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%
<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
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