Combustion Analysis Calculator
Determine elemental composition and empirical formula from combustion data
Combustion Analysis Calculator
Analysis: Burn organic compound, measure CO₂ and H₂O produced
From CO₂: %C = (12/44) × (mCO₂ / msample) × 100
From H₂O: %H = (2/18) × (mH₂O / msample) × 100
What is Combustion Analysis?
Combustion analysis is a classical analytical technique used to determine the elemental composition of organic compounds. The method involves burning a known mass of the compound in excess oxygen and measuring the masses of combustion products (CO₂, H₂O, and sometimes N₂).
This powerful technique allows chemists to calculate the empirical formula of unknown organic compounds, which is essential for structure determination in organic chemistry, pharmaceutical analysis, and materials science.
How Combustion Analysis Works
Step 1: Combustion Reaction
The organic compound is burned completely in excess O₂:
CxHyOzNw + O₂ → CO₂ + H₂O + N₂
Step 2: Product Collection
- • CO₂ is absorbed by NaOH or Mg(ClO₄)₂ and weighed
- • H₂O is absorbed by anhydrous CaCl₂ or Mg(ClO₄)₂ and weighed
- • N₂ (if present) is collected and measured separately
Step 3: Calculate Element Masses
- • Carbon: mass C = (12.01/44.01) × mass CO₂
- • Hydrogen: mass H = (2.016/18.015) × mass H₂O
- • Nitrogen: mass N = mass N₂ (if measured)
- • Oxygen: mass O = mass sample - mass C - mass H - mass N
Formula Derivations
Carbon Percentage Calculation
mass of C = (12.01 g/mol / 44.01 g/mol) × mass of CO₂
mass of C = 0.2729 × mass of CO₂
%C = (mass of C / mass of sample) × 100
Molar mass ratio: 12.01 g C per 44.01 g CO₂
Hydrogen Percentage Calculation
mass of H = (2.016 g/mol / 18.015 g/mol) × mass of H₂O
mass of H = 0.1119 × mass of H₂O
%H = (mass of H / mass of sample) × 100
Molar mass ratio: 2.016 g H per 18.015 g H₂O
Oxygen by Difference
mass of O = mass sample - mass C - mass H - mass N
%O = (mass of O / mass of sample) × 100
Oxygen is calculated by difference since it's not directly measured
Detailed Example: Glucose Analysis
Problem: A 1.50 mg sample of an unknown carbohydrate produces 2.20 mg CO₂ and 0.90 mg H₂O upon complete combustion. No nitrogen is detected. Determine the empirical formula.
Step 1: Calculate mass of carbon
mass C = (12.01/44.01) × 2.20 mg = 0.600 mg
%C = (0.600/1.50) × 100 = 40.0%
Step 2: Calculate mass of hydrogen
mass H = (2.016/18.015) × 0.90 mg = 0.101 mg
%H = (0.101/1.50) × 100 = 6.7%
Step 3: Calculate mass of oxygen (by difference)
mass O = 1.50 - 0.600 - 0.101 = 0.799 mg
%O = (0.799/1.50) × 100 = 53.3%
Step 4: Convert to moles
moles C = 0.600 mg / 12.01 mg/mmol = 0.0500 mmol
moles H = 0.101 mg / 1.008 mg/mmol = 0.100 mmol
moles O = 0.799 mg / 16.00 mg/mmol = 0.0499 mmol
Step 5: Find mole ratios (divide by smallest)
C: 0.0500/0.0499 ≈ 1
H: 0.100/0.0499 ≈ 2
O: 0.0499/0.0499 ≈ 1
Empirical Formula: CH₂O
(This is the empirical formula of glucose and other simple sugars)
Applications in Chemistry
🔬 Organic Structure Determination
Identify unknown organic compounds, verify synthesis products, and confirm molecular structures in research laboratories.
💊 Pharmaceutical Quality Control
Verify drug purity, confirm active ingredient composition, and ensure batch-to-batch consistency in pharmaceutical manufacturing.
🧪 Polymer Analysis
Characterize polymer composition, determine monomer ratios, and analyze degradation products in materials science.
📚 Educational Chemistry
Teach stoichiometry, empirical formula determination, and quantitative analysis techniques in undergraduate chemistry courses.
Quick Reference
Carbon from CO₂
mC = 0.2729 × mCO₂
Hydrogen from H₂O
mH = 0.1119 × mH₂O
Oxygen by Difference
mO = msample - mC - mH - mN
Related Calculators
Where It's Used
- •Organic chemistry laboratories
- •Pharmaceutical R&D
- •Quality control labs
- •University chemistry courses
- •Materials characterization