How this instrument works
Combustion analysis is a classic technique for figuring out what an organic compound is made of without ever seeing its structure directly. Burn a sample completely in oxygen, and every carbon atom it contained ends up as CO2, while every hydrogen atom ends up as H2O — nothing else forms from those two elements under complete combustion. Weigh the CO2 and H2O produced, and you can work backward to the exact mass of carbon and hydrogen that must have been present in the original sample, since every mole of CO2 came from exactly one mole of carbon and every mole of H2O came from exactly two moles of hydrogen atoms.
The arithmetic rests on molar mass ratios, not guesswork: carbon makes up 12.011 out of every 44.01 grams of CO2 (its molar mass), so multiplying the measured CO2 mass by that ratio, 12.011/44.01, recovers the mass of carbon alone. The same logic applies to H2O: 2.016 out of every 18.015 grams of H2O is hydrogen, so multiplying measured H2O mass by 2.016/18.015 recovers the mass of hydrogen alone. Neither calculation depends on knowing the original compound's structure — only on the fixed atomic weights of carbon, hydrogen and oxygen.
Historically this was chemistry's primary tool for determining empirical formulas before modern spectroscopic methods existed, and it's still taught for exactly that reason: once you know the mass of carbon and hydrogen recovered, subtracting both from the original sample's total mass reveals any oxygen (or other element) that didn't show up as CO2 or H2O, and converting each element's mass to moles gives the whole-number ratio that defines the compound's empirical formula.
- Enter the measured mass of carbon dioxide produced into Mass of CO2 produced (g).
- Enter the measured mass of water produced into Mass of H2O produced (g).
- Read the recovered carbon mass off Mass of carbon recovered (g) and the recovered hydrogen mass off Mass of hydrogen recovered (g).
- Both readouts only account for carbon and hydrogen — if your original sample's total mass exceeds the sum of these two, the difference is oxygen or another element that didn't leave as CO2 or H2O.
- To find an empirical formula, convert each recovered mass to moles (divide by 12.011 for carbon, by 1.008 for hydrogen) and compare the mole ratios.
Worked example — 3.667 g CO2 and 1.5 g H2O
Enter 3.667 into Mass of CO2 produced (g) and 1.5 into Mass of H2O produced (g). Mass of carbon recovered (g) reads 1.00078 g and Mass of hydrogen recovered (g) reads 0.16786 g.
By hand: mass C = 3.667 × (12.011/44.01) = 3.667 × 0.27295 ≈ 1.0008 g, and mass H = 1.5 × (2.016/18.015) = 1.5 × 0.11190 ≈ 0.16786 g. If the original burned sample weighed, say, 1.5 g, subtracting both recovered masses (1.0008 + 0.16786 ≈ 1.169 g) from that total would leave roughly 0.33 g unaccounted for — most likely oxygen bound in the original compound, since oxygen from the air used to burn the sample can't be told apart from oxygen already present in it.
Questions
Why does every carbon atom become CO2 and every hydrogen atom become H2O?
Complete combustion in excess oxygen fully oxidizes both elements — carbon has nowhere else to go but CO2, and hydrogen has nowhere else to go but H2O, provided the sample burns completely rather than partially (which would produce soot or carbon monoxide instead). Because those are the only two products carbon and hydrogen can form under these conditions, their masses can be traced back unambiguously.
Can this method tell me if my sample contained oxygen?
Indirectly, yes. Combustion analysis alone can't detect oxygen the way it detects carbon and hydrogen, but if you know the original sample's total mass, subtracting the recovered carbon and hydrogen masses from that total reveals any remaining mass — which is typically oxygen, since combustion products can't distinguish oxygen already in the compound from oxygen supplied by the air used to burn it.
Where do the numbers 12.011, 44.01, 2.016 and 18.015 come from?
They're standard molar masses built from IUPAC atomic weights: carbon is 12.011 g/mol, so CO2 (one carbon plus two oxygens at 15.999 each) is 44.01 g/mol; hydrogen gas is 2.016 g/mol (two atoms at 1.008 each), so H2O (two hydrogens plus one oxygen) is 18.015 g/mol. These ratios are fixed by atomic physics, not by the specific compound being analyzed.
How do I turn recovered carbon and hydrogen masses into an empirical formula?
Convert each recovered mass to moles by dividing by its atomic weight — carbon mass ÷ 12.011, hydrogen mass ÷ 1.008 — then divide both mole counts by whichever is smaller to get a simple whole-number ratio. That ratio is the compound's empirical formula's subscripts for carbon and hydrogen; if oxygen or another element is also present, its moles are found the same way after subtracting known masses from the sample's total.
Does the amount of oxygen used for burning affect the result?
No — as long as combustion is complete, the excess oxygen supplied simply reacts and leaves as more CO2 and H2O without changing the ratio of carbon to hydrogen recovered from the original sample. What matters is that every carbon and hydrogen atom originally in the sample ends up fully accounted for in the measured product masses, regardless of how much extra oxygen was used to get there.