Combustion Analysis: Empirical Formula Calculations
Combustion analysis is the foundational quantitative method taught in Class 11 chemistry for finding empirical and molecular formulas of organic compounds. Questions on this topic appear regularly in CBSE board exams, state boards, JEE Main, NEET, and CUET-UG. By burning a known mass of an organic substance in excess oxygen and measuring the masses of carbon dioxide and water produced, students determine the percentages of carbon, hydrogen, and oxygen. Learning this clear procedure ensures full accuracy in organic stoichiometry numericals.
Principle of Combustion Analysis
In combustion analysis (Liebig's method), a weighed sample of a dry organic compound containing carbon, hydrogen, and sometimes oxygen is placed in a combustion tube. The compound is heated strongly in a continuous stream of pure oxygen over copper(II) oxide (CuO), which acts as an oxidising catalyst.
Carbon oxidises completely to carbon dioxide gas:
C + O₂ → CO₂
Hydrogen oxidises completely to water vapour:
2H + ½ O₂ → H₂O
The produced gases pass sequentially through two pre-weighed absorption tubes:
- First U-tube: Contains anhydrous calcium chloride (CaCl₂), which absorbs water vapour (H₂O). The mass increase of this tube gives the mass of water formed.
- Second U-tube: Contains concentrated potassium hydroxide (KOH) solution, which absorbs carbon dioxide (CO₂). The mass increase gives the mass of carbon dioxide formed.
Water must be absorbed first because potassium hydroxide absorbs both water vapour and carbon dioxide.
Formulas for Percentage Composition and Empirical Formula
Standard IUPAC atomic masses are used: C = 12.011 g/mol (taken as 12.0 g/mol in school exams), H = 1.008 g/mol (taken as 1.0 g/mol), and O = 15.999 g/mol (taken as 16.0 g/mol). Molar mass of CO₂ is 44.01 g/mol (taken as 44.0 g/mol) and H₂O is 18.015 g/mol (taken as 18.0 g/mol).
Key Formulas:
1. Mass of Carbon: m_C = (12 / 44) × m_CO₂
2. Percentage of Carbon: % C = [ (12 × m_CO₂) / (44 × m_sample) ] × 100
3. Mass of Hydrogen: m_H = (2 / 18) × m_H₂O
4. Percentage of Hydrogen: % H = [ (2 × m_H₂O) / (18 × m_sample) ] × 100
5. Percentage of Oxygen (by difference):
% O = 100% - (% C + % H)
Mass of Oxygen: m_O = m_sample - (m_C + m_H)
6. Atomic Molar Ratio: Moles = (% of element) / (atomic mass)
7. Whole-number ratio: Divide each mole value by the smallest mole value. If fractions appear (such as 0.5), multiply by 2.
8. Molecular formula = (Empirical formula)_n, where n = Molar mass / Empirical formula mass.
Worked Numerical Examples
Question: Complete combustion of 0.200 g of an organic hydrocarbon yielded 0.628 g of carbon dioxide and 0.257 g of water. Determine the empirical formula of the hydrocarbon. (Atomic masses: C = 12.0 g/mol, H = 1.0 g/mol).
Step 1: Calculate mass and percentage of carbon.
Mass of C = (12 / 44) × 0.628 g = 0.1713 g
% C = (0.1713 g / 0.200 g) × 100 = 85.65%
Step 2: Calculate mass and percentage of hydrogen.
Mass of H = (2 / 18) × 0.257 g = 0.0286 g
% H = (0.0286 g / 0.200 g) × 100 = 14.30%
Step 3: Check mass balance.
Total percentage = 85.65% + 14.30% = 99.95% ≈ 100.0%. The substance contains only carbon and hydrogen.
Step 4: Determine the atomic mole ratio.
Moles of C = 85.65 / 12 = 7.138 mol
Moles of H = 14.30 / 1 = 14.30 mol
Divide by smallest value (7.138):
C = 7.138 / 7.138 = 1.00
H = 14.30 / 7.138 = 2.00
Ratio C : H = 1 : 2.
Route 2 verification (Direct stoichiometric moles):
Moles of C = 0.628 / 44.01 = 0.01427 mol
Moles of H = 2 × (0.257 / 18.015) = 0.02854 mol
Ratio H / C = 0.02854 / 0.01427 = 2.00. Confirms empirical formula CH₂.
Answer: CH₂
Question: A 0.450 g sample of an organic compound containing carbon, hydrogen, and oxygen was burned in excess oxygen. It produced 0.660 g of carbon dioxide and 0.270 g of water. Its molar mass is 60.05 g/mol. Find the empirical and molecular formulas. (Atomic masses: C = 12.0, H = 1.0, O = 16.0 g/mol).
Step 1: Calculate masses of carbon and hydrogen.
Mass of C = (12 / 44) × 0.660 g = 0.180 g
Mass of H = (2 / 18) × 0.270 g = 0.030 g
Step 2: Calculate mass and percentage of oxygen.
Mass of O = 0.450 g - (0.180 g + 0.030 g) = 0.240 g
% C = (0.180 / 0.450) × 100 = 40.00%
% H = (0.030 / 0.450) × 100 = 6.67%
% O = (0.240 / 0.450) × 100 = 53.33%
Step 3: Calculate the atomic mole ratio.
Moles of C = 40.00 / 12 = 3.333
Moles of H = 6.67 / 1 = 6.670
Moles of O = 53.33 / 16 = 3.333
Divide by 3.333: C = 1.00, H = 2.00, O = 1.00
Empirical formula = CH₂O
Step 4: Find the molecular formula.
Empirical formula mass = 12 + 2 + 16 = 30.0 g/mol
n = 60.05 / 30.0 = 2
Molecular formula = (CH₂O)₂ = C₂H₄O₂
Route 2 verification (Direct mole basis):
Moles C = 0.180 / 12 = 0.015 mol
Moles H = 0.030 / 1 = 0.030 mol
Moles O = 0.240 / 16 = 0.015 mol
Ratio C : H : O = 0.015 : 0.030 : 0.015 = 1 : 2 : 1.
Answer: Empirical formula: CH₂O; Molecular formula: C₂H₄O₂
Common Mistakes Students Make
- Assuming no oxygen: If % C + % H is less than 100%, the difference is oxygen. Always check the sum.
- Using 1 instead of 2 for hydrogen: Water is H₂O. Using 1/18 instead of 2/18 cuts the calculated hydrogen mass in half.
- Reversing absorption tubes: Placing KOH before CaCl₂ ruins the test because KOH absorbs water vapour as well as CO₂.
- Incorrect rounding: Never round 1.5 to 1 or 2; multiply by 2 to get integer ratios.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, JEE Main, NEET, State Engineering and Medical Entrances, CUET-UG
- Topic: Quantitative Elemental Analysis of Organic Compounds (Combustion Analysis)
- Difficulty: Medium
Exam Relevance and Testing Patterns
Combustion analysis questions test understanding of stoichiometry and empirical formula derivations.
| Exam | Question Style | Core Concepts Tested |
|---|---|---|
| CBSE Class 11 & State Boards | Structured descriptive problems and short numericals | Apparatus order, calculation of % C and % H, and finding molecular formula from molar mass |
| JEE Main | Numerical value questions | Oxygen determination by difference, handling fractional ratios, and linking with unsaturation |
| NEET | Direct calculation MCQs | Standard percentage formulas, rapid identification of empirical formulas |
| CUET-UG | Objective questions | Functions of anhydrous CaCl₂ and KOH, stoichiometric calculations from combustion data |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
More Concepts articles