In a Carius determination, 0.55g 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]

Explanation

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

### Step 1: Identify the given data
- Mass of the organic substance = 0.55 g
- Mass of barium sulfate (BaSO₄) = 0.44 g
- Molar mass of barium sulfate (BaSO₄) = 233 g/mol
- Molar mass of sulfur (S) = 32 g/mol

### Step 2: Calculate the mass of sulfur in barium sulfate
The mass of sulfur in barium sulfate can be calculated using the ratio of the molar mass of sulfur to the molar mass of barium sulfate.

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

Substituting the values:

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

Calculating this gives:

\[ \text{Mass of S in BaSO₄} = \left(\frac{32}{233}\right) \times 0.44 \approx 0.060 g \]

### Step 3: Calculate the percentage of sulfur in the organic compound
Now, we can find the percentage of sulfur in the organic compound using the formula:

\[ \text{Percentage of S} = \left(\frac{\text{Mass of S in BaSO₄}}{\text{Mass of organic substance}}\right) \times 100 \]

Substituting the values:

\[ \text{Percentage of S} = \left(\frac{0.060 \, \text{g}}{0.55 \, \text{g}}\right) \times 100 \]

Calculating this gives:

\[ \text{Percentage of S} \approx 10.9\% \]

### Final Answer:
The percentage of sulfur in the given organic compound is approximately **10.9%**.

---