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)

Solution

Chemistry ma’am · 4 steps

All the bromine ends up in AgBr, so find how much of that 0.36 g is bromine and compare it with 0.45 g.

1. Where Br goes. In the Carius method the compound is heated with fuming HNO₃ and AgNO₃. Every bromine atom in the compound is caught as AgBr, so the bromine in the precipitate equals the bromine in the sample.
2. Mass fraction. One mole of AgBr (188 g) contains one mole of Br (80 g). So bromine makes up 80/188 of the precipitate's mass.

   % Br = 80/188 × m_(AgBr)/m_(compound) × 100

3. Put in numbers. Mass of Br = 80/188 × 0.36 = 28.8/188 = 0.1532 g. Then 0.1532/0.45 × 100 = 34.04.
4. So. The compound contains 34.04% bromine, which is option (a).

The idea to keep. For any precipitate method, the element's share of the precipitate is (atomic mass of element) divided by (molar mass of precipitate). Multiply by the precipitate mass, then divide by the sample mass.

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