In Carius method of estimation of bromine, 1.53 g of an organic compound gives 1 g of AgBr.
Atomic masses:
Find the percentage of bromine in the organic compound.
Solution:
Molar mass of AgBr:
= 108 + 80
= 188 g mol-1
Mass of bromine present in 188 g AgBr:
= 80 g
Therefore, bromine present in 1 g AgBr:
= 80/188
= 0.4255 g
Percentage of bromine in the organic compound:
%Br = (0.4255/1.53) × 100
= 27.81%
Answer:
27.81%