Logo
Jeeify
MCQEasyJEE 2026Stoichiometry & Calculations

JEE Chemistry 2026 Question with Solution

By usual analysis, 1.00 g1.00 \, \text{g} of compound (X) gave 1.79 g1.79 \, \text{g} of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : \hspace{1cm} (nearest integer) (Given, molar mass in g mol−1\text{g mol}^{-1} ; O = 1616, Mg = 2424, P = 3131)

  • A

    2020

  • B

    4040

  • C

    3030

  • D

    5050

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Mass of compound (X) = 1.00 g1.00 \, \text{g} and mass of magnesium pyrophosphate formed = 1.79 g1.79 \, \text{g}.

Find: Percentage of phosphorus in compound (X).

Phosphorus is estimated by converting it into magnesium pyrophosphate, Mg2P2O7Mg_2P_2O_7. The phosphorus content is obtained from the stoichiometry of this product.

Molar mass of Mg2P2O7Mg_2P_2O_7 is

2(24)+2(31)+7(16)=48+62+112=222 g/mol2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 \, \text{g/mol}

Mass of phosphorus present in 222 g222 \, \text{g} of Mg2P2O7Mg_2P_2O_7 is

2×31=62 g2 \times 31 = 62 \, \text{g}

Therefore,

%P=62222×1.791.00×100\%P = \frac{62}{222} \times \frac{1.79}{1.00} \times 100

Now evaluating,

%P≈0.2792×1.79×100≈49.99%\%P \approx 0.2792 \times 1.79 \times 100 \approx 49.99\%

The nearest integer is 5050. Therefore, the correct option is D.

Stoichiometric Factor Approach

Given: The stoichiometric factor for phosphorus in Mg2P2O7Mg_2P_2O_7 is

62222\frac{62}{222}

Find: Percentage of phosphorus in 1.00 g1.00 \, \text{g} of compound (X).

Mass of phosphorus in 1.79 g1.79 \, \text{g} of Mg2P2O7Mg_2P_2O_7 is

1.79×622221.79 \times \frac{62}{222}

Hence percentage of phosphorus in the compound is

1.79×622221.00×100≈49.99%\frac{1.79 \times \frac{62}{222}}{1.00} \times 100 \approx 49.99\%

So the percentage of phosphorus is 50%50\%, and the correct option is D.

Common mistakes

  • Using the molar mass of phosphorus as 3131 instead of 2×312 \times 31 is incorrect because Mg2P2O7Mg_2P_2O_7 contains two phosphorus atoms. Always account for the full formula of the precipitate.

  • Taking the molar mass of Mg2P2O7Mg_2P_2O_7 incorrectly leads to a wrong percentage. Use the given atomic masses and calculate 2(24)+2(31)+7(16)=2222(24) + 2(31) + 7(16) = 222 carefully.

  • Forgetting to divide by the mass of the original compound is wrong because percentage composition must be based on the sample taken. Here the denominator is 1.00 g1.00 \, \text{g}, not 1.79 g1.79 \, \text{g}.

Practice more Stoichiometry & Calculations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Stoichiometry & Calculations

Related questions