Combustion Analysis Calculator

Find a compound's empirical and molecular formula from its combustion products.

Handles C-H and C-H-O samples in micrograms, milligrams, grams and ounces.

Updated October 18, 2026

CreatorFrank Zhao

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Introduction

Burn a weighed organic sample in plenty of oxygen, catch the carbon dioxide and the water it gives off, and weigh them. Every carbon atom in the sample has moved into the CO₂, every hydrogen atom into the H₂O, so those two masses carry the compound's composition back with them. This page does that bookkeeping: product masses in, empirical formula out, and a molecular formula as soon as you add the compound's molar mass.

It is the calculation that sits between a finished combustion run and the next thing you want to do with the compound, and it is usually done in a hurry at the end of a lab session. Students check a lab report with it, technicians use it to see whether an instrument run makes sense before it goes into a notebook, and anyone who has a set of product masses but no formula can get to one from here.

Two of the inputs are conditional, and it is worth knowing why before you look for them. The sample mass only appears for a C-H-O compound, because that is the only case where oxygen has to be found by subtraction — a hydrocarbon gives it up directly through the two products. The molar mass is optional and changes nothing until you fill it in: the empirical formula is complete without it, and the molecular formula needs it.

The page also works in the other direction. Any of the elemental masses or the sample mass can be typed in directly, and the one you left out is solved for — useful when you weighed the water differently, or when you have a carbon mass from another method.

C-H and C-H-O

Hydrocarbon or oxygen-containing samples

Molar mass optional

Adds the molecular formula

µg, mg, g, oz

On every mass box, sample included

Quick start

Pick the family your sample belongs to

Hydrocarbons (C-H) is the opening choice. If your compound also contains oxygen, switch to CHO compounds (C-H-O) — that adds the sample-mass box and the oxygen line, because oxygen is then the part of the sample that the two products do not account for.

Type the two product masses

Carbon dioxide and water, in whichever unit your balance reads. The examples below use 18.942 g of CO₂ and 7.749 g of H₂O; a µg, mg or oz reading is converted to grams before anything else happens, so the arithmetic never mixes units.

For a C-H-O sample, add the mass you burnt

12.915 g in this example. The moment both product masses and the sample mass are in, the carbon, hydrogen and oxygen masses fill themselves in below the “Show atom masses from sample” switch. Type over any of the three and its partner is back-solved: carbon from CO₂ or CO₂ from carbon, hydrogen from water or water from hydrogen, and the oxygen through the sample balance.

Add the molar mass if you have one

90.0779 g/mol here. This is the whole molecule, not the empirical unit, and it normally comes from a separate measurement — a mass spectrum, freezing point depression, or a value on the sample label. If you only have the formula, the molar mass calculator turns a formula into that number.

Read the three result boxes

The result area shows the empirical formula, its mass, and the molecular formula with the multiple that connects them. For this run:

Elemental masses

m_{\mathrm{C}}

=

18.942 \times \frac{12.0107}{44.0095}

=

5.1695\ \mathrm{g}

Hydrogen

m_{\mathrm{H}}

=

7.749 \times \frac{2 \times 1.00784}{18.01508}

=

0.8670\ \mathrm{g}

Oxygen, by difference

m_{\mathrm{O}}

=

12.915 - 5.1695 - 0.8670

=

6.8785\ \mathrm{g}

Mole ratio, then the formula

n_{\mathrm{C}}:n_{\mathrm{H}}:n_{\mathrm{O}}

=

0.43041 : 0.86028 : 0.42992

=

1.001 : 2.001 : 1

\rightarrow

\mathrm{CH_2O}

The empirical mass is 30.0258 g/mol, so the supplied molar mass is

n = \frac{90.0779}{30.0258} = 3.0000 \approx 3

\rightarrow

\mathrm{C_3H_6O_3}

A ratio of 1.001 rather than a clean 1 is normal: the three elemental masses come from weighed products, so the counts sit a thousandth of an atom away from whole numbers. The method below says how much slack is allowed.

How the formula is worked out

Three relations do all the work. Two of them move carbon and hydrogen out of the products; the third closes the mass balance of the sample, which is where oxygen comes from.

