In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)
[Atomic mass : Ag = 108, Br = 80]

Solution

Chemistry ma’am · 4 steps

All the bromine ends up inside AgBr, so find what fraction of that precipitate is bromine.

1. Where bromine goes. In the Carius method the compound is heated with fuming HNO₃ and AgNO₃. Every bromine atom ends up in the AgBr precipitate, so the bromine in 0.188 g of AgBr is all the bromine the sample had.
2. Bromine in AgBr. Molar mass of AgBr = 108 + 80 = 188 g mol⁻¹. So 80/188 of the precipitate's mass is bromine.

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

3. Put in numbers. Mass of Br = 80/188 × 0.188 g = 0.08 g. Then % Br = 0.08/0.2 × 100 = 40.
4. So. The compound contains 40% bromine.

The common slip: 94. Dividing the whole mass of AgBr by the sample mass, without taking the 80/188 bromine fraction, gives 94, not 40.

The idea to keep. In any gravimetric estimation, multiply the precipitate's mass by (atomic mass of the element ÷ molar mass of the precipitate) before dividing by the sample mass.

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