Question

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g mol⁻¹). Molar mass of barium sulphate is 233 g mol⁻¹.

• A. 16.48\%
• B. 10.30\%
• C. 21.97\%
• D. 4.55\%

Hint:

For quantitative analysis problems (like Carius, Dumas, Kjeldahl methods), memorizing the final formula is key to saving time. The core principle is always based on stoichiometry: relating the mass of the final product (like BaSO₄ or AgX) to the mass of the element of interest in the original sample.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:

The question asks us to calculate the percentage of sulphur in an organic compound using data obtained from the Carius method. In this method, sulphur present in the compound is converted completely into barium sulphate (BaSO₄), which is then weighed to determine the sulphur content.

Step 2: Formula Used:

The percentage of sulphur is calculated using the formula: \[ \% \text{S} = \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4 \text{ formed}}{\text{Mass of organic compound taken}} \times 100 \]

Step 3: Substitution of Given Values:

Given data:
Mass of organic compound = 0.75 g
Mass of BaSO₄ formed = 1.2 g
Atomic mass of sulphur (S) = 32 g mol⁻¹
Molar mass of BaSO₄ = 233 g mol⁻¹

Substituting into the formula: \[ \% \text{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100 \]

First, evaluate each term: \[ \frac{32}{233} \approx 0.13734 \] \[ \frac{1.2}{0.75} = 1.6 \]

Now multiply: \[ \% \text{S} = 0.13734 \times 1.6 \times 100 \] \[ \% \text{S} = 21.97\% \]

Step 4: Final Answer:

The percentage of sulphur present in the given organic compound is approximately 21.97%.
This corresponds to option (C).