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Combustion Analysis Calculator

Ready to calculate
Liebig Mass-Balance Method.
C-H + C-H-O Compounds.
Empirical + Molecular.
100% Free.
No Data Stored.

How it Works

01Pick Substance Type

Hydrocarbon (C-H only) or CHO compound (C-H-O) — determines whether oxygen is found by mass balance

02Enter CO₂ and H₂O Masses

Mass of carbon dioxide and water collected from complete combustion in the absorption traps

03Mass Balance for Oxygen

For CHO mode, also enter sample mass — m(O) = m(sample) − m(C) − m(H) by conservation

04Empirical + Molecular Formula

Mole ratios → empirical formula; molar mass scales it to the molecular formula with % composition

What is a Combustion Analysis Calculator?

Combustion analysis is the 190-year-old workhorse of organic chemistry — the technique Justus von Liebig perfected in the 1830s for determining the empirical formula of any organic compound. The principle is elegantly simple: burn a known mass of compound in excess oxygen, trap the CO₂ and H₂O products, and weigh them. Carbon mass comes from the CO₂; hydrogen mass comes from the H₂O; oxygen mass (if present) comes from mass balance. Our Combustion Analysis Calculator implements this method in two modes: Hydrocarbons (C-H only), where you only need CO₂ and H₂O masses, and CHO compounds (C-H-O), where you also enter the original sample mass so oxygen is recovered by conservation: m(O) = m(sample) − m(C) − m(H). Output includes the empirical formula by mole-ratio reduction, the molecular formula scaled by the compound's molar mass, the percent composition for every element, and a complete step-by-step calculation breakdown.

Just enter the masses you measured: CO₂ from the absorption trap (typically NaOH or Ascarite), H₂O from the desiccant trap (Mg(ClO₄)₂ or P₂O₅), the original sample mass (CHO mode only), and the molar mass of your compound (from mass spectrometry, freezing-point depression, or another method). The calculator converts CO₂ and H₂O masses to moles of C and H using the universal mole ratios — m(C) = m(CO₂) × (12.011 / 44.009) ≈ 0.273 × m(CO₂) and m(H) = m(H₂O) × (2 × 1.008 / 18.015) ≈ 0.112 × m(H₂O) — divides by atomic weights to get moles, normalizes to the smallest mole count to obtain the mole ratio, and tries small integer multipliers (1-6) to scale to whole-number subscripts.

Designed for organic chemistry students learning structural elucidation, biochemistry students working with natural-product characterization, instrumental analysis labs running CHN combustion analyzers (Perkin-Elmer 2400, LECO TruSpec, Elementar vario MICRO), and historians of chemistry studying the foundations of modern analytical methods, the tool runs entirely in your browser — no data is stored or transmitted.

Pro Tip: Pair this with our Molecular Weight Calculator to compute molar masses for known formulas, or our Combustion Reaction Calculator to balance the corresponding combustion equation.

How to Use the Combustion Analysis Calculator?

Pick the Substance Type: "Hydrocarbon (C-H only)" if your sample contains just carbon and hydrogen (alkanes, alkenes, aromatics). "CHO compound (C-H-O)" if it also contains oxygen (alcohols, ethers, carbonyls, carboxylic acids, sugars). The mode determines whether oxygen is computed by mass balance.
Enter the Sample Molar Mass: The molecular weight of your unknown, in g/mol. Get it from mass spectrometry (M⁺ peak), freezing-point depression, vapor-density measurements, or another method. The calculator needs this to scale empirical → molecular formula.
Enter Sample Mass (CHO mode only): The original mass of compound burned, in g (or μg/mg/kg via the unit selector). For pure hydrocarbon mode, this is not needed — the C and H masses sum to the sample mass automatically.
Enter CO₂ and H₂O Masses: The masses collected in the absorption traps after combustion in pure O₂. CO₂ in the basic trap (NaOH/Ascarite); H₂O in the desiccant trap (Mg(ClO₄)₂ or P₂O₅). Modern CHN analyzers report these directly.
Press Calculate: Get the empirical formula by mole-ratio reduction, the molecular formula by molar-mass scaling, percent composition for each element, and a calculation breakdown showing every intermediate step (moles of C, moles of H, moles of O, ratios, multipliers).

How does combustion analysis work?

