How to Balance Complete and Incomplete Hydrocarbon Combustion

A gas burner can use the same fuel and produce either a clean blue flame or a yellow flame that leaves black soot behind. If the fuel hasn’t changed, what’s different?

The oxygen supply and how well it mixes with the fuel. That changes how far the carbon in the fuel is oxidised, which changes the products of combustion and the equation you need to balance.

The answer to the title question is straightforward:

• Complete combustion: a hydrocarbon reacts with sufficient oxygen to produce carbon dioxide and water.
• Incomplete combustion: limited oxygen causes carbon monoxide, solid carbon (soot), or a mixture of carbon-containing products to form, alongside water.

To balance either type, count carbon atoms first, hydrogen atoms second, and oxygen atoms last.

There’s one important complication: incomplete combustion does not have just one possible balanced equation. You must know which carbon-containing products are being formed before you can balance it.

01What does the flame tell us?

Start with a familiar piece of laboratory equipment: the Bunsen burner.

A Bunsen burner allows you to change how much air mixes with the fuel before combustion. This gives us a useful way to investigate the relationship between oxygen supply and combustion products.

The Bunsen burner experiment

Follow the path of the gas through the apparatus.

1. Gas enters through the inlet at the base of the burner.
2. It passes through a narrow jet, drawing air into the burner through adjustable holes near the base.
3. The fuel and air travel upwards through the barrel.
4. The mixture exits the top and burns when ignited.

The adjustable collar controls how much air enters the burner before combustion.

Imagine closing the air holes. What would you predict happens to the flame?

Less air mixes with the gas before it burns. The flame becomes yellow and luminous, and a cool porcelain surface held above it can collect a black deposit.

That deposit is soot: tiny particles of solid carbon produced by incomplete combustion. They become hot and glow, giving the flame its yellow appearance.

Now open the air holes.

More air mixes with the fuel before ignition. The flame becomes blue and non-luminous, and considerably less soot forms.

This is evidence that combustion is more complete.

What did the experiment actually establish?

The original burner comparison gives us useful evidence.

The prediction was that increasing the available air would improve combustion. The observations support that prediction: the flame changes colour, and soot formation decreases.

However, the experiment has a limitation.

A yellow flame does not tell us exactly how much carbon monoxide has formed. Similarly, a blue flame does not prove that every fuel molecule has been completely converted into carbon dioxide and water.

Carbon monoxide, \(\ce{CO}\), is particularly important here. It is colourless and odourless, so you cannot identify it by looking at the flame or smelling the exhaust. It is also poisonous. See NSW Health’s carbon monoxide safety information.

A more precise investigation would control the fuel and air flow, then analyse the exhaust using an appropriate gas analyser to measure \(\ce{CO2}\), \(\ce{CO}\), and remaining \(\ce{O2}\). A suitable filter could also be used to collect solid carbon.

This revised design measures the actual products rather than relying on flame appearance.

The key conclusion is that oxygen availability affects the products of combustion, but visual observations alone cannot establish the exact composition of the exhaust.

02Complete versus incomplete combustion

Before balancing equations, make sure you understand which products belong on the right-hand side.

A hydrocarbon contains only carbon and hydrogen. During combustion, these elements react with oxygen.

In the standard school-level model, hydrogen forms water. The carbon-containing product depends on the oxygen supply.

Complete combustion

With sufficient oxygen, carbon is fully oxidised to carbon dioxide.

Hydrogen forms water.

For example, complete combustion of methane is:

\[ \ce{CH4 + 2O2 -> CO2 + 2H2O} \]

Check the atoms:

• Carbon: 1 on each side.
• Hydrogen: 4 on each side.
• Oxygen: 4 on each side.

The equation is balanced.

Incomplete combustion

When oxygen is limited, carbon may form carbon monoxide instead of carbon dioxide.

