A 0.4657 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 0.8878 g. What is the mass percentage of bromine in the original compound?

Question

Chemistry

Dominique Collier

3 November, 16:05

A 0.4657 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 0.8878 g. What is the mass percentage of bromine in the original compound?

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Answers (1)

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Roland Lowery · Answer 1

The mass percentage of bromine in the original compound is 81,12%

Explanation:

Step 1: Calculate moles AgBr

moles AgBr = mass AgBr / molar mass AgBr

= 0.8878 g / 187.77 g/mol

= 0.00472812 moles AgBr

⇒

Since 1 mol AgBr contains 1 mol Br-

Then the amount of moles Br - in the original sample must also have been 0.00472812 moles

Step 2: Calculating mass Br-

mass Br - = molar mass Br x moles Br-

= 79.904 g/mol x 0.00472812 mol

= 0.377796 g Br-

⇒

There were 0.377796 g Br - in the original sample

Step 3: Calculating mass percentage Br-

⇒mass percentage = actual mass Br - / total mass x 100%

% mass Br = 0.377796 g / 0.4657 g x 100 %

= 81.12%