pb+ 



Mg + 
Mn + 



Hg+ 



Pt + 



F- 



H 2 



I- 

ror 



Fe0 4 ~ 
Pb0 8 - 



MnO*~ 
MnO 4 



N" 1 

NO 2 - 

N0 3 - 

0" 
pur 

H 2 P0 2 - 
H 8 P0 3 
POi 

PO,- 

H 2 P0 4 



yellow 
yellow 



green 
yellow- 
brown 
red 



si. pink 
si. green 
green 
purple 



green 
(unstable) 



yellow 
orange 



106 



APPENDIX 



Name of element, ions or 


Valence 




Symbol of 




Color of 


radicals 


number 


Element 


Cation 


Anion 


ions 


Silicon . . 


o 


Si 








silicic! e . 


4 






Si IV 




silicate 


4 






SiO4 




metasilicate 


4 






SiOs 




Silver . ... 





Aff 








silver 


1 




Ag + 






Sodium 





Na 








sodium. 


1 




Na+ 






Strontium 





Sr 








strontium 


2 




Sr ++ 






Sulphur 


o 


g 








sulphide 
hydrosulnde 


-2 
-2 






s 

HS- 




sulfite 


4 






SOs 




bisulfite 


4 






HSO 3 ~ 




sulfate 


6 






SO4 




bisulf ate 


6 






HSO4~ 




Tin 
stannous 



2 


Sn 


Sn ++ 






stannite 


2 






SnOs 




stannic . . 


4 




Sn ++ ++ 






stannate 


4 






SnOa 




Zinc 


o 


Zn 








zinc 


2 




Zn ++ 






zincate . 


2 






ZnO2 




zinc ammonium 


2 




Zn(NHa)4++ 






Uranium 





u 








uranous 


2 




U++ 




light green 


uranyl 


6 




UOa++ 




yellow 


uranate 


6 






UOi" 


yellow 


Common Organic Radicals 
acetates (CHsCOQ-) ..... 
cyanides 


1 
1 






Ac- 

CN- 




cyanates 


1 






CNO- 




ferrocyanides 


4 






Fe(CN) 6 




ferricyanides .... 


3 






Fe(CN)e 






2 






Ox~ 




rhodanates .... 


1 






CNS- 

















of a distinctive color in the test tube will show the 
respective ion or compound. Thus, if during a chemical 
reaction, the yellow solution of a chromate turns green- 
chromic ion has been formed, if the purple permanganate 
solution turns green manganate ion has been produced, 
and so on. Where no color is given the substance is 
colorless or white. 

If necessary the student may enlarge the list to meet 
special requirements. A frequent use of this key in the 
laboratory is recommended. 



APPENDIX II 
DISPLACEMENT SERIES 

(Elements printed in capitals to be memorized by the 

student) 

In the displacement series the elements are arranged 
according to their electro-motive force, that is, the 
capacity of holding the ionic charges more or less firmly. 

NEGATIVE tantalum columbium 

palladium cadmium 

fluorine ruthenium IRON 

CHLORINE rhodium ZINC 

OXYGEN ANTIMONY manganese 

NITROGEN BISMUTH uranium 

BROMINE ARSENIC gadolinium 

IODINE MERCURY indium 

SULFUR SILVER gallium 

selenium COPPER ALLUMINIUM 

tellurium silicon rare-earth metals 

PHOSPHORUS titanium beryllium 
chromium scandium 

vanadium HYDROGEN yttrium 

tungsten MAGNESIUM 

molybdenum TIN lithium 

CARBON LEAD CALCIUM 

boron germanium strontium 

GOLD zirconium BARIUM 

osmium cerium SODIUM 

platinum nickel POTASSIUM 

iridium cobalt 

thallium POSITIVE 
107 



108 APPENDIX 

At the top of the list are the most negative elements 
which displace all following anions. The elements near 
the end of the list are the most positive and will displace 
each cation above them. Thus Br will displace I and S, 
but F will displace Cl and Br. Ca will displace Mg, Al, 
and Fe, but K will displace Na and Ca, etc. 

The farther apart the elements, the more stable their 
compounds. Thus KF is the most stable compound, 
while a compound of and F is so unstable that it 
cannot exist. 



APPENDIX III 

THE PERIODIC SYSTEM AND THE 
CLASSIFICATION OF THE ELEMENTS 

A systematical arrangement of all chemical elements 
is essential for an understanding of their correlation, 
their similarity or difference. The periodic system is 
such a classification in which all elements are logically 
tabulated. The position occupied by an element in this 
system is an indication of its chemical and physical 
properties. 

The periodic system was not suddenly discovered as is 
often stated, but has gradually developed and taken 
form with increasing chemical knowledge during the 
last 100 years. One of the earliest attempts at classifi- 
cation was made in 1829 by Doebereiner who arranged 
some elements in triads or groups-of-three, such as 
Ca-Sr-Ba, Li-Na-K, Cl-Br-I, etc.; because the properties 
and the atomic weight in each triad show a certain 
relationship. The same idea was further developed by 
Cooke in 1854 and by DeChancourtois in 1865. The 
next noteworthy step was taken in 1867 by Newlands in 
his octaves or groups-of-eight in which he pointed out, 
that if all elements are arranged according to increasing 
atomic weights, each eighth element has analogous 
properties, for elements of similar properties are sepa- 
rated by seven others, just as is the case in a harmonious 
musical scale. This law of octaves contained the nucleus 
of the periodic system and in 1869 Mendeleeff and 
Lothar Meyer, working independently of each other, 

109 



110 APPENDIX 

presented the first table of a system of elements which 
was generally accepted. Mendeleeff predicted the 
properties of three unknown elements from three empty 
spaces occuring in his table, and when these elements 
were later discovered, it was found that his predictions 
came true with an astounding accuracy. Since that 
time numerous discoveries have been made and the 
system has gradually been completed. Today we may 
assume with great confidence that only five elements 
remain undiscovered and that the total number of 
elements in the series from hydrogen to uranium inclusive 
is 92. The five unknown elements with atomic numbers 
43, 61, 75, 85, and 87 would complete the table. 