For methane:

\[ \ce{2CH4 + 3O2 -> 2CO + 4H2O} \]

Notice that the carbon-containing product has changed from \(\ce{CO2}\) to \(\ce{CO}\).

Another possibility is the formation of solid carbon:

\[ \ce{CH4 + O2 -> C(s) + 2H2O} \]

Here, the carbon is present as solid elemental carbon rather than a carbon-containing gas.

These are three different possible combustion equations for the same hydrocarbon.

A useful comparison

One misconception is worth correcting.

Incomplete combustion does not always produce only carbon monoxide. It can produce solid carbon, carbon monoxide, or a mixture of carbon dioxide, carbon monoxide, and solid carbon.

For an exam question, use the products specified in the question. If they aren’t specified, you cannot assume that there is one unique incomplete combustion equation.

03How to balance hydrocarbon combustion equations

The balancing method is the same for complete and incomplete combustion.

The only thing that changes is the products you write.

Use this sequence:

1. Write the correct reactants and products.
2. Balance carbon atoms.
3. Balance hydrogen atoms.
4. Balance oxygen atoms last.
5. Multiply all coefficients if necessary to obtain the smallest whole-number ratio.
6. Check every atom on both sides.

Why leave oxygen until last?

Because oxygen appears in several products, including carbon dioxide, carbon monoxide, and water. Balancing it first often means having to change it again.

A general formula for complete combustion

Consider a hydrocarbon with the molecular formula \(\ce{C_xH_y}\).

Here:

• \(x\) represents the number of carbon atoms.
• \(y\) represents the number of hydrogen atoms.

Complete combustion follows this general equation:

\[ \ce{C_xH_y + (x + y/4)O2 -> xCO2 + (y/2)H2O} \]

This formula comes directly from balancing the atoms.

• There are \(x\) carbon atoms, so we need \(x\) molecules of \(\ce{CO2}\).
• There are \(y\) hydrogen atoms, so we need \(y/2\) molecules of \(\ce{H2O}\).
• We then count the oxygen atoms in the products and divide by two to find the number of \(\ce{O2}\) molecules.

For incomplete combustion producing carbon monoxide:

\[ \ce{C_xH_y + (x/2 + y/4)O2 -> xCO + (y/2)H2O} \]

For incomplete combustion producing solid carbon:

\[ \ce{C_xH_y + (y/4)O2 -> xC + (y/2)H2O} \]

These are useful templates, but they assume that all carbon follows the specified pathway. A mixture of carbon products requires a separate atom balance.

04Don’t confuse addition, substitution, and combustion

These are three different types of hydrocarbon reactions.

This distinction matters because addition and substitution involve different changes to the bonds within a hydrocarbon. They are not alternative ways of writing combustion equations.

Addition reactions

Addition reactions commonly occur with alkenes, which contain a carbon-carbon double bond, \(\ce{C=C}\).

The double bond contains one sigma bond and one pi bond. During addition, the pi bond is broken, allowing new atoms or groups to bond to the two carbon atoms.

For example, ethene reacts with bromine:

\[ \ce{CH2=CH2 + Br2 -> BrCH2-CH2Br} \]

The product is 1,2-dibromoethane.

The important changes are:

• The carbon-carbon double bond becomes a single bond.
• The bromine-bromine bond is broken.
• Two new carbon-bromine bonds form.

Conditions and observations: Bromine can add across the double bond at room temperature. In the usual bromine-water test, the orange-brown colour disappears when bromine reacts with an alkene.

Substitution reactions

Substitution reactions commonly occur with alkanes.

Instead of adding atoms across a double bond, an existing atom is replaced by another atom.

For example, methane reacts with chlorine under ultraviolet light:

\[ \ce{CH4 + Cl2 ->[UV\ light] CH3Cl + HCl} \]

The products are chloromethane and hydrogen chloride.

The important changes are:

• One carbon-hydrogen bond is broken.
• The chlorine-chlorine bond is broken.
• A carbon-chlorine bond forms.
• A hydrogen-chlorine bond forms.

