CHEMICAL EQUATIONS AND CALCULATIONS
~Formulas.~ Since the molecule of any chemical compound consists of a
definite number of atoms, and this number never changes without
destroying the identity of the compound, it is very convenient to
represent the composition of a compound by indicating the composition of
its molecules. This can be done very easily by using the symbols of the
atoms to indicate the number and the kind of the atoms which constitute
the molecule. HgO will in this way represent mercuric oxide, a molecule
of which has been found to contain 1 atom each of mercury and oxygen.
H_{2}O will represent water, the molecules of which consist of 1 atom of
oxygen and 2 of hydrogen, the subscript figure indicating the number of
the atoms of the element whose symbol precedes it. H_{2}SO_{4} will
stand for sulphuric acid, the molecules of which contain 2 atoms of
hydrogen, 1 of sulphur, and 4 of oxygen. The combination of symbols
which represents the molecule of a substance is called its _formula_.
~Equations.~ When a given substance undergoes a chemical change it is
possible to represent this change by the use of such symbols and
formulas. In a former chapter it was shown that mercuric oxide
decomposes when heated to form mercury and oxygen. This may be expressed
very briefly in the form of the equation
(1) HgO = Hg + O.
When water is electrolyzed two new substances, hydrogen and oxygen, are
formed from it. This statement in the form of an equation is
(2) H_{2}O = 2H + O.
The coefficient before the symbol for hydrogen indicates that a single
molecule of water yields two atoms of hydrogen on decomposition.
In like manner the combination of sulphur with iron is expressed by the
equation
(3) Fe + S = FeS.
The decomposition of potassium chlorate by heat takes place as
represented by the equation
(4) KClO_{3} = KCl + 3O.
~Reading of equations.~ Since equations are simply a kind of shorthand way
of indicating chemical changes which occur under certain conditions, in
reading an equation the full statement for which it stands should be
given. Equation (1) should be read, "Mercuric oxide when heated gives
mercury and oxygen"; equation (2) is equivalent to the statement, "When
electrolyzed, water produces hydrogen and oxygen"; equation (3), "When
heated together iron and sulphur unite to form iron sulphide"; equation
(4), "Potassium chlorate when heated yields potassium chloride and
oxygen."
~Knowledge required for writing equations.~ In order to write such
equations correctly, a considerable amount of exact knowledge is
required. Thus, in equation (1) the fact that red oxide of mercury has
the composition represented by the formula HgO, that it is decomposed by
heat, that in this decomposition mercury and oxygen are formed and no
other products,--all these facts must be ascertained by exact experiment
before the equation can be written. An equation expressing these facts
will then have much value.
Having obtained an equation describing the conduct of mercuric oxide on
being heated, it will not do to assume that other oxides will behave in
like manner. Iron oxide (FeO) resembles mercuric oxide in many respects,
but it undergoes no change at all when heated. Manganese dioxide, the
black substance used in the preparation of oxygen, has the formula
MnO_{2}. When this substance is heated oxygen is set free, but the metal
manganese is not liberated; instead, a different oxide of manganese
containing less oxygen is produced. The equation representing the
reaction is
3MnO_{2} = Mn_{3}O_{4} + 2O.
~Classes of reactions.~ When a chemical change takes place in a substance
the substance is said to undergo a reaction. Although a great many
different reactions will be met in the study of chemistry, they may all
be grouped under the following heads.
1. _Addition._ This is the simplest kind of chemical action. It consists
in the union of two or more substances to produce a new substance. The
combination of iron with sulphur is an example:
Fe + S = FeS.
2. _Decomposition._ This is the reverse of addition, the substance
undergoing reaction being parted into its constituents. The
decomposition of mercuric oxide is an example: HgO = Hg + O.
3. _Substitution._ It is sometimes possible for an element in the free
state to act upon a compound in such a way that it takes the place of
one of the elements of the compound, liberating it in turn. In the study
of the element hydrogen it was pointed out that hydrogen is most
conveniently prepared by the action of sulphuric or hydrochloric acid
upon zinc. When sulphuric acid is used a substance called zinc sulphate,
having the composition represented by the formula ZnSO_{4}, is formed
together with hydrogen. The equation is
Zn + H_{2}SO_{4} = ZnSO_{4} + 2H.
When hydrochloric acid is used zinc chloride and hydrogen are the
products of reaction:
Zn + 2HCl = ZnCl_{2} + 2H.
When iron is used in place of zinc the equation is
Fe + H_{2}SO_{4} = FeSO_{4} + 2H.
These reactions are quite similar, as is apparent from an examination of
the equations. In each case 1 atom of the metal replaces 2 atoms of
hydrogen in the acid, and the hydrogen escapes as a gas. When an element
in the free state, such as the zinc in the equations just given, takes
the place of some one element in a compound, setting it free from
chemical combination, the act is called _substitution_.
Other reactions illustrating substitution are the action of sodium on
water,
Na + H_{2}O = NaOH + H;
and the action of heated iron upon water,
3Fe + 4H_{2}O = Fe_{3}O_{4} + 8H.
