When 0.15 g of an organic compound was analyzed using carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is . (Nearest integer)

$$\begin{gathered} \text{Bromine }=\frac{0.2397\times 80}{188}=0.102 \\ Percent of \text{Br }=\frac{{\displaystyle 0.102}}{{\displaystyle 0.15}}\times 100=68 \end{gathered}$$

OR

We have mass of Bromine which is 80u

We have mass of AgBr which is 188u

We have weight of AgBr which is 0.12g

We have weight of organic compound which is 0.15g

Now, equation which follows is:

Percentage of Bromine = Molar Mass of AgBr,

Molar Mass of Bromine​

×Weight of organic compound Weight of AgBr​×100

Therefore, Percentage of the bromine

=18880​×0.150.12​×100

=68%