1.6 g of an organic compound gave 2.6 g of magnesium pyrophosphate. The percentage of phosphorus in the compound is

Options

A. 0.4538
B. 0.5438
C. 0.3776
D. 0.1902

Explanation

To calculate the percentage of phosphorus in the organic compound, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the mass of magnesium pyrophosphate (Mg2P2O7)**:
- Given mass of magnesium pyrophosphate = 2.6 g.

2. **Determine the molar mass of magnesium pyrophosphate (Mg2P2O7)**:
- Molar mass of magnesium (Mg) = 24 g/mol.
- Molar mass of phosphorus (P) = 31 g/mol.
- Molar mass of oxygen (O) = 16 g/mol.
- Molar mass of Mg2P2O7 = (2 × 24) + (2 × 31) + (7 × 16)
- = 48 + 62 + 112
- = 222 g/mol.

3. **Calculate the mass of phosphorus in magnesium pyrophosphate**:
- Since one mole of Mg2P2O7 contains 2 moles of phosphorus, the mass of phosphorus in one mole of Mg2P2O7 = 2 × 31 g = 62 g.

4. **Calculate the percentage of phosphorus in the organic compound**:
- The formula for percentage of phosphorus is:
\[ \text{Percentage of P} = \left( \frac{\text{mass of P in Mg}_2\text{P}_2\text{O}_7}{\text{molar mass of Mg}_2\text{P}_2\text{O}_7} \right) \times \text{mass of Mg}_2\text{P}_2\text{O}_7 \div \text{mass of organic compound} \times 100 \]
- Substituting the values:
\[ \text{Percentage of P} = \left( \frac{62 \, \text{g}}{222 \, \text{g/mol}} \right) \times 2.6 \, \text{g} \div 1.6 \, \text{g} \times 100 \]

5. **Perform the calculations**:
- Calculate the fraction:
\[ \frac{62}{222} \approx 0.279 \]
- Now, substitute and calculate:
\[ \text{Percentage of P} = 0.279 \times 2.6 \div 1.6 \times 100 \]
- Calculate \(0.279 \times 2.6 \approx 0.726\).
- Now divide by 1.6:
\[ \frac{0.726}{1.6} \approx 0.45375 \]
- Finally, multiply by 100:
\[ 0.45375 \times 100 \approx 45.375\% \]

6. **Final Result**:
- The percentage of phosphorus in the organic compound is approximately **45.38%**.