A. F. Pollard - "Henry VIII."

Various - "Scientific American Supplement, No. 286"

Emile Zola - "La Conquete De Plassans"

D. G. Hogarth - "The Ancient East"

An Elementary Study of Chemistry

W >> William McPherson >> An Elementary Study of Chemistry




[Illustration: Fig. 66]

One ton of good gas coal yields approximately 10,000 cu. ft. of
gas, 1400 lb. of coke, 120 lb. of tar, and 20 gal. of
ammoniacal liquor.

Not only is the ammonia obtained in the manufacture of the gas
of great importance, but the coal tar also serves as the source
of many very useful substances, as will be explained in Chapter
XXXII.

~Water gas.~ Water gas is essentially a mixture of carbon monoxide and
hydrogen. It is made by passing steam over very hot anthracite coal,
when the reaction shown in the following equation takes place:

C + H_{2}O = CO + 2H.

When required merely to produce heat the gas is at once ready for use.
When made for illuminating purposes it must be enriched, that is,
illuminants must be added, since both carbon monoxide and hydrogen burn
with non-luminous flames. This is accomplished by passing it into
heaters containing highly heated petroleum oils. The gas takes up
hydrocarbon gases formed in the decomposition of the petroleum oils,
which make it burn with a luminous flame.

Water gas is very effective as a fuel, since both carbon monoxide and
hydrogen burn with very hot flames. It has little odor and is very
poisonous. Its use is therefore attended with some risk, since leaks in
pipes are very likely to escape notice.

~Natural gas.~ This substance, so abundant in many localities, varies much
in composition, but is composed principally of methane. When used for
lighting purposes it is usually burned in a burner resembling an open
Bunsen, the illumination being furnished by an incandescent mantle. This
is the case in the familiar Welsbach burner. Contrary to statements
frequently made, natural gas contains no free hydrogen.


TABLE SHOWING COMPOSITION OF GASES

=====================+================+========+========+==========
| PENNSYLVANIA | COAL | WATER | ENRICHED
| NATURAL | GAS | GAS | WATER
| GAS | | | GAS
---------------------+----------------+--------+--------+----------
Hydrogen | | 41.3 | 52.88 | 30.00
Methane | 90.64 | 43.6 | 2.16 | 24.00
Illuminants | | 3.9 | | 12.05
Carbon monoxide | | 6.4 | 36.80 | 29.00
Carbon dioxide | 0.30 | 2.0 | 3.47 | 0.30
Nitrogen | 9.06 | 1.2 | 4.69 | 2.50
Oxygen | | 0.3 | | 1.50
Hydrocarbon vapors | | 1.5 | | 1.50
=====================+================+========+========+==========

These are analyses of actual samples, and may be taken as about
the average for the various kinds of gases. Any one of these
may vary considerably. The nitrogen and oxygen in most cases is
due to a slight admixture of air which is difficult to exclude
entirely in the manufacture and handling of gases.

~Fuels.~ A variety of substances are used as fuels, the most important of
them being wood, coal, and the various gases mentioned above. Wood
consists mainly of compounds of carbon, hydrogen, and oxygen. The
composition of coal and the fuel gases has been given. Since these fuels
are composed principally of carbon and hydrogen or their compounds, the
chief products of combustion are carbon dioxide and water. The practice
of heating rooms with portable gas or oil stoves with no provision for
removing the products of combustion is to be condemned, since the carbon
dioxide is generated in sufficient quantities to render the air unfit
for breathing. Rooms so heated also become very damp from the large
amount of water vapor formed in the combustion, and which in cold
weather condenses on the window glass, causing the glass to "sweat."
Both coal and wood contain a certain amount of mineral substances which
constitute the ashes.

~The electric furnace.~ In recent years electric furnaces have come into
wide use in operations requiring a very high temperature. Temperatures
as high as 3500 deg. can be easily reached, whereas the hottest oxyhydrogen
flame is not much above 2000 deg.. These furnaces are constructed on one of
two general principles.

[Illustration: Fig. 67]

1. _Arc furnaces._ In the one type the source of heat is an electric arc
formed between carbon electrodes separated a little from each other, as
shown in Fig. 67. The substance to be heated is placed in a vessel,
usually a graphite crucible, just below the arc. The electrodes and
crucible are surrounded by materials which fuse with great difficulty,
such as magnesium oxide, the walls of the furnace being so shaped as to
reflect the heat downwards upon the contents of the crucible.

