Question

1 g of an organic compound produces 1.49 g of \( \mathrm{Mg_2P_2O_7} \). Determine the percentage of phosphorus (P).

Hint:

For gravimetric estimation problems: \[ %\text{Element} = \frac{\text{Mass of precipitate} \times \text{Mass of element in 1 mole of precipitate}} {\text{Molar mass of precipitate} \times \text{Mass of sample}} \times 100 \]

Answer 1

Correct Answer: 41.6

Concept: This problem is based on gravimetric analysis of phosphorus. In gravimetric estimation, an element present in a compound is converted quantitatively into a stable compound of known composition. Here, phosphorus present in the organic compound is finally obtained as magnesium pyrophosphate, \( \mathrm{Mg_2P_2O_7} \). Once the mass of this compound is known, the mass of phosphorus can be calculated using stoichiometry.
Step 1: Write the formula and calculate molar mass of \( \mathrm{Mg_2P_2O_7} \) \[ \mathrm{Mg_2P_2O_7} \] Molar 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: Determine how much phosphorus is present in one mole of \( \mathrm{Mg_2P_2O_7} \) From the formula, each 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: Find the mass fraction of phosphorus in \( \mathrm{Mg_2P_2O_7} \) \[ \text{Fraction of P} = \frac{\text{Mass of P}}{\text{Molar mass of } \mathrm{Mg_2P_2O_7}} = \frac{62}{222} \]
Step 4: Calculate the mass of phosphorus in 1.49 g of \( \mathrm{Mg_2P_2O_7} \) \[ \text{Mass of P} = 1.49 \times \frac{62}{222} \] \[ = 1.49 \times 0.2793 \] \[ = 0.416 \ \text{g} \]
Step 5: Calculate the percentage of phosphorus in the given sample Given mass of organic compound = 1 g \[ %\text{P} = \frac{\text{Mass of P}}{\text{Mass of sample}} \times 100 \] \[ %\text{P} = \frac{0.416}{1} \times 100 \] \[ %\text{P} = 41.6% \] Final Answer: \[ \boxed{41.6%} \]