In a Carius determination, 0.25g of an organic substance gave 0.44g of barium sulfate. Calculate the percentage of sulphur in the given compound. [Ba=137,S=32,O=16]

Options

A. 24%
B. 15%
C. 19%
D. 22%

Explanation

To calculate the percentage of sulfur in the given organic compound based on the Carius determination, we can follow these steps:

### Step 1: Identify the given data
- Mass of organic substance = 0.25 g
- Mass of barium sulfate (BaSO₄) = 0.44 g
- Molar mass of barium sulfate (BaSO₄) = 233 g/mol
- Mass of sulfur (S) in barium sulfate = 32 g

### Step 2: Calculate the mass of sulfur in the barium sulfate
Using the ratio of sulfur in barium sulfate, we can find the mass of sulfur present in 0.44 g of barium sulfate.

\[ \text{Mass of S in BaSO₄} = \left(\frac{\text{Mass of S}}{\text{Molar mass of BaSO₄}}\right) \times \text{Mass of BaSO₄} \]

Substituting the values:

\[ \text{Mass of S in BaSO₄} = \left(\frac{32 \text{ g}}{233 \text{ g/mol}}\right) \times 0.44 \text{ g} \]

Calculating this gives:

\[ \text{Mass of S in BaSO₄} = \frac{32 \times 0.44}{233} = 0.060 g \]

### Step 3: Calculate the percentage of sulfur in the organic compound
To find the percentage of sulfur in the original organic compound, we use the formula:

\[ \text{Percentage of S} = \left(\frac{\text{Mass of S}}{\text{Mass of organic substance}}\right) \times 100 \]

Substituting the values:

\[ \text{Percentage of S} = \left(\frac{0.060 \text{ g}}{0.25 \text{ g}}\right) \times 100 \]

Calculating this gives:

\[ \text{Percentage of S} = 0.24 \times 100 = 24\% \]

### Final Answer
The percentage of sulfur in the given compound is **24%**.

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