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JEE Chemistry 2025 Question with Solution

On complete combustion 1.0 g1.0 \, \text{g} of an organic compound (X) gave 1.46 g1.46 \, \text{g} of CO2\text{CO}_2 and 0.567 g0.567 \, \text{g} of H2O\text{H}_2\text{O}. The empirical formula mass of compound (X) is:

(Given molar mass in g mol−1\text{g mol}^{-1}: C: 1212, H: 11, O: 1616)

  • A

    3030

  • B

    4545

  • C

    6060

  • D

    1515

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mass of compound =1.0 g= 1.0 \, \text{g}, mass of CO2=1.46 g\text{CO}_2 = 1.46 \, \text{g}, mass of H2O=0.567 g\text{H}_2\text{O} = 0.567 \, \text{g}.

Find: The empirical formula mass of compound (X).

Use combustion analysis: all carbon goes to CO2\text{CO}_2, all hydrogen goes to H2O\text{H}_2\text{O}, and oxygen is obtained by difference.

Step 1: Mass of carbon

Mass of C=(1244)×1.46=0.39818 g≈0.398 g\text{Mass of C} = \left(\frac{12}{44}\right) \times 1.46 = 0.39818 \, \text{g} \approx 0.398 \, \text{g}

Step 2: Mass of hydrogen

Mass of H=(218)×0.567=0.063 g\text{Mass of H} = \left(\frac{2}{18}\right) \times 0.567 = 0.063 \, \text{g}

Step 3: Mass of oxygen by difference

Mass of O=1.0−(0.398+0.063)=0.539 g\text{Mass of O} = 1.0 - (0.398 + 0.063) = 0.539 \, \text{g}

Step 4: Convert masses into moles

Moles of C=0.39812≈0.0332\text{Moles of C} = \frac{0.398}{12} \approx 0.0332 Moles of H=0.0631=0.063\text{Moles of H} = \frac{0.063}{1} = 0.063 Moles of O=0.53916≈0.0337\text{Moles of O} = \frac{0.539}{16} \approx 0.0337

Step 5: Find the simplest ratio Divide by the smallest value 0.03320.0332:

C:H:O=1:1.89:1.01≈1:2:1\text{C} : \text{H} : \text{O} = 1 : 1.89 : 1.01 \approx 1 : 2 : 1

So the empirical formula is CH2O\text{CH}_2\text{O}.

Step 6: Empirical formula mass

(1×12)+(2×1)+(1×16)=30(1 \times 12) + (2 \times 1) + (1 \times 16) = 30

Therefore, the empirical formula mass is 30 g mol−130 \, \text{g mol}^{-1}. The correct option is A.

The solution also states directly that the empirical formula is CH2O\text{CH}_2\text{O} and its empirical formula mass is 3030.

Direct Mole-Ratio Route

Given: Mass of CO2\text{CO}_2 and H2O\text{H}_2\text{O} formed on combustion.

Find: Empirical formula mass.

A faster route is to calculate moles of atoms directly from the combustion products:

Moles of C=1.4644≈0.033\text{Moles of C} = \frac{1.46}{44} \approx 0.033 Moles of H=2×0.56718=0.063\text{Moles of H} = 2 \times \frac{0.567}{18} = 0.063

Then mass of oxygen in the sample is found by difference, giving oxygen moles approximately 0.0330.033. Hence the ratio is

C:H:O=1:2:1\text{C} : \text{H} : \text{O} = 1 : 2 : 1

So empirical formula =CH2O= \text{CH}_2\text{O}, and empirical formula mass

12+2+16=3012 + 2 + 16 = 30

Therefore, the correct option is A.

Common mistakes

  • Finding the mass of carbon or hydrogen incorrectly from CO2\text{CO}_2 or H2O\text{H}_2\text{O}. The product mass is not the element mass; use the mass fraction of the element in that compound, such as 1244\frac{12}{44} for carbon in CO2\text{CO}_2 and 218\frac{2}{18} for hydrogen in H2O\text{H}_2\text{O}.

  • Using masses directly as the atomic ratio. Empirical formula requires the mole ratio, so each elemental mass must be divided by its atomic mass before simplifying.

  • Forgetting to calculate oxygen by difference. In combustion analysis, oxygen in the original compound is obtained from sample mass minus masses of carbon and hydrogen, not from the combustion products directly.

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