Last updated: July 3, 2026
Combustion Analysis Calculator
Creators
Dharmendra SinghReviewers
Creators
Dharmendra SinghReviewers
Quick Answer
A combustion analysis calculator determines carbon from CO2 mass, hydrogen from H2O mass, and oxygen by subtracting those masses from the original CHO sample mass. It then converts masses to moles and normalizes the mole amounts to estimate the empirical formula.
Combustion analysis finds carbon from carbon dioxide, hydrogen from water, and oxygen by difference. Convert each element mass to moles and divide by the smallest mole value to get the empirical formula ratio.
Key Takeaways
- Combustion analysis converts CO2 mass into carbon mass and H2O mass into hydrogen mass.
- For CHO compounds, oxygen mass is the sample mass minus carbon and hydrogen masses.
- Element masses become moles by dividing by atomic masses C = 12.011, H = 1.008, and O = 16.00.
- Dividing each mole amount by the smallest mole amount gives the empirical-formula ratio.
- A molecular formula needs an independent molar mass in addition to the empirical formula.
Creators
Dharmendra SinghReviewers
Creators
Dharmendra SinghReviewers
Formula
mass_C = mass_CO2 × (12.011/44.01); mass_H = mass_H2O × (2×1.008/18.015); mass_O = sample_mass − mass_C − mass_H
Where:
- m_C=Mass of carbon in the original sample(g)
- m_H=Mass of hydrogen in the original sample(g)
- m_O=Mass of oxygen by difference(g)
- m_CO2=Mass of carbon dioxide produced(g)
- m_H2O=Mass of water produced(g)
- m_sample=Original sample mass(g)
Worked Examples
CHO compound from 0.500 g sample
A 0.500 g sample produces 1.500 g CO2 and 0.409 g H2O.
- 1Carbon: 1.500 × (12.011 ÷ 44.01) = 0.4094 g C.
- 2Hydrogen: 0.409 × (2 × 1.008 ÷ 18.015) = 0.0458 g H.
- 3Oxygen by difference: 0.500 − 0.4094 − 0.0458 ≈ 0.0448 g O.
- 4Convert each mass to moles and divide by the smallest mole amount to estimate the empirical ratio.
Small CO2 product check
A sample that gives 0.220 g CO2 contains about 0.0600 g carbon.
- 1Carbon fraction in CO2 is 12.011 ÷ 44.01 = 0.27292.
- 2Mass carbon = 0.220 × 0.27292 = 0.0600 g.
- 3Hydrogen and oxygen are calculated from the water product and remaining sample mass.
Water product hydrogen check
A combustion run producing 0.900 g H2O contains about 0.1007 g hydrogen.
- 1Hydrogen fraction in H2O is (2 × 1.008) ÷ 18.015 = 0.11190.
- 2Mass hydrogen = 0.900 × 0.11190 = 0.1007 g.
- 3Carbon is found from CO2, and oxygen is the sample mass left after C and H are subtracted.
Introduction
Combustion analysis converts the products of burning an organic compound into the elemental composition of the original sample. Carbon in the sample becomes CO2, hydrogen becomes H2O, and oxygen in a CHO compound is found by difference. This calculator uses measured product masses to report carbon, hydrogen, oxygen, and an empirical formula. It pairs well with the percent composition calculator and molar mass calculator. Atomic masses follow common values consistent with IUPAC and analytical stoichiometry references such as LibreTexts.
What is combustion analysis?
Combustion analysis is a classical elemental-analysis method for compounds containing carbon, hydrogen, and sometimes oxygen. A precisely weighed sample is burned completely in excess oxygen. The CO2 and H2O formed are trapped and weighed, allowing the carbon and hydrogen in the unknown compound to be calculated from conservation of atoms.
Every mole of CO2 contains one mole of carbon atoms.
Every mole of H2O contains two moles of hydrogen atoms.
For a CHO compound, oxygen is usually calculated by difference from the original sample mass.
The mole ratio of C:H:O gives the empirical formula.
Combustion analysis formula
The carbon mass is the CO2 mass multiplied by carbon's mass fraction in CO2: 12.011/44.01. The hydrogen mass is the water mass multiplied by hydrogen's mass fraction in H2O: 2×1.008/18.015. Any remaining sample mass is assigned to oxygen for a compound known to contain only C, H, and O.
