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MCQEasyJEE 2025Quantitative Analysis (C, H, N…)

JEE Chemistry 2025 Question with Solution

On combustion 0.210 g0.210 \, \text{g} of an organic compound containing C, H and O gave 0.127 g0.127 \, \text{g} H2OH_2O and 0.307 g0.307 \, \text{g} CO2CO_2. The percentages of hydrogen and oxygen in the given organic compound respectively are:

  • A

    6.72,39.876.72, 39.87

  • B

    6.72,53.416.72, 53.41

  • C

    7.55,43.857.55, 43.85

  • D

    53.41,39.653.41, 39.6

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Mass of organic compound = 0.210 g0.210 \, \text{g}. On combustion it gives 0.127 g0.127 \, \text{g} of H2OH_2O and 0.307 g0.307 \, \text{g} of CO2CO_2.

Find: Percentage of hydrogen and oxygen in the compound.

All hydrogen of the compound appears in H2OH_2O and all carbon appears in CO2CO_2. Oxygen is then found by difference.

From water formed,

Mass of H=218×0.127=0.0141 g\text{Mass of H} = \frac{2}{18} \times 0.127 = 0.0141 \, \text{g}

Hence,

%H=(0.01410.210)×100=6.72%\%\text{H} = \left( \frac{0.0141}{0.210} \right) \times 100 = 6.72\%

From carbon dioxide formed,

Mass of C=1244×0.307=0.0837 g\text{Mass of C} = \frac{12}{44} \times 0.307 = 0.0837 \, \text{g}

So,

%C=(0.08370.210)×100=39.87%\%\text{C} = \left( \frac{0.0837}{0.210} \right) \times 100 = 39.87\%

Now oxygen percentage is obtained by difference,

%O=100−(%C+%H)=100−(39.87+6.72)=53.41%\%\text{O} = 100 - (\%\text{C} + \%\text{H}) = 100 - (39.87 + 6.72) = 53.41\%

Therefore, the percentages of hydrogen and oxygen are 6.72%6.72\% and 53.41%53.41\% respectively. The correct option is B.

Direct Mass-Fraction Method

Given: Sample mass = 0.210 g0.210 \, \text{g}, water formed = 0.127 g0.127 \, \text{g}, carbon dioxide formed = 0.307 g0.307 \, \text{g}.

Find: Percentages of H and O.

Use direct mass fractions from combustion products:

Mass of H in H2O=218×0.127=0.01412 g\text{Mass of H in } H_2O = \frac{2}{18} \times 0.127 = 0.01412 \, \text{g} Mass of C in CO2=1244×0.307=0.08376 g\text{Mass of C in } CO_2 = \frac{12}{44} \times 0.307 = 0.08376 \, \text{g}

Then mass of oxygen in the compound is

0.210−(0.01412+0.08376)=0.11212 g0.210 - (0.01412 + 0.08376) = 0.11212 \, \text{g}

So,

%H=0.014120.210×100=6.72%\%\text{H} = \frac{0.01412}{0.210} \times 100 = 6.72\% %O=0.112120.210×100=53.41%\%\text{O} = \frac{0.11212}{0.210} \times 100 = 53.41\%

This works because combustion converts all H into H2OH_2O and all C into CO2CO_2, so their masses can be read directly from the products. Hence the correct option is B.

Common mistakes

  • Using the entire mass of H2OH_2O as the mass of hydrogen is incorrect because water contains both hydrogen and oxygen. Use only the hydrogen fraction, 218\frac{2}{18} of the mass of H2OH_2O.

  • Calculating oxygen percentage directly from the oxygen present in the combustion products is wrong because that oxygen may come from atmospheric oxygen during combustion. First find masses of C and H in the original sample, then subtract from the sample mass.

  • Using an incorrect carbon fraction in CO2CO_2 causes error. Carbon contributes only 1244\frac{12}{44} of the mass of CO2CO_2, not the whole mass.

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