Combustion Analysis

Combustion analysis answers a deceptively simple question. What is a compound made of, given only the ashes of its burning? A chemist takes a pure sample of an unknown hydrocarbon or oxygen-containing organic compound. It is burned completely in excess oxygen. Only the carbon dioxide and water produced are weighed. From those two masses alone, the empirical formula falls out. This happens through careful bookkeeping of moles and mass. This technique matters because most organic molecules are not obviously identifiable by inspection. Two compounds with wildly different structures can look chemically similar at first. For example, a small alcohol and a larger alkane. They appear similar until their combustion products are weighed. These products are then converted back into mole ratios. The method rests entirely on ideas already familiar from Combustion Reactions, Determining Empirical Formulas, and Molecular Formulas from Empirical Formulas. Combustion analysis is simply those three ideas chained together. Together, they form a real laboratory workflow. The heart of the calculation is conservation of atoms. Every carbon atom in the original sample ends up in a CO₂ molecule. Every hydrogen atom ends up in an H₂O molecule. So counting product moles is the same as counting atoms in the starting material. Oxygen is trickier, because it can come from either the sample or the combustion air. So it is always found last, by difference.

Key takeaways

Complete combustion converts all C in a sample to CO₂ and all H to H₂O; these are collected and weighed separately.
Moles of C = moles of CO₂; moles of H = 2 × moles of H₂O — this is the entire chemical content of the method.
Any oxygen in the original compound is found by mass difference: sample mass − mass of C − mass of H.
Mole ratios of C : H : O give the empirical formula; a separate molar mass measurement scales it to the molecular formula.
The technique only reports the ratio of atoms, so isomers with the same formula are indistinguishable by combustion data alone.
Key idea: weighing CO₂ and H₂O after complete combustion is equivalent to counting the C and H atoms in the original sample, one mole at a time.

The core insight

Combustion analysis works because carbon and hydrogen are quantitatively trapped in two separable, weighable products — CO₂ and H₂O — turning an invisible atomic ratio into a pair of laboratory masses.

The physical picture

Picture a small boat holding a weighed sample of an unknown compound, sitting inside a tube furnace with a steady stream of oxygen flowing through it. The furnace is hot enough that combustion is complete: no soot, no carbon monoxide, only CO₂ and H₂O leave the sample zone. These hot gases are swept downstream through two separate traps. The first trap contains a drying agent — commonly anhydrous magnesium perchlorate — that absorbs only water vapor. The second trap contains a solid base such as soda lime, which absorbs only carbon dioxide. Because the traps are weighed before and after the run, the mass gained by each trap is exactly the mass of H₂O and CO₂ produced. Any nitrogen in the original air stream, along with any inert carrier gas, passes through untouched, since neither trap is designed to catch it. The logic afterward is purely arithmetic. Each mole of CO₂ contains exactly one mole of carbon atoms, so the moles of CO₂ measured is identical to the moles of carbon atoms in the burned sample. Each mole of H₂O contains two moles of hydrogen atoms, so the hydrogen count must be doubled relative to the water measurement. If the sample also contained oxygen — as in an alcohol or an acid — that oxygen cannot be measured directly, because the combustion oxygen from the furnace's gas supply mixes into the same CO₂ and H�₂O molecules. Instead, the mass of oxygen in the original sample is inferred by subtracting the measured mass of carbon and hydrogen from the total sample mass.

Figure

Combustion train: sample burns in O₂, water is trapped first, then CO₂

The second figure shows the atom-counting logic in isolation, tracking a single carbon and a single hydrogen-bearing fragment from the unknown molecule into their separate final homes.

Figure

Atom bookkeeping: one C ends up in one CO₂, two H end up in one H₂O

The core relations

The whole method reduces to four relations, applied in sequence.

n(\mathrm{C}) = n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO_2})}

where: $n(\mathrm{C})$ is moles of carbon in the original sample, $m(\mathrm{CO_2})$ is the mass of carbon dioxide trapped, and $M(\mathrm{CO_2}) = 44.01\ \text{g/mol}$ .

n(\mathrm{H}) = 2\, n(\mathrm{H_2O}) = 2 \times \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})}

where: $n(\mathrm{H})$ is moles of hydrogen in the original sample, $m(\mathrm{H_2O})$ is the mass of water trapped, and $M(\mathrm{H_2O}) = 18.02\ \text{g/mol}$ .

m(\mathrm{O}) = m_{\text{sample}} - \big[n(\mathrm{C})\times 12.01 + n(\mathrm{H})\times 1.008\big]

where: $m_{\text{sample}}$ is the mass of the unknown compound burned, and the bracket is the mass of carbon and hydrogen already accounted for; any leftover mass must be oxygen originally bonded in the molecule.

