Question

In an estimation of sulphur by Carius method \(0.2 \text{ g}\) of the substance gave \(0.6 \text{ g}\) of \(BaSO_4\). The percentage of sulphur in the substance is ____ %.
(Given molar mass in \(\text{g mol}^{-1}\) S : 32, \(BaSO_4 : 231\))}

Answer 1

Correct Answer: 42

Step 1: Understanding the Question:
In the Carius method, all sulfur in an organic compound is converted to sulfuric acid, which is then precipitated as barium sulfate (\(BaSO_4\)). We calculate the mass of sulfur in the precipitate to find its percentage in the original compound.
Step 2: Key Formula or Approach:
\[ \text{% of Sulfur} = \frac{\text{Atomic mass of S}}{\text{Molar mass of } BaSO_4} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of substance}} \times 100 \]
Step 3: Detailed Explanation:
Given:
Mass of substance = \(0.2 \text{ g}\)
Mass of \(BaSO_4\) = \(0.6 \text{ g}\)
Molar mass of \(BaSO_4\) = \(231 \text{ g/mol}\)
Atomic mass of \(S\) = \(32 \text{ g/mol}\)
Calculation:
\[ \text{% S} = \frac{32}{231} \times \frac{0.6}{0.2} \times 100 \]
\[ \text{% S} = \frac{32}{231} \times 3 \times 100 \]
\[ \text{% S} = \frac{9600}{231} \]
\[ \text{% S} = 41.558 \dots \approx 42 \]
Step 4: Final Answer:
The percentage of sulfur is 42.