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

To determine the percentage of phosphorus in the compound (X), we need to follow a systematic approach using stoichiometry. The given data tells us that 1.00 g of compound X gives 1.79 g of magnesium pyrophosphate (Mg₂P₂O₇).

The molecular formula of magnesium pyrophosphate is Mg₂P₂O₇. Let's calculate its molar mass:

• Mg = 24 g/mol. For 2 magnesium atoms: \(2 \times 24 = 48 \, \text{g/mol}\) 
• P = 31 g/mol. For 2 phosphorus atoms: \(2 \times 31 = 62 \, \text{g/mol}\)
• O = 16 g/mol. For 7 oxygen atoms: \(7 \times 16 = 112 \, \text{g/mol}\)

Using the mass of Mg₂P₂O₇ obtained (1.79 g), calculate the moles:

\(\text{Moles of Mg}_{2}\text{P}_{2}\text{O}_{7} = \frac{1.79 \, \text{g}}{222 \, \text{g/mol}} \approx 0.00806 \, \text{mol}\)

The molecular formula of magnesium pyrophosphate, Mg₂P₂O₇, contains 2 moles of phosphorus for every mole of Mg₂P₂O₇. Thus, the moles of phosphorus in the sample:

\(2 \times 0.00806 \, \text{mol} = 0.01612 \, \text{mol}\)

Determine the mass of phosphorus:

\(\text{Mass of P} = 0.01612 \, \text{mol} \times 31 \, \text{g/mol} = 0.4997 \, \text{g}\)

Calculate the percentage of phosphorus in compound (X):

\(\text{Percentage of P} = \left( \frac{0.4997 \, \text{g}}{1.00 \, \text{g}} \right) \times 100 \approx 50\%\)

Thus, the percentage of phosphorus in compound (X) is 50%. The correct answer is rounded to the nearest integer as required.