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 × 16.03
g. of sulphur. 65.4 g. of zinc combine with 70.9 g. of chlorine and 2 ×
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
× 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 × 0.9046 × 32;
or, still more briefly,
M. = D. × 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. × 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 × 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
3O_{2} = 2O_{3}.
~Other methods of determining molecular weights.~ It will be noticed that
Avogadro's law gives us a method by which we can determine the relative
weights of the molecules of two gases because it enables us to tell when
we are dealing with an equal number of the two kinds of molecules. If by
any other means we can get this information, we can make use of the
knowledge so gained to determine the molecular weights of the two
substances.
~Raoult's laws.~ Two laws have been discovered which give us just such
information. They are known as Raoult's laws, and can be stated as
follows:
1. _When weights of substances which are proportional to their molecular
weights are dissolved in the same weight of solvent, the rise of the
boiling point is the same in each case._
2. _When weights of substances which are proportional to their molecular
weights are dissolved in the same weight of solvent, the lowering of the
freezing point is the same in each case._
By taking advantage of these laws it is possible to determine when two
solutions contain the same number of molecules of two dissolved
substances, and consequently the relative molecular weights of the two
substances.
~Law of Dulong and Petit.~ In 1819 Dulong and Petit discovered a very
interesting relation between the atomic weight of an element and its
specific heat, which holds true for elements in the solid state. If
equal weights of two solids, say, lead and silver, are heated through
the same range of temperature, as from 10° to 20°, it is found that very
different amounts of heat are required. The amount of heat required to
change the temperature of a solid or a liquid by a definite amount
compared with the amount required to change the temperature of an equal
weight of water by the same amount is called its specific heat. Dulong
and Petit discovered the following law: _The specific heat of an element
in the solid form multiplied by its atomic weight is approximately equal
to the constant 6.25._ That is,
at. wt. × sp. ht. = 6.25.
Consequently,
6.25
at. wt. = --------
sp. ht.
This law is not very accurate, but it is often possible by means of it
to decide upon what multiple of the equivalent is the real atomic
weight. Thus the specific heat of iron is found by experiment to be
0.112, and its equivalent is 27.95. 6.25 ÷ 0.112 = 55.8. We see,
therefore, that the atomic weight is twice the equivalent, or 55.9.
~How formulas are determined.~ It will be well in connection with
molecular weights to consider how the formula of a compound is decided
upon, for the two subjects are very closely associated. Some examples
will make clear the method followed.
The molecular weight of a substance containing hydrogen and chlorine was
36.4. By analysis 36.4 parts of the substance was found to contain 1
part of hydrogen and 35.4 parts of chlorine. As these are the simple
atomic weights of the two elements, the formula of the compound must be
HCl.
A substance consisting of oxygen and hydrogen was found to have a
molecular weight of 34. Analysis showed that in 34 parts of the
substance there were 2 parts of hydrogen and 32 parts of oxygen.
Dividing these figures by the atomic weights of the two elements, we get
2 ÷ 1 = 2 for H; 32 ÷ 16 = 2 for O. The formula is therefore H_{2}O_{2}.
A substance containing 2.04% H, 32.6% S, and 65.3% O was found to have a
molecular weight of 98. In these 98 parts of the substance there are 98
× 2.04% = 2 parts of H, 98 × 32.6% = 32 parts of S, and 98 × 65.3% = 64
parts of O. If the molecule weighs 98, the hydrogen atoms present must
together weigh 2, the sulphur atoms 32, and the oxygen atoms 64.
Dividing these figures by the respective atomic weights of the three
elements, we have, for H, 2 ÷ 1 = 2 atoms; for S, 32 ÷ 32 = 1 atom; for
O, 64 ÷ 16 = 4 atoms. Hence the formula is H_{2}SO_{4}.
We have, then, this general procedure: Find the percentage composition
of the substance and also its molecular weight. Multiply the molecular
weight successively by the percentage of each element present, to find
the amount of the element in the molecular weight of the compound. The
figures so obtained will be the respective parts of the molecular weight
due to the several atoms. Divide by the atomic weights of the respective
elements, and the quotient will be the number of atoms present.
~Avogadro's hypothesis and chemical calculations.~ This law simplifies
many chemical calculations.
1. _Application to volume relations in gaseous reactions._ Since equal
volumes of gases contain an equal number of molecules, it follows that
when an equal number of gaseous molecules of two or more gases take part
in a reaction, the reaction will involve equal volumes of the gases. In
the equation
C_{2}H_{2}O_{4} = H_{2}O + CO_{2} + CO,
since 1 molecule of each of the gases CO_{2} and CO is set free from
each molecule of oxalic acid, the two substances must always be set free
in equal volumes.
Acetylene burns in accordance with the equation
2C_{2}H_{2} + 5O_{2} = 4CO_{2} + 2H_{2}O.
Hence 2 volumes of acetylene will react with 5 volumes of oxygen to form
4 volumes of carbon dioxide and 2 volumes of steam. That the volume
relations may be correct a gaseous element must be given its molecular
formula. Thus oxygen must be written O_{2} and not 2O.
2. _Application to weights of gases._ It will be recalled that the
molecular weight of a gas is determined by ascertaining the weight of
22.4 l. of the gas. This weight in grams is called the _gram-molecular
weight_ of a gas. If the molecular weight of any gas is known, the
weight of a liter of the gas under standard conditions may be determined
by dividing its gram-molecular weight by 22.4. Thus the gram-molecular
weight of a hydrochloric acid gas is 36.458. A liter of the gas will
therefore weigh 36.458 ÷ 22.4 = 1.627 g.
EXERCISES
1. From the following data calculate the atomic weight of sulphur. The
equivalent, as obtained by an analysis of sulphur dioxide, is 16.03. The
densities and compositions of a number of compounds containing sulphur
are as follows:
NAME DENSITY COMPOSITION BY PERCENTAGE
Hydrosulphuric acid 1.1791 S = 94.11 H = 5.89
Sulphur dioxide 2.222 S = 50.05 O = 49.95
Sulphur trioxide 2.74 S = 40.05 O = 59.95
Sulphur chloride 4.70 S = 47.48 Cl = 52.52
Sulphuryl chloride 4.64 S = 23.75 Cl = 52.53 O = 23.70
Carbon disulphide 2.68 S = 84.24 C = 15.76
2. Calculate the formulas for compounds of the following compositions:
MOLECULAR
WEIGHT
(1) S = 39.07% O = 58.49% H = 2.44% 81.0
(2) Ca = 29.40 S = 23.56 O = 47.04 136.2
(3) K = 38.67 N = 13.88 O = 47.45 101.2
3. The molecular weight of ammonia is 17.06; of sulphur dioxide is
64.06; of chlorine is 70.9. From the molecular weight calculate the
weight of 1 l. of each of these gases. Compare your results with the
table on the back cover of the book.
4. From the molecular weight of the same gases calculate the density of
each, referred to air as a standard.
5. A mixture of 50 cc. of carbon monoxide and 50 cc. of oxygen was
exploded in a eudiometer, (a) What gases remained in the tube after
the explosion? (b) What was the volume of each?
6. In what proportion must acetylene and oxygen be mixed to produce the
greatest explosion?
7. Solve Problem 18, Chapter XVII, without using molecular weights.
Compare your results.
8. Solve Problem 10, Chapter XVIII, without using molecular weights.
Compare your results.
9. The specific heat of aluminium is 0.214; of lead is 0.031. From these
specific heats calculate the atomic weights of each of the elements.