A 0.8838-g sample of an ionic compound containing bromide ions and an unknown metal cation is dissolved in water and treated with an excess of AgNO₃. If 1.573 g of a AgBr precipitate forms, what is the percent by mass of Br in the original compound? A) 62.13% B) 75.72% C) 56.19% D) 42.55% E) 37.87%

Question

Quizplus · Accepted Answer

First, we need to determine how many moles of AgBr were formed in the reaction:
moles of AgBr = mass / molar mass = 1.573 g / (107.87 g/mol) = 0.0146 mol AgBr

Since AgBr is a 1:1 ratio compound, we know that moles of AgBr = moles of Br in the original compound.

Next, we can use the mass of the original compound and the moles of Br to find the percent by mass of Br:
moles of Br = 0.0146 mol AgBr
mass of original compound = 0.8838 g

mass percent of Br = (moles of Br x molar mass of Br / mass of original compound) x 100%
mass percent of Br = (0.0146 mol x 79.90 g/mol / 0.8838 g) x 100%
mass percent of Br = 75.72%

Therefore, the answer is B) 75.72%.

A 08838-G Sample of an Ionic Compound Containing Bromide Ions and and an Unknown