[Illustration: Fig. 68]

2. _Resistance furnaces._ In the other type of furnace the heat is
generated by the resistance offered to the current in its passage
through the furnace. In its simplest form it may be represented by Fig.
68. The furnace is merely a rectangular box built up of loose bricks.
The electrodes E, each consisting of a bundle of carbon rods, are
introduced through the sides of the furnace. The materials to be heated,
C, are filled into the furnace up to the electrodes, and a layer of
broken coke is arranged so as to extend from one electrode to the other.
More of the charge is then placed on top of the coke. In passing through
the broken coke the electrical current encounters great resistance. This
generates great heat, and the charge surrounding the coke is brought to
a very high temperature. The advantage of this type of furnace is that
the temperature can be regulated to any desired intensity.


EXERCISES

1. Why does charcoal usually burn with no flame? How do you account for
the flame sometimes observed when it burns?

2. How do you account for the fact that a candle burns with a flame?

3. What two properties must the mantle used in the Welsbach lamp
possess?

4. (a) In what respects does the use of the Welsbach mantle resemble
that of lime in the calcium light? (b) If the mantle were made of
carbon, would it serve the same purpose?

5. Would anthracite coal be suitable for the manufacture of coal gas?

6. How could you prove the formation of carbon dioxide and water in the
combustion of illuminating gases?

7. Suggest a probable way in which natural gas has been formed.

8. Coal frequently contains a sulphide of iron. (a) What two sulphur
compounds are likely to be formed when gas is made from such coal? (b)
Suggest some suitable method for the removal of these compounds.

9. Why does the use of the bellows on the blacksmith's forge cause a
more intense heat?

10. What volume of oxygen is necessary to burn 100 l. of marsh gas and
what volume of carbon dioxide would be formed, all of the gases being
measured under standard conditions?

11. Suppose a cubic meter of Pennsylvania natural gas, measured under
standard conditions, were to be burned. How much water by weight would
result?




CHAPTER XIX

MOLECULAR WEIGHTS, ATOMIC WEIGHTS, FORMULAS


~Introduction.~ In the chapter on The Atomic Theory, it was shown that if
it were true that two elements uniting to form a compound always
combined in the ratio of one atom of one element to one atom of the
other element, it would be a very easy matter to decide upon figures
which would represent the relative weights of the different atoms. It
would only be necessary to select some one element as a standard and
determine the weight of every element which combines with a definite
weight (say 1 g.) of the standard element. The figures so obtained would
evidently represent the relative weights of the atoms.

But the law of multiple proportion at once reminds us that two elements
may unite in several proportions; and there is no simple way to
determine the number of atoms present in the molecule of any compound.
Consequently the problem of deciding upon the relative atomic weights is
not an easy one. To the solution of this problem we must now turn.

~Dalton's method of determining atomic weights.~ When Dalton first
advanced the atomic theory he attempted to solve this problem by very
simple methods. He thought that when only one compound of two elements
is known it is reasonable to suppose that it contains one atom of each
element. He therefore gave the formula HO to water, and HN to ammonia.
When more than two compounds were known he assumed that the most
familiar or the most stable one had the simple formula. He then
determined the atomic weight as explained above. The results he
obtained were contradictory and very far from satisfactory, and it was
soon seen that some other method, resting on much more scientific
grounds, must be found to decide what compounds, if any, have a single
atom of each element present.

~Determination of atomic weights.~ Three distinct steps are involved in
the determination of the atomic weight of an element: (1) determination
of the equivalent, (2) determination of molecular weights of its
compounds, and (3) deduction of the exact atomic weight from the
equivalent and molecular weights.

~1. Determination of the equivalent.~ By the equivalent of an element is
meant the weight of the element which will combine with a fixed weight
of some other element chosen as a standard. It has already been
explained that oxygen has been selected as the standard element for
atomic weights, with a weight of 16. This same standard will serve very
well as a standard for equivalents. _The equivalent of an element is the
weight of the element which will combine with 16 g. of oxygen._ Thus 16
g. of oxygen combines with 16.03 g. of sulphur, 65.4 g. of zinc, 215.86
g. of silver, 70.9 g. of chlorine. These figures, therefore, represent
the equivalent weights of these elements.

