Question

In Carius method of estimation of halogen, 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound.
(Molar masses: AgBr = 188 g mol^–1; Br = 80 g mol^–1)

• A. 34.04%
• B. 40.04%
• C. 36.03%
• D. 38.04%

Answer 1

The Correct Option is A

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

1. First, we determine the mass of bromine present in the 0.36 g of AgBr formed.
2. The molar mass of AgBr is given as 188 g/mol. The reaction for the formation of AgBr from bromine is: Br + Ag \rightarrow AgBr
3. From this equation, 1 mole of AgBr (188 g) contains 1 mole of Br (80 g).
4. We use a ratio to find the mass of bromine in 0.36 g of AgBr:
   \(\text{Mass of Br} = \frac{80}{188} \times 0.36\, \text{g}\)
5. Calculating the above expression:
   \(\text{Mass of Br} = \frac{80 \times 0.36}{188}\)
6. Solving, we find:
   \(\text{Mass of Br} = \frac{28.8}{188} = 0.15319\, \text{g}\)
7. Next, we calculate the percentage of bromine in the original organic compound (0.45 g):
   \(\text{Percentage of bromine} = \left(\frac{0.15319}{0.45}\right) \times 100\%)\)
8. Performing the calculation:
   \(\text{Percentage of bromine} = 34.04\%\)

Thus, the percentage of bromine in the organic compound is 34.04%.