If all elements are arranged in the order of increasing 
atomic weights, then the 92 elements between H (At. No. 
1) and U (At. No. 92) inclusive are divided into six periods, 
the terminals of each period being the noble gases. Thus 
there are in : 

Period I, extending from He to Ne 8 elements ( = 2 X 2 2 ) 

Period II, extending from Ne to Ar 8 elements ( = 2 X 2 2 ) 

Period III, extending from Ar to Kr 18 elements ( = 2 X 3 2 ) 

Period IV, extending from Kr to Xe 18 elements ( = 2 X 3 2 ) 

Period V, extending from Xe to Nt 32 elements (= 2 X 4 2 ) 

Period VI, extending from Nt to U 7 elements 

Each period has the following characteristics : 

The terminals of each period are electrically neutral 
(noble gases). 

The first three elements of each period are strongly 
electropositive (light metals). 

The last three elements of each period are strongly 
electronegative (non-metals). 

The central membets of each period are amphoteric 
(heavy metals). 

The members of the carbon family are the framework 



APPENDIX 



111 



of the system and form a transition between the different 
groups of elements (C the life-element, Si the rock- 
forming element). 

Besides the horizontal division into periods there is 
the vertical division into groups. The general rule of 
the periodic chart is : 

The similarity among the elements in the upper half 



The Periodic System 



NON-METALS INERT ELEMENTS 

Group 4 5a 6a 7a la 



111! It If 



LIGHT MEf ALS 

2a 3a 4a 



-ft a 



Period 
Vb 



IVb 

Illb 

lib 

Ib 



Iron period 
III' 



Silver period 



er pi 
IV 



Rare earth 
metals 

V" 

Gold 
period 

Radioactive 

elements 

VI ' 

Group 



82 
Pb 


83 
Bi 


84 
Po 


85 


86 

Nt 


87 


88 
Ra 


89 
Ac 


90 
Th 


50 

Sn 


51 
Sb 


52 
Te 


53 


54 

Xe 


55 
Cs 


56 
Ba 


57 
La 


58 
Ce 


32 
Ge 


33 

As 


34 

Se 


35 

Br 


36 
Kr 


37 
Rb 


38 

Sr 


39 
Y 


40 
Zr 


14 
Si 


15 
P 


16 
S 


17 
Cl 


18 
Ar 


19 
K 


20 
Ca 


21 

Sc 


22 
Ti 


6 
C 


N 


8 



9 
F 


10 

Ne 

2 
He 


11 
Na 

3 
Li 


12 
Mg 

4 
Be 


13 
Al 

5 
B 


14 
Si 

6 
C 


H 


22 23 24 25 
Ti V Cr Mn 


26 27 28 
Fe Co Ni 


29 30 31 32 
Cu Zn Ga Ge 


40 41 42 43 
Zr Cb Mo 


44 45 46 
Ru Rh Pd 


47 48 49 50 
Ag Cd In Sn 


58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 
Ce Pr Nd Sm Eu Gd Tb Dy Ho Er Db Tm Yb Lu 


72 73 74 75 
Lu Ta W 


76 77 78 
Os Ir Pt 


79 80 81 82 
Au Hg Tl Pb 


90 91 92 
Th Bv U 



VI 

Va 

IVa 

Ilia 

Ila 

la 

Iron period 
III' 



Silver 

period 

IV 

Rare earth 

metals 

V" 

Gold 
period 



5b 6b 7b 888 
HEAVY METALS 



Ib 2b 3b 



112 



APPENDIX 



of the table is most pronounced in the vertical direction, 
therefore these elements possess group-analogy. 

The similarity among the elements of the lower half 
of the table is most pronounced in the horizontal direct- 
ion, therefore these elements possess period-analogy. 

ATOMIC NUMBERS AND ATOMIC WEIGHTS 

Importance of the elements is indicated by * and o. The atomic numbers serve as 
n index to the periodic table. 



Actinium 


89 
13 
51 
18 
33 

56 
4 
83 
5 

, 91 

35 

48 
55 
20 
6 

58 
17 
24 
27 
41 

29 
69 
66 
68 
63 

9 
64 
31 
32 
79 

2 
1 
49 
53 

77 

26 
36 
57 

82 
3 

72 
12 
25 
80 
42 


Ac 
Al 
Sb 
A 
As 

Ba 
Be 
Bi 
B 
Bv 

Br 
Cd 
Cs 
Ca 
C 

Ce 
Cl 
Cr 
Co 
Cb 

Cu 
Db 
Dy 
Er 
Eu 

F 
Gd 
Ga 
Ge 
Au 

He 
H 
In 
I 
Ir 

Fe 
Kr 
La 
Pb 
Li 

Lu 

Mg 
Mn 
Hg 
Mo 


27.1 
120.2 
39.88 
74.96 

137.37 
9.1 
208.0 
11.0 

79.92 
112.4 
132.81 
40.07 
12.00 

140 .25 
35.46 
52.0 
58.97 
93.1 

63.57 
168.5 
162.5 
167.5 
152.0 

19.0 
157.3 
69.9. 
72.5 
197.2 

4.00 
1.008 
114.8 
126.96 
193.1 

55.84 
82.92 
139.0 
207.2 
6.49 

17.50 
24.32 
54.93 
200.6 
96.0 


Neodymium .... 


60 
10 
28 
7 
76 

8 
46 
15 

78 
84 

19 
59 
88 
45 
37 

44 
62 
21 
34 
14 

47 
11 
38 
16 
73 

52 
65 
81 
90 
70 

50 
22 
74 
92 
23 

54 
71 
39 
30 
40 

tBeryl 
(Colun 
,isY. 
jod) is J 


Nd 

Ne 
Ni 
N 
Os 


Pd 
P 
Pt 
Po 

K 
Pr 
Ra 
Rh 
Rb 

Ru 

Sm 
Sc 
Se 
Si 

Ag 
Na 
Sr 
S 
Ta 

Te 
Tb 
Tl 
Th 
Tm 

Sn 
Ti 
W 
U 
V 

X 

Yb 
Y 
Zn 
Zr 

ium) 
ibiuir 

. 


