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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

During the elemental analysis of a newly synthesized drug molecule,...

During the elemental analysis of a newly synthesized drug molecule, 0.45 g0.45\text{ g} of the sample was subjected to Liebig's combustion method. The experiment yielded 0.66 g0.66\text{ g} of carbon dioxide and 0.27 g0.27\text{ g} of water. What are the respective percentages of carbon and hydrogen in the drug sample?

Options

  1. A40%40\% and 3.33%3.33\%
  2. B27.27%27.27\% and 11.11%11.11\%
  3. C40%40\% and 6.67%6.67\%
  4. D6.67%6.67\% and 40%40\%

Correct answer

C. 40%40\% and 6.67%6.67\%

Step-by-step solution

Mass of organic compound, W=0.45 gW = 0.45\text{ g}

Mass of CO2CO_2 formed =0.66 g= 0.66\text{ g}

Mass of H2OH_2O formed =0.27 g= 0.27\text{ g}

Percentage of carbon is given by:

%C=1244×Mass of CO2Mass of compound×100\% C = \dfrac{12}{44} \times \dfrac{\text{Mass of } CO_2}{\text{Mass of compound}} \times 100

%C=1244×0.660.45×100=1244×6645×100=40%\% C = \dfrac{12}{44} \times \dfrac{0.66}{0.45} \times 100 = \dfrac{12}{44} \times \dfrac{66}{45} \times 100 = 40\%

Percentage of hydrogen is given by:

%H=218×Mass of H2OMass of compound×100\% H = \dfrac{2}{18} \times \dfrac{\text{Mass of } H_2O}{\text{Mass of compound}} \times 100

%H=218×0.270.45×100=19×2745×100=6.67%\% H = \dfrac{2}{18} \times \dfrac{0.27}{0.45} \times 100 = \dfrac{1}{9} \times \dfrac{27}{45} \times 100 = 6.67\%

Answer: 40%40\% and 6.67%6.67\%
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