Determine the empirical formula of an organic compound from combustion analysis data. Enter the mass of your sample, CO₂ produced, and H₂O produced to instantly get carbon, hydrogen, and oxygen percentages, molar ratios, and the empirical formula.
✓ Verified: IUPAC atomic masses & standard combustion analysis method — April 2026
Mass of organic compound burnedEnter a valid sample mass greater than 0.
Mass of carbon dioxide collectedEnter a valid CO₂ mass (0 or more).
Mass of water collectedEnter a valid H₂O mass (0 or more).
Known molar mass to find molecular formula
Empirical Formula
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Sources & Methodology
🛡️Atomic masses from IUPAC 2021 standard. Method follows standard organic combustion analysis procedure as described in Atkins' Physical Chemistry and NIST chemistry resources.
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IUPAC 2021 Atomic Weights of the Elements
Standard atomic masses: C = 12.011, H = 1.008, O = 15.999, CO₂ = 44.010, H₂O = 18.015. iupac.org
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Atkins' Physical Chemistry, 12th Edition — Atkins & de Paula
Standard university textbook covering combustion analysis methodology, empirical formula determination, and stoichiometry (Chapter 1).
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NIST Chemistry WebBook
Reference thermochemical data for combustion reactions of organic compounds. webbook.nist.gov
Step-by-step method:
1. Mass C = mass CO₂ × (12.011 / 44.010)
2. Mass H = mass H₂O × (2.016 / 18.015)
3. Mass O = sample mass − mass C − mass H
4. Moles C = mass C / 12.011 | Moles H = mass H / 1.008 | Moles O = mass O / 15.999
5. Divide all moles by the smallest value → molar ratios
6. Round to nearest whole number (multiply if needed for .5, .33, .25)
Example: 1.000 g sample produces 1.466 g CO₂ and 0.600 g H₂O
Mass C = 1.466×(12.011/44.010) = 0.400 g | Mass H = 0.600×(2.016/18.015) = 0.0672 g
Mass O = 1.000−0.400−0.0672 = 0.533 g
Moles: C=0.0333, H=0.0667, O=0.0333 → ratio 1:2:1 → Empirical formula: CH₂O
How Combustion Analysis Works: Finding Empirical Formulas
Combustion analysis is one of the oldest and most reliable techniques in organic chemistry for determining the elemental composition of unknown compounds. The method works by burning a precisely weighed sample of an organic compound in a stream of pure oxygen, then capturing and weighing the combustion products — primarily carbon dioxide (CO₂) and water (H₂O).
Because all carbon in the original compound ends up in CO₂, and all hydrogen ends up in H₂O, measuring these products allows exact back-calculation of how much C and H were in the original sample. Any remaining mass (after subtracting C and H) is attributed to oxygen, though other elements like nitrogen or sulfur require separate analysis methods.
The Six-Step Process
The calculation follows a systematic six-step process. First, convert CO₂ mass to carbon mass using the mass ratio 12.011/44.010 = 0.2729. Second, convert H₂O mass to hydrogen mass using 2.016/18.015 = 0.1119. Third, calculate oxygen mass by subtracting C and H from the total sample mass. Fourth, convert each elemental mass to moles by dividing by the respective atomic mass. Fifth, divide all moles by the smallest mole value to get the simplest ratio. Sixth, round to whole numbers — multiplying the entire set if needed when fractions like 1.5 or 1.33 appear.
Step
Operation
Example (glucose, 1.800 g)
1
Mass C from CO₂
2.640 g CO₂ × 0.2729 = 0.720 g C
2
Mass H from H₂O
1.080 g H₂O × 0.1119 = 0.121 g H
3
Mass O = sample − C − H
1.800 − 0.720 − 0.121 = 0.959 g O
4
Convert to moles
C: 0.0599 | H: 0.1200 | O: 0.0599
5
Divide by smallest
C: 1.00 | H: 2.00 | O: 1.00
6
Empirical formula
CH₂O (molar mass 30 g/mol)
From Empirical to Molecular Formula
The empirical formula is the simplest whole-number ratio, but the molecular formula may be a multiple of it. To find the molecular formula, divide the known molar mass of the compound by the empirical formula mass. The result (n) is the multiplier: molecular formula = empirical formula × n.
For glucose: empirical formula CH₂O has mass 30.026 g/mol. The actual molar mass of glucose is 180.16 g/mol. So n = 180.16 / 30.026 = 6. Molecular formula = C₆H₁₂O₆.
