0.2595g of an organic compound yielded quantitatively 0.35g of BaSO4. Find the percentage of sulphur in the compound. (Ba= 137.3, S= 32, O= 16)

Answer

To find the percentage of sulfur in the organic compound based on the amount of barium sulfate (BaSO₄) produced, we can follow these steps:

### Step 1: Calculate the molar mass of BaSO₄
The molar mass of BaSO₄ can be calculated using the atomic masses provided:
- Barium (Ba) = 137.3 g/mol
- Sulfur (S) = 32.0 g/mol
- Oxygen (O) = 16.0 g/mol

The molar mass of BaSO₄ is calculated as follows:
\[ \text{Molar mass of BaSO₄} = \text{Ba} + \text{S} + 4 \times \text{O} = 137.3 + 32.0 + (4 \times 16.0) = 137.3 + 32.0 + 64.0 = 233.3 \, \text{g/mol} \]

### Step 2: Calculate the moles of BaSO₄ produced
Using the mass of BaSO₄ produced (0.35 g), we can calculate the number of moles of BaSO₄:
\[ \text{Moles of BaSO₄} = \frac{\text{mass of BaSO₄}}{\text{molar mass of BaSO₄}} = \frac{0.35 \, \text{g}}{233.3 \, \text{g/mol}} \approx 0.0015 \, \text{mol} \]

### Step 3: Determine the moles of sulfur in the compound
From the stoichiometry of the reaction, we know that 1 mole of BaSO₄ contains 1 mole of sulfur. Therefore, the moles of sulfur in the compound is the same as the moles of BaSO₄ produced:
\[ \text{Moles of S} = 0.0015 \, \text{mol} \]

### Step 4: Calculate the mass of sulfur in the compound
Now, we can calculate the mass of sulfur using its molar mass:
\[ \text{Mass of S} = \text{moles of S} \times \text{molar mass of S} = 0.0015 \, \text{mol} \times 32.0 \, \text{g/mol} = 0.048 \, \text{g} \]

### Step 5: Calculate the percentage of sulfur in the organic compound
Finally, we can find the percentage of sulfur in the original organic compound:
\[ \text{Percentage of S} = \left( \frac{\text{mass of S}}{\text{mass of organic compound}} \right) \times 100 = \left( \frac{0.048 \, \text{g}}{0.2595 \, \text{g}} \right) \times 100 \approx 18.46\% \]

### Final Answer
The percentage of sulfur in the organic compound is approximately **18.46%**.

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