The Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of $\mathrm{AgBr}$. The percentage of bromine in the compound is: (atomic mass, $\mathrm{Ag}=108\mathrm{u},\mathrm{Br}=80\mathrm{u})$

The percentage of Br is $\left(\frac{M_{\mathrm{Br}}}{M_{\mathrm{AgBr}}}\right)\left(\frac{m_{\mathrm{AgBr}}}{m_{\text{compound }}}\right)\left(100\right)=\left(\frac{80\mathrm{u}}{188\mathrm{u}}\right)\left(\frac{141\mathrm{mg}}{250\mathrm{mg}}\right)\left(100\right)=24\mathrm{\%}$The molar mass of AgBr is 108+80=188g/mol .

Hence, 141 mg of AgBr will contain $141\mathrm{mg}\times \frac{80\mathrm{g}}{188\mathrm{g}}=60\mathrm{mg}$ of bromine.

Hence, the percentage of bromine in the compound is $\frac{60\mathrm{mg}}{250\mathrm{mg}}\times 100=24\mathrm{\%}$.