144.3 
20.2 
58.68 
14.01 
190.9 

16.000 
106.7 
31.04 
195.2 

39.10 
140.9 
226.0 
102.9 

85.45 

101.7 
150.4 
44.1 
79.2 
28.3 

107.88 
23.00 
87.63 
32.06 
181.5 

127.5 
159.2 
204.0 
232.4 
170.5 

118.7 
48.1 
184.0 
238.2 
51.0 

130.2 
173.5 
88.7 
65.37 
90.6 

r 


* Aluminum 
Antimony 
Argon 
Arsenic 


Neon 
o Nickel 


* Nitrogen 
Osmium 

* Oxygen 
Palladium 


o Barium . . 


Beryllium 
o Bismuth 


* Phosphorus 
o Platinum 


o Boron 


Brevium 


Polonium 

* Potassium 
Praseodymium.. . 
o Radium 


o Bromine 


Cadmium 
Caesium . . 


* Calcium 
* Carbon 


Rhodium 


Rubidium 

Ruthenium 
Samarium 
Scandium 


Cerium 
* Chlorine 


o Chromium 


o Cobalt 


o Selenium 
* Silicon 


Columbium 


o Copper .... 


o Silver . . . 


Denebium 


* Sodium 


Dysprosium .... 


o Strontium 




* Sulfur 
Tantalum 

Tellurium 


Europium . . 


o Fluorine 


Gadolinium 
Gallium 


Terbium 
Thallium 
o Thorium 
Thulium 


Germanium 


Gold 


o Helium 


o Tin 


* Hydrogen 
Indium 
. Iodine 


o Titanium ..... . 
Tungsten 
o Uranium 
Vanadium 

Xenon 


Iridium 
*Iron 


Krypton . . . 


Ytterbium 
Yttrium 


Lanthanum 
o Lead 


o Zinc 


Lithium 

Lutecium 
* Magnesium 
o Manganese 


Zirconium 


Gl (glucinum) is Be 
Nb (niobium) is Cb 
Flis,F; Xeis X; Y1 
Az (azote) is N ; J ( 


o Mercury 


Molybdenum 



APPENDIX 113 

Accordingly there are group-relations and period-rela- 
tions. E.g. the group-relation of Au refers to its proper- 
ties which are similar to those of Ag and Cu, while the 
period-relation of Au refers to those properties which 
resemble Hg and Pt; in other words, gold has a group 
similarity to silver and copper, and a period similarity 
to mercury and platinum. 

OUTLINE OF PROPERTIES OF ELEMENTS AS DEDUCED 
FROM THE PERIODIC TABLE 

In the UPPER half of the table are the elements of 
the strongest e.m.f., the simplest spectra, low density, 
and colorless ions. 

In the LOWER half of the chart are the elements of 
the weakest e.m.f., complex spectra, high density, 
and colored ions. 

On the LEFT of the table are the electro-negative 
elements which form the acids. The solid elements are 
brittle and paramagnetic. 

On the RIGHT of the table are the electro-positive 
elements forming the bases. The solid elements are 
malleable. 

In the LEFT UPPER HALF and RIGHT LOWER 
HALF are located the non-metals and heavy-metals 
which are found in their native form in nature (free 
state). Their sulfides are common and they are readily 
reduced to the free state. 

In the RIGHT UPPER HALF and LEFT LOWER 
HALF are the light metals and heavy metals which 
always occur in combination and never in their free 
state in nature. Their oxides and oxysalts are common, 
and it is with difficulty that they are reduced to^free 
elements. Not all elements are of equal importance. 
Fourteen elements are of primary importance (marked 



114 APPENDIX 

with bold type and *) for they constitute 99 per cent of 
the known earth surface and are abundant in the litho- 
sphere, hydrosphere, atmosphere. These elements are 
essential to all living matter (biosphere) and are of 
fundamental importance in every industry. About 40 
elements are of secondary importance (marked o), for 
they are less abundant and are used only for specific 
purposes in industry and technic. About 33 elements are 
of more or less scientific interest only for they very rarely 
occur and are of little or no present use. 

The student must be familiar with the fourteen im- 
portant and fundamental elements, know their position 
in the table, and be familiar with their chemistry, for 
they are most common on the surface of our earth and 
their compounds are used in every industry, in every 
material science, and in every-day life. Elementary 
chemistry confines itself mainly to these elements as 
they are the basis for a study of the other elements. 
Thus, e.g. with chlorine as type or example, the proper- 
ties of fluorine, bromine, and iodine can readily be 
memorized; while with sulfur as type or example the 
characteristics of selenium and tellurium are easily 
learned; potassium and sodium is the type for lithium, 
rubidium, and cesium; manganese is similar to chromium, 
iron similar to cobalt and nickel and so on. 

REFERENCES 

Classification of chemical elements, Scientific American Supplement, 
vol. 87, p. 146, 1919. 

Modification of the periodic table, Am. Jour, of Science, vol. 46, p. 481, 
1918. 

Distribution of chemical elements, Science Progress, vol. 14, p. 602, 
1920. 
BOOKS ON THE SUBJECT: 

Rudorf, G., The Periodic System. 

Garrett, A. E., The Periodic Law. 



APPENDIX IV 

SOLUBILITY TABLE OF COMPOUNDS 

(To be Memorized by Students) 

SOLUBLE in water: 

ALL CHLORIDES, except AgCl, PbCl 2 , HgCl; 
ALL NITRATES, except bismuth, subnitrate; 
ALL SULFATES, except barium sulfate, strontium 

sulfate, calcium sulfate, lead sulfate; 
ALL ACETATES, except basic ferric acetate and basic 

aluminum acetate ; 
ALL ALKALI SALTS, except acid potassium tartrate, 

acid ammonium tartrate, potassium platin chloride, 

ammonium platin chloride, and sodium pyro- 

antimoniate. 

INSOLUBLE in water: 

ALL OXIDES ] 

ATT QTTT T^TTvira except those of the alkalies 

AL/lj oU-LrlJJJiib . 

