Combustion Analysis Calculator
Determine empirical formula from combustion analysis data. Find mole ratios of C, H, and O from masses of CO2 and H2O produced when burning an organic compound.
Formula & Reference
| Variable | Symbol | Formula | Units |
|---|---|---|---|
| Combustion Analysis Calculator | — | C from CO2, H from H2O, O by difference | empirical formula |
Step-by-Step Examples
3.000 g compound gives 4.501 g CO2 and 1.232 g H2O.
- n(C) = 4.501/44.010 = 0.1023 mol, m(C) = 0.1023x12.011 = 1.228 g
- n(H) = 1.232/18.015 x 2 = 0.1369 mol, m(H) = 0.1369x1.008 = 0.138 g
- m(O) = 3.000-1.228-0.138 = 1.634 g, n(O) = 1.634/15.999 = 0.1021
- Ratios: C:H:O = 0.1023:0.1369:0.1021 divided by 0.1021 = 1:1.34:1
- Multiply by 3: C3H4O3... simplify to C3H4O3
3.000 g compound gives 10.33 g CO2 and 2.12 g H2O.
- n(C) = 10.33/44.01 = 0.2347 mol
- n(H) = 2.12/18.02 x 2 = 0.2353 mol
- m(C)+m(H) = 2.818+0.237 = 3.055 g > 3.000 g... O<0
- Compound is pure hydrocarbon (CH ratio ~1:1)
Ethanol: 2.300 g gives 4.399 g CO2, 2.700 g H2O.
- n(C) = 4.399/44.01 = 0.09994 mol
- n(H) = 2.700/18.015 x 2 = 0.2998 mol
- m(O) = 2.300 - 1.200 - 0.302 = 0.798 g, n(O) = 0.798/16 = 0.04988
- Ratio C:H:O = 0.09994:0.2998:0.04988 / 0.04988 = 2:6:1
Real-World Applications
Common Mistakes to Avoid
Moles C = moles CO2. All C in compound converts to CO2 in complete combustion.
mol H = 2 x mol H2O. Each water contains 2 hydrogen atoms from the organic compound.
m(O) = m_sample - m(C) - m(H) - m(other). Cannot be measured directly; calculated from mass balance.
Frequently Asked Questions
Related Chemistry Calculators
Formula Explorer connections
Interpretation: This relationship converts chemical amount, mass, composition or balanced-equation ratios into a reaction quantity. Assumption: Use a balanced reaction, consistent units and the correct molar mass. Purity, side reactions and limiting reagents can change experimental results.