Barrons SAT Subject Test Chemistry, 13th Edition

the correct formula mass. For example, if your empirical formula is CH 2 (which

has a formula mass of 14) and the true formula mass is given as 28, you can see
that you must double the empirical formula by doubling all the subscripts. The
true formula is C 2 H 4.

LAWS OF DEFINITE COMPOSITION AND MULTIPLE

PROPORTIONS

In the problems involving percent composition, we have depended on two things:
each unit of an element has the same atomic mass, and every time the particular
compound forms, it forms in the same percent composition. That this latter
statement is true no matter the source of the compound is the Law of Definite
Composition. There are some compounds formed by the same two elements in
which the mass of one element is constant, but the mass of the other varies. In
every case, however, the mass of the other element is present in a small-whole-
number ratio to the weight of the first element. This is called the Law of Multiple
Proportions. An example is H 2 O and H 2 O 2.

In H 2 O the proportion of H : O = 2 : 16 or 1 : 8

In H 2 O 2 the proportion of H : O = 2 : 32 or 1 : 16

The ratio of the mass of oxygen in each is 8 : 16 or 1 : 2 (a small-whole-
number ratio).

WRITING AND BALANCING SIMPLE EQUATIONS

An equation is a simplified way of recording a chemical change. Instead of
words, chemical symbols and formulas are used to represent the reactants and
the products. Here is an example of how this can be done. The following is the
word equation of the reaction of burning hydrogen with oxygen:

Hydrogen + oxygen yields water.

Replacing the words with the chemical formulas, we have

H 2 + O 2 → H 2 O

We replaced hydrogen and oxygen with the formulas for their diatomic
molecular states and wrote the appropriate formula for water based on the
respective oxidation (valence) numbers for hydrogen and oxygen. Note that the
word yields was replaced with the arrow.
Although the chemical statement tells what happened, it is not an equation