ALL HYDROXIDES | and earth alkali metals ; 

ALL CARBONATES ^ 

ATT PHOSPHATFS exce P^ those 01 the alkali 

ALL SILICATES metals ; 

ALL CHROMATES, except those of the alkalies; 

Ca, Sr. 

SOLUBILITY TABLE OF COMPOUNDS 
, (for reference) 

ACETATES are SOLUBLE, except AgAc, HgAc, 
HgAc 2 . 

115 



116 



APPENDIX 



ARSENITES are INSOLUBLE, except those of the 

alkalies. 
ARSENATES are INSOLUBLE, except those of the 

alkalies and the acid arsenates of the earth alkali 

metals. 

BORAXES are SOLUBLE. 
BROMATES are SOLUBLE, least soluble are 

HgBrO,, AgBr0 3 . 
BROMIDES are SOLUBLE, except CuBr, AgBr, 

HgBr, PbBr 2 . 
CARBONATES are insoluble, except those of alkalies 

and Tl. 

CHLORATES are SOLUBLE, except HgClO 3 . 
CHLORIDES are SOLUBLE, except AgCl, HgCl; 

CuCl, AuCl, PtCl 2 , PbCl 2 . 
CHROMATES are INSOLUBLE, except those of 

alkalies and CaCr0 4 , SrCrO 4 and MgCrO 4 . 
CYANIDES are IN SOLUBLE, except those of alkalies 

and earth alkalies. 
FERROCYANIDES are INSOLUBLE, except those 

of alkalies and earth alkalies. 
FERRICYANIDES are INSOLUBLE, except those of 

alkalies and earth alkalies. 
FLUORIDES are INSOLUBLE, except those of 

alkalies and AgF, FeF 2 , SnF 2 , SrF 2 and CdF 2 . 
IODATES are INSOLUBLE, except those of alkalies. 
IODIDES are SOLUBLE, except Agl, Cul, HgI 2 , 

PbI 2 , Bil 3 , SbI 3 , Pt metals. 
MANGANATES are INSOLUBLE, except those of 

the alkalies. 
MOLYBDATES are INSOLUBLE, except those of 

the alkalies. 

NITRATES are SOLUBLE. 
NITRITES are SOLUBLE, except AgNO 2 . 



APPENDIX 117 

OXALATES are INSOLUBLE, except those of the 

alkalies. 

PERMANGANATES are SOLUBLE, except AgMnO 4 . 
PHOSPHATES are INSOLUBLE, except those of the 

alkalies. 
PHOSPHITES are INSOLUBLE, except those of the 

alkalies, and Tl. 
SILICATES are INSOLUBLE, except those of the 

alkalies. 
SULFATES are SOLUBLE, except BaSO 4 , SrSO 4 , 

PbS0 4 , slightly soluble are Ca, Ag, Hg. 
SULFIDES are INSOLUBLE, except those of the 

alkalies. 



APPENDIX V 
PREPARATION OF SALTS 

A Metallic Salt May be Prepared in Any One of the 
Following Ways (To be Memorized by the Student) 

Reaction between: 

1. METAL and HALOGEN: 

Zn + C1 2 = ZnCl 2 

la. METAL and ACID: 

Zn H- H 2 SO 4 = ZnSO 4 + H 2 

2. METAL-OXIDE and ACID-ANHYDRIDE : 

ZnO + SO 3 = ZnS0 4 

2a. METAL-OXIDE and ACID: 

ZnO + H 2 S0 4 = ZnSO 4 + H 2 O 

3. METAL HYDROXIDE and ACID ANHYDRIDE: 

Zn(OH) 2 + SO 3 = ZnSO 4 + H 2 O 

3a. METAL HYDROXIDE and ACID: 

Zn(OH) 2 + H 2 SO 4 = ZnSO 4 + 2H 2 O 

4. METAL CARBONATE and ACID: 

ZnCO 3 + H 2 S0 4 = ZnSO 4 + CO 2 + H 2 

5. METAL SULFIDE and ACID: 

ZnS + H 2 SO 4 = ZnS0 4 + H 2 S 

6. Interaction-of two METAL SALTS: 

Zn(NO 3 ) 2 + Na 2 SO 4 = ZnSO 4 + 2NaNO 3 

118 



APPENDIX 119 

KEY TO THE EQUATIONS 

The following is a systematic index of the equations 
used in this book. The numbers to the left of the 
formula refer to the reacting substances (which stand 
at the left of an equation), and the numbers to the 
right refer to reaction products (which stand at the 
right side of an equation). 

This index makes it possible to find at a glance how 
some of the substances are prepared thus under oxygen 
there is a score of equations showing how this gas may 
be made. It also gives the analytical tests for many 
compounds, that is, the reactions used for their detection. 
Furthermore it will facilitate the construction of similar 
equations for reactions taking place with related ele- 
ments. Thus many reactions given under sulfur occur 
also with selenium and tellurium, likewise those of 
arsenic compounds have a relation to those of antimony 
and bismuth compounds. 

It is well for students to make a comparative study 
of reactions by consulting the periodic system and the 
displacement series and extend this index to the labora- 
tory and lecture note book for further reference. 