~Relation of atomic weights to equivalents.~ According to the atomic
theory combination always takes place between whole numbers of atoms.
Thus one atom unites with one other, or with two or three; or two atoms
may unite with three, or three with five, and so on.

When oxygen combines with zinc the combination must be between definite
numbers of the two kinds of atoms. Experiment shows that these two
elements combine in the ratio of 16 g. of oxygen to 65.4 g. of zinc. If
one atom of oxygen combines with one atom of zinc, then this ratio must
be the ratio between the weights of the two atoms. If one atom of oxygen
combines with two atoms of zinc, then the ratio between the weights of
the two atoms will be 16: 32.7. If two atoms of oxygen combine with one
atom of zinc, the ratio by weight between the two atoms will be 8: 65.4.
It is evident, therefore, that the real atomic weight of an element must
be some multiple or submultiple of the equivalent; in other words, the
equivalent multiplied by 1/2, 1, 2, or 3 will give the atomic weight.

~Combining weights.~ A very interesting relation holds good between the
equivalents of the various elements. We have just seen that the figures
16.03, 65.4, 215.86, and 70.9 are the equivalents respectively of
sulphur, zinc, silver, and chlorine. These same figures represent the
ratios by weight in which these elements combine among themselves. Thus
215.86 g. of silver combine with 70.9 g. of chlorine and with 2 x 16.03
g. of sulphur. 65.4 g. of zinc combine with 70.9 g. of chlorine and 2 x
16.03 g. of sulphur.

By taking the equivalent or some multiple of it a value can be obtained
for each element which will represent its combining value, and for this
reason is called its _combining weight_. It is important to notice that
the fact that a combining weight can be obtained for each element is not
a part of a theory, but is the direct result of experiment.

~Elements with more than one equivalent.~ It will be remembered that
oxygen combines with hydrogen in two ratios. In one case 16 g. of oxygen
combine with 2.016 g. of hydrogen to form water; in the other 16 g. of
oxygen combine with 1.008 g. of hydrogen to form hydrogen dioxide. The
equivalents of hydrogen are therefore 2.016 and 1.008. Barium combines
with oxygen in two proportions: in barium oxide the proportion is 16 g.
of oxygen to 137.4 g. of barium; in barium dioxide the proportion is 16
g. of oxygen to 68.7 g. of barium.

In each case one equivalent is a simple multiple of the other, so the
fact that there may be two equivalents does not add to the uncertainty.
All we knew before was that the true atomic weight is some multiple of
the equivalent.

~2. The determination of molecular weights.~ To decide the question as to
which multiple of the equivalent correctly represents the atomic weight
of an element, it has been found necessary to devise a method of
determining the molecular weights of compounds containing the element in
question. Since the molecular weight of a compound is merely the sum of
the weights of all the atoms present in it, it would seem to be
impossible to determine the molecular weight of a compound without first
knowing the atomic weights of the constituent atoms, and how many atoms
of each element are present in the molecule. But certain facts have been
discovered which suggest a way in which this can be done.

~Avogadro's hypothesis.~ We have seen that the laws of Boyle, Charles, and
Gay-Lussac apply to all gases irrespective of their chemical character.
This would lead to the inference that the structure of gases must be
quite simple, and that it is much the same in all gases.

In 1811 Avogadro, an Italian physicist, suggested that if we assume all
gases under the same conditions of temperature and pressure to have the
same number of molecules in a given volume, we shall have a probable
explanation of the simplicity of the gas laws. It is difficult to prove
the truth of this hypothesis by a simple experiment, but there are so
many facts known which are in complete harmony with this suggestion that
there is little doubt that it expresses the truth. Avogadro's hypothesis
may be stated thus: _Equal volumes of all gases under the same
conditions of temperature and pressure contain the same number of
molecules._

~Avogadro's hypothesis and molecular weights.~ Assuming that Avogadro's
hypothesis is correct, we have a very simple means for deciding upon the
relative weights of molecules; for if equal volumes of two gases contain
the same number of molecules, the weights of the two volumes must be in
the same ratio as the weights of the individual molecules which they
contain. If we adopt some one gas as a standard, we can express the
weights of all other gases as compared with this one, and the same
figures will express the relative weights of the molecules of which the
gases are composed.

