Question

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : _________ (nearest integer)
(Given, molar mass in g mol\(^{-1}\) ; O = 16, Mg = 24, P = 31)}

• A. 20
• B. 40
• C. 30
• D. 50

Hint:

The stoichiometric factor for phosphorus in magnesium pyrophosphate is \( 62/222 \). Always use the exact molar masses provided in the question.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Phosphorus estimation is done by converting it into magnesium pyrophosphate (\( Mg_2P_2O_7 \)). The mass of phosphorus in the compound is derived from the stoichiometry of this product.

Step 2: Key Formula or Approach:
\[ %P = \frac{2 \times \text{At. mass of P}}{\text{Molar mass of } Mg_2P_2O_7} \times \frac{\text{Mass of } Mg_2P_2O_7 \text{ formed}}{\text{Mass of compound taken}} \times 100 \]

Step 3: Detailed Explanation:
Molar mass of \( Mg_2P_2O_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 \text{ g/mol} \).
Mass of phosphorus in 222 g of \( Mg_2P_2O_7 \) is 62 g.
Mass of organic compound = 1.00 g.
Mass of magnesium pyrophosphate = 1.79 g.
\[ %P = \frac{62}{222} \times \frac{1.79}{1.00} \times 100 \]
\[ %P \approx 0.2792 \times 1.79 \times 100 \approx 49.99% \]
The nearest integer is 50.

Step 4: Final Answer:
The percentage of phosphorus is 50.