Principle of Carius method for bromine

In Carius method, bromine present in the organic compound is converted into silver bromide, $\mathrm{AgBr}$.

So, from the mass of $\mathrm{AgBr}$ obtained, we first find the mass of bromine present.

Molar mass of $\mathrm{AgBr}$

Given:

• Atomic mass of Ag $= 108$
• Atomic mass of Br $= 80$

Therefore, $M(\mathrm{AgBr}) = 108 + 80 = 188$

Mass of bromine in $0.2397\,\text{g}$ of $\mathrm{AgBr}$

In $188\,\text{g}$ of $\mathrm{AgBr}$, mass of bromine $= 80\,\text{g}$.

Hence in $0.2397\,\text{g}$ of $\mathrm{AgBr}$, $\text{mass of Br} = \frac{80}{188} \times 0.2397$

$\text{mass of Br} = 0.102\,\text{g} \quad (\text{approximately})$

Percentage of bromine in the organic compound

Mass of organic compound taken $= 0.15\,\text{g}$

Therefore, $\%\text{Br} = \frac{0.102}{0.15} \times 100$

$\%\text{Br} = 68\%$

Nearest integer

$\boxed{68}$