A 12.5843 g sample of ZrBr4 was dissolved and, after several chemical steps, all of the combined Bromine was precipitated as AgBr. The silver content of the AgBr was found to be 13.2160 g. Assume the atomic weights of silver and bromine to be 107.870 and 79.909 respectively. What value was obtained for the atomic weight of Zr for this experiment?

Answer 1

107.870 + 79.909 = 187.779
%Ag = [(107.870 / 187.779](100) = 57.4452%
%Br = [(79.909/187.779)](100) = 42.5548%
(13.2160 g AgBr)(0.42.5548) = 5.624g Br in AgBr

[(5.624g Br)/(12.5843g ZrBr4)](100) = 44.691% Br in ZrBr4
100.000 - 44.691 = 55.309% Zr in ZrBr4
Let the atomic mass of Zr = x
The formula mass of ZrBr4 = x + (4)(79.909), or
ZrBr4 = x + 319.636

Set up the equation:
x / (x + 319.636) = 0.44691
Solve for x to get the atomic mass of Zr based on the information given here.