ALUMINUM 

442, 446 Al 443 

21, 338 Al +++ 

162, 335, 337, 342, 345, 443, A1C1 3 113-5, 442, 445 

444 

113 A1 2 O 3 424 

114 A1 2 S 3 127.446 

115 ' Al 2 Se 3 

116 Al 2 Te 3 

131 A1 4 C 3 

9,15,127,343,344,426,445. A1(OH) 3 20, 21, 131, 162, 335, 337 

338, 342, 444 

H 3 A10 3 343 

HA1O 2 426 



120 APPENDIX 

, A1(N0 3 ) 3 9 

20, 425 A1 2 (S0 4 ) 3 

424 A1 2 (C0 3 ) 3 

A1P0 4 15 

AMMONIUM 

86,88,89,92,233,333,374.. NH 3 101, 106, 168, 282, 294 

46 NH 4 + 

90, 91 NH 4 OH 46, 86, 374 

92, 101 NH 4 C1 88, 233, 333, 360, 362 

289 NH 4 N0 3 

94, 95, 96, 362 (NH 4 ) 2 S 361 

360, 361 (NH 4 ) 2 AsS 3 94 

(NH 4 ) 2 SnS 3 95, 96 

289 (NH 4 ) 2 MoO 3 

ANTIMONY 

185 Sb 

165, 421 SbCl 3 

Sb 4 O 6 421 

80, 81 Sb 2 S 3 

Sb 2 S 5 150 

SbOCl 165 

ARSENIC 

186 As 420 

416 AsH 3 118, 169 

420 As 4 O 6 

94 . As 2 S 3 321, 322, 360 

321. H 3 AsO 3 272, 320 

320, 322 H 3 AsO 4 

BARIUM 

18, 139, 140 Ba ++ . . 151 

17, 372 BaCl 2 407 

BaO 243, 244, 246 

151, 407 BaO 2 

157, 243 BaS 

BaC 2 247 

7, 13 Ba(OH) 2 157, 158 

158 Ba(SH) 2 157 

Ba(N0 3 ) 2 7 

243,^246 BaSO 4 17, 18, 139 

244/247 BaCO 8 



APPENDIX 121 

BaCrO 4 ... 140, 372 

Ba 3 (PO 4 ) 3 .13 



BERYLLIUM 

390 BeO 

Be(OH) 2 390 

BISMUTH 

184 Bi 

160 .' BiCl 3 

'.'. BiI 3 184 

Bi(OH) 3 161 

161 BiOCl 160 

BORON 

, B 2 O 3 133, 404 

402 B 2 S 2 

168 BN 106 

404 BO(OH) 168, 403 

133, 403 B(OH) 3 402 

106 B(NH 2 ) 3 

BROMINE 

54, 213, 221, 263, 264, 285. .. Br 27, 31, 32, 33, 53, 214, 220, 

412 

27, 33, 53 Br- 54 

31, 32, 383, 412 HBr 221, 263, 264 

CADMIUM 
388 CdCl 2 



CALCIUM 

73, 177 Ca 70 

399 CaF 2 123, 124 

75, 77 CaCl 2 .... .6, 76, 110, 111, 112, 177 

70, 74, 79, 84, 110, 120, 123. CaO 69, 73 

CaS 120, 121 

112 Ca 3 P 2 

169 Ca 3 As 2 

130, 245 CaC 2 

6, 12, 76 Ca(OH) 2 77, 84, 130, 169 

. CaCN 2 245 

CaSO 4 12, 399 

69, 121 CaCO 3 74, 75, 79 



122 APPENDIX 



124... 


CaSiO 3 


, 274, 275 


273, 274 


Ca(PO 3 ) 2 




275 


Ca 3 (P0 4 ) 2 


273 




CARBON 




244, 246, 247, 258, 273, 276, 


C 


245 


420 








C 2 H 2 


130 


234,400 


CH 4 


131 


167 


CC1 4 




192,242 


CO 


246, 244, 258, 273, 274, 275 


71, 79 


CO 2 


20, 21, 69, 134, 242, 288, 






400, 401, 420, 430, 431 


127, 401 


CS 2 




145-7, 334 ... 


CN- 






C 2 N 2 


254 




CH 3 C1 


... 234 


288 


CH 2 O 






COC1 2 


167 




COC1 


276 




CHLORINE 




27, 53, 177, 182, 183, 214, 


C1 2 


196, 198, 200, 201, 215, 218, 


234, 276, 280, 292 




219, 251, 253, 371, 410, 411, 






415, 437, 443 


387, 391 


ci- 


27, 53 


2, 4, 6, 8, 19, 110, 118, 196, 


HC1 


101, 119, 234, 235, 283, 284, 


223, 287, 371 




372, 421, 443 


262 


C1O 2 






HOC1 


235, 282 




CHROMIUM 




298, 305 


Cr ++ 






Q r +++ 


35, 306, 310, 312, 317 


303 


CrOr 


; 302 


140, 382, 433 


CrO 4 ~ 


25, 303, 305 


25 35 281 306 310 312 


Cr 2 7 




317, 372 








CrCl 3 


. . ; 253 




Cr 2 O 3 


201, 252, 268 281 


252 253 


CrO 3 




301, 302, 432 


Cr(OH) 3 


298 


201 


Cr(OCl) 2 






cwsa) a 


..432 



APPENDIX 123 

COBALT 

Co 430 

142, 297 Co ++ 

CoO 431 

191 CoO 3 

285 Co(OH) 2 142 

255 Co(OH) 3 285, 297 

331 Co(NO 2 ) 2 

100 Co(NO 2 ) 3 331 

CoSo 4 255 

431 CoCO 3 

430 Co(OOC) 2 

MgCoO 3 * 191 

COPPER 

51, 207, 324, 328, 330 Cu . . 26, 50, 56, 208, 229, 240,241 

26, 50, 56, 90, 141, 148, 153. Cu ++ 51, 324, 329, 330 

134 CuCl 

135, 208 ( CuCl 2 19, 207 

Cu 2 O ..134 

19, 229, 240, 329 CuO 103, 104, 328 

81, 240, 241 CuS 153 

103, 149 Cu(OH) 2 141, 159 

CuCN 254 

104 Cu(NO 3 ) 2 

1086, 159, 241, 254 CuSO 4 108a 

CuCO 3 135 

Cu(NH 3 ) 4 ++ 90, 149 

Cu 3 SbS 3 81 

Cu 2 Fe(CN 6 ) . , 148 

FLUORINE 

181, 215, 216, 217, 415, 418. . F 2 

124 HF ., . .181, 216, 217, 415, 418 

IF 5 236 

GALLIUM 

193 Ga 

GaCl 2 193 

193 GaCl 3 

, Ga 2 O 3 422 

422 Ga(NO 3 ) 3 



124 



APPENDIX 





GOLD 




266 


Au 


367, 370 




AuCl 


437 


367, 368 


AuCl 3 


438 


137, 370 


Au 2 O 3 


369 


369 


Au(OH) 3 


368 


437, 438 


HAuCl 4 


266 




HYDROGEN 




40, 44, 176, 181, 228-232, 


H 2 


42, 43, 41, 222-227, 416, 442 


277, 281, 293, 384 






16, 33, 34, 35, 45, 306 to 315, 


H+ 


23, 332 


323 to 33 f 






16, 24, 23, 46, 295 to 298, 302, 


OH- 




304-8, 440 






20,21,41-3,216,335,338... 