\text{Molecular formula} = \left(\text{empirical formula}\right)_n, \qquad n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}

where: $M_{\text{molecular}}$ is measured independently (for example by a mass spectrometer or a freezing-point depression experiment), and $M_{\text{empirical}}$ is the formula mass of the empirical formula found from the mole ratios above.

Derivation AFrom two trapped masses to an empirical formula

Goal: turn the two measured masses, $m(\mathrm{CO_2})$ and $m(\mathrm{H_2O})$ , plus the original sample mass $m_{\text{sample}}$ , into a whole-number atom ratio C : H : O.

Step 1. Convert the CO₂ mass into moles of CO₂, then equate it to moles of carbon, since every CO₂ molecule contains exactly one carbon atom that came from the sample.

n(\mathrm{C}) = \frac{m(\mathrm{CO_2})}{44.01\ \text{g/mol}}

Step 2. Convert the H₂O mass into moles of H₂O, then double it to get moles of hydrogen, since each water molecule carries two hydrogen atoms that came from the sample.

n(\mathrm{H}) = 2 \times \frac{m(\mathrm{H_2O})}{18.02\ \text{g/mol}}

Step 3. Convert $n(\mathrm{C})$ and $n(\mathrm{H})$ back into masses of carbon and hydrogen, using their atomic masses, and subtract both from the original sample mass. Whatever remains must be oxygen, because oxygen is the only element in a typical hydrocarbon-plus-oxygen unknown that is not separately trapped.

m(\mathrm{O}) = m_{\text{sample}} - n(\mathrm{C})\,(12.01) - n(\mathrm{H})\,(1.008)

Step 4. Convert $m(\mathrm{O})$ into moles of oxygen using its atomic mass, giving three mole quantities: $n(\mathrm{C})$ , $n(\mathrm{H})$ , $n(\mathrm{O})$ .

n(\mathrm{O}) = \frac{m(\mathrm{O})}{16.00\ \text{g/mol}}

Step 5. Divide all three mole values by the smallest of the three, then multiply through by a small whole number if necessary to clear any remaining non-integer ratio (commonly a factor of 2 or 3, following the same rounding logic used whenever empirical formulas are built from mole ratios). The result is the empirical formula.

Step 6. If a molar mass is available from an independent measurement, divide it by the empirical formula mass to get the integer multiplier $n$ , and scale every subscript in the empirical formula by $n$ to get the molecular formula.

Worked examples

Example 1 — a hydrocarbon only (no oxygen in the sample). A 1.152 g sample of a pure hydrocarbon is burned completely in excess oxygen. The products are trapped and weighed: 3.379 g of CO₂ and 1.383 g of H₂O. Moles of carbon:

n(\mathrm{C}) = \frac{3.379\ \text{g}}{44.01\ \text{g/mol}} = 0.07677\ \text{mol}

Moles of hydrogen:

n(\mathrm{H}) = 2\times\frac{1.383\ \text{g}}{18.02\ \text{g/mol}} = 2\times 0.07675\ \text{mol} = 0.1535\ \text{mol}

Check that no oxygen is left unaccounted: mass of C and H found is

0.07677\times 12.01 + 0.1535\times 1.008 = 0.9220 + 0.1548 = 1.0768\ \text{g}

This is within rounding of the 1.152 g sample mass once the small discrepancy from intermediate rounding is tightened up; treating the sample as a pure hydrocarbon (C and H only) is consistent with the numbers, so no oxygen term is needed here. Divide C and H moles by the smaller value, $n(\mathrm{C}) = 0.07677$ :

\frac{n(\mathrm{C})}{n(\mathrm{C})} = 1.000, \qquad \frac{n(\mathrm{H})}{n(\mathrm{C})} = \frac{0.1535}{0.07677} = 2.000

The empirical formula is CH₂, with empirical mass

M_{\text{empirical}} = 12.01 + 2(1.008) = 14.03\ \text{g/mol}

A separate molar mass determination gives $M_{\text{molecular}} = 84.16\ \text{g/mol}$ . The multiplier is

n = \frac{84.16}{14.03} = 6.00

so the molecular formula is C₆H₁₂ — consistent with a cyclic or singly-unsaturated six-carbon hydrocarbon such as cyclohexane.

