In an estimation of bromine by Carius method, $$1.6\,\text{g}$$ of an organic compound gave $$1.88\,\text{g}$$ of AgBr. The mass percentage of bromine in the compound is........ (Atomic mass, Ag = 108, Br = 80 $$\text{g mol}^{-1}$$)

We have to find the mass percentage of bromine present in the organic compound whose mass is given as $$1.6\;\text{g}$$. During the Carius estimation the bromine present in the compound is completely converted into silver bromide, $$\text{AgBr}$$, whose mass is reported to be $$1.88\;\text{g}$$. Hence the whole problem boils down to: how much bromine is contained in $$1.88\;\text{g}$$ of $$\text{AgBr}$$, and what fraction of the original $$1.6\;\text{g}$$ sample does that bromine represent?

First we need the molar mass of silver bromide. Using the given atomic masses

$$M_{\text{Ag}} = 108\;\text{g mol}^{-1}, \qquad M_{\text{Br}} = 80\;\text{g mol}^{-1}$$

we write

$$M_{\text{AgBr}} = M_{\text{Ag}} + M_{\text{Br}} = 108 + 80 = 188\;\text{g mol}^{-1}.$$

Now, the number of moles of $$\text{AgBr}$$ actually formed is obtained from the basic relation

$$n = \dfrac{m}{M},$$

where $$m$$ is mass and $$M$$ is molar mass. Substituting the known quantities,

$$n_{\text{AgBr}} = \dfrac{1.88\;\text{g}}{188\;\text{g mol}^{-1}} = 0.01\;\text{mol}.$$

In the silver bromide crystal lattice each formula unit contains exactly one bromine atom, so the amount (in moles) of bromine captured is numerically equal to the number of moles of $$\text{AgBr}$$ formed. Thus

$$n_{\text{Br}} = n_{\text{AgBr}} = 0.01\;\text{mol}.$$

To get the actual mass of bromine present we again employ

$$m = n \times M.$$

Hence, substituting,

$$m_{\text{Br}} = 0.01\;\text{mol} \times 80\;\text{g mol}^{-1} = 0.8\;\text{g}.$$

We now know that $$0.8\;\text{g}$$ of bromine came from the original $$1.6\;\text{g}$$ sample. The percentage composition by mass is therefore computed from the standard formula

$$\%\text{Br} = \dfrac{m_{\text{Br}}}{m_{\text{sample}}}\times 100.$$

Substituting the values,

$$\%\text{Br} = \dfrac{0.8\;\text{g}}{1.6\;\text{g}}\times 100 = 50.$$

Hence, the correct answer is Option 50.