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Combustion Analysis Calculator

Chemistry

Determine percent composition and empirical formula of organic compounds from combustion analysis data: mass of CO₂, H₂O produced, and original sample mass.

Reviewed by the thecalcu.com team · Last updated July 5, 2024

10
29.3
12
0

Carbon (%)

79.97
Hydrogen (%)
13.43
Oxygen (%)
6.61
Nitrogen (%)
0
Empirical Formula
C16H32O

Breakdown

How the total splits

Carbon (%)
79.97
Hydrogen (%)
13.43
Oxygen (%)
6.61
Nitrogen (%)
0

This calculator computes your Carbon (%), Hydrogen (%), Oxygen (%), Nitrogen (%), Empirical Formula from the values you enter.

Inputs
Sample Mass (mg)CO₂ Produced (mg)H₂O Produced (mg)Nitrogen Present?Nitrogen Content (%)
Outputs
Carbon (%)Hydrogen (%)Oxygen (%)Nitrogen (%)Empirical Formula

What is a Combustion Analysis?

The Combustion Analysis Calculator determines the percent composition (C, H, O, N) and empirical formula of an organic compound from combustion data: the mass of sample burned, CO₂ produced, H₂O produced, and optionally nitrogen content. Enter the measured masses in milligrams.

Combustion analysis is the primary technique for confirming the elemental composition of newly synthesised organic compounds. When a sample burns completely in excess O₂: all C → CO₂, all H → H₂O. Measuring these products back-calculates the %C and %H; %O (and other elements) are obtained by difference. The resulting percent composition converts to mole ratios and then to the empirical formula.

The Empirical Formula Calculator takes percent composition directly and outputs the empirical formula, a complementary tool when you already have %C, %H, %O from a lab report. The Percent Composition Calculator works in the forward direction (from molecular formula to %composition), while this calculator works in reverse (from combustion data to empirical formula).

Why Use a Combustion Analysis Calculator?

The combustion analysis calculation requires: (1) computing mass C from CO₂ (multiply by 12.011/44.009), (2) computing mass H from H₂O (multiply by 2.016/18.015), (3) subtracting from sample mass to get mass O, (4) converting to percent, (5) dividing by atomic masses to get mole ratios, (6) finding smallest ratio and dividing, (7) rounding to integers. Seven sequential steps where one error at any stage gives the wrong empirical formula. This calculator does all seven steps at once.

Who Should Use This Calculator?

Organic chemistry laboratory students characterising synthesised compounds: combustion analysis results are required for compound characterisation in JEE Advanced lab sections and university organic chemistry courses.

Research chemists at IITs, IISc, CSIR labs, and pharmaceutical R&D departments who need to verify CHN analyser output and convert raw mass data to percent composition and empirical formula.

JEE Advanced aspirants working combustion analysis calculation problems, a standard numerical problem type in the organic chemistry section.

Pharmaceutical and quality control analysts verifying elemental composition of drug substances and intermediates per ICH Q6A guidelines and Indian Pharmacopoeia requirements.

What Insights Does the Combustion Analysis Calculator Give You?

Carbon (%) is the primary output, the mass percentage of carbon in the compound, derived from CO₂ mass.

Hydrogen (%), the mass percentage of hydrogen, derived from H₂O mass.

Oxygen (%), computed by difference (100 − %C − %H − %N). If oxygen is actually absent in the compound, this should be ≈ 0; any non-zero value indicates oxygen presence or experimental error.

Nitrogen (%), entered directly from a separate measurement (Kjeldahl or Dumas N₂ detection).

Empirical Formula, the simplest whole-number atom ratio, the starting point for identifying the compound.

