A compound has a percent composition of 15.24% sodium (molar mass = 22.99

Chemistry

1 answer:

RUDIKE [14]2 years ago

8 0

The empirical formula of the compound obtained from the question given is NaBrO₃

<h3>Data obtained from the question </h3>

Sodium (Na) = 15.24%
Bromine (Br) = 52.95%
Oxygen (O) = 31.81%

Empirical formula =?

<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Na = 15.24 / 22.99 = 0.663

Br = 52.95 / 79.90 = 0.663

O = 31.81 / 16 = 1.988

Divide by the smallest

Na = 0.663 / 0.663 = 1

Br = 0.663 / 0.663 = 1

O = 1.988 / 0.663 = 3

Thus, the empirical formula of the compound is NaBrO₃

Learn more about empirical formula:

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