~Oxygen as the standard.~ It is important that the same standard should be
adopted for the determination of molecular weights as has been decided
upon for atomic weights and equivalents, so that the three values may be
in harmony with each other. Accordingly it is best to adopt oxygen as
the standard element with which to compare the molecular weights of
other gases, being careful to keep the oxygen atom equal to 16.

~The oxygen molecule contains two atoms.~ One point must not be
overlooked, however. We desire to have our unit, the oxygen _atom_,
equal to 16. The method of comparing the weights of gases just suggested
compares the molecules of the gases with the _molecule_ of oxygen. Is
the molecule and the atom of oxygen the same thing? This question is
answered by the following considerations.

We have seen that when steam is formed by the union of oxygen and
hydrogen, two volumes of hydrogen combine with one volume of oxygen to
form two volumes of steam. Let us suppose that the one volume of oxygen
contains 100 molecules; then the two volumes of steam must, according
to Avogadro's hypothesis, contain 200 molecules. But each of these 200
molecules must contain at least one atom of oxygen, or 200 in all, and
these 200 atoms came from 100 molecules of oxygen. It follows that each
molecule of oxygen must contain at least two atoms of oxygen.

Evidently this reasoning merely shows that there are _at least_ two
atoms in the oxygen molecule. There may be more than that, but as there
is no evidence to this effect, we assume that the molecule contains two
atoms only.

It is evident that if we wish to retain the value 16 for the atom of
oxygen we must take twice this value, or 32, for the value of the oxygen
molecule, when using it as a standard for molecular weights.

~Determination of the molecular weights of gases from their weights
compared with oxygen.~ Assuming the molecular weight of oxygen to be 32,
Avogadro's hypothesis gives us a ready means for determining the
molecular weight of any other gas, for all that is required is to know
its weight compared with that of an equal volume of oxygen. For example,
1 l. of chlorine is found by experiment to weigh 2.216 times as much as
1 l. of oxygen. The molecular weight of chlorine must therefore be 2.216
x 32, or 70.91.

If, instead of comparing the relative weights of 1 l. of the two gases,
we select such a volume of oxygen as will weigh 32 g., or the weight in
grams corresponding to the molecular weight of the gas, the calculation
is much simplified. It has been found that 32 g. of oxygen, under
standard conditions, measure 22.4 l. This same volume of hydrogen weighs
2.019 g.; of chlorine 70.9 g.; of hydrochloric acid 36.458 g. The
weights of these equal volumes must be proportional to their molecular
weights, and since the weight of the oxygen is the same as the value of
its molecular weight, so too will the weights of the 22.4 l. of the
other gases be equal to the value of their molecular weights.

As a summary we can then make the following statement: _The molecular
weight of any gas may be determined by calculating the weight of 22.4 l.
of the gas, measured under standard conditions._

~Determination of molecular weights from density of gases.~ In an actual
experiment it is easier to determine the density of a gas than the
weight of a definite volume of it. The density of a gas is usually
defined as its weight compared with that of an equal volume of air.
Having determined the density of a gas, its weight compared with oxygen
may be determined by multiplying its density by the ratio between the
weights of air and oxygen. This ratio is 0.9046. To compare it with our
standard for atomic weights we must further multiply it by 32, since the
standard is 1/32 the weight of oxygen molecules. The steps then are
these:

1. Determine the density of the gas (its weight compared with air).

2. Multiply by 0.9046 to make the comparison with oxygen molecules.

3. Multiply by 32 to make the comparison with the unit for atomic
weights.

We have, then, the formula:

molecular weight = density x 0.9046 x 32;

or, still more briefly,

M. = D. x 28.9.

The value found by this method for the determination of molecular
weights will of course agree with those found by calculating the weight
of 22.4 l. of the gas, since both methods depend on the same principles.

[Illustration: Fig. 69]

~Determination of densities of gases.~ The relative weights of
equal volumes of two gases can be easily determined. The
following is one of the methods used. A small flask, such as is
shown in Fig. 69, is filled with one of the gases, and after
the temperature and pressure have been noted the flask is
sealed up and weighed. The tip of the sealed end is then broken
off, the flask filled with the second gas, and its weight
determined. If the weight of the empty flask is subtracted from
these two weighings, the relative weights of the gases is
readily found.