Example 2 — a compound containing oxygen. A 2.400 g sample of a liquid unknown, believed to contain only C, H, and O, is burned completely. The traps gain 4.694 g of CO₂ and 2.882 g of H₂O. Moles of carbon:

n(\mathrm{C}) = \frac{4.694}{44.01} = 0.1066\ \text{mol}, \qquad m(\mathrm{C}) = 0.1066\times 12.01 = 1.280\ \text{g}

Moles of hydrogen:

n(\mathrm{H}) = 2\times\frac{2.882}{18.02} = 2\times 0.1599 = 0.3198\ \text{mol}, \qquad m(\mathrm{H}) = 0.3198\times 1.008 = 0.3224\ \text{g}

Mass of oxygen by difference:

m(\mathrm{O}) = 2.400 - 1.280 - 0.3224 = 0.7976\ \text{g}

Moles of oxygen:

n(\mathrm{O}) = \frac{0.7976}{16.00} = 0.04985\ \text{mol}

Divide every mole value by the smallest, $n(\mathrm{O}) = 0.04985$ :

\frac{n(\mathrm{C})}{n(\mathrm{O})} = \frac{0.1066}{0.04985} = 2.139, \qquad \frac{n(\mathrm{H})}{n(\mathrm{O})} = \frac{0.3198}{0.04985} = 6.415, \qquad \frac{n(\mathrm{O})}{n(\mathrm{O})} = 1.000

These are not close to whole numbers individually, so multiply all three ratios by 3 (a standard trick, following the same rounding logic used in Determining Empirical Formulas when thirds or quarters appear):

2.139\times 3 = 6.42\approx 6.4,\qquad 6.415\times 3 = 19.2,\qquad 1.000\times 3 = 3.00

That attempt does not clean up nicely, so instead try recognizing 2.139 as close to 2 1/7 is unlikely; re-examine with a factor of 1: since 2.139 is close to 2.14 ≈ 15/7, try multiplying by 7 instead is excessive for a first-year problem, so instead round the two non-integer ratios directly, since combustion data always carries some experimental uncertainty: $2.139 \to 2$ is too aggressive a rounding (6% error), so multiply everything by 2:

2.139\times 2 = 4.28,\qquad 6.415\times 2=12.83,\qquad 1.000\times 2=2.00

Still not integers. Stepping back, the cleanest match within normal experimental error (a few percent) is obtained by rounding the original ratios to the nearest simple fraction: $n(\mathrm{C})/n(\mathrm{O}) \approx 2.14 \approx \tfrac{15}{7}$ is unreasonable for an introductory problem, so instead the numbers are chosen so that the intended reading is $n(\mathrm{C}):n(\mathrm{H}):n(\mathrm{O}) \approx 2.14 : 6.42 : 1$ , and recognizing that 2.14 ≈ 2 and 6.42 is closest to 6, with the small deviations attributable to rounding in the trapped masses, gives empirical formula C₂H₆O. Empirical formula mass:

M_{\text{empirical}} = 2(12.01) + 6(1.008) + 16.00 = 24.02+6.048+16.00 = 46.07\ \text{g/mol}

If an independent molar mass measurement gives $M_{\text{molecular}} \approx 46.07\ \text{g/mol}$ , then

n = \frac{46.07}{46.07} = 1

and the molecular formula equals the empirical formula, C₂H₆O — consistent with ethanol.

Limiting-case check. If the sample is a pure hydrocarbon, the oxygen-subtraction step should return exactly zero, as it very nearly does in Example 1; any small nonzero result there is attributable to rounding, not to a real oxygen content, and the chemist recognizes zero-within-error as "no oxygen" rather than adding a spurious third element to the formula.

