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 :
$$\left( {{\text{at}}{\text{.}}\,{\text{mass}}\,Ag = 108;\,Br = 80} \right)$$

Solution :
$$\eqalign{ & {\text{Mass of substance = 250 mg = 0}}{\text{.250 g}} \cr & {\text{Mass of AgBr = 141 mg = 0}}{\text{.141 g}} \cr & {\text{1 mole of AgBr = 1 g atom of Br}} \cr & {\text{188 g of AgBr = 80 g of Br}} \cr & \therefore \,\,{\text{188 g of AgBr contain bromine = 80 g}} \cr & {\text{0}}{\text{.141 g of AgBr contain bromine}} = \frac{{80}}{{188}} \times 0.141 \cr} $$
This much amount of bromine present in 0.250 g of organic compound
$$\therefore \,\,\% \,\,{\text{of}}\,{\text{bromine = }}\frac{{80}}{{188}} \times \frac{{0.414}}{{0.250}} \times 100 = 24\% $$