Common Empirical Formulas from Combustion Analysis
Compound
%C
%H
%O
Empirical Formula
Molecular Formula
Glucose
40.00%
6.71%
53.29%
CH₂O
C₆H₁₂O₆
Acetic acid
40.00%
6.71%
53.29%
CH₂O
C₂H₄O₂
Ethanol
52.14%
13.13%
34.73%
C₂H₆O
C₂H₆O
Benzene
92.26%
7.74%
0%
CH
C₆H₆
Aspirin
60.00%
4.44%
35.56%
C₉H₈O₄
C₉H₈O₄
💡 Note on Nitrogen & Sulfur: Standard combustion analysis only directly determines C, H, and O. Nitrogen is measured separately using the Dumas method (converting N to N₂ gas). Sulfur requires precipitation or ion chromatography. Modern CHNS elemental analyzers can measure C, H, N, and S simultaneously in a single automated run within 10 minutes.
When Molar Ratios Are Not Whole Numbers
In practice, molar ratios from combustion analysis are rarely perfect whole numbers due to measurement uncertainty. Ratios within 0.05 of a whole number can be rounded directly. Ratios near 0.5 (e.g., 1.50, 2.50) indicate you need to multiply all values by 2. Ratios near 0.33 or 0.67 require multiplication by 3. Ratios near 0.25 or 0.75 require multiplication by 4. If ratios don't simplify to small whole numbers, recheck your measurements for errors.
Frequently Asked Questions
Combustion analysis is a technique for determining the elemental composition of an organic compound. A known mass of sample is burned completely in excess oxygen. The masses of CO₂ and H₂O produced are measured, allowing calculation of carbon, hydrogen, and (by difference) oxygen percentages in the original compound.
Step 1: Calculate mass C from CO₂ (mass C = mass CO₂ × 12.011/44.010). Step 2: Calculate mass H from H₂O (mass H = mass H₂O × 2.016/18.015). Step 3: Mass O = sample mass − mass C − mass H. Step 4: Convert masses to moles. Step 5: Divide all moles by the smallest. Step 6: Round to whole numbers for the empirical formula.
The empirical formula gives the simplest whole-number ratio of atoms (e.g., CH₂O). The molecular formula gives the actual atom count per molecule (e.g., C₆H₁₂O₆ for glucose). To find the molecular formula, divide the compound's molar mass by the empirical formula mass to get multiplier n, then multiply the empirical formula by n.
Carbon mass = mass of CO₂ × (12.011 / 44.010). This ratio equals the fraction of CO₂ that is carbon by mass. For example, 2.200 g CO₂ gives: mass C = 2.200 × 0.2729 = 0.600 g carbon.
Hydrogen mass = mass of H₂O × (2.016 / 18.015) = mass H₂O × 0.1119. For example, 0.900 g H₂O gives: mass H = 0.900 × 0.1119 = 0.1007 g hydrogen.
Non-whole molar ratios mean fractions need to be multiplied to reach whole numbers. If a ratio is ~1.5, multiply all by 2. If ~1.33, multiply by 3. If ~1.25, multiply by 4. Always apply the same multiplier to all elements. If ratios still don't simplify, recheck measurements for errors.
Standard combustion analysis only determines C, H, and O. Nitrogen requires the Dumas method (measuring N₂ gas). Sulfur requires separate precipitation or ion chromatography. Modern CHNS elemental analyzers can measure C, H, N, and S simultaneously in one automated run.
Incomplete combustion produces CO instead of CO₂, so not all carbon is captured in the CO₂ trap. This gives falsely low carbon percentages. To ensure complete combustion, excess oxygen is used, CuO catalyst wire is often added, and the tube is heated above 800°C.
Glucose (C₆H₁₂O₆) contains 40.00% C, 6.71% H, and 53.29% O. Burning 1.800 g produces 2.640 g CO₂ and 1.080 g H₂O. Molar ratios C:H:O = 1:2:1, giving empirical formula CH₂O. The molecular formula requires knowing molar mass = 180 g/mol, so n = 6, giving C₆H₁₂O₆.
The empirical formula mass is compared to the compound's known molar mass to find n: n = molar mass / empirical formula mass. The molecular formula = empirical formula × n. For example, empirical CH₂O (mass 30) with molar mass 180 gives n = 6, so molecular formula = C₆H₁₂O₆.