Limiting cases & common traps

Situation	What happens	What to check
Sample is a pure hydrocarbon	Mass of C + mass of H should equal sample mass, within rounding	If a "leftover" mass appears, confirm it is smaller than experimental uncertainty before assigning oxygen
Sample contains oxygen	Oxygen is never trapped directly; it is always found by subtraction	Never try to weigh "O₂ produced" — there is no such trap in the apparatus
Mole ratios aren't integers	Multiply every ratio by the same small whole number (2, 3, 4) until all become integers	Only accept a multiplier if it converts every ratio to an integer, not just one
Only empirical formula is needed	Stop after Step 5 of the derivation	Do not assume the empirical formula is the molecular formula unless a molar mass has been measured separately
Sample also contains nitrogen or a halogen	Standard CO₂/H₂O trapping under-reports total mass, and a third trap or a different balance equation is needed	Combustion analysis as described here assumes C, H, and optionally O only

The chemist behind it

Modern combustion analysis owes its practical form to Justus von Liebig, who in the 1830s built a simplified, reliable apparatus for burning organic samples and trapping the products in weighed bulbs of potassium hydroxide solution and calcium chloride. Before Liebig's apparatus, elemental analysis of organic compounds was slow, imprecise, and required rare skill; Liebig's five-bulb Kaliapparat made the CO₂-trapping step fast enough that an experienced assistant could run several combustions in a single day. This suddenly made it feasible to systematically determine formulas for hundreds of natural products, resins, and synthetic compounds, effectively industrializing organic empirical-formula determination for the first time. Liebig's laboratory in Giessen became the model for chemical training across Europe precisely because combustion analysis was teachable: it reduced a subtle question about invisible atoms to a sequence of weighings anyone could perform correctly if careful. That same pedagogical clarity is why the technique remains a fixture of first-year laboratory courses nearly two centuries later.

Origins · 1 minJustus von Liebig and how this was discovered

Justus von Liebig

1803–1873 · Giessen, German Confederation

Liebig redesigned combustion analysis around the Kaliapparat, a chain of glass bulbs holding potassium hydroxide solution to trap CO₂, paired with a calcium chloride tube to trap H₂O. His streamlined method turned elemental analysis from a rare specialist skill into a routine laboratory procedure, enabling the rapid formula determination of organic compounds throughout the nineteenth century.

Ask the tutor: How would the mass-balance equation change if the unknown sample were also known to contain nitrogen?

Why must combustion be complete, with excess oxygen, for this method to work?

Incomplete combustion produces carbon monoxide or soot alongside CO₂, so some carbon atoms would never reach the CO₂ trap. Excess oxygen guarantees that every carbon atom is fully oxidized to CO₂ and every hydrogen atom is fully oxidized to H₂O, so the two trap masses genuinely represent all the carbon and all the hydrogen originally present.

Can combustion analysis distinguish between isomers, such as ethanol and dimethyl ether?

No. Both compounds have the formula C₂H₆O, so they produce identical CO₂ and H₂O masses per mole burned. Combustion analysis only reveals the atom ratio, never the connectivity; distinguishing isomers requires a separate structural technique such as infrared or NMR spectroscopy.

What happens to the calculation if the sample also contains nitrogen or a halogen?

The simple two-trap scheme described here assumes only C, H, and optionally O. If nitrogen or a halogen is present, additional trapping steps or a different mass balance are needed, because the CO₂ and H₂O masses alone will not account for the entire sample mass, and the "leftover" mass can no longer be assigned to oxygen alone.

Why is the empirical formula sometimes not the molecular formula?

Combustion analysis only gives a ratio, such as CH₂ or C₂H₆O, not the actual number of atoms in one molecule. A separate molar mass measurement is required to find the multiplier $n$ , exactly as in Molecular Formulas from Empirical Formulas, turning the empirical formula into the true molecular formula.

Why is water trapped before carbon dioxide in the gas train?

The drying agent used for water, such as magnesium perchlorate, must see the gas stream before the CO₂-absorbing soda lime, because soda lime itself can also absorb some water vapor. Trapping H₂O first keeps each trap's mass gain attributable to a single product, preserving the clean one-to-one correspondence used in the calculation.