How to use this Combustion Analysis calculator

  1. Enter Sample Mass (mg), the precisely weighed mass of compound burned.
  2. Enter CO₂ Produced (mg), the mass of CO₂ absorbed in the CO₂ absorption tube.
  3. Enter H₂O Produced (mg), the mass of H₂O absorbed in the moisture absorption tube.
  4. Select Nitrogen Present, choose Yes if N was measured and enter %N.
  5. Read %C, %H, %O, %N and Empirical Formula.
Show formula & methodology ↓Show less ↑

Formula & Methodology

Percent composition from combustion:

%C = (mass_CO₂ × 12.011 / 44.009) / mass_sample × 100 %H = (mass_H₂O × 2 × 1.008 / 18.015) / mass_sample × 100 %O = 100 − %C − %H − %N  (by difference)

Empirical formula:

Mole ratios: nC = %C/12.011;  nH = %H/1.008;  nO = %O/15.999 Divide by smallest: ratio = n / min(nC, nH, nO) Round to nearest integer (or multiply if fractional: e.g. ×2 if ratio ≈ 0.5)

Worked example, aspirin (C₉H₈O₄):

Sample = 10.00 mg; CO₂ produced = 22.01 mg; H₂O = 1.80 mg; no nitrogen.

%C = 22.01 × (12.011/44.009) / 10.00 × 100 = 6.00/10.00 × 100 = 60.00% %H = 1.80 × (2×1.008/18.015) / 10.00 × 100 = 0.4479/10.00 × 100 = 4.48% %O = 100 − 60.00 − 4.48 = 35.52%  Mole ratios: C = 60.00/12.011 = 4.996; H = 4.48/1.008 = 4.444; O = 35.52/15.999 = 2.220 Divide by 2.220: C = 2.25; H = 2.00; O = 1.00 Multiply by 4: C = 9, H = 8, O = 4 → Empirical formula: C₉H₈O₄

The empirical formula C₉H₈O₄ matches aspirin's molecular formula (M=180.16 g/mol). Aspirin is the world's most widely consumed pharmaceutical; India produces ~1,000 tonnes/year of bulk aspirin API (active pharmaceutical ingredient) in Hyderabad's pharma cluster.

