Combustion Analysis Calculator
Enter the masses of your combustion products and sample to find the empirical formula and, when a molar mass is known, the molecular formula. The calculator handles pure hydrocarbons, oxygenated organics, and nitrogen-containing compounds. Every step of the stoichiometric calculation is shown below the result.
Formula
Worked example
A 12.915 g sample of an unknown C/H/O compound produces 18.942 g CO2 and 7.749 g H2O. m_C = 18.942 x (12.011/44.009) = 5.170 g; m_H = 7.749 x (2.016/18.015) = 0.867 g; m_O = 12.915 - 5.170 - 0.867 = 6.878 g. Moles: C = 0.4305, H = 0.8601, O = 0.4299. Ratio (divide by 0.4299): C1.0 H2.0 O1.0 -> empirical formula CH2O. With molar mass 90.08 g/mol: n = 90.08 / 30.026 = 3, so molecular formula = C3H6O3.
What is combustion analysis?
Combustion analysis is the oldest and most widely used method for determining the elemental composition of an organic compound. A small, precisely weighed sample is burned in excess oxygen inside a furnace. The resulting gases are passed through two sequential absorption traps: a desiccant (such as magnesium perchlorate) captures all the water produced, and an alkaline absorbent (such as ascarite or concentrated NaOH) captures all the carbon dioxide. The mass gained by each trap equals the mass of H2O and CO2 generated by the combustion. Because every carbon atom in the original compound ends up in one CO2 molecule, and every hydrogen atom ends up in one H2O molecule, the trapped masses let chemists calculate exactly how much carbon and hydrogen were present. Oxygen is found by subtraction: whatever mass remains after removing carbon and hydrogen from the total sample must be oxygen. Nitrogen can be measured with the Dumas modification, where NO2 is also captured and weighed. From these element masses the molar ratios are found, reduced to the smallest whole numbers, and written as the empirical formula. If the molar mass of the compound is also known (from mass spectrometry or colligative methods), the molecular formula follows directly.
How to use this calculator
Select your compound type from the dropdown: pure hydrocarbon (C and H only), oxygenated organic (C, H, and O), or a nitrogen-containing compound (C, H, O, and N). Enter the mass of your combustion sample in grams, then the mass gained by the CO2 trap and the H2O trap. For nitrogen-containing compounds, also enter the NO2 trap mass. If you already know the molar mass of your compound (for example from a mass spectrum), enter it in the optional molar mass field and the calculator will determine the molecular formula as well as the empirical formula. The step-by-step panel below the result shows every stoichiometric conversion with your actual numbers. If the molar mass you enter is inconsistent with the empirical formula (meaning the calculated n value is not close to a whole number), the calculator tells you so and suggests checking your trap masses or molar mass value.
Empirical formula vs. molecular formula
The empirical formula is the simplest whole-number ratio of atoms in the compound. Acetic acid (CH3COOH) and glucose (C6H12O6) both have the empirical formula CH2O, because their C:H:O atom ratios are the same even though the molecules are very different sizes. Combustion analysis alone gives only the empirical formula. The molecular formula shows the actual count of each atom in one molecule: glucose is C6H12O6, which is 6 times the empirical unit CH2O. To get from the empirical formula to the molecular formula you need the molar mass. The ratio n = molar mass / empirical formula mass must be a whole number (or very close to one within experimental error). Multiply each subscript in the empirical formula by n to get the molecular formula.
Accuracy, mass balance, and error sources
Combustion analysis is accurate to about 0.3 percent under controlled laboratory conditions when the traps are fully absorbing and the sample is dry and pure. Common sources of error include: incomplete combustion (leaving carbonaceous residue), moisture in the sample or reagents inflating the H2O reading, contamination of the CO2 trap with other acidic gases such as SO2 or NOx, and weighing errors. A mass balance check is built into this calculator: for C/H-only compounds the sum of carbon mass and hydrogen mass should equal the sample mass within a small tolerance; for oxygenated compounds the derived oxygen mass should be positive. If the oxygen by difference is negative, the CO2 or H2O reading is likely too high, or the sample contained an element not accounted for (such as sulfur). For compounds containing halogen, sulfur, or phosphorus, additional traps and a modified combustion train are required before the standard formulas apply.