Combustion analysis converts masses you can weigh (CO₂, H₂O, sample) into moles you can ratio via the unchanging mole stoichiometry of CO₂ (1 C per molecule) and H₂O (2 H per molecule). Here's the complete derivation:


The whole method is built on conservation of mass — every carbon in the original sample ends up in exactly one CO₂ molecule; every two hydrogens end up in exactly one H₂O molecule. Knowing that, you can run the masses backwards.

Step 1 — Mass of Carbon


In CO₂, the carbon mass fraction is M(C)/M(CO₂) = 12.011/44.009 ≈ 0.2729. So:


m(C) = m(CO₂) × (12.011 / 44.009)

Step 2 — Mass of Hydrogen


In H₂O, the hydrogen mass fraction is 2·M(H)/M(H₂O) = 2.016/18.015 ≈ 0.1119. So:


m(H) = m(H₂O) × (2 × 1.008 / 18.015)

Step 3 — Mass of Oxygen (CHO mode only)


By conservation of mass — every gram of compound that wasn't C or H must be O:


m(O) = m(sample) − m(C) − m(H)


For pure hydrocarbon mode, this step is skipped (oxygen mass is 0 by definition).

Step 4 — Convert Masses to Moles


Divide each element mass by its atomic weight:


n(C) = m(C) / 12.011, n(H) = m(H) / 1.008, n(O) = m(O) / 15.999

Step 5 — Mole Ratios → Empirical Formula


Divide every mole count by the smallest one to get a normalized ratio. If the ratios aren't already integers (within ~5% rounding tolerance), try multiplying by 2, 3, 4, 5, or 6 until they all become integers. The result gives the empirical-formula subscripts.


Example: ratios C:H:O = 1.000 : 2.000 : 1.000 → CH₂O. Or 1.000 : 1.500 : 1.000 → multiply by 2 → C₂H₃O₂.

Step 6 — Empirical → Molecular Formula


Compute the empirical-formula molar mass M(empirical), then divide the known molecular molar mass by it:


n = M(molecular) / M(empirical)  →  molecular formula = (empirical)ₙ


For example, empirical CH₂O has M = 30.026 g/mol. If the compound's molar mass is 180.16 g/mol, then n = 180.16/30.026 ≈ 6, so molecular formula = C₆H₁₂O₆ (glucose). If n ≈ 1, the molecular formula equals the empirical formula.

Real-World Example

Combustion Analysis Calculator – Liebig's Method In Practice

Consider a classic textbook problem: a 0.255 g sample of an unknown CHO compound is burned in excess O₂, producing 0.561 g of CO₂ and 0.306 g of H₂O. Mass spectrometry shows the molar mass is 60.10 g/mol. What is the molecular formula?
  • Step 1 — Mass of carbon: m(C) = 0.561 × (12.011/44.009) = 0.561 × 0.2729 = 0.1531 g.
  • Step 2 — Mass of hydrogen: m(H) = 0.306 × (2.016/18.015) = 0.306 × 0.1119 = 0.0342 g.
  • Step 3 — Mass of oxygen by balance: m(O) = 0.255 − 0.1531 − 0.0342 = 0.0677 g.
  • Step 4 — Moles: n(C) = 0.1531/12.011 = 0.01275; n(H) = 0.0342/1.008 = 0.03393; n(O) = 0.0677/15.999 = 0.00423.
  • Step 5 — Mole ratio (divide by smallest, 0.00423): C : H : O = 3.014 : 8.020 : 1.000 ≈ 3 : 8 : 1. Empirical formula: C₃H₈O.
  • Step 6 — Empirical molar mass: M(C₃H₈O) = 3 × 12.011 + 8 × 1.008 + 15.999 = 36.033 + 8.064 + 15.999 = 60.096 g/mol.
  • Step 7 — Scale factor: n = 60.10 / 60.096 ≈ 1.00 → molecular formula = empirical formula.
  • Step 8 — Result: C₃H₈O. This is propan-1-ol or propan-2-ol (isopropanol) — combustion analysis can't distinguish isomers, but it nails the formula.
  • Percent composition: %C = 60.0%, %H = 13.4%, %O = 26.6%.

