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

Given:

• 1 g of an organic compound produces 1.49 g of \( \text{Mg}_2\text{P}_2\text{O}_7 \) (magnesium pyrophosphate).
• We are asked to determine the percentage of phosphorus (P) in the compound.

Step 1: Write the molecular formula of magnesium pyrophosphate

The formula for magnesium pyrophosphate is \( \text{Mg}_2\text{P}_2\text{O}_7 \), where: - There are 2 atoms of magnesium (Mg) - 2 atoms of phosphorus (P) - 7 atoms of oxygen (O) The molar masses of the elements are: - Molar mass of \( \text{Mg} \) = 24.305 g/mol - Molar mass of \( \text{P} \) = 30.974 g/mol - Molar mass of \( \text{O} \) = 15.999 g/mol So, the molar mass of \( \text{Mg}_2\text{P}_2\text{O}_7 \) is: \[ \text{Molar mass of } \text{Mg}_2\text{P}_2\text{O}_7 = (2 \times 24.305) + (2 \times 30.974) + (7 \times 15.999) \] \[ = 48.610 + 61.948 + 111.993 = 222.551 \, \text{g/mol} \]

Step 2: Calculate the mass of phosphorus in 1.49 g of \( \text{Mg}_2\text{P}_2\text{O}_7 \)

From the molecular formula, there are 2 moles of phosphorus (P) in 1 mole of \( \text{Mg}_2\text{P}_2\text{O}_7 \). The molar mass of phosphorus is 30.974 g/mol, so the mass of phosphorus in 1 mole of \( \text{Mg}_2\text{P}_2\text{O}_7 \) is: \[ \text{Mass of phosphorus in 1 mole of } \text{Mg}_2\text{P}_2\text{O}_7 = 2 \times 30.974 = 61.948 \, \text{g} \] Now, we find the mass of phosphorus in 1.49 g of \( \text{Mg}_2\text{P}_2\text{O}_7 \) using the ratio: \[ \text{Mass of phosphorus} = \left( \frac{61.948}{222.551} \right) \times 1.49 = 0.414 \, \text{g} \]

Step 3: Calculate the percentage of phosphorus in the compound

The percentage of phosphorus in the organic compound is given by the ratio of the mass of phosphorus to the total mass of the compound, multiplied by 100: \[ \text{Percentage of phosphorus} = \left( \frac{0.414}{1} \right) \times 100 = 41.6\% \]

Final Answer:

The percentage of phosphorus in the organic compound is \( \boxed{41.6\%} \).