In Carius method of estimation of halogen, 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which of these is the correct structure of the compound?

Solution

Chemistry ma’am · 3 steps

Turn the masses into a bromine percentage, then find the structure whose formula gives the same percentage.

1. Percentage of bromine. The Carius result gives the bromine mass directly. Divide it by the sample mass.

   % Br = 0.08/0.172 × 100 = 46.5%

2. Check each option. For each structure, find the molar mass and the mass of bromine in one mole (80 g per Br atom). Then compare its bromine percentage with 46.5%.

   Option	Formula	Molar mass	% Br
   (a) bromoethane	C₂H₅Br	109 g mol⁻¹	80/109 × 100 = 73.4%
   (b) bromomethane	CH₃Br	95 g mol⁻¹	80/95 × 100 = 84.2%
   (c) 4-bromoaniline	C₆H₆BrN	172 g mol⁻¹	80/172 × 100 = 46.5%
   (d) 2,4-dibromoaniline	C₆H₅Br₂N	251 g mol⁻¹	160/251 × 100 = 63.7%

3. So. Only 4-bromoaniline matches. Its molar mass is 172 g mol⁻¹, so 0.172 g is 0.001 mol, and that holds 0.001 mol of Br, which is exactly 0.08 g. The answer is (c).

The idea to keep. In Carius questions, compute the halogen percentage first. Then match it against the halogen percentage of each candidate formula. Count every halogen atom when you do.

Answer from the NTA final answer key. Solution by PrepWiser.