Frequently Asked Questions

What is combustion analysis?
Combustion analysis (elemental analysis) is an analytical technique for determining the carbon, hydrogen, oxygen, and nitrogen content of organic compounds by burning a known mass of the compound in excess oxygen and measuring the CO₂ and H₂O produced. From the masses: %C = mass_CO₂ × (12.011/44.009) / mass_sample × 100; %H = mass_H₂O × (2.016/18.015) / mass_sample × 100; %O = 100 − %C − %H (by difference for C,H,O compounds). The resulting percent composition is used to determine the empirical formula.
How do I calculate percent carbon from combustion analysis data?
%C = (mass of CO₂ produced / molar mass of CO₂) × (molar mass of C) / mass of sample × 100 = (mass_CO₂ × 12.011 / 44.009) / mass_sample × 100. Every carbon atom in the compound appears as one CO₂ molecule in the combustion products. For example: 10 mg sample produces 29.3 mg CO₂ → mass C = 29.3 × 12.011/44.009 = 7.994 mg → %C = 7.994/10 × 100 = 79.9% (approximately, matching benzene C₆H₆).
What is the empirical formula and how does combustion analysis give it?
The empirical formula is the simplest whole-number ratio of atoms in a compound. From combustion analysis: (1) compute %C, %H, %O (and %N if measured). (2) Divide each percentage by the element's atomic mass to get mole ratios: C: %C/12.011; H: %H/1.008; O: %O/15.999. (3) Divide all mole ratios by the smallest ratio. (4) Round to whole numbers. If result is CH₂O, the molecular formula could be CH₂O (formaldehyde, M=30), C₂H₄O₂ (acetic acid, M=60), C₆H₁₂O₆ (glucose, M=180). Molecular mass data is needed to distinguish these. The [Empirical Formula Calculator](/empirical-formula-calculator/) automates steps 1–4.
How do I use the Combustion Analysis Calculator?
Enter Sample Mass (mg), CO₂ Produced (mg), and H₂O Produced (mg). Select whether nitrogen is present, if yes, enter %N separately (measured from a separate Kjeldahl or Dumas analysis). The calculator computes %C, %H, %O (by difference), and the empirical formula. Default: 10 mg sample, 29.3 mg CO₂, 12.0 mg H₂O (approximately benzene C₆H₆: 80.0% C, 6.7% H, 13.3% remaining).
What is the 'by difference' method for oxygen?
In combustion analysis, oxygen (and other elements) in the compound are not directly measured, they are calculated by subtracting from 100%: %O = 100 − %C − %H − %N − %S − %halogens. This works because elements other than C and H appear in different combustion products (SO₂, HX, N₂ or NOₓ) that are typically not measured directly in simple combustion analysis setups. Modern instruments (CHN analysers like those from Elementar or LECO) can measure N simultaneously by thermal conductivity detection; S and halogens require additional methods.
What is the minimum sample mass needed for combustion analysis?
Modern CHN elemental analysers (automated combustion analysis) require 1–5 mg of sample for reliable results, smaller samples can be used with high-sensitivity instruments. The classic wet chemistry approach using absorption tubes requires 5–20 mg. Very volatile compounds may need to be analysed at low temperature. Indian research institutions (IITs, IISc, NCL Pune, IICT Hyderabad) use automated CHN analysers for characterising newly synthesised organic compounds, a requirement for publication in organic chemistry journals (each compound must be ≥95% pure by combustion analysis).
What is the CHNS analyser and how does it work?
A CHN or CHNS analyser (Dumas method) burns the sample at ~900–1100°C in oxygen, then passes the combustion gases over catalyst beds to convert all C to CO₂, H to H₂O, N to N₂, and S to SO₂. The gases are separated by chromatography and detected by thermal conductivity. Results: %C (from CO₂), %H (from H₂O), %N (from N₂), %S (from SO₂), all in a single automated run in 3–5 minutes. The classic Pregl-Dumas method, which won the 1923 Nobel Prize in Chemistry (Fritz Pregl), originally required 5–10 mg; modern analysers handle sub-milligram samples.
How does combustion analysis give the ratio C:H:O but not the molecular formula?
Combustion analysis gives elemental mass percentages → mole ratios → empirical formula, which is the simplest ratio. The molecular formula is an integer multiple of the empirical formula: molecular formula = n × (empirical formula). To find n, you need the molecular mass: n = M_molecular / M_empirical. Example: empirical formula CH₂O (M=30.026); if mass spectrometry shows M=60.052, then n=2, molecular formula C₂H₄O₂ (acetic acid). If M=180, n=6, molecular formula C₆H₁₂O₆ (glucose). Mass spectrometry is now the standard complement to combustion analysis for molecular formula determination.
What are the combustion analysis results for common organic compounds?
Benzene (C₆H₆): 92.26% C, 7.74% H. Glucose (C₆H₁₂O₆): 40.00% C, 6.71% H, 53.29% O. Aspirin (C₉H₈O₄): 60.00% C, 4.48% H, 35.52% O. Urea (CH₄N₂O): 20.00% C, 6.71% H, 46.65% N, 26.64% O. Ethanol (C₂H₆O): 52.14% C, 13.13% H, 34.73% O. These are standard reference results for verifying combustion analysis instrument calibration.
Does combustion analysis work for inorganic compounds?
Classic combustion analysis (CHN/CHNS analysis) is designed for organic compounds containing C, H, N, S, O. For inorganic materials: metals cannot be burned to measurable gaseous products; metal salts produce residues. Inorganic elemental analysis uses different techniques: ICP-OES (inductively coupled plasma optical emission spectrometry) for metals, X-ray fluorescence (XRF) for bulk composition, and titrimetry for specific ions. Some inorganic compounds with organic ligands (metal-organic frameworks, coordination compounds) can be partially characterised by CHN analysis to confirm the organic portion.