Question

In Carius method of estimation of halogens $250\, mg$ of an organic compound gave $141\, mg$ of $AgBr$. The percentage of bromine in the compound is (atomic mass $Ag\, =\, 108, Br \,= \,80$)

• A. 24
• B. 36
• C. 48
• D. 60

Answer 1

The Correct Option is A

To determine the percentage of bromine in the compound using the Carius method, we need to follow these steps:

1. First, understand that in the Carius method, halogens in an organic compound are converted to their corresponding silver halides (here AgBr: silver bromide) by heating with fuming nitric acid.

2. Given: The mass of the organic compound is 250 \, \text{mg} and it produced 141 \, \text{mg} of \text{AgBr}.

3. The molar masses are: Ag = 108 \, \text{g/mol} and Br = 80 \, \text{g/mol}.

   The molar mass of \text{AgBr} = 108 + 80 = 188 \, \text{g/mol}.

4. Determine the moles of \text{AgBr} formed:

   \text{Moles of AgBr} = \dfrac{141 \, \text{mg}}{188 \, \text{mg/mmol}} = \dfrac{141}{188} \times 10^{-3} \, \text{mol}

5. Since the moles of bromine match those of \text{AgBr} from the reaction stoichiometry (1:1), the mass of bromine is:

   \text{Mass of Br} = \dfrac{141}{188} \times 80 \, \text{mg}

6. Convert the mass of bromine to grams for percentage calculation:

   \text{Mass of Br} = \dfrac{141}{188} \times 80 \, \text{mg} \approx 60 \, \text{mg}

7. Finally, calculate the percentage of bromine in the compound:

   \text{Percentage of Br} = \left( \dfrac{60}{250} \right) \times 100\% = 24\%

The percentage of bromine in the compound is therefore 24%. This matches the correct answer.