Carbon and hydrogen come out of the products

Each CO₂ molecule carries one carbon and each H₂O molecule carries two hydrogens, so the elemental mass is the product mass scaled by the fraction of the product that is that element:

m_{\mathrm{C}} = m_{\mathrm{CO_2}} \times \frac{M_{\mathrm{C}}}{M_{\mathrm{CO_2}}}

\qquad

m_{\mathrm{H}} = m_{\mathrm{H_2O}} \times \frac{2\,M_{\mathrm{H}}}{M_{\mathrm{H_2O}}}

With water, the factor of two matters: a mole of water weighs about 18 g and holds only about 2 g of hydrogen, so the hydrogen mass is roughly a ninth of the water mass. Forgetting that two doubles the hydrogen you claim and turns CH₄ into CH₈.

Oxygen is whatever is left of the sample

A C-H-O compound contributes no marker of its own to the products, so its oxygen cannot be weighed directly. It is the remainder:

m_{\mathrm{O}} = m_{\mathrm{sample}} - m_{\mathrm{C}} - m_{\mathrm{H}}

This is also the mass balance the page keeps enforcing: type an oxygen mass and the sample mass becomes the quantity that is solved for, so the four boxes can never disagree on screen.

Moles, then the smallest whole-number ratio

Divide each elemental mass by its atomic mass to get moles, then divide all three moles by the smallest of them. The resulting ratios are close to small whole numbers, and the page searches for the smallest multiplier that clears them all at once:

r_i = \frac{n_i}{n_{\min}}, \qquad \text{counts} = \mathrm{round}(k\,r_i)

Two limits keep the search honest. The tolerance is 0.05 atom per count, so a ratio of 1.5986 does not get rounded to 2 — it needs a multiplier of five to become 7.99, which is close enough to 8. And the search stops at a multiplier of twelve, with no element allowed more than 30 atoms, because combustion data that needs a larger factor no longer identifies a small molecule: at that point a formula would be invented rather than found.

Ratios that appear out of real weighing data, and the multiplier they need:

Ratio seen (smallest = 1)	Multiplier	Counts
1 : 2.001	×1	1 : 2
1 : 1.500	×2	2 : 3
1 : 1.333	×3	3 : 4
1 : 1.250	×4	4 : 5
1 : 1.599	×5	5 : 8
1 : 1.167	×6	6 : 7

Empirical mass, then the molecular multiple

The empirical mass is what one formula unit weighs: the atom counts multiplied by their atomic masses and added up.

M_{\mathrm{emp}} = a\,M_{\mathrm{C}} + b\,M_{\mathrm{H}} + c\,M_{\mathrm{O}}

\qquad

n = \frac{M_{\mathrm{molar}}}{M_{\mathrm{emp}}}

A real molecule is a whole number of empirical units, so the page accepts a factor that is within 0.1 of a whole number and at least 1. If the molar mass you gave does not divide that way, no molecular formula is printed — the page says so instead of rounding 1.5 or 2.4 into a molecule that does not exist.

The atomic masses behind those numbers

Carbon is taken as 12.0107 g/mol, hydrogen as 1.00784 g/mol and oxygen as 15.9994 g/mol. The two product molar masses are then built from the same three numbers rather than typed in separately, which is what keeps the arithmetic self-consistent:

M_{\mathrm{CO_2}} = 12.0107 + 2 \times 15.9994 = 44.0095\ \mathrm{g/mol}

M_{\mathrm{H_2O}} = 2 \times 1.00784 + 15.9994 = 18.01508\ \mathrm{g/mol}

All three sit inside the uncertainty intervals of the current IUPAC standard atomic weights, quoted at the precision the source tables print them rather than the five-digit abridged values (12.011, 1.0080 and 15.999). The difference lives in the fifth significant figure, so it cannot change a formula for clean data — but it is why a textbook answer may print an empirical mass of 30.03 g/mol where this page prints 30.0258 g/mol. The CIAAW abridged table of standard atomic weights lists the current five-digit values with their uncertainties.

Every box can drive the others

Because each relation is two-sided, the page solves whichever quantity you left blank: type a carbon mass and the CO₂ mass follows, type the water mass and the hydrogen mass follows, type the sample mass and the oxygen is the difference. The molar mass box is the only one that never feeds back into the empirical part — filling it in adds the molecular formula and stops there.

