Question

In Carius method, of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (atomic mass Ag = 108; Br =80)
A. 24
B. 36
C. 48
D. 60

Answer 1

Hint:. Carius method is used to determine the amount of halogen in the chemical compounds quantitatively. There is a formula to calculate the percentage of the halogen in the organic compound and it is as follows.
\[\text{Percentage of halogen (X)=}\dfrac{\text{Atomic weight of X }\!\!\times\!\!\text{ weight of AgX }\!\!\times\!\!\text{ 100}}{\text{Molecular weight of AgX }\!\!\times\!\!\text{ Weight of the organic compound}}\]

Complete step by step answer:
- We have to calculate the percentage of bromine which is liberated from 250 mg of the organic compound.
 - Weight of the given compound = 250 mg
- Weight of the AgBr obtained from organic compound = 141 mg
- Molecular weight of the bromine (X) = 80
- Molecular weight of the silver bromide (AgBr) = 188
- Now substitute all the known values in the below equation to get the percentage of Bromine in 141 mg of AgBr.
\[\begin{align}
 & \text{Percentage of Bromine (X)=}\dfrac{\text{Atomic weight of bromine }\!\!\times\!\!\text{ weight of AgBr }\!\!\times\!\!\text{ 100}}{\text{Molecular weight of AgBr }\!\!\times\!\!\text{ Weight of the organic compound}} \\
& \text{Percentage of Bromine (X)=}\dfrac{\text{80 }\!\!\times\!\!\text{ 141 }\!\!\times\!\!\text{ 100}}{\text{188 }\!\!\times\!\!\text{ 250}} \\
 & \text{Percentage of Bromine (X)=}\dfrac{\text{1128000}}{\text{47000}}\text{=24 %}\!\!\!\!\text{ } \\
\end{align}\]
- Therefore the percentage of bromine present in 141 mg silver bromide (AgBr) which is obtained by 250 mg of an organic compound is 24%.
So, the correct answer is “Option A”.

Note: We can calculate the percentage of any halogen present in the respective compounds by using the Carius method. In the Carius method a known mass of organic compound is heated by adding nitric acid and silver nitrate chemicals in a glass tube called Carius tube. The carbon and hydrogen present in the organic compound are converted into carbon dioxide and water and the halogen released from the organic compound reacts with silver nitrate and forms respective silver halide.