H 2 O 


. .4-15, 82-86, 218, 300-331 


295-309, 363, 367, 370, 373. . 


H 2 O 2 


407 




IODINE 




55, 184, 236, 286, 414 


li 


54, 199, 213, 287, 299, 312- 






316, 376-379, 419 


54, 434 


I- 


55 


199, 312-316 


HI 


286, 375, 379, 414 




IRON 




26, 41, 42, 43, 56, 178, 179, 


Fe 


40, 44, 205, 209 


195, 206, 208, 212 






34, 145, 296, 317-319 


Fe ++ 


....26, 56 


22 ... 


Fe+++ 


34, 317-319 


209 


FeCl 2 


117, 208 


3 


FeCl 3 


8 




FeBra 


383 


40, 44, 205 


Fe 2 O 3 


41, 42, 43, 179, 206 


117, 187 


FeS 


178, 212 


8, 14, 383 


Fe(OH) 3 


279,296 




FeSO 4 


187, 195 


195 


Fe-zfSOJi 


14 




Fe 2 (CO 3 ) 3 


22 


97 


Fe(CN) 2 


145 




Fe(CNS) 3 


3 


148 


Fe(GN) 6 ~~ 




280 


K 4 Fe(CN) 6 


97 




Fe(CN) 8 ~~ 






K 3 Fe(CN) 6 


. .280 






APPENDIX 125 

LEAD 

50, 59, 423 Pb 49, 65, 239 

23, 49, 61, 65, 433, 434 Pb ++ 50, 59, 64, 327 

357, 359 PbCl 2 410 

435, 410 PbCl 4 408 

PbI 2 . 1, 434 

239 PbO 204, 237, 423 

60, 267 PbO 2 435 

62, 68, 408 PbO 2 ~ 60, 349 

80, 237, 239, 327 PbS 22, 358, 359 

1, 204, 346, 349 Pb(NO 3 ) 2 267 

64, 66 PbSO 4 

348 Pb(OH) 2 .346, 347 

PbCr0 4 433 

. PbSb 2 S 4 80 

MAGNESIUM 

224, 227, 247 Mg 

174, 219 MgCl 2 397 

191 MgO 174, 219, 227, 247 

Mg(N0 3 ) 2 224 

MgCoO 3 191 

MANGANESE 

295, 304, 387 Mn ++ 33, 34, 308, 309, 311, 313, 

318 

34, 265, 307, 309, 311, 313 MnOr 

318 

MnO 4 ~ 259 

373, 385 MnCl 2 . 371, 387, 411 

411 MnCl 4 409 

' MnO 293 

32, 33, 259, 293, 308, 371, MnO 2 300, 304, 307 

392, 409 

366 Mn(OH) 2 365 

203, 293, 300 Mn(OH) 3 295 

HMriO 2 203 

365 HMnO 4 364 

H 2 MnO 3 373 

MnSO 4 32, 265, 363, 366, 392 

MnCO 3 385 



MERCURY 

52, 194, 325, 389 Hg 28, 51, 197, 207, 269, 319, 

332, 334 



126 



APPENDIX 



332, 334, 382 

51, 319, 389 

333 

207, 233, 270, 354, 356. 

269 

28, 197 



Hg+ 



...339 
52, 325 
..270 



194 

268.. 



HgCl 

HgCl 2 

Hg 2 O 268 

HgO 

HgS 332, 355, 356 

HgNO 3 194 

Hg(N0 3 ) 2 

Hg 2 CrO 4 382 

Hg(CN) 2 334 

Hg(NH 2 )Cl 233,333 



136. 



MOLYBDENUM 
Mo0 3 

NICKEL 



.294 



89.... 


Ni ++ 




99.... 


Ni(CN) 2 






NUNHjOe* 4 - 


. .89 




K 2 Ni(CN) 4 


99 


180, 245..., 


NITROGEN 

N 2 


248, 252, 287 


287 


N 2 H 4 






NH 3 


(see ammonia) . . . 


189, 190, 265 


N 2 O 2 


266, 323, 325, 327, 331 




N 2 O 3 


30 




NO 2 


180, 189, 204, 261, 288, 289, 


82 


N 2 O 6 


290 
396, 422 


290 


H 2 N 2 O 2 


291 


291 . 


HNO 2 




5, 7, 9, 30 


HNO 3 


82, 190 


35, 311, 310 






323,327 


NOr 


35, 310, 311 


282 


NC1 3 




291 


NH 2 OH 




29, 176, 179, 180, 187, 191, 


OXYGEN 
2 


28, 47, 197, 201, 202, 205, 



206, 218, 219, 220, 237, 249, 
259, 294 

419.. 



3 



217, 255, 257, 268, 277, 279, 
306-9, 363, 366, 367, 370, 
386, 393, 400, 401, 419 
216, 418 



APPENDIX 127 





PHOSPHORUS 




182, 263, 271..... 


P 


273, 274, 275 




PH 3 


132, 271 


163 


PC1 3 


182, 251 


164, 251, 87, 133 


PC1 5 




171 220 


PBr 3 




172 


PBr 5 




87, 128 


P 2 5 


220 




HPO 3 


128, 263 




H 3 P0 3 


. . . . 163, 171 


11, 13, 15 


H 3 P0 4 


164, 173 




POC1 


87 


173 


POC1 3 


133 




POBr 3 


172 


132 


PHJ 






PLATINUM 






Pt 


200,384 


384 


PtS . . . . 




152 


ptci 6 






POTASSIUM 




192, 210, 248 


K 


226, 227 


152 


K + 






KC1 


....3, 210, 256, 257 


1, 299, 375-9, 419... 


. . . . KI 




413. 