m_{\mathrm{CO_2}},\ m_{\mathrm{H_2O}}

Masses of carbon dioxide and water weighed after the burn

m_{\mathrm{sample}}

Mass of the compound that was burnt; needed for a C-H-O sample

m_{\mathrm{C}},\ m_{\mathrm{H}},\ m_{\mathrm{O}}

Elemental masses inside the sample; oxygen is the remainder

M_{\mathrm{C}},\ M_{\mathrm{H}},\ M_{\mathrm{O}}

Atomic masses: 12.0107, 1.00784 and 15.9994 g/mol

n_i

Moles of an element: its mass divided by its atomic mass

r_i

Mole ratio: each mole count divided by the smallest one

k

Smallest multiplier that turns every ratio into a whole number

M_{\mathrm{emp}}

Mass of one empirical formula unit, in g/mol

n

Molar mass divided by the empirical mass

Two worked examples

The first run is a hydrocarbon, where the ratio needs a multiplier and the molar mass then confirms it. The second is the same kind of data with one mass typed in wrong, to show what a failing ratio looks like before you decide the chemistry is wrong.

A hydrocarbon that only resolves at ×5

Situation: A 204.35 g/mol hydrocarbon gives 33.057 g of CO₂ and 10.816 g of H₂O. The sample mass is not needed, because a C-H compound has no oxygen to find by subtraction.

Entered

Carbon dioxide33.057 g

Water10.816 g

Molar mass204.35 g/mol

What the page returns

Elemental masses

m_{\mathrm{C}} = 33.057 \times \frac{12.0107}{44.0095} = 9.0216\ \mathrm{g}

\qquad

m_{\mathrm{H}} = 10.816 \times \frac{2 \times 1.00784}{18.01508} = 1.2102\ \mathrm{g}

Mole ratio

n_{\mathrm{C}} : n_{\mathrm{H}} = 0.75113 : 1.20077 = 1 : 1.5986

Multiplier and formula

1.5986 \times 5 = 7.993 \approx 8

\rightarrow

\mathrm{C_5H_8}

\quad M_{\mathrm{emp}} = 68.1162\ \mathrm{g/mol}

n = \frac{204.35}{68.1162} = 3.0000

\rightarrow

\mathrm{C_{15}H_{24}}

In practice: Five carbon to eight hydrogen is a real formula (C₅H₈ is the empirical unit of several cyclic and unsaturated hydrocarbons), and the molar mass confirms it: 3 × 68.1162 is 204.35, so the molecule is C₁₅H₂₄.

A sample mass that is one gram short

Situation: The same 18.942 g of CO₂ and 7.749 g of H₂O, but the sample mass is typed as 12.000 g instead of 12.915 g. Everything else is identical, and the page refuses to name a formula.

Entered

Carbon dioxide18.942 g

Water7.749 g

Sample mass12.000 g (mistyped)

Molar mass90.0779 g/mol

What the page returns

Where the damage lands

m_{\mathrm{C}} = 5.1695\ \mathrm{g}

\qquad m_{\mathrm{H}} = 0.8670\ \mathrm{g}

\qquad m_{\mathrm{O}} = 12.000 - 5.1695 - 0.8670 = 5.9635\ \mathrm{g}

Mole ratio

n_{\mathrm{C}} : n_{\mathrm{H}} : n_{\mathrm{O}} = 1.1547 : 2.3080 : 1

No multiplier up to twelve brings those three within 0.05 of whole numbers, so the result area reports that the masses do not reduce to a small whole-number ratio instead of printing a formula. The Mass of oxygen box also stays grey-blue rather than turning into a plausible-looking number.

In practice: Correct the sample mass rather than the products. The carbon and hydrogen lines are pinned by their own weighed products, so a sample that is one gram light shows up entirely in the oxygen line — 5.9635 g instead of 6.8785 g — and from there in the ratio.

Telling a good answer from a shaky one

A formula that appears on screen is not automatically right. These four checks take a few seconds and catch almost every bad combustion result.

The oxygen line absorbs the mistakes

Carbon and hydrogen are each locked to their own weighed product, so an error in the CO₂ mass hurts carbon and nothing else. Oxygen is a subtraction, which means the errors of the sample mass, the carbon and the hydrogen all land on it together. A sample mass that is 1% low moves the oxygen count by 1% while the other two counts stay put.

So if one element is off, look at the oxygen first; if all three are off, the problem is upstream — an incomplete burn, a leak, or the wrong sample mass on the report.

Look at the multiplier before the formula

Clean data lands on the smallest multipliers. Ratios that resolve at ×1 or ×2 are the normal case; ×3 and ×4 show up in real compounds; ×7 and above are a sign that something in the weighing is off, even if the arithmetic manages to find a fit.

A fit that only just squeezes inside the 0.05 atom window is a warning, not a triumph.

