Question

A 0.5781 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 1.1166 g. What is the mass percentage of bromine in the original compound? %

Answer 1

Answer:

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

Explanation:

Mass of AgBr = 1.1166 g

Moles of AgBr = $\frac{1.1166 g}{187.77 g/mol}=0.0059466 mol$

$AgNO_3(aq)+Br^-(aq)\rightarrow AgBr(s)+NO_3^{-} (aq)$

According to reaction, 1 mole of AgBr is obtained from 1 mole of bromide ions , then 0.0059466 moles of AgBr will be formed from :

$\frac{1}{1}\times 0.0059466 mol=0.0059466 mol$ of bromide ions

Mass of 0.005939 moles of bromide ions :

0.0059466 mol × 79.90 g/mol = 0.4751 g

Mass of the sample = 0.5781 g

Mass percentage of bromine in sample :

$=\frac{0.4751 g}{0.5781 g}\times 100=82.18\%$

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