A 0.4505 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 0.7514 g. What is the mass percentage of bromine in the original compound

70.98% is the mass percentage of bromine in the original compound

Explanation:

First, we need to convert mass of AgBr to moles with its molar mass (AgBr: 187.77g/mol). Moles of AgBr = Moles of Br. With molar mass of Br-: 79.904g/mol we can find mass of bromide and mass percentage as follows:

0.7514g AgBr * (1mol / 187.77g) = 0.00400 moles AgBr = Moles of Br-

0.00400 moles Br- * (79.904g / mol) = 0.320g Br- in the sample

Mass percentage is:

0.320g Br- / 0.4505g sample * 100

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