Formula from combustion data: Class 11 practice
When an organic compound burns, all its carbon ends up in CO₂ (12/44 of the mass) and its hydrogen in H₂O (2/18 of the mass). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Empirical formula
The idea behind formula from combustion data
When an organic compound burns, all its carbon ends up in CO₂ (12/44 of the mass) and its hydrogen in H₂O (2/18 of the mass). From these masses find % C and % H, then the empirical formula.
Step-by-step method
- Mass of C = (12/44) × mass of CO₂.
- Mass of H = (2/18) × mass of H₂O.
- Convert to moles and find the simplest ratio.
Worked example 1 (Basic)
A hydrocarbon gives 44 g CO₂ on burning. What mass of carbon did it contain?
- 12/44 × 44
- Answer: 12 g
Worked example 2 (Standard)
A hydrocarbon burns to give 8.8 g CO₂ and 5.4 g H₂O. Find its empirical formula.
- C: 2.4 g = 0.2 mol
- H: 0.6 g = 0.6 mol
- Ratio 1 : 3
- Answer: CH3
Worked example 3 (Challenge)
A hydrocarbon burns to give 52.8 g CO₂ and 10.8 g H₂O. Find its empirical formula.
- C: 14.4 g = 1.2 mol
- H: 1.2 g = 1.2 mol
- Ratio 1 : 1
- Answer: CH1
Common mistakes and misconceptions
- Using 12/18
- Forgetting oxygen by difference
Practice questions
- (Basic) A hydrocarbon gives 66 g CO₂ on burning. What mass of carbon did it contain?
- (Basic) A hydrocarbon gives 8.8 g CO₂ on burning. What mass of carbon did it contain?
- (Standard) A hydrocarbon burns to give 22 g CO₂ and 18 g H₂O. Find its empirical formula.
- (Standard) A hydrocarbon burns to give 26.4 g CO₂ and 16.2 g H₂O. Find its empirical formula.
- (Challenge) A hydrocarbon burns to give 17.6 g CO₂ and 9 g H₂O. Find its empirical formula.
- (Challenge) A hydrocarbon burns to give 26.4 g CO₂ and 14.4 g H₂O. Find its empirical formula.
Answer key with full working
- Q1: 12/44 × 66 → Answer: 18 g
- Q2: 12/44 × 8.8 → Answer: 2.4 g
- Q3: C: 6 g = 0.5 mol → H: 2 g = 2 mol → Ratio 1 : 4 → Answer: CH4
- Q4: C: 7.2 g = 0.6 mol → H: 1.8 g = 1.8 mol → Ratio 1 : 3 → Answer: CH3
- Q5: C: 4.8 g = 0.4 mol → H: 1 g = 1 mol → Ratio 2 : 5 → Answer: C2H5
- Q6: C: 7.2 g = 0.6 mol → H: 1.6 g = 1.6 mol → Ratio 3 : 8 → Answer: C3H8
Syllabus note
This skill is listed under "Some Basic Concepts of Chemistry" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
- Some Basic Concepts of Chemistry chapter notes (Class 11)
- Molarity of solutions
- Mass–mass stoichiometry
- Limiting reagent
- Empirical and molecular formula
- Theoretical and percentage yield
- Quantum numbers and orbitals
- Hydrogen spectrum and Rydberg formula
- Hybridisation and molecular shape (VSEPR)
- Hess's law and enthalpy of reaction
- All Class 11 practice
Frequently asked questions
Which chapter is formula from combustion data in?
"Some Basic Concepts of Chemistry", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.