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)
[Atomic mass : Silver = 108, Bromine = 80]

In the Carius method for estimation of bromine, the organic compound is burnt in excess silver nitrate, and all the bromine is converted to silver bromide (AgBr), which is weighed.

The molar mass of AgBr = 108 + 80 = 188 g mol$$^{-1}$$.

Moles of AgBr formed: $$n_{\text{AgBr}} = \frac{0.2397}{188} = 1.275 \times 10^{-3}$$ mol

Since each mole of AgBr contains one mole of Br, the mass of bromine in the sample is:

$$m_{\text{Br}} = 1.275 \times 10^{-3} \times 80 = 0.1020 \text{ g}$$

The percentage of bromine in the organic compound is:

$$\% \text{Br} = \frac{m_{\text{Br}}}{m_{\text{compound}}} \times 100 = \frac{0.1020}{0.15} \times 100 = 68.0\%$$

The percentage of bromine in the organic compound is $$\mathbf{68}$$%.