4. _Double decomposition._ When barium dioxide (BaO_{2}) is treated with
sulphuric acid two compounds are formed, namely, hydrogen dioxide
(H_{2}O_{2}) and barium sulphate (BaSO_{4}). The equation is
BaO_{2} + H_{2}SO_{4} = BaSO_{4} + H_{2}O_{2}.
In this reaction it will be seen that the two elements barium and
hydrogen simply exchange places. Such a reaction is called a _double
decomposition_. We shall meet with many examples of this kind of
chemical reactions.
~Chemical equations are quantitative.~ The use of symbols and formulas in
expressing chemical changes has another great advantage. Thus, according
to the equation
H_{2}O = 2H + O,
1 molecule of water is decomposed into 2 atoms of hydrogen and 1 atom of
oxygen. But, as we have seen, the relative weights of the atoms are
known, that of hydrogen being 1.008, while that of oxygen is 16. The
molecule of water, being composed of 2 atoms of hydrogen and 1 atom of
oxygen, must therefore weigh relatively 2.016 + 16, or 18.016. The
amount of hydrogen in this molecule must be 2.016/18.016, or 11.18% of
the whole, while the amount of oxygen must be 16/18.018, or 88.82% of
the whole. Now, since any definite quantity of water is simply the sum
of a great many molecules of water, it is plain that the fractions
representing the relative amounts of hydrogen and oxygen present in a
molecule must likewise express the relative amounts of hydrogen and
oxygen present in any quantity of water. Thus, for example, in 20 g. of
water there are 2.016/18.016 × 20, or 2.238 g. of hydrogen, and
16/18.016 × 20, or 17.762 g. of oxygen. These results in reference to
the composition of water of course agree exactly with the facts obtained
by the experiments described in the chapter on water, for it is because
of those experiments that the values 1.008 and 16 are given to hydrogen
and oxygen respectively.
It is often easier to make calculations of this kind in the form of a
proportion rather than by fractions. Since the molecule of water and
the two atoms of hydrogen which it contains have the ratio by weight of
18.016: 2.016, any mass of water has the same ratio between its total
weight and the weight of the hydrogen in it. Hence, to find the number
of grams (x) of hydrogen in 20 g. of water, we have the proportion
18.016 : 2.016 :: 20 g. : x (grams of hydrogen).
Solving for x, we get 2.238 for the number of grams of hydrogen.
Similarly, to find the amount (x) of oxygen present in the 20 g. of
water, we have the proportion
18.016 : 16 :: 20 : x
from which we find that x = 17.762 g.
Again, suppose we wish to find what weight of oxygen can be obtained
from 15 g. of mercuric oxide. The equation representing the
decomposition of mercuric oxide is
HgO = Hg + O.
The relative weights of the mercury and oxygen atoms are respectively
200 and 16. The relative weight of the mercuric oxide molecule must
therefore be the sum of these, or 216. The molecule of mercuric oxide
and the atom of oxygen which it contains have the ratio 216: 16. This
same ratio must therefore hold between the weight of any given quantity
of mercuric oxide and that of the oxygen which it contains. Hence, to
find the weight of oxygen in 15 g. of mercuric oxide, we have the
proportion
216 : 16 :: 15 : x (grams of oxygen).
On the other hand, suppose we wish to prepare, say, 20 g. of oxygen. The
problem is to find out what weight of mercuric oxide will yield 20 g. of
oxygen. The following proportion evidently holds
216 : 16 :: x (grams of mercuric oxide) : 20;
from which we get x = 270.
In the preparation of hydrogen by the action of sulphuric acid upon
zinc, according to the equation,
Zn + H_{2}SO_{4} = ZnSO_{4} + 2 H,
suppose that 50 g. of zinc are available; let it be required to
calculate the weight of hydrogen which can be obtained. It will be seen
that 1 atom of zinc will liberate 2 atoms of hydrogen. The ratio by
weight of a zinc to an hydrogen atom is 65.4: 1.008; of 1 zinc atom to 2
hydrogen atoms, 65.4: 2.016. Zinc and hydrogen will be related in this
reaction in this same ratio, however many atoms of zinc are concerned.
Consequently in the proportion
65.4 : 2.016 :: 50 : x,
x will be the weight of hydrogen set free by 50 g. of zinc. The weight
of zinc sulphate produced at the same time can be found from the
proportion
65.4 : 161.46 :: 50 : x;
where 161.46 is the molecular weight of the zinc sulphate, and x the
weight of zinc sulphate formed. In like manner, the weight of sulphuric
acid used up can be calculated from the proportion
65.4 : 98.076 :: 50 : x.
These simple calculations are possible because the symbols and formulas
in the equations represent the relative weights of the substances
concerned in a chemical reaction. When once the relative weights of the
atoms have been determined, and it has been agreed to allow the symbols
to stand for these relative weights, an equation or formula making use
of the symbols becomes a statement of a definite numerical fact, and
calculations can be based on it.
~Chemical equations not algebraic.~ Although chemical equations are
quantitative, it must be clearly understood that they are not algebraic.
