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._

~De