Now consider a hydrocarbon: 1.000 g of unknown gives 3.149 g CO₂ and 1.290 g H₂O. Molar mass = 86.18 g/mol. m(C) = 3.149 × 0.2729 = 0.8595 g → n(C) = 0.07156 mol. m(H) = 1.290 × 0.1119 = 0.1444 g → n(H) = 0.1432 mol. Ratio C:H = 1 : 2.001 → empirical CH₂. M(CH₂) = 14.027 g/mol. Scale factor = 86.18 / 14.027 ≈ 6.14 ≈ 6 → molecular formula C₆H₁₂ (cyclohexane, hexenes, or methylcyclopentane — same formula).

For glucose verification: a 0.500 g sample yields 0.733 g CO₂ and 0.300 g H₂O. M = 180.16 g/mol. m(C) = 0.2001, m(H) = 0.0336, m(O) = 0.500 − 0.2001 − 0.0336 = 0.2664 g. Moles: 0.01666 : 0.03331 : 0.01665 = 1 : 2 : 1 → CH₂O. M(CH₂O) = 30.026. n = 180.16/30.026 = 6.000 → C₆H₁₂O₆. Glucose confirmed.

Who Should Use the Combustion Analysis Calculator?

1
Organic Chemistry Students: Solve textbook problems on empirical/molecular formula determination — the bread and butter of structural elucidation chapters.
2
Natural Product Chemists: Confirm molecular formulas of newly isolated alkaloids, terpenes, and other natural products before NMR/MS structure elucidation.
3
Pharmaceutical QC Labs: Use CHN combustion analysis (Perkin-Elmer 2400, LECO TruSpec, Elementar vario MICRO) to verify identity and purity of drug substances against specifications.
4
Forensic Chemists: Identify unknown organic residues from crime scenes, fire investigations, or environmental contamination using elemental composition.
5
Polymer Chemists: Determine the empirical formula of a copolymer or polymerization product when the repeating unit is unknown.
6
History of Chemistry Educators: Demonstrate the Liebig method that founded modern analytical organic chemistry — including the brilliant Kaliapparat absorption bulb design from 1831.

Technical Reference

Atomic Weights (IUPAC 2021): M(C) = 12.011 g/mol, M(H) = 1.008 g/mol, M(O) = 15.999 g/mol. Derived: M(CO₂) = 44.009 g/mol, M(H₂O) = 18.015 g/mol. The all-important mass-fraction constants: carbon fraction in CO₂ = 12.011/44.009 = 0.2729, hydrogen fraction in H₂O = 2.016/18.015 = 0.1119.

The Liebig Apparatus (1831). Justus von Liebig's Kaliapparat — a five-bulb glass apparatus filled with concentrated KOH solution — absorbs CO₂ from the combustion gas stream. A separate calcium chloride tube absorbs H₂O. The apparatus is weighed before and after combustion; mass gains give CO₂ and H₂O directly. This design dropped analysis time from several days to a few hours and made organic chemistry as a rigorous quantitative discipline possible. The Kaliapparat is the symbol of the American Chemical Society.

Modern CHN Analyzers. Today's instruments (Perkin-Elmer 2400 Series II, LECO TruSpec Micro, Elementar vario MICRO cube) automate the Liebig method: tin capsule containing 1-3 mg of sample drops into a 950-1050 °C combustion tube under O₂; products pass through reduction copper to scrub excess O₂ and reduce NOₓ to N₂; CO₂, H₂O, and N₂ are separated on a chromatographic column or by selective traps; quantified by thermal-conductivity detector. Typical precision: ±0.3% absolute on each element.

Reference Mass-Fraction Compositions:

  • CH₄ (methane): %C = 74.87, %H = 25.13
  • C₂H₆ (ethane): %C = 79.89, %H = 20.11
  • C₆H₆ (benzene): %C = 92.26, %H = 7.74
  • C₈H₁₈ (octane): %C = 84.12, %H = 15.88
  • CH₃OH (methanol): %C = 37.48, %H = 12.58, %O = 49.93
  • C₂H₅OH (ethanol): %C = 52.14, %H = 13.13, %O = 34.73
  • CH₃COOH (acetic acid): %C = 39.99, %H = 6.71, %O = 53.29
  • C₆H₁₂O₆ (glucose): %C = 40.00, %H = 6.71, %O = 53.29 (same as acetic acid! both are CH₂O empirical)
  • C₁₂H₂₂O₁₁ (sucrose): %C = 42.11, %H = 6.48, %O = 51.42

Sources of Error. (1) Incomplete combustion — visible as carbon residue in the boat or low %C result; mitigated with more O₂ and combustion catalyst. (2) Hygroscopic samples — absorbed water inflates apparent H content; dry to constant mass first. (3) Heteroatoms in the sample (N, S, halogens, P) — distort the mass balance unless trapped separately or accounted for. (4) Sample weighing precision — micro-balance precision (~1 μg) is the practical limit. (5) Boat tare-weight drift — modern analyzers use calibration standards (acetanilide, sulfanilamide) every 10-20 samples.

