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Purification and Characterisation of Organic Compounds
On complete combustion,
0.492 g of an organic compound gave
0.792 g of
CO2.
The
% of carbon in the organic compound is (Nearest integer)
Hint 1 of 3
What formula relates the mass of CO2 produced to the mass of carbon present in the organic sample?
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Step-by-step solutionView
Correct answer
The percentage of carbon in the organic compound is 44%.
Option analysis
Why each option works or fails
StepWorking
01given
Mass of organic compound w=0.492 g, mass of CO2 formed m=0.792 g.
02find
Percentage of carbon in the organic compound, rounded to the nearest integer.
03strategise
Since 1 mol of CO2 (44 g) contains 12 g of carbon, the mass of carbon in 0.792 g of CO2 is 4412×mCO2. Then, %C=wmC×100=4412×wmCO2×100.
04execute
Calculate the mass of carbon: mC=4412×0.792=12×0.018=0.216 g. Then calculate %C=0.4920.216×100=43.902%≈44%.
✓verify
0.792/44=0.018 mol of CO2. Carbon mass =0.018×12=0.216 g. Fraction in compound =0.216/0.492=216/492≈0.439. Nearest integer is 44.
Hints that build this answer step by step
What formula relates the mass of CO2 produced to the mass of carbon present in the organic sample?
Mass of C=4412×Mass of CO2What is the mass of carbon contained in 0.792 g of CO2?
0.216 gWhat is the percentage of carbon in the 0.492 g sample of the organic compound?
44%
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✓ Source and academic review↓
- Question type
- Numerical
- Exam relevance
- JEE Main · Chemistry
- Concepts assessed
- Chemistry
- Academic status
- Reviewed by official_key
- Source
- pyq
- Editorial review
- 9 September 2026
Quick checks
Students also ask
Is 0.792 completely divisible by 44?
Yes, 792/44=18, so 0.792/44=0.018, which simplifies the arithmetic considerably.