er 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