sider 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
x 2.04% = 2 parts of H, 98 x 32.6% = 32 parts of S, and 98 x 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.

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