In a Carius determination, 0.15g 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. 20.5%
B. 21.9%
C. 25.6%
D. 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 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 (BaSO4).
The molar mass of barium sulfate 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)

So, the molar mass of BaSO4 is:
\[ \text{Molar mass of BaSO}_4 = 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \, \text{g/mol} \]

### Step 3: Determine the mass of sulfur in the barium sulfate.
Using the ratio of the mass of sulfur to the molar mass of barium sulfate, we can find the mass of sulfur in the 0.24 g of BaSO4:
\[ \text{Mass of S} = \left( \frac{32 \, \text{g}}{233 \, \text{g}} \right) \times 0.24 \, \text{g} \]
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.
Now, we can find the percentage of sulfur in the original organic compound:
\[ \text{Percentage of S} = \left( \frac{\text{Mass of S}}{\text{Mass of organic compound}} \right) \times 100 \]
Substituting the values we have:
\[ \text{Percentage of S} = \left( \frac{0.0329 \, \text{g}}{0.15 \, \text{g}} \right) \times 100 \approx 21.93\% \]

### Final Answer:
The percentage of sulfur in the given compound is approximately **21.9%**.

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