Empirical vs Molecular Formula. Combustion analysis returns the empirical formula — the smallest integer mole ratio. Multiple compounds share the same empirical formula: glucose (C₆H₁₂O₆), fructose (C₆H₁₂O₆), formaldehyde (CH₂O), acetic acid (C₂H₄O₂) all have empirical CH₂O. To go from empirical → molecular, you need an independent molar-mass measurement (mass spec, freezing-point depression, vapor density, osmometry).

Key Takeaways

Combustion analysis converts CO₂ and H₂O masses into mole counts of C and H (and via mass balance, O), then mole ratios into the empirical formula, then molar-mass scaling into the molecular formula. The two governing constants you'll use forever: m(C) = 0.273 × m(CO₂) and m(H) = 0.112 × m(H₂O). Use the ToolsACE Combustion Analysis Calculator to solve any C-H or C-H-O empirical-formula problem instantly, see every intermediate step, get percent composition for every element, and verify the molecular formula. Bookmark it for organic chemistry homework, qualitative analysis labs, natural-product characterization, or any time you have CO₂/H₂O combustion data and need to find the formula.

Frequently Asked Questions

What is the Combustion Analysis Calculator?
It implements the classic Liebig combustion-analysis method for determining the empirical and molecular formula of an organic compound from CO₂ and H₂O combustion-product masses. Two modes: Hydrocarbons (C-H only), where you enter only CO₂ and H₂O masses plus the molar mass; CHO compounds (C-H-O), where you also enter the original sample mass so oxygen is recovered by mass balance: m(O) = m(sample) − m(C) − m(H). Output: empirical formula, molecular formula, percent composition for every element, and a complete step-by-step breakdown showing every conversion (mass → mole ratios → integer subscripts → molar-mass scaling).

Designed for organic chemistry students, instrumental analysis labs running CHN analyzers, natural-product chemists confirming structures, and history-of-chemistry educators teaching the Liebig method. Runs entirely in your browser — no data is stored or transmitted.

Pro Tip: Pair this with our Molecular Weight Calculator for fast empirical-mass computation.