~3. Deduction of atomic weights from molecular weights and equivalents.~
We have now seen how the equivalent of an element and the molecular
weight of compounds containing the element can be obtained. Let us see
how it is possible to decide which multiple of the equivalent really is
the true atomic weight. As an example, let us suppose that the
equivalent of nitrogen has been found to be 7.02 and that it is desired
to obtain its atomic weight. The next step is to obtain the molecular
weights of a large number of compounds containing nitrogen. The
following will serve:

==================+============+=============+================+==============
| | APPROXIMATE | PERCENTAGE OF | PART OF
| DENSITY BY | MOLECULAR | NITROGEN BY | MOLECULAR
| EXPERIMENT | WEIGHT | EXPERIMENT | WEIGHT DUE
| | (D. x 28.9) | | TO NITROGEN
------------------+------------+-------------+----------------+--------------
Nitrogen gas | 0.9671 | 27.95 | 100.00 | 27.95
Nitrous oxide | 1.527 | 44.13 | 63.70 | 27.11
Nitric oxide | 1.0384 | 30.00 | 46.74 | 14.02
Nitrogen peroxide | 1.580 | 45.66 | 30.49 | 13.90
Ammonia | 0.591 | 17.05 | 82.28 | 14.03
Nitric acid | 2.180 | 63.06 | 22.27 | 14.03
Hydrocyanic acid | 0.930 | 26.87 | 51.90 | 13.94
==================+============+=============+================+==============

~Method of calculation.~ The densities of the various gases in the first
column of this table are determined by experiment, and are fairly
accurate but not entirely so. By multiplying these densities by 28.9 the
molecular weights of the compounds as given in the second column are
obtained. By chemical analysis it is possible to determine the
percentage composition of these substances, and the percentages of
nitrogen in them as determined by analysis are given in the third
column. If each of these molecular weights is multiplied in turn by the
percentage of nitrogen in the compound, the product will be the weight
of the nitrogen in the molecular weight of the compound. This will be
the sum of the weights of the nitrogen atoms in the molecule. These
values are given in the fourth column in the table.

If a large number of compounds containing nitrogen are studied in this
way, it is probable that there will be included in the list at least one
substance whose molecule contains a single nitrogen atom. In this case
the number in the fourth column will be the approximate atomic weight of
nitrogen. On comparing the values for nitrogen in the table it will be
seen that a number which is approximately 14 is the smallest, and that
the others are multiples of this. These compounds of higher value,
therefore, contain more than one nitrogen atom in the molecule.

~Accurate determination of atomic weights.~ Molecular weights cannot be
determined very accurately, and consequently the part in them due to
nitrogen is a little uncertain, as will be seen in the table. All we can
tell by this method is that the true weight is very near 14. The
equivalent can however be determined very accurately, and we have seen
that it is some multiple or submultiple of the true atomic weight.
Since molecular-weight determinations have shown that in the case of
nitrogen the atomic weight is near 14, and we have found the equivalent
to be 7.02, it is evident that the true atomic weight is twice the
equivalent, or 7.02 x 2 = 14.04.

~Summary.~ These, then, are the steps necessary to establish the atomic
weight of an element.

1. Determine the equivalent accurately by analysis.

2. Determine the molecular weight of a large number of compounds of the
element, and by analysis the part of the molecular weight due to the
element. The smallest number so obtained will be approximately the
atomic weight.

3. Multiply the equivalent by the small whole number (usually 1, 2, or
3), which will make a number very close to the approximate atomic
weight. The figure so obtained will be the true atomic weight.

~Molecular weights of the elements.~ It will be noticed that the molecular
weight of nitrogen obtained by multiplying its density by 28.9 is 28.08.
Yet the atomic weight of nitrogen as deduced from a study of its gaseous
compounds is 14.04. The simplest explanation that can be given for this
is that the gaseous nitrogen is made up of molecules, each of which
contains two atoms. In this respect it resembles oxygen; for we have
seen that an entirely different line of reasoning leads us to believe
that the molecule of oxygen contains two atoms. When we wish to indicate
molecules of these gases the symbols N_{2} and O_{2} should be used.
When we desire to merely show the weights taking part in a reaction this
is not necessary.

The vapor densities of many of the elements show that, like oxygen and
nitrogen, their molecules consist of two atoms. In other cases,
particularly among the metals, the molecule and the atom are identical.
Still other elements have four atoms in their molecules.

While oxygen contains two atoms in its molecules, a study of ozone has
led to the conclusion that it has three. The formation of ozone from
oxygen can therefore be represented by the equation

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