The oxygen-by-difference step is valid only when the original sample contains no other elements or when those elements have been independently corrected.
How to calculate an empirical formula
After finding elemental masses, convert each to moles: nC = mC/12.011, nH = mH/1.008, and nO = mO/16.00. Divide all mole amounts by the smallest positive value. If the resulting ratio is near whole numbers, write the empirical formula directly; if not, multiply all ratios by 2, 3, 4, or another small integer until whole numbers are obtained.
Measure sample, CO2, and H2O masses carefully.
Calculate C and H masses from product stoichiometry.
Calculate O by mass difference for CHO compounds.
Convert grams to moles using atomic masses.
Normalize the mole ratio to the smallest mole amount.
Verified reference values
Useful checks are the mass fractions built into the calculation. Carbon is 12.011/44.01 = 0.27292 of CO2 by mass. Hydrogen is 2.016/18.015 = 0.11190 of H2O by mass. Therefore 0.220 g CO2 contains about 0.0600 g carbon, and 0.900 g water contains about 0.1007 g hydrogen.
| Product measured | Conversion | Example product mass | Element mass |
|---|---|---|---|
| CO2 | × 12.011/44.01 | 0.220 g | 0.0600 g C |
| CO2 | × 12.011/44.01 | 1.500 g | 0.4094 g C |
| H2O | × 2.016/18.015 | 0.900 g | 0.1007 g H |
| H2O | × 2.016/18.015 | 0.409 g | 0.0458 g H |
Uses and limitations
Combustion analysis is widely used for organic unknowns, purity checks, and validating empirical formulas. It is strongest when combustion is complete and products are quantitatively collected. It cannot by itself distinguish molecular formulas with the same empirical formula; for that, combine it with molar mass from mass spectrometry or a mole calculator.
If you know the molecular molar mass, divide it by the empirical-formula mass to scale the empirical formula to the molecular formula.
Common mistakes to avoid
Do not treat the mass of CO2 as the mass of carbon, or the mass of H2O as the mass of hydrogen. Only a fraction of each product is the element you need. Another common mistake is assigning a negative oxygen mass when measurements are inconsistent; recheck collection losses, drying agents, and sample purity before interpreting the formula.
Use product masses in grams consistently.
Use the original sample mass for oxygen by difference.
Do not round until the final displayed values.
Use empirical ratios as estimates when product masses have limited significant figures.
Quick Reference Card
Combustion Analysis — Quick Reference
Quick reference • Combustion Analysis Calculator
mC = mCO2×12.011/44.01; mH = mH2O×2.016/18.015; mO = msample−mC−mHValid range: Positive product and sample masses; best for compounds known to contain only C, H, and O
Common Values
⚠ Watch Out
- •Incomplete combustion underestimates CO2 and can distort the formula.
- •Moisture in traps can overstate H2O mass and hydrogen content.
- •Oxygen by difference is invalid if other elements are present but unmeasured.
- •Do not round intermediate masses before converting to moles.
Pro Tips
- →Carry at least four significant figures through mass and mole calculations.
- →If ratios end in about 0.5, multiply all ratios by 2 before writing the formula.
- →Use a molar mass measurement to convert empirical formula to molecular formula.
- →Check percent composition totals as a sanity check after finding the formula.
FAQs
How does combustion analysis find carbon?
All carbon in the original organic sample is assumed to become CO2. The carbon mass is therefore mass_CO2 × 12.011/44.01.
How does the calculator find hydrogen?
Hydrogen is calculated from water because each H2O molecule contains two hydrogen atoms. The mass is mass_H2O × (2×1.008/18.015).
Why is oxygen calculated by difference?
CO2 and H2O directly reveal carbon and hydrogen. For a compound known to contain only C, H, and O, any sample mass not accounted for by carbon and hydrogen is assigned to oxygen.
Can this calculator handle nitrogen or halogens?
Not directly. If the sample contains N, S, halogens, metals, or other elements, those must be measured separately before oxygen by difference is meaningful.
What is the difference between empirical and molecular formula?
The empirical formula is the simplest whole-number atom ratio. The molecular formula is the actual atom count and requires the empirical formula plus the compound's molar mass.
Why might my empirical formula look unusual?
Small product-mass errors are magnified when ratios are divided by a small mole amount. Check that combustion was complete, product traps were dry, and the sample contains only C, H, and O.