A 0.4708 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.9093 g. What is the mass percentage of bromine in the original compound

The correct answer is 82.18%.

Explanation:

Based on the given information, the mass of the pure soluble bromide compound is 0.4708 grams, and the mass of the formed silver bromide is found to be 0.9093 grams.

The molecular weight of AgBr is 187.77 g/mol, and the molecular weight of Br is 79.9 g/mol.

So, 0.9093 grams of AgBr comprise (79.9 * 0.9093)/187.77 = 0.3869 grams of bromine.

Now, 0.4708 grams of sample comprise 0.3869 grams of bromine. Therefore, 100 grams of sample comprise (0.3869*100) / 0.4708 g = 82.18 grams of bromine.

The mass % of bromine in the original compound is 82.18%.

answer

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