Question

In phosphorus estimation, 0.5 g of an organic compound gives 0.75 g of \( \mathrm{Mg_2P_2O_7} \). The percentage of phosphorus (P) in the compound is (nearest integer):

Hint:

In gravimetric estimation problems, always remember: \[ %\text{Element} = \frac{\text{Mass of precipitate} \times \text{Mass of element in 1 mole of precipitate}} {\text{Molar mass of precipitate} \times \text{Mass of sample}} \times 100 \] Round off only at the final step if the question asks for the nearest integer.

Answer 1

Correct Answer: 42

Concept: This question is based on the gravimetric estimation of phosphorus. In this method, phosphorus present in the organic compound is quantitatively converted into a stable compound, magnesium pyrophosphate \( \mathrm{Mg_2P_2O_7} \). Knowing the mass of this precipitate allows us to calculate the mass and percentage of phosphorus using stoichiometry.
Step 1: Calculate the molar mass of \( \mathrm{Mg_2P_2O_7} \) Atomic masses: \[ \mathrm{Mg} = 24,\quad \mathrm{P} = 31,\quad \mathrm{O} = 16 \] \[ \text{Molar mass of } \mathrm{Mg_2P_2O_7} = 2(24) + 2(31) + 7(16) \] \[ = 48 + 62 + 112 = 222 \ \text{g mol}^{-1} \]
Step 2: Calculate the mass of phosphorus in one mole of \( \mathrm{Mg_2P_2O_7} \) From the formula, one mole of \( \mathrm{Mg_2P_2O_7} \) contains: \[ 2 \text{ atoms of phosphorus} \] \[ \text{Mass of phosphorus} = 2 \times 31 = 62 \ \text{g} \]
Step 3: Determine the fraction of phosphorus in \( \mathrm{Mg_2P_2O_7} \) \[ \text{Fraction of P} = \frac{62}{222} \]
Step 4: Calculate the mass of phosphorus in 0.75 g of \( \mathrm{Mg_2P_2O_7} \) \[ \text{Mass of P} = 0.75 \times \frac{62}{222} \] \[ = 0.75 \times 0.2793 \] \[ = 0.209 \ \text{g} \]
Step 5: Calculate the percentage of phosphorus in the organic compound Given mass of organic compound = 0.5 g \[ %\text{P} = \frac{\text{Mass of P}}{\text{Mass of sample}} \times 100 \] \[ %\text{P} = \frac{0.209}{0.5} \times 100 \] \[ %\text{P} = 41.8% \]
Step 6: Nearest integer value \[ %\text{P} \approx \boxed{42%} \]