Question

A 5.18-g sample of a mixture of MgBr2(s) and sucrose (C12H22O11) is dissolved in water and all the bromide is precipitated as AgBr(s). If the AgBr(s) precipitate has a mass of 7.65 g, what is the mass percent of MgBr2(s) in the mixture?

Answer 1

Answer: 62.8%

Explanation:

step 1: Calculate the number of moles (n) for AgBr.

n(AgBr) = mass(AgBr) / Molar mass(AgBr)

             = (7.65g) / (187.77g/mol)

             = 0.0407 mol

step 2: Determine the number of moles of Br in AgBr.

1 mol AgBr = 1 mol of Br

therefore; n(Br) = 0.0407mol

step 3: Calculate the mass of Br precipitated.

m(Br) = n(Br) × Molar mass(Br)

          = (0.0407 mol) / (79.904g/mol)

          = 3.255g

step 4: Calculate the mass percent of MgBr₂.

%MgBr₂ = mass(Br) / mass(MgBr₂ + C₁₂H₂₂O₁₁) × 100

              = (3.225g / 5.18g) × 100

              = 62.8%