An Elementary Study of Chemistry

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~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 x 20, or 2.238 g. of hydrogen, and
16/18.016 x 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 deg. 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)?

CHAPTER VII

NITROGEN AND THE RARE ELEMENTS: ARGON, HELIUM, NEON, KRYPTON, XENON

~Historical.~ Nitrogen was discovered by the English chemist Rutherford in
1772. A little later Scheele showed it to be a constituent of air, and
Lavoisier gave it the name _azote_, signifying that it would not support
life. The name _nitrogen_ was afterwards given it because of its
presence in saltpeter or niter. The term azote and symbol Az are still
retained by the French chemists.

~Occurrence.~ Air is composed principally of oxygen and nitrogen in the
free state, about 78 parts by volume out of every 100 parts being
nitrogen. Nitrogen also occurs in nature in the form of potassium
nitrate (KNO_{3})--commonly called saltpeter or niter--as well as in
sodium nitrate (NaNO_{3}). Nitrogen is also an essential constituent of
all living organisms; for example, the human body contains about 2.4% of
nitrogen.

~Preparation from air.~ Nitrogen can be prepared from air by the action of
some substance which will combine with the oxygen, leaving the nitrogen
free. Such a substance must be chosen, however, as will combine with the
oxygen to form a product which is not a gas, and which can be readily
separated from the nitrogen. The substances most commonly used for this
purpose are phosphorus and copper.

1. _By the action of phosphorus._ The method used for the preparation of
nitrogen by the action of phosphorus is as follows:

The phosphorus is placed in a little porcelain dish, supported on a cork
and floated on water (Fig. 26). It is then ignited by contact with a hot
wire, and immediately a bell jar or bottle is brought over it so as to
confine a portion of the air. The phosphorus combines with the oxygen to
form an oxide of phosphorus, known as phosphorus pentoxide. This is a
white solid which floats about in the bell jar, but in a short time it
is all absorbed by the water, leaving the nitrogen. The withdrawal of
the oxygen is indicated by the rising of the water in the bell jar.

[Illustration: Fig. 26]

2. _By the action of copper._ The oxygen present in the air may also be
removed by passing air slowly through a heated tube containing copper.
The copper combines with the oxygen to form copper oxide, which is a
solid. The nitrogen passes on and may be collected over water.

~Nitrogen obtained from air is not pure.~ Inasmuch as air, in
addition to oxygen and nitrogen, contains small amounts of
other gases, and since the phosphorus as well as the copper
removes only the oxygen, it is evident that the nitrogen
obtained by these methods is never quite pure. About 1% of the
product is composed of other gases, from which it is very
difficult to separate the nitrogen. The impure nitrogen so
obtained may, however, be used for a study of most of the
properties of nitrogen, since these are not materially affected
by the presence of the other gases.

~Preparation from compounds of nitrogen.~ Pure nitrogen may be obtained
from certain compounds of the element. Thus, if heat is applied to the
compound ammonium nitrite (NH_{4}NO_{2}), the change represented in the
following equation takes place:

NH_{4}NO_{2} = 2H_{2}O + 2N.

~Physical properties.~ Nitrogen is similar to oxygen and hydrogen in that
it is a colorless, odorless, and tasteless gas. One liter of nitrogen
weighs 1.2501 g. It is almost insoluble in water. It can be obtained in
the form of a colorless liquid having a boiling point of -195 deg. at
ordinary pressure. At -214 deg. it solidifies.

~Chemical properties.~ Nitrogen is characterized by its inertness. It is
neither combustible nor a supporter of combustion. At ordinary
temperatures it will not combine directly with any of the elements
except under rare conditions. At higher temperatures it combines with
magnesium, lithium, titanium, and a number of other elements. The
compounds formed are called _nitrides_, just as compounds of an element
with oxygen are called _oxides_. When it is mixed with oxygen and
subjected to the action of electric sparks, the two gases slowly combine
forming oxides of nitrogen. A mixture of nitrogen and hydrogen when
treated similarly forms ammonia, a gaseous compound of nitrogen and
hydrogen. Since we are constantly inhaling nitrogen, it is evident that
it is not poisonous. Nevertheless life would be impossible in an
atmosphere of pure nitrogen on account of the exclusion of the necessary
oxygen.

