In a Carius tube, 0.25 g of an organic compound gave 0.699 g of barium sulphate. What is the percentage of sulphur in the compound? (Atomic weight of Ba = 137)

Options

A. 0.425
B. 0.355
C. 0.452
D. 0.384

Answer

To find the percentage of sulfur in the organic compound based on the amount of barium sulfate produced, we can follow these steps:

### Step 1: Calculate the molar mass of barium sulfate (BaSO₄)

The molar mass of barium sulfate can be calculated as follows:
- Atomic weight of Barium (Ba) = 137 g/mol
- Atomic weight of Sulfur (S) = 32 g/mol
- Atomic weight of Oxygen (O) = 16 g/mol

The formula for barium sulfate is BaSO₄, which consists of:
- 1 Ba = 137 g/mol
- 1 S = 32 g/mol
- 4 O = 4 × 16 g/mol = 64 g/mol

Adding these together:
\[ \text{Molar mass of BaSO₄} = 137 + 32 + 64 = 233 \text{ g/mol} \]

### Step 2: Determine the amount of sulfur in the barium sulfate produced

From the question, we know that 0.699 g of barium sulfate was produced. We need to find out how much sulfur is in this amount.

Using the ratio of sulfur's molar mass to the molar mass of barium sulfate:
\[ \text{Mass of S in BaSO₄} = \left( \frac{32 \text{ g/mol}}{233 \text{ g/mol}} \right) \times 0.699 \text{ g} \]

Calculating this:
\[ \text{Mass of S} = \frac{32}{233} \times 0.699 \approx 0.096 \text{ g} \]

### Step 3: Calculate the percentage of sulfur in the organic compound

Now, we need to find the percentage of sulfur in the original organic compound. We know that the mass of the organic compound is 0.25 g.

Using the formula for percentage:
\[ \text{Percentage of S} = \left( \frac{\text{Mass of S}}{\text{Mass of organic compound}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage of S} = \left( \frac{0.096 \text{ g}}{0.25 \text{ g}} \right) \times 100 \]

Calculating this:
\[ \text{Percentage of S} = 0.384 \times 100 = 38.4\% \]

### Final Answer:
The percentage of sulfur in the organic compound is **38.4%**.

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