KBr 




78, 259 


K 2 O 


248 


5, 11 226, 227 


KOH 


299 


, 


KSH 


398 


97, 98, 99 


KCN 


254 


248, 250 


KNO 3 


1,5 


100 


KN0 2 


250, 261 




K 2 SO 4 


78,375-379 





K 3 PO 4 


.11 


256.. 


KC1O 












KC10 2 


262 


257, 413 


KC10 3 


256, 262 




KC1O 4 


257 




KBr0 3 ........ 


413 


397 


K 2 MgCl 3 




279 


. . . . KoFeO 4 




281,372 


K 2 Cr 2 O 7 






K 2 Mn0 4 


259 



130 APPENDIX 

TIN 

30,326 Sn 

119, 129, 350, 353 SnCl 2 

95 ...-. SnS 119 

96 SnS 2 

352 Sn(OH) 2 129, 350 351 

H 2 SnO 3 107, 326 

107 H 4 SnO 4 30 

TELLURIUM 

436 Te 260 

TeH 2 116 

TeO 436 

TeO 3 105 

105 Te(OH) 6 

260 H 2 TeCl 6 

THALLIUM 

391 .. Tl* 

292 T1C1 391, 392 

,V T1C1 3 292 

392 -,.. T1NO 3 

THORIUM 

ThOs . . . 396 

396 Th(N0 3 ) 4 

TITANIUM 

276 TiO 2 

395 TiCl 4 ...... ? 276 

H 4 TiO 4 395 

TUNGSTEN 

183 W 232 

170 WC1 6 183 

125, 232 WO 3 

WO(OH) 4 126, 170 

ZINC 
49, 185, 186, 209 

222, 225, 323 Zn 198, 210 

143, 147 Zn* 4 - 49, 323, 428, 429 

198, 210. . .. ZnCl, .211 



APPENDIX 



131 



188. 



ZnS 



.......................... Zn 3 Sb 2 

24, 92, 144 ................ Zn(OH) 2 

.......................... ZnO 2 

98 ........ . ............... Zn(CN), 

427, 428 ................... ZnSO 4 



186 

185 

143 

. 24, 144, 225 

147 

188, 222 



K 2 Zn(CN) 4 4, 98 

Zn(NH 3 ) 6 Cl 92 



134 INDEX AND GLOSSARY 

CLASSIFICATION OF ELEMENTS is derived from the periodic 

system 109 

COLLOIDS are particles of ultra-microscopic size 51 

COLOR often indicates the stage of oxidation 106 

change indicates a chemical reaction 23 

COMBINATION is a chemical reaction in which two elements unite 
and form a compound, one element being reduced, the other 

oxidized 87 

COMPLEX-ION is a radicle or group of atoms which is electrically 

charged, either positively or negatively 3 

COMPLEX REACTION results from chemical changes in which 

two or more types of reaction simultaneously take place 96 

COMPOUNDS are molecules with unlike atoms 6 

the sum of valence numbers in stable compounds is zero 14 

the weight relation of their constituents is shown in a formula ... 6 

the percentage composition is calculated from formula 6, 19 

the same element may form different series of compounds 13 

- the names of different series depend upon valence number 17, 103 

CONCENTRATION is the amount of atoms or molecules in a cer- 
tain volume; cone, of gases is expressed in terms of pressure; 

cone, of substances in solutions is expressed in moles per liter ... 53 

change in concentration affects the chemical equilibrium 56 

DECOMPOSITION is the reverse reaction of addition 82 

DISPLACEMENT is a chemical reaction in which one element 
exchanges charges with another element it is either oxidation 

or reduction 69, 89 

DISPLACEMENT SERIES is an arrangement of the elements in 

the order in which they hold their electric charges 70, 107 

indicates the electro motive force of elements 68 

DISSOCIATION is the breaking apart of molecules 63 

thermal dissociation takes place under the influence of heat .... 64 
electrical dissociation or ionization is produced by electricity ... 3, 65 
DIVISION is the reverse reaction of combination in which a com- 
pound breaks apart, one element being oxidized, the other 
reduced 88 

E. M. F. = electro-motive force 68 

EARTH ALKALI METALS are the elements of the second group 

of the periodic system: Mg, Ca, Sr, Ba Ill 

they are bivalent but univalent 13, 110 

EARTH METALS are the elements of the third group of the periodic 

system e.g., Al, Sc Ill 

they are trivalent, but univalent 13, 110 



INDEX AND GLOSSARY 135 

ELEMENTS are chemically indivisable and indestruc table sub- 
stances whose molecules consist of like atoms 2 

in their free state they have a valence number of zero* 12 

those having in their compounds but one valence number are 

univalent, those with two or more are polyvalent 13 

ELECTRIC BATTERY is formed when two elments are connected 
by a conducting media but are locally separated so that a dis- 
placement reaction can take place 71 

ELECTRIC DISSOCIATION is ionization 3, 65 

ELECTRO AFFINITY or e.m.f. is the force with which the atoms 

hold their equally large ionic charges 68 

ELECTRODE is the positive or negative pole 65, 71 

ELECTROLYSIS is the spacial or local separation of ions (cations 

and anions) by an electric current 65 

ELECTRO-MOTIVE FORCE is the power of an atom to 

retain electrical charges as expressed in the displacement series 68 
ELECTRO-POTENTIAL is the tension between two different sets 

of atoms and their ions, expressed in volts 71 

ELEMENT is an aggregation of matter consisting of one type of 

atoms 2 

ENDOTHERMIC COMPOUND requires heat for its formation, 

, and liberates heat in its decomposition 62 

ENDOTHERMIC REACTION always proceeds slowly and 

requires heat which it absorbs 62 

EQUATION is the expression of a chemical reaction 23 

there are non-ionic or molecular and ionic equations 28 

balancing of equations < 32 

finishing of incomplete equations 29 

the universal equation of neutralization is H+ -f OH~ = H 2 O . . 27 
EQUILIBRIUM is the balanced state reached in a chemical reaction 
when the concentration between the reacting substances has 
become such, that decomposition and re-combination proceeds 

with equal speed 55 

the removal of one reaction-product shifts the equilibrium and the 

reaction may proceed to completion 56 

EXOTHERMIC COMPOUND liberates heat in its formation, and 

requires or absorbs heat in its decomposition 62 

EXOTHERMIC REACTION proceeds rapidly and liberates heat. 62 
EXPLOSIONS are exothermic reactions 62 