~Argon, helium, neon, krypton, xenon.~ These are all rare
elements occurring in the air in very small quantities. Argon,
discovered in 1894, was the first one obtained. Lord Rayleigh,
an English scientist, while engaged in determining the exact
weights of various gases, observed that the nitrogen obtained
from the air is slightly heavier than pure nitrogen obtained
from its compounds. After repeating his experiments many times,
always with the same results, Rayleigh finally concluded that
the nitrogen which he had obtained from the air was not pure,
but was mixed with a small amount of some unknown gas, the
density of which is greater than that of nitrogen. Acting on
this assumption, Rayleigh, together with the English chemist
Ramsay, attempted to separate the nitrogen from the unknown
gas. Knowing that nitrogen would combine with magnesium, they
passed the nitrogen obtained from the air and freed from all
known substances through tubes containing magnesium heated to
the necessary temperature. After repeating this operation, they
finally succeeded in obtaining from the atmospheric nitrogen a
small volume of gas which would not combine with magnesium and
hence could not be nitrogen. This proved to be a new element,
to which they gave the name _argon_. As predicted, this new
element was found to be heavier than nitrogen, its density as
compared with hydrogen as a standard being approximately 20,
that of nitrogen being only 14. About 1% of the atmospheric
nitrogen proved to be argon. The new element is characterized
by having no affinity for other elements. Even under the most
favorable conditions it has not been made to combine with any
other element. On this account it was given the name argon,
signifying lazy or idle. Like nitrogen, it is colorless,
odorless, and tasteless. It has been liquefied and solidified.
Its boiling point is -187 deg..

Helium was first found in the gases expelled from certain
minerals by heating. Through the agency of the spectroscope it
had been known to exist in the sun long before its presence on
the earth had been demonstrated,--a fact suggested by the name
helium, signifying the sun. Its existence in traces in the
atmosphere has also been proven. It was first liquefied by
Onnes in July, 1908. Its boiling point, namely -269 deg., is the
lowest temperature yet reached.

The remaining elements of this group--neon, krypton, and
xenon--have been obtained from liquid air. When liquid air is
allowed to boil, the constituents which are the most difficult
to liquefy, and which therefore have the lowest boiling points,
vaporize first, followed by the others in the order of their
boiling points. It is possible in this way to make at least a
partial separation of the air into its constituents, and Ramsay
thus succeeded in obtaining from liquid air not only the known
constituents, including argon and helium, but also the new
elements, neon, krypton, and xenon. These elements, as well as
helium, all proved to be similar to argon in that they are
without chemical activity, apparently forming no compounds
whatever. The percentages present in the air are very small.
The names, neon, krypton, xenon, signify respectively, new,
hidden, stranger.

EXERCISES

1. How could you distinguish between oxygen, hydrogen, and nitrogen?

2. Calculate the relative weights of nitrogen and oxygen; of nitrogen
and hydrogen.

3. In the preparation of nitrogen from the air, how would hydrogen do as
a substance for the removal of the oxygen?

4. What weight of nitrogen can be obtained from 10 l. of air measured
under the conditions of temperature and pressure which prevail in your
laboratory?

5. How many grams of ammonium nitrite are necessary in the preparation
of 20 l. of nitrogen measured over water under the conditions of
temperature and pressure which prevail in your laboratory?

6. If 10 l. of air, measured under standard conditions, is passed over
100 g. of hot copper, how much will the copper gain in weight?

[Illustration: WILLIAM RAMSAY (Scotch) (1855-)

Has made many studies in the physical properties of substances;
discovered helium; together with Lord Rayleigh and others he discovered
argon, krypton, xenon, and neon; has contributed largely to the
knowledge of radio-active substances, showing that radium gradually
gives rise to helium; professor at University College, London]

CHAPTER VIII

THE ATMOSPHERE

~Atmosphere and air.~ The term _atmosphere_ is applied to the gaseous
envelope surrounding the earth. The term _air_ is generally applied to a
limited portion of this envelope, although the two words are often used
interchangeably. Many references have already been made to the
composition and properties of the atmosphere. These statements must now
be collected and discussed somewhat more in detail.

~Air formerly regarded as an element.~ Like water, air was at first
regarded as elementary in character. Near the close of the eighteenth
century Scheele, Priestley, and Lavoisier showed by their experiments
that it is a mixture of at least two gases,--those which we now call
oxygen and nitrogen. By burning substances in an inclosed volume of air
and noting the contraction in volume due to the removal of the oxygen,
they were able to determine with some accuracy the relative volumes of
oxygen and nitrogen present in the air.

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