In a Carius determination, 0.25g of an organic substance gave 0.34g of barium sulfate. Calculate the percentage of sulphur in the given compound. [Ba=137,S=32,O=16]

Options

A. 18.68%
B. 21%
C. 22%
D. 16.8%

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 organic substance = 0.25 g
- Mass of barium sulfate (BaSO₄) = 0.34 g

### Step 2: Calculate the molar mass of barium sulfate (BaSO₄)
The molar mass of BaSO₄ can be calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (there are 4 oxygen atoms)

Calculating the total molar mass:
\[ \text{Molar mass of BaSO₄} = 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \text{ g/mol} \]

### Step 3: Determine the amount of sulfur in the barium sulfate
In 233 g of BaSO₄, there are 32 g of sulfur. To find the amount of sulfur in 0.34 g of BaSO₄, use a proportion:
\[ \text{Mass of sulfur} = \left(\frac{32 \text{ g S}}{233 \text{ g BaSO₄}}\right) \times 0.34 \text{ g BaSO₄} \]
Calculating this gives:
\[ \text{Mass of sulfur} = \frac{32}{233} \times 0.34 \approx 0.0467 \text{ g S} \]

### Step 4: Calculate the percentage of sulfur in the organic compound
To find the percentage of sulfur in the organic compound, use the formula:
\[ \text{Percentage of sulfur} = \left(\frac{\text{Mass of sulfur}}{\text{Mass of organic substance}}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage of sulfur} = \left(\frac{0.0467 \text{ g}}{0.25 \text{ g}}\right) \times 100 \approx 18.68\% \]

### Final Answer
The percentage of sulfur in the given compound is approximately **18.68%**.

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