Question

A 0.4042 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 0.6797 g.
What is the mass percentage of bromine in the original compound?
%

Answer 1

70.80 % is the mass percentage of bromine in the original compound.

Mass of AgBr = 0.6797 g

Moles of AgBr = 0.6797 g / 187.77 g/mol

= 0.0036198 mol

Silver Nitrate + Bromine --------->  Silver Bromide + Nitrate

According to reaction, 1 mole of AgBr is obtained from 1 mole of bromide ions , then 0.0036198 moles of AgBr will be formed from :

1/1 * 0.0036198  mol = 0.0036198 mole of bromide ions

Mass of 0.0036198 moles of bromide ions :

0.0036198 moles * 79.09 g/ mol

= 0.2862 g

Mass of the sample = 0.4042 g

Mass percentage = 0.2862 * 100 / 0.4041

Mass percentage (%) = 70.80 %

Hence, the mass percentage of bromine in the original compound is 70.80%

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