Combustion Analysis Calculator — Empirical Formula
Find the empirical formula of an organic compound from combustion analysis. Enter the sample mass and the masses of CO₂, H₂O (and optionally N₂ and SO₂) collected from the combustion, and the calculator derives the mole ratios and empirical formula.
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- 1
Moles of carbon
26.67 ÷ 44.009 = 0.60601 molEach CO₂ molecule contains exactly one carbon atom. - 2
Moles of hydrogen
2 × 10.91 ÷ 18.015 = 1.21121 molEach H₂O molecule contains two hydrogen atoms. - 3
Carbon mass in sample
0.60601 mol × 12.011 g/mol = 7.2788 g - 4
Carbon percentage (%C)
(7.2788 ÷ 10) × 100 = 72.79 %
How does this calculator work?
Burn the compound, collect CO₂ and H₂O. Moles of C = mass CO₂ ÷ 44.01; moles of H = 2 × mass H₂O ÷ 18.02; oxygen by difference. Divide all moles by the smallest to get whole-number subscripts — that's the empirical formula. Molar masses used: IUPAC 2021 standard values.
Formula
How this is calculated
Combustion analysis burns a known mass of an organic compound in excess oxygen and collects the gaseous products. Because carbon is fully converted to CO₂ and hydrogen to H₂O, the mass of each product directly gives the moles of C and H in the original sample. Similarly, nitrogen exits as N₂ and sulfur as SO₂ in some setups. Oxygen in the original compound is found by subtracting the masses of all other elements from the sample mass (the "by difference" method).
Once the moles of each element are known, dividing by the smallest mole count gives a whole-number (or near whole-number) ratio. The calculator tries multipliers from 1 to 8 to find the simplest integer subscripts within 5% rounding tolerance. The result is the empirical formula — the simplest ratio of atoms. To get the molecular formula you also need the molar mass from a separate measurement (mass spectrometry or vapour density).
The molar masses used are: C = 12.011, H = 1.008, O = 15.999, N = 14.007, S = 32.06 g/mol (IUPAC 2021 standard atomic weights). Combustion analysis assumes complete combustion with no side reactions. Incomplete combustion, presence of halogens, or phosphorus will give incorrect results without additional corrections.
Frequently asked questions
Oxygen is found by difference: subtract the masses of C, H, N, and S (all calculated from their combustion products) from the total sample mass. What remains is attributed to oxygen in the original compound. This works only if no other elements are present.
The empirical formula is the simplest integer ratio of atoms (e.g. CH₂O for glucose). The molecular formula shows the actual number of each atom (C₆H₁₂O₆ for glucose). To go from empirical to molecular you need the molar mass from a separate experiment and divide: n = M_molecular / M_empirical.
Small measurement errors in the collected masses can give slightly off ratios (e.g. C₁H₂.₁O₀.₉). The calculator tries multipliers up to 8 to find a whole-number approximation. If the result still looks non-integer, check your measured masses for errors, or the compound may contain elements this calculator doesn't model (halogens, phosphorus, metals).
Also known as
TG we-Calculate Editorial Team. (2026). Combustion Analysis Calculator — Empirical Formula [Online calculator]. TG we-Calculate. https://we-calculate.com/calculator/combustion-analysis-calculator
TG we-Calculate Editorial Team. "Combustion Analysis Calculator — Empirical Formula." TG we-Calculate. 2026. https://we-calculate.com/calculator/combustion-analysis-calculator.
TG we-Calculate Editorial Team, "Combustion Analysis Calculator — Empirical Formula," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/calculator/combustion-analysis-calculator
@misc{wecalculate_combustion_analysis_calculator, title = {Combustion Analysis Calculator — Empirical Formula}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/calculator/combustion-analysis-calculator}}, year = {2026}, note = {TG we-Calculate} }
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