0 . 400   g of an organic compound ( X ) gave 0 . 376   g of AgBr in Carius method for estimation…

The percentage composition of element $=\frac{\mathrm{mass} \mathrm{of} \mathrm{element}}{\mathrm{mass} \mathrm{of} \mathrm{organic} \mathrm{compound}}\times 100$

Moles of $\mathrm{AgBr}=\frac{0.376}{188}$

Moles of $\mathrm{Br}=\frac{0.376}{188}$

Mass of $\mathrm{Br}=\frac{0.376}{188}\times 80$

$\%\text{ of }\mathrm{Br}=\frac{0.376\times 80}{188\times (0.400)}\times 100$

$= 40\%$