What's the formula for combustion analysis?
Two universal relations: m(C) = m(CO₂) × (12.011/44.009) ≈ 0.273 × m(CO₂) and m(H) = m(H₂O) × (2.016/18.015) ≈ 0.112 × m(H₂O). For CHO compounds, oxygen comes from mass balance: m(O) = m(sample) − m(C) − m(H). Convert each element mass to moles by dividing by atomic weight, then normalize to the smallest mole count to get the mole ratio, multiply by 1-6 to get integer subscripts, and you have the empirical formula.
How do I get from empirical to molecular formula?
You need an independent measurement of the compound's molar mass — typically from mass spectrometry (M⁺ peak), freezing-point depression, vapor-density measurement, or osmometry. Then divide: n = M(molecular) / M(empirical). The molecular formula is the empirical formula multiplied through by n. Example: empirical CH₂O has M(empirical) = 30.026 g/mol. If the compound has M = 180.16 g/mol, then n = 6, and the molecular formula is C₆H₁₂O₆ (glucose).
What if my sample contains nitrogen, sulfur, or halogens?
This calculator handles only C, H, and O. For nitrogen-containing samples, use a CHN analyzer that quantifies N₂ separately and gives you nitrogen mass directly — then n(N) = m(N)/14.007. For sulfur, use a CHN-S analyzer (the S goes to SO₂ and is trapped). For halogens, use Schöniger combustion or oxygen-flask combustion followed by ion chromatography. Once you have all element masses, the mole-ratio step is the same.
Why does the calculator multiply ratios by 2, 3, 4, 5, or 6?
Because empirical formulas must have integer subscripts. After dividing all moles by the smallest one, ratios may come out as 1.5, 2.33, 1.25 — clearly not integers. Multiplying through by 2 turns 1.5 into 3 (giving subscripts like C₃H₃ instead of C₁.₅H₁.₅). Multiplying by 3 turns 1.33 into 4. The calculator tries multipliers 1 through 6 and picks the smallest that produces integers within ~5% tolerance.
What happens if mass(C) + mass(H) > sample mass?
Either there's an experimental error (CO₂ or H₂O mass measured too high, sample mass too low) or the sample doesn't contain only C, H, O. Common culprits: water of hydration absorbed by hygroscopic sample (inflates H), incomplete drying of the absorption traps before tare weighing (inflates baseline), or heteroatoms (N/S/halogens) in the sample. The calculator flags this case as an error.
How precise are CHN combustion measurements?
Modern instruments (Perkin-Elmer 2400, LECO TruSpec, Elementar vario MICRO) deliver ±0.3% absolute precision on C, H, N when run with proper calibration standards (acetanilide, sulfanilamide) every 10-20 samples. For a compound that's 60.0% C, that means ±0.3% absolute = the reported value falls in 59.7-60.3%. Sample sizes are 1-3 mg in tin capsules; the analysis takes 4-8 minutes. Older manual Pregl micro-combustion methods get ±0.5% with practiced technique.
Why is my empirical formula CH₂O for both glucose and acetic acid?
Because both have the same C:H:O mole ratio of 1:2:1 — they share the same empirical formula. Glucose (C₆H₁₂O₆) is six times the empirical unit; acetic acid (C₂H₄O₂) is two times. Combustion analysis can't distinguish them — you need an independent molar-mass measurement to find the multiplier. This is why combustion analysis must be paired with mass spectrometry or another molar-mass method to fully identify a compound.
Can I use this calculator for unknowns containing only carbon (like graphite)?
If your sample is pure carbon (graphite, coal, diamond dust), then m(H₂O) = 0 and the H mass-balance step is trivial — you'd only see %C in the result. Combustion analysis is overkill for pure C; just weigh the sample and the CO₂ — the C:O₂ ratio will be 1:1, and m(C)/m(sample) should equal 1.000. If it doesn't, your sample isn't pure C — there's some hydrogen, oxygen, or ash content.
Who invented combustion analysis?
Antoine Lavoisier introduced the principle of quantitative combustion in the 1780s as part of his chemical revolution that overthrew phlogiston theory. Joseph Louis Gay-Lussac and Louis Jacques Thénard (early 1800s) refined the apparatus. Justus von Liebig in 1831 invented the Kaliapparat — a five-bulb glass tube filled with KOH solution that absorbs CO₂ — which dropped analysis time from days to hours and made the method routine for the first time. The Kaliapparat is so important to chemistry that it's the central element of the American Chemical Society's logo.
What's the difference between empirical and molecular formula?
Empirical formula: the smallest integer mole ratio of atoms — e.g., CH₂O for formaldehyde, glycoaldehyde, glucose, fructose, sucrose (well, almost — sucrose is C₁₂H₂₂O₁₁, slightly different). Molecular formula: the actual count of each element in one molecule — e.g., glucose is C₆H₁₂O₆ (six empirical units), formaldehyde is just CH₂O (one). Combustion analysis returns the empirical; you need molar mass (MS, freezing-point depression, etc.) to scale it to molecular. Structural formula goes further and shows connectivity — that requires NMR, IR, X-ray crystallography, etc.

Author Spotlight

The ToolsACE Team - ToolsACE.io Team

The ToolsACE Team

Our chemistry tools team implements the classical Liebig combustion-analysis method that has been the gold standard for organic elemental analysis since the 1830s. The calculator handles two cases: pure hydrocarbons (C-H), where mass of carbon and hydrogen come directly from CO₂ and H₂O masses, and CHO compounds (C-H-O), where oxygen mass is recovered by conservation: m(O) = m(sample) − m(C) − m(H). Output includes the empirical formula by mole-ratio reduction, the molecular formula scaled by molar mass, percent composition for every element, and a full step-by-step calculation breakdown so students can verify each line.

Organic ChemistryQuantitative Elemental AnalysisSoftware Engineering Team

Disclaimer

The calculator assumes complete combustion of a sample containing only C, H, and optionally O. Heteroatoms (N, S, halogens, P) will distort the oxygen mass-balance result — use a CHN(S) analyzer for those. Atomic weights use IUPAC 2021 standard values. Empirical formula is determined to within ~5% rounding tolerance on mole ratios; if your data has experimental error larger than this, the rounding may pick the wrong integers.