termination 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:

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