Common organic compound benchmarks
| Compound | Formula | Molar mass (g/mol) | % C | % H | % O |
|---|---|---|---|---|---|
| Methane | CH4 | 16.04 | 74.9 | 25.1 | - |
| Ethanol | C2H6O | 46.07 | 52.1 | 13.1 | 34.7 |
| Acetic acid | C2H4O2 | 60.05 | 40.0 | 6.7 | 53.3 |
| Benzene | C6H6 | 78.11 | 92.3 | 7.7 | - |
| Glucose | C6H12O6 | 180.16 | 40.0 | 6.7 | 53.3 |
| Aspirin | C9H8O4 | 180.16 | 60.0 | 4.4 | 35.5 |
| Caffeine | C8H10N4O2 | 194.19 | 49.5 | 5.2 | 16.5 |
| Naphthalene | C10H8 | 128.17 | 93.7 | 6.3 | - |
Typical percent compositions from combustion analysis for reference compounds.
Frequently asked questions
What inputs are required for combustion analysis?
You need the mass of the sample burned, the mass gained by the CO2 absorption trap, and the mass gained by the H2O absorption trap. The CO2 mass gives you carbon content, the H2O mass gives you hydrogen content, and oxygen (if present) is found by subtracting both from the sample mass. A molar mass value (optional) from mass spectrometry or a colligative property measurement lets the calculator go further and determine the molecular formula.
How is oxygen determined in combustion analysis?
Oxygen is not directly measured in the standard method. Instead it is found by difference: oxygen mass = sample mass - carbon mass - hydrogen mass. This works because every element in the compound must go somewhere after combustion, and if the only elements are C, H, and O, any mass not accounted for by carbon and hydrogen must be oxygen. For this reason the sample must be pure and dry, and all other elements must be absent or separately accounted for.
Why do glucose and acetic acid give the same empirical formula?
Glucose (C6H12O6) and acetic acid (C2H4O2) both reduce to the ratio C1H2O1, written CH2O, because the ratios of their atom counts are identical even though the totals differ. Combustion analysis measures mass ratios and atom ratios, not the absolute size of the molecule. This is why the molecular formula requires an independent molar mass determination. Two isomers with the same molecular formula cannot be distinguished by combustion analysis at all.
What is the Dumas method for nitrogen?
In the Dumas modification of combustion analysis, a nitrogen-containing compound is burned in a copper oxide furnace. Nitrogen oxides produced during combustion are reduced over hot copper metal to N2 gas, or alternatively the NO2 is absorbed in a second trap. By weighing the NO2 collected and applying the ratio M_N / M_NO2, the mass of nitrogen in the original sample can be calculated. This calculator uses the NO2 trap approach: enter the mass gained by the NO2 trap and the calculator adds nitrogen to the analysis.
What compounds cannot be analyzed by standard combustion analysis?
Standard combustion analysis is limited to compounds containing only C, H, O, and N. Compounds containing sulfur produce SO2, which can enter the CO2 trap and inflate the carbon reading unless a scrubber is added. Halogen-containing compounds produce HX gases. Phosphorus and silicon form solid oxides. Metals and inorganic ions require a modified furnace train and different absorption chemistry. The method also fails if the compound does not burn completely.
How accurate is combustion analysis?
Under good laboratory conditions, combustion analysis is accurate to within about 0.3 percent by mass for each element. This is precise enough to determine empirical formulas reliably, but it means that the molecular formula derived from the empirical formula and a molar mass can still be ambiguous if the molar mass measurement is imprecise. For large molecules where n is 10 or more, a 1 g/mol error in the molar mass will still give the correct integer ratio, but errors of 5-10 g/mol can shift n by one unit and give the wrong molecular formula.
How do I find the molecular formula if the molar mass is not a whole multiple of the empirical mass?
In practice, the n value is never perfectly integer because of measurement uncertainties in both the combustion data and the molar mass. The standard convention is to round n to the nearest integer if the deviation is within about 5-10 percent. If the deviation is larger, either the molar mass source is inaccurate, the combustion data contains a systematic error, or the compound contains an element not included in the analysis such as a halogen. Check the mass balance first: for a C/H compound, the sum of carbon and hydrogen masses should be close to the total sample mass.