Combustion Analysis Calculator
Determine the empirical formula of an organic compound from combustion analysis data. Enter sample mass, CO2 mass, and H2O mass for step-by-step solutions.
Combustion Analysis Calculator
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Formula: Moles C = mass CO2 / 44.01 | Moles H = 2 * mass H2O / 18.015
Worked example โ Empirical formula determination from C, H, O ratios
Formula
Moles C = mass CO2 / 44.01 | Moles H = 2 * mass H2O / 18.015
Combustion analysis burns an organic sample in excess O2. Each mole of CO2 contains one mole of C, and each mole of H2O contains two moles of H. Mass of O is found by difference. Mole ratios give the empirical formula.
Worked Examples
Example 1: Combustion of an Unknown Hydrocarbon
Problem:0.500 g of an organic compound produces 1.467 g CO2 and 0.600 g H2O. Find the empirical formula.
Solution:Moles CO2 = 1.467/44.01 = 0.03334 mol = moles C Moles H2O = 0.600/18.015 = 0.03331 mol, moles H = 0.06662 Mass C = 0.03334 * 12.011 = 0.4003 g Mass H = 0.06662 * 1.008 = 0.0671 g Mass O = 0.500 - 0.4003 - 0.0671 = 0.0326 g Moles O = 0.0326/15.999 = 0.00204 Ratio C:H:O = 16.3:32.6:1 = ~16:33:1
Result:Empirical formula determination from C, H, O ratios
Example 2: Simple Combustion - Ethanol
Problem:1.000 g sample produces 1.913 g CO2 and 1.174 g H2O.
Solution:Moles CO2 = 1.913/44.01 = 0.04347 mol C Moles H2O = 1.174/18.015 = 0.06517 mol, H = 0.1303 mol Mass C = 0.5222 g, Mass H = 0.1314 g Mass O = 1.000 - 0.5222 - 0.1314 = 0.3464 g C:H:O ratio = 2:6:1 => C2H6O
Result:Empirical formula: C2H6O (ethanol)
Frequently Asked Questions
What is combustion analysis in chemistry?
Combustion analysis is an analytical technique used to determine the empirical formula of an organic compound by burning it in excess oxygen and measuring the masses of combustion products. Carbon in the sample becomes CO2, hydrogen becomes H2O, and nitrogen becomes N2. By weighing these products, you can calculate the moles of each element in the original sample. Any remaining mass is attributed to oxygen. This technique was developed by Justus von Liebig in the 1830s and remains a standard method for elemental analysis of organic compounds.
How do you find oxygen content from combustion analysis?
Oxygen content is determined by difference. First, calculate the mass of carbon from CO2 and the mass of hydrogen from H2O. If nitrogen is present, calculate its mass from N2. Subtract all these masses from the total sample mass to find the mass of oxygen. This indirect method is necessary because oxygen from the sample mixes with the excess oxygen used for combustion, making it impossible to measure directly. If the sum of C, H, and N masses exceeds the sample mass, recheck your calculations as this indicates an error.
What is the difference between empirical and molecular formulas?
The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms. For example, glucose has an empirical formula of CH2O and a molecular formula of C6H12O6. Combustion analysis directly gives the empirical formula. To find the molecular formula, you need the molar mass from another technique like mass spectrometry. The molecular formula is always a whole-number multiple of the empirical formula: molecular formula = n * empirical formula, where n = molar mass / empirical mass.
What are common sources of error in combustion analysis?
Common errors include incomplete combustion leaving unburned carbon (soot), moisture contamination of the sample or absorption tubes, air leaks in the apparatus allowing atmospheric gases to interfere, and incomplete absorption of CO2 or H2O by the traps. Halogen-containing compounds can produce HCl or other acids that interfere with water measurement. Sulfur compounds produce SO2 which must be trapped separately. Modern automated analyzers minimize these errors but require careful calibration with standard compounds of known composition.
Why is the oxygen figure the least reliable number in the analysis?
Because it is never measured, only left over, so every error made anywhere else accumulates in it. The sample weighing, the CO2 mass and the H2O mass each carry their own uncertainty, and the subtraction hands all of it to oxygen. The effect is worst when oxygen is a small part of the compound. In the 0.500 g example on this page the oxygen works out at 0.0326 g, about 6.5 percent of the sample, so a one percent error in the carbon mass, roughly 0.004 g, shifts the oxygen figure by more than 12 percent. A negative oxygen mass is the extreme form of the same arithmetic and always means a measurement is wrong rather than that the compound contains no oxygen.
Why does the calculation double the moles of water but not the moles of carbon dioxide?
It simply follows the stoichiometry of the two products. Every CO2 molecule carries exactly one carbon atom, so moles of carbon equal moles of CO2. Every H2O molecule carries two hydrogen atoms, so moles of hydrogen are twice the moles of water. For 0.600 g of water, 0.600 / 18.015 = 0.03331 mol of H2O, which is 0.06662 mol of H, or 0.06662 x 1.008 = 0.0672 g. Dropping that factor of two halves the hydrogen count and quietly turns a CH3 group into CH1.5.
What should I do when the mole ratios do not land near whole numbers?
Divide every mole count by the smallest one, then read the remainder. A value near .5 means multiplying the whole set by 2, so 1 : 1.5 becomes 2 : 3 and the formula is C2H3. A remainder near .33 or .67 means multiplying by 3. Ratios that stay stubbornly awkward, such as 1 : 2.2, are a data problem rather than an arithmetic one, typically adsorbed moisture, residual solvent of crystallisation or incomplete combustion, and the sample should be dried and reburned.
How is a nitrogen-containing compound handled in the analysis?
Nitrogen leaves the furnace as N2 once the nitrogen oxides formed during combustion are reduced over hot copper, so each mole of N2 collected stands for two moles of nitrogen atoms, exactly as one mole of water stands for two hydrogens. Entering the N2 mass subtracts that contribution before oxygen is taken by difference. This is the Dumas principle, and it differs from the Kjeldahl method, which digests the sample in sulfuric acid and titrates the ammonia released, reporting nitrogen alone.
Can combustion analysis distinguish between two isomers?
No. The technique reports composition rather than connectivity, so any two compounds sharing an atom ratio return identical data. Ethanol and dimethyl ether both analyse as C2H6O; glucose, fructose and galactose all give CH2O as the empirical formula and C6H12O6 as the molecular one. Telling them apart requires a structural method: NMR spectroscopy for the carbon-hydrogen framework, infrared for functional groups, or the fragmentation pattern from mass spectrometry. Elemental analysis instead confirms that a prepared sample has the intended composition and purity.
How do I check a proposed structure against the percentages reported here?
Run the calculation backwards. Sum the atomic masses for the candidate formula, then express each element's share as its total mass over the formula mass. For ethanol, C2H6O at 46.07 g/mol, carbon is 2 x 12.011 / 46.07 = 52.1 percent and hydrogen is 6 x 1.008 / 46.07 = 13.1 percent. Setting these calculated figures beside the found percentages is the standard reporting convention, and agreement to within a few tenths of a percent absolute is normally treated as satisfactory.
References
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Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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