In a Carius determination, 0.25g of an organic substance gave 0.24g of barium sulfate. Calculate the percentage of sulphur in the given compound. [Ba=137,S=32,O=16]

Options

A. 10%
B. 9.6%
C. 13%
D. 14%

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 mass of barium sulfate (BaSO₄) produced
The mass of barium sulfate given is:
\[ \text{Mass of BaSO}_4 = 0.24 \, \text{g} \]

### Step 2: Calculate the molar mass of barium sulfate
The molar mass of barium sulfate (BaSO₄) can be calculated using the atomic masses:
- Barium (Ba) = 137 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol (there are 4 oxygen atoms in sulfate)

Calculating the molar mass:
\[ \text{Molar mass of BaSO}_4 = 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \, \text{g/mol} \]

### Step 3: Determine the mass of sulfur in the barium sulfate
From the molar mass, we know that 233 g of BaSO₄ contains 32 g of sulfur. To find the mass of sulfur in 0.24 g of BaSO₄, we can set up a proportion:
\[ \text{Mass of S} = \left(\frac{32 \, \text{g S}}{233 \, \text{g BaSO}_4}\right) \times 0.24 \, \text{g BaSO}_4 \]
Calculating this gives:
\[ \text{Mass of S} = \frac{32}{233} \times 0.24 \approx 0.0329 \, \text{g} \]

### Step 4: Calculate the percentage of sulfur in the organic compound
To find the percentage of sulfur in the original organic substance, 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:
\[ \text{Percentage of S} = \left(\frac{0.0329 \, \text{g}}{0.25 \, \text{g}}\right) \times 100 \]
Calculating this gives:
\[ \text{Percentage of S} = 13.16 \approx 13\% \]

### Final Answer
The percentage of sulfur in the given compound is approximately **13%**.

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