In a Carius determination, 0.15g 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, we can follow these steps:

### Step 1: Identify the given data
- Mass of the organic substance = 0.15 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 amount of sulfur in the barium sulfate
We know that in 233 g of barium sulfate, there are 32 g of sulfur. We can set up a proportion to find out how much sulfur is present in 0.44 g of barium sulfate.

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

Calculating this gives:

\[ \text{Mass of sulfur} = \frac{32}{233} \times 0.44 \approx 0.060 g \]

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

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

Substituting the values we have:

\[ \text{Percentage of sulfur} = \left(\frac{0.060 \text{ g}}{0.15 \text{ g}}\right) \times 100 \approx 40\% \]

### Final Answer
The percentage of sulfur in the given organic compound is **40%**.

---