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: 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%