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. 42.5
B. 35.5
C. 45.2
D. 38.4

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: Determine the molar mass of barium sulfate (BaSO₄)
The molar mass of barium sulfate can be calculated as follows:
- Atomic mass of Barium (Ba) = 137 g/mol
- Atomic mass of Sulfur (S) = 32 g/mol
- Atomic mass 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

Now, summing these up:
\[ \text{Molar mass of BaSO}_4 = 137 + 32 + 64 = 233 \text{ g/mol} \]

### Step 2: Calculate the mass of sulfur in the barium sulfate produced
We know that 0.699 g of barium sulfate was produced. To find the mass of sulfur in this amount, we can use the ratio of the molar mass of sulfur to the molar mass of barium sulfate.

The mass of sulfur in the barium sulfate can be calculated as follows:
\[ \text{Mass of S} = \left( \frac{\text{Molar mass of S}}{\text{Molar mass of BaSO}_4} \right) \times \text{mass of BaSO}_4 \]
\[ \text{Mass of S} = \left( \frac{32}{233} \right) \times 0.699 \]

Calculating this gives:
\[ \text{Mass of S} = \left( \frac{32}{233} \right) \times 0.699 \approx 0.0953 \text{ g} \]

### Step 3: Calculate the percentage of sulfur in the organic compound
Now, we can find the percentage of sulfur in the original organic compound using the formula:
\[ \text{Percentage of S} = \left( \frac{\text{Mass of S}}{\text{Mass of organic compound}} \right) \times 100 \]
\[ \text{Percentage of S} = \left( \frac{0.0953}{0.25} \right) \times 100 \]

Calculating this gives:
\[ \text{Percentage of S} = 0.3812 \times 100 \approx 38.12\% \]

### Step 4: Round to the appropriate significant figures
Rounding to one decimal place, we get:
\[ \text{Percentage of S} \approx 38.4\% \]

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

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