IB Chemistry HL Topic 4 — Energy from Fuels Paper 1 & 2 Core idea ~10 min read

Incomplete Combustion

Starve a flame of oxygen and it does not simply burn less. It burns differently — producing a yellow, sooty flame, less energy, and a gas that has killed people in their sleep. The chemistry is straightforward once you know the pecking order: hydrogen always gets the oxygen first.

📘 What you need to know

The pecking order for oxygen

Here is the single idea that makes every incomplete combustion equation predictable. When oxygen is scarce, it is not shared out evenly. Hydrogen is oxidised first, every time. Only whatever oxygen is left over goes to the carbon.

That gives a clean three-step ladder as the oxygen supply drops:

As the oxygen runs out, the carbon gets less oxidised oxygen supply falling plenty of O₂ CO₂ + H₂O complete combustion limited O₂ CO + H₂O carbon monoxide forms very little O₂ C + H₂O soot (carbon) forms Hydrogen gets the oxygen first, so water appears in every row. Hydrogen gas is never a product of incomplete combustion.
In a real faulty boiler you get a mixture of all three carbon products at once. Exam questions usually ask for one specific product and will tell you which.
Incomplete combustion — two versions fuel + oxygen → carbon monoxide + water
fuel + oxygen → carbon + water

Reading the flame

You already have a piece of laboratory equipment that demonstrates all of this: the Bunsen burner. The air hole controls the oxygen supply, and the flame tells you which reaction is running.

The Bunsen burner tells you which reaction is running AIR HOLE CLOSED incomplete combustion yellow, luminous, sooty AIR HOLE OPEN complete combustion blue, non-luminous, clean A yellow, sooty flame is the visible warning of incomplete combustion. The yellow glow is white-hot soot particles, not the gas itself.
That last point is a nice one to know: a luminous flame glows because it is full of incandescent carbon particles. No soot, no glow — which is why the clean flame is so hard to see.
FeatureComplete combustionIncomplete combustion
Oxygen supplyExcess / plentifulLimited or restricted
Carbon-containing productCO2CO and / or C
Other productH2OH2O, always
Flame appearanceBlue, non-luminousYellow, luminous, sooty
Energy released per gramMaximumLower
HazardCO2, a greenhouse gasCO toxicity and soot particulates

Why carbon monoxide is so dangerous

Carbon monoxide is the reason gas appliances need servicing and why you should never run an engine in a closed garage. The danger comes from a combination of properties:

Soot is a hazard too. Carbon particulates cause respiratory problems, blacken buildings, and reflect sunlight back into space — an effect known as global dimming. Soot deposits on the bottom of a calorimeter can also ruin an enthalpy of combustion experiment, since the energy that should have been released was never released at all.

Which fuels soot the most?

The more carbon a molecule contains relative to hydrogen, the harder it is to supply enough oxygen for every carbon atom, so the sootier the flame. A useful quantitative measure is the percentage of carbon by mass:

Percentage carbon by mass % C = (total mass of carbon ÷ Mr) × 100

Benzene and its relatives burn with famously smoky flames for exactly this reason; alcohols, with an oxygen atom already built in, burn far more cleanly.

WORKED EXAMPLE

Calculate the percentage of carbon by mass in hexane (C6H14, Mr = 86.17), benzene (C6H6, Mr = 78.11) and ethanol (C2H5OH, Mr = 46.07). Predict which burns with the sootiest flame.

Step 1: Mass of carbon in one mole of each hexane and benzene: 6 × 12.01 = 72.06 g; ethanol: 2 × 12.01 = 24.02 g Step 2: Divide by Mr and multiply by 100 hexane: (72.06 ÷ 86.17) × 100 = 83.6% benzene: (72.06 ÷ 78.11) × 100 = 92.3% ethanol: (24.02 ÷ 46.07) × 100 = 52.1% Step 3: Predict Benzene sootiest, ethanol cleanest same six carbons in hexane and benzene, but benzene has far less hydrogen to dilute them
WORKED EXAMPLE

Write a balanced equation for the incomplete combustion of butane, C4H10, where carbon monoxide is the only carbon-containing product. Use whole-number coefficients.

Step 1: Carbon and hydrogen, exactly as before 4 carbons → 4CO; 10 hydrogens → 5H₂O Step 2: Count oxygen on the right (4 × 1) + 5 = 9 oxygen atoms CO has only one oxygen each, which is the whole point Step 3: Halve to get O₂ 9 ÷ 2 = 4½, so double everything 2C₄H₁₀(g) + 9O₂(g) → 8CO(g) + 10H₂O(l) complete combustion would need 13O₂ for the same 2 butane − noticeably more
WORKED EXAMPLE

Write a balanced equation for the combustion of ethanol, C2H5OH, in a very poor oxygen supply, where soot is the only carbon-containing product.

Step 1: Carbon and hydrogen 2 carbons → 2C(s); 6 hydrogens in total → 3H₂O Step 2: Oxygen on the right soot has no oxygen, so right-hand total = 3 atoms Step 3: Subtract the alcohol’s own oxygen 3 − 1 = 2 atoms from O₂, so just 1O₂ C₂H₅OH(l) + O₂(g) → 2C(s) + 3H₂O(l) one third of the oxygen that complete combustion needed, and far less energy out

💡 Exam tip

⚠ Common mix-up

Up next: Carbon Dioxide from Burning Fuels — how to calculate exactly how much CO2 a fuel produces, and why methane is the cleanest hydrocarbon there is.

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