Mass of organic compound (m1) = 0.20 g
Mass of silver bromide formed (m2) = 0.15 g
188(108 + 80)g of AgBr contains 80g of bromine
∴ 0.20 g of AgBr contains
= \(\frac{80\times 0.15}{188}g\) = 0.0638g of Br
Percentage of bromine
= \(\frac{80\times 15\times 100}{188\times 20}\) = 31.92
The percentage of bromine in the given compound = 31.92