Question

0.20 g of an organic compound gave 0.12 g of AgBr. By using Carius method, the % of bromine in the compound will be:

• A. 34.06%
• B. 44.04%
• C. 54%
• D. 25%

Hint:

In the Carius method, the % of an element in an organic compound is calculated by finding the moles of the halide (AgBr in this case) formed and relating it to the moles and mass of the element in the compound.

Answer 1

The Correct Option is D

Given,
Mass of an organic compound \(= 0.20\, \text{g}\)
Mass of AgBr \(= 0.12\, \text{g}\)
Molecular mass of AgBr \(= 188\, \text{g mol}^{-1}\)
\(188\, \text{g}\) of AgBr contains \(80\, \text{g}\) of bromine.
\[ \therefore 0.12\, \text{g of AgBr will contain } = \frac{80}{188} \times 0.12 \] \[ = 0.05\, \text{g of bromine} \] \[ \therefore \text{Percentage of bromine} = \frac{0.05}{0.20} \times 100 \] \[ = 25\% \]