
In Carius' method of estimation of halogens, 0.15g of an organic compound gave 0.12g of AgBr. What is the percentage of bromine in the compound?
[A] 68.08%
[B] 34.04%
[C] 42.1%
[D] 50%
Answer
228.3k+ views
Hint: The Carius method is carried out for the estimation of the halogen presence in organic compounds. This method is a quantitative analysis and is used in analytical chemistry. There is a formula which we can use to estimate the percentage of halogens through this method. The formula is-\[Percentage\text{ }of\text{ halogen }=\frac{molar\text{ }mass\text{ }of\text{ halogen}}{molar\text{ }mass\text{ }of\text{ silver halide}}\times \frac{weight\text{ }of\text{ silver halide obtained}}{weight\text{ }of\text{ }organic\text{ }compound}\times 100\]
Complete Step by Step Solution: The Carius halogen method is generally used by us to determine the quantity of halogens in a chemical compound.
In this method, we heat a known mass of an organic compound with fuming nitric acid in presence of silver nitrate in a hard glass tube. The carbon and hydrogen present in the compound is oxidised forming carbon dioxide and water and the halogen compound with the silver nitrate forms a precipitate of silver halide which is dried and weighed and is used for the quantitative calculation of the halogen.
Here, the weight of the organic compound = 0.15g
Weight of the silver bromide precipitate = 0.12g
We know that, atomic mass of bromine = 80
Atomic mass of silver bromide = 188
Therefore, according to the Carius method, we can write that-
Percentage of bromine =$\frac{molar\text{ }mass\text{ }of\text{ }bromine}{molar\text{ }mass\text{ }of\text{ AgBr}}\times \frac{weight\text{ }of\text{ AgBr}}{weight\text{ }of\text{ }organic\text{ }compound}\times 100$
Putting the values in the above formula, we will get-
Percentage of bromine =$\frac{80}{188}\times \frac{0.12}{0.15}\times 100=34.04%$
Therefore, the correct answer is option [B] 34.04%.
Note: In this method, we are adding nitric acid for the oxidation of carbon and hydrogen into carbon dioxide, which will evolve as gas and water, which can be removed upon heating. Nitric acid does not affect any halogen. However, concentrated nitric acid may oxidise iodine to iodic acid but a very high temperature is needed for that.
Complete Step by Step Solution: The Carius halogen method is generally used by us to determine the quantity of halogens in a chemical compound.
In this method, we heat a known mass of an organic compound with fuming nitric acid in presence of silver nitrate in a hard glass tube. The carbon and hydrogen present in the compound is oxidised forming carbon dioxide and water and the halogen compound with the silver nitrate forms a precipitate of silver halide which is dried and weighed and is used for the quantitative calculation of the halogen.
Here, the weight of the organic compound = 0.15g
Weight of the silver bromide precipitate = 0.12g
We know that, atomic mass of bromine = 80
Atomic mass of silver bromide = 188
Therefore, according to the Carius method, we can write that-
Percentage of bromine =$\frac{molar\text{ }mass\text{ }of\text{ }bromine}{molar\text{ }mass\text{ }of\text{ AgBr}}\times \frac{weight\text{ }of\text{ AgBr}}{weight\text{ }of\text{ }organic\text{ }compound}\times 100$
Putting the values in the above formula, we will get-
Percentage of bromine =$\frac{80}{188}\times \frac{0.12}{0.15}\times 100=34.04%$
Therefore, the correct answer is option [B] 34.04%.
Note: In this method, we are adding nitric acid for the oxidation of carbon and hydrogen into carbon dioxide, which will evolve as gas and water, which can be removed upon heating. Nitric acid does not affect any halogen. However, concentrated nitric acid may oxidise iodine to iodic acid but a very high temperature is needed for that.
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