In one experiment, a 0.8024 g sample of the mixture resulted in 0.9978 g of AgBr. Determine the percent (by mass) of Zn in the mixture.

Question

A mixture of ZnBr2 and inert material is analyzed to determine the Zn content. First the mixture is dissolved in water. Then all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate.

ZnBr2(aq) + 2AgNO3(aq) 2AgBr(s) + Zn(NO3)2(aq)

In one experiment, a 0.8024 g sample of the mixture resulted in 0.9978 g of AgBr. Determine the percent (by mass) of Zn in the mixture.

Answer 1

ZnBr₂(aq) + 2AgNO₃(aq) ==> 2AgBr(s) + Zn(NO₃)₂(aq) .. balanced equation

All of the Br in AgBr must have come from the ZnBr₂ since that's the only source of Br. So, first we'll find the moles of ZnBr₂ present.

Molar mass AgBr = 187.8 g / mol

0.9978 g AgBr x 1 mol AgBr / 187.8 g = 5.313x10^-3 moles AgBr

5.313x10^-3 mols AgBr x 1 mol ZnBr₂ / 2 mols AgBr = 2.657x10^-3 mols ZnBr₂

Now that we have the mols of ZnBr₂, we can convert this to grams of Zn

2.657x10^-3 mols ZnBr₂ x 1 mol Zn / 1 mol ZnBr₂ = 2.657x10^-3 mols Zn

2.657x10^-3 mols Zn x 65.39 g Zn / mol = 0.1737 g Zn

Finally we can determine the percent (by mass) of Zn in the original sample that weighed 0.8024 g

0.1737 g Zn / 0.8024 g (x100%) = 21.65% Zn