A scale slip looks like a chemistry problem

The mass boxes accept µg, mg, g and oz independently, and one box left in the wrong unit is a factor of a thousand. A CO₂ mass entered as 18.942 mg when it was grams makes carbon a thousand times too small, and the ratio falls apart rather than looking obviously wrong.

Before rethinking the chemistry, check that the three masses are the same order of magnitude as the sample.

Cross-check in percent, not in grams

Take the formula the page gives you, work out what mass percentage of each element it implies, and compare that with your own data: carbon is $m_{\mathrm{C}} / m_{\mathrm{sample}}$ , and likewise for hydrogen and oxygen. For the first example the formula CH₂O implies 40.0% carbon, 6.7% hydrogen and 53.3% oxygen, which is much easier to eyeball against a report than grams of CO₂. The percent composition calculator does that conversion for any formula you paste in.

Round the answer, not the working

The result boxes show the empirical mass to four decimals and the molar-mass multiple to three, which is more than the input data usually deserves. Quote the formula as the answer and the masses as supporting arithmetic; do not carry 30.0258 g/mol into a report as if it were measured to six figures.

Questions

The page says CH₂O but the molecule is C₃H₆O₃. Which one is right?

Both, at different levels. Combustion analysis measures the ratio between the elements, which is the lowest whole-number ratio — the empirical formula. It cannot tell you how many of those units are stacked into the real molecule; only the molar mass can. That is why the molar mass box changes the molecular formula and never the empirical one.

\mathrm{CH_2O} \times 3 = \mathrm{C_3H_6O_3}

\qquad 30.0258 \times 3 = 90.0779\ \mathrm{g/mol}

Why does it sometimes refuse to give an empirical formula?

Because the ratio does not reduce. The page needs a multiplier of at most twelve that brings every mole count within 0.05 of a whole number, and no element count above 30. When that fails, the usual causes, in order of likelihood, are:

A sample mass that is wrong by more than a few tenths of a percent, which shifts the oxygen line.
One mass entered in a different unit from the others.
An incomplete burn, so not all the carbon reached the CO₂ trap.
A compound that contains nitrogen, sulfur or a halogen, whose products were not weighed.

Can it handle a compound with nitrogen, sulfur or chlorine?

No. This page covers carbon, hydrogen and oxygen only. A nitrogen-containing compound puts some of its nitrogen into the gas stream, and a halogen ends up as a silver halide in the trap rather than in the CO₂ or the water, so the oxygen-by-difference step would quietly absorb that mass and give a wrong formula.

For those samples, work from a measured composition instead — the empirical formula calculator turns mass or percentage data for up to ten elements into a formula without assuming what the combustion products were.

Why did typing the oxygen mass change my sample mass?

Because the four masses are one balance, not four independent facts: $m_{\mathrm{sample}} = m_{\mathrm{C}} + m_{\mathrm{H}} + m_{\mathrm{O}}$ . When you type an oxygen mass, the page treats it as the measurement you trust and solves for whichever of the other three is then unknown, so the sample mass becomes a calculated value.

Clearing the oxygen box hands the difference back and the sample mass stops being derivable — you would need to type it again. If you want the sample mass to stay exactly as you entered it, leave the oxygen box alone and let the page fill it in.

What this analysis cannot tell you

It assumes complete combustion. If any carbon leaves as soot, carbon monoxide or an unburnt fragment, the CO₂ mass is short and the formula is wrong — with no warning from the arithmetic.
It only knows carbon, hydrogen and oxygen. Nitrogen, sulfur and the halogens are outside its model, and their mass would be counted as oxygen.
The sample must be pure and dry. Absorbed water inflates the hydrogen line and can look exactly like a real difference in composition.
The atomic masses are conventional values, so the empirical mass carries uncertainty in the fifth significant figure. A difference in the last digits between this page and a textbook is not an error in either.
The molar mass is an input, not a measurement this page makes. An incorrect molar mass changes only the molecular formula — and if it is not close to a whole multiple of the empirical mass, the page withholds that formula rather than guessing.
The result is for teaching, lab bookkeeping and estimation. It is not a substitute for instrument-verified identification (mass spectrometry, NMR, elemental analysis by a certified laboratory) when a composition has to be reported formally.

When the formula has to be right

Combustion analysis narrows down the formula; it does not confirm a structure. Two isomers share one empirical formula, and the molecular formula alone still leaves the connectivity unknown. Treat this page as the step that turns product masses into a candidate formula, and confirm the candidate with the techniques your lab already trusts.

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