A glance at the equations
7 + 4 = 11, 8 + 5 = 9 + 4
will show at once that they are true. The equations
HgO = Hg + O, FeO = Fe + O
are equally true in an algebraic sense, but experiment shows that only
the first is true chemically, for iron oxide (FeO) cannot be directly
decomposed into iron and oxygen. Only such equations as have been found
by careful experiment to express a real chemical transformation, true
both for the kinds of substances as well as for the weights, have any
value.
_Chemical formulas and equations, therefore, are a concise way of
representing qualitatively and quantitatively facts which have been
found by experiment to be true in reference to the composition of
substances and the changes which they undergo._
~Formulas representing water of crystallization.~ An examination of
substances containing water of crystallization has shown that in every
case the water is present in such proportion by weight as can readily be
represented by a formula. For example, copper sulphate (CuSO_{4}) and
water combine in the ratio of 1 molecule of the sulphate to 5 of water;
calcium sulphate (CaSO_{4}) and water combine in the ratio 1: 2 to form
gypsum. These facts are expressed by writing the formulas for the two
substances with a period between them. Thus the formula for crystallized
copper sulphate is CuSO_{4}·5H_{2}O; that of gypsum is CaSO_{4}·2H_{2}O.
~Heat of reaction.~ Attention has frequently been directed to the fact
that chemical changes are usually accompanied by heat changes. In
general it has been found that in every chemical action heat is either
absorbed or given off. By adopting a suitable unit for the measurement
of heat, the heat change during a chemical reaction can be expressed in
the equation for the reaction.
Heat cannot be measured by the use of a thermometer alone, since the
thermometer measures the intensity of heat, not its quantity. The
easiest way to measure a quantity of heat is to note how warm it will
make a definite amount of a given substance chosen as a standard. Water
has been chosen as the standard, and the unit of heat is called a
_calorie. A calorie is defined as the amount of heat required to raise
the temperature of one gram of water one degree._
By means of this unit it is easy to indicate the heat changes in a given
chemical reaction. The equation
2H + O = H_{2}O + 68,300 cal.
means that when 2.016 g. of hydrogen combine with 16 g. of oxygen,
18.016 g. of water are formed and 68,300 cal. are set free.
C + 2S = CS_{2} - 19,000 cal.
means that an expenditure of 19,000 cal. is required to cause 12 g. of
carbon to unite with 64.12 g. of sulphur to form 76.12 g. of carbon
disulphide. In these equations it will be noted that the symbols stand
for as many grams of the substance as there are units in the weights of
the atoms represented by the symbols. This is always understood to be
the case in equations where the heat of reaction is given.
~Conditions of a chemical action are not indicated by equations.~
Equations do not tell the conditions under which a reaction will take
place. The equation
HgO = Hg + O
does not tell us that it is necessary to keep the mercuric oxide at a
high temperature in order that the decomposition may go on. The equation
Zn + 2HCl = ZnCl_{2} + 2H
in no way indicates the fact that the hydrochloric acid must be
dissolved in water before it will act upon the zinc. From the equation
H + Cl = HCl
it would not be suspected that the two gases hydrogen and chlorine will
unite instantly in the sunlight, but will stand mixed in the dark a long
time without change. It will therefore be necessary to pay much
attention to the details of the conditions under which a given reaction
occurs, as well as to the expression of the reaction in the form of an
equation.
EXERCISES
1. Calculate the percentage composition of the following substances:
(a) mercuric oxide; (b) potassium chlorate; (c) hydrochloric acid;
(d) sulphuric acid. Compare the results obtained with the compositions
as given in Chapters II and III.
2. Determine the percentage of copper, sulphur, oxygen, and water in
copper sulphate crystals. What weight of water can be obtained from 150
g. of this substance?
3. What weight of zinc can be dissolved in 10 g. of sulphuric acid? How
much zinc sulphate will be formed?
4. How many liters of hydrogen measured under standard conditions can be
obtained from the action of 8 g. of iron on 10 g. of sulphuric acid? How
much iron sulphate (FeSO_{4}) will be formed?
5. 10 g. of zinc were used in the preparation of hydrogen; what weight
of iron will be required to prepare an equal volume?
6. How many grams of barium dioxide will be required to prepare 1 kg. of
common hydrogen dioxide solution? What weight of barium sulphate will be
formed at the same time?
7. What weight of the compound Mn_{3}O_{4} will be formed by strongly
heating 25 g. of manganese dioxide? What volume of oxygen will be given
off at the same time, measured under standard conditions?
8. (a) What is the weight of 100 l. of hydrogen measured in a
laboratory in which the temperature is 20° and pressure 750 mm.? (b)
What weight of sulphuric acid is necessary to prepare this amount of
hydrogen? (c) The density of sulphuric acid is 1.84. Express the acid
required in (b) in cubic centimeters.
9. What weight of potassium chlorate is necessary to furnish sufficient
oxygen to fill four 200 cc. bottles in your laboratory (the gas to be
collected over water)?