FORMULA is a combination of symbols showing the composition of 

a molecule 6, 10 

There are four types of formula: empirical, constitutional, 

rational, and structural 8 



136 INDEX AND GLOSSARY 

Summary of the information contained in formula: 18 

GAS formation indicates a chemical reaction 23 

density, weight of 1 liter, specific gravity, and volume relation 

of a gas is shown in the formula 19 

GOLD PERIOD is composed of the elements of the fifth sub-period 

of the periodic system Ill 

GRAM MOLECULE is the molecular weight of a compound in 

grams 19 

HEAT increases the speed of reactions 61 

produces thermal dissociation, 62 

is absorbed in endothermic reactions and liberated in exothermic 

reactions, : 63 

is absorbed in the formation of endothermic compounds and 

liberated in the formation of exothermic compounds, 62 

HEAVY METALS are the elements of the sub-periods of the 

periodic system (lower part of chart) 110, 113 

are all polyvalent, therefore have different series of compounds 

(-ous and -ic compounds) 13, 17 

may be amphoteric, and form many complex compounds 113 

HYDROGEN is the basis of valency and valence numbers 11 

Elements combining with H have a negative valence number. ... 12 

Elements replacing H have a positive valence number 12 

HYDROLYSIS is the reverse reaction of neutralisation 86 

INERT ELEMENTS (or noble gases) do not combine with any 

other element Ill, 113 

IONIC REACTIONS occur only when the substance is in solution 57 
proceed very rapidly, and depend upon the formation of one of 

7 compounds 57 

IONIZATION or electric dissociation is the breaking apart of 
molecules (electrically neutral) into electrified atoms (charged 
positively or negatively) taking place when certain substances 

are dissolved in water 3 

IONS are electrically charged atoms or groups of atoms 3 

They may have one or more positive (cations) or negative (anions) 

charges 4 

Ionic charges are balanced before the atoms are balanced. 31 

and after oxidation and reduction is balanced 41 

IRON-PERIOD, the elements of the third sub-period of the periodic 

system Cr, Mn, Fe, Co, Ni, Cu Ill, 113 

LIGHT METALS are the elements of the a-sub-periods of the 

periodic system Ill, 113 



INDEX AND GLOSSARY 137 

METATHESIS is a chemical reaction with an even exchange 83 

MOLAR SOLUTIONS contain the gram molecular weight of sub- 
stance in 1 liter 53 

MOLE is the molecular weight of an element or compound in grams 

( = gram molecule) 19, 53 

MOLECULE is the chemical combination of two or more like or 

unlike atoms 2 

MOLECULAR WEIGHT is the relative mass of a molecule which 

is found by adding the atomic weights 6 

NEUTRALIZATION is a metathesical reaction between an acid and 
a base which gives a salt and water the universal equation 

is H + + OH- = H 2 O 27, 85 

NOBLE GASES are the inert elements of the zero group Ill 

NOBLE METALS are the elements of group 8 and Ib of the silver- 

and gold period Ill, 113 

NOMENCLATURE of compounds depends mainly on the valence 

number , 17 

table and general rules 103 

NON-METALS are the elements of the b-sub-periods of the periodic 

system : Ill 

are strongly electro-negative, form acids, are polyvalent 113 

NORMAL SOLUTIONS contain the gram equivalent weight of the 

substance in 1 liter solution 53 

OXIDATION is the increase or augmentation of the valence 
number 15, 37 

POLAR NUMBER is the same as valence number 12 

PERCENTAGE of constituents in a compound is calculated from 

the formula 6 

PERIODIC SYSTEM is a logical and natural classification of all 

elements according to their properties 109 

aids to memorize the properties of elements, and enables the pre- 
diction of properties of unknown elements . HO 

PHOSPHOR GROUP are the elements of the fifth group of the 

periodic system Ill 

PRECIPITATE formation is due to an insoluble compound and 

indicates a chemical reaction 23 

PREPARATION OF SALTS is tabulated on page 118 

RADICLE is a group of atoms having one or more free valencies. 

It reacts like an atom and cannot exist in the free state. 26 



138 INDEX AND GLOSSARY 

RADIOACTIVE ELEMENTS are the elements of highest atomic 

weight which spontaneously disintegrate or break apart Ill 

RARE EARTH METALS are the elements of the V" sub-period. . Ill 

REACTION is a chemical change or interaction of molecules 23 

Control of reaction depends on concentration, velocity, and the 

e.m.f 48, 75 

Control of ionic reactions depends on the formation of one of 

seven substances 57, 75 

REDUCTION is the decrease or diminution of the valence number 15, 37 
RESTITUTION is the reverse reaction of substitution 91 

SALTS are generally formed from acids and bases (neutralization) 26 
They may be prepared from metal, metaloxides, metalhydroxides, 

metalcarbonates, metal sulfides 118 

SILVER PERIOD are the elements of the fourth sub-period. . 111 

SOLUBILITY TABLE 115 

SUBSTITUTION is a chemical reaction in which an element is 

either oxidised or reduced and combines with the displaced 

element 91 

SULFUR GROUP are the elements of the sixth group of the periodic 

system: O, S, Se, Te Ill 

SYMBOL is one or two letters representing an atom and its relative 

mass 1 

is used to represent an atom, molecule, or ion 4 

VALENCE NUMBER is the arithmetical expression of valency. 

It can be either + or , and is based upon H = 1, andO = 2.12, 14 
of the same element may be different and then it indicates different 

series of compounds 13 

of free elements is always zero 12 

the sum in a stable compound is always zero 14 

VALENCY is the capacity of an atom to combine with other atoms 

in a definite proportion 11 

is measured with regard to H and can be any integer from 18 12 

when unknown may be calculated from the formula 14 

WATER formation is the characteristic earmark of neutralization . 27 
WATER OF CRYSTALLIZATION is the water contained in 

crystals 7 



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