Molar mass of AgBr = 108 + 80 = 188 g mol-1
188 g AgBr contains 80 g bromine
0.12 g AgBr contains= 80 X 0.12/188 g bromine
Percentage of bromine=80 X0.12 X100/188 X 0.15
= 34.04%
In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (At. Mass Ag = 108; Br = 80)