Conditions and observations: Ultraviolet light initiates the reaction. The pale green chlorine colour fades as chlorine is consumed.

The key differences

When you see a hydrocarbon reaction question, identify the reaction type before writing the products. That prevents you from applying the wrong reaction pattern.

05Questions and solutions

The following questions progress from straightforward balancing to more challenging applications involving experimental evidence and multiple combustion products.

Question 1

Write the balanced chemical equation for the complete combustion of methane, \(\ce{CH4}\).

Solution 1

The balanced equation is:

\[ \boxed{\ce{CH4 + 2O2 -> CO2 + 2H2O}} \]

Complete combustion produces carbon dioxide and water.

Balance carbon first: one carbon atom requires one \(\ce{CO2}\) molecule.

Next, balance hydrogen: four hydrogen atoms require two \(\ce{H2O}\) molecules.

Finally, there are four oxygen atoms on the product side, requiring two \(\ce{O2}\) molecules.

Both sides now contain one carbon, four hydrogen, and four oxygen atoms.

Question 2

A sample of 3.00 mol of ethene, \(\ce{C2H4}\), undergoes incomplete combustion. Assume that the only products containing carbon are carbon monoxide and that all hydrogen forms water.

(a) Write the balanced chemical equation.

(b) Calculate the amount, in moles, of oxygen consumed.

(c) Calculate the amount, in moles, of carbon monoxide produced.

(d) Calculate how much oxygen would be required if the same amount of ethene underwent complete combustion instead.

Solution 2

The balanced incomplete combustion equation is:

\[ \boxed{\ce{C2H4 + 2O2 -> 2CO + 2H2O}} \]

For 3.00 mol of ethene, 6.00 mol of oxygen is consumed and 6.00 mol of carbon monoxide is produced. Complete combustion would require 9.00 mol of oxygen.

Step 1: Establish the mole ratios from the balanced equation.

\[ \ce{1C2H4 : 2O2 : 2CO} \]

Step 2: Calculate the oxygen consumed.

The equation gives a ratio of one mole of ethene to two moles of oxygen.

\[ n(\ce{O2})=3.00\times2 \]

\[ \boxed{n(\ce{O2})=6.00\ \mathrm{mol}} \]

Step 3: Calculate the carbon monoxide produced.

One mole of ethene produces two moles of carbon monoxide.

\[ n(\ce{CO})=3.00\times2 \]

\[ \boxed{n(\ce{CO})=6.00\ \mathrm{mol}} \]

Step 4: Compare with complete combustion.

The complete combustion equation is:

\[ \ce{C2H4 + 3O2 -> 2CO2 + 2H2O} \]

One mole of ethene now requires three moles of oxygen.

\[ n(\ce{O2})=3.00\times3 \]

\[ \boxed{n(\ce{O2})=9.00\ \mathrm{mol}} \]

The difference is:

\[ 9.00-6.00=3.00\ \mathrm{mol} \]

Complete combustion requires an additional 3.00 mol of oxygen.

This happens because carbon monoxide contains less oxygen per carbon atom than carbon dioxide.

Question 3

A 5.80 g sample of butane, \(\ce{C4H10}\), undergoes incomplete combustion. Assume all carbon forms solid carbon and all hydrogen forms water.

Given:

• Molar mass of butane = \(58.0\ \mathrm{g\,mol^{-1}}\)
• Molar mass of oxygen = \(32.0\ \mathrm{g\,mol^{-1}}\)
• Molar mass of carbon = \(12.0\ \mathrm{g\,mol^{-1}}\)

(a) Write the balanced chemical equation.

(b) Calculate the amount of oxygen required, in moles.

(c) Calculate the mass of oxygen required.

(d) Calculate the mass of solid carbon produced.

Solution 3

The balanced equation is:

\[ \boxed{\ce{2C4H10 + 5O2 -> 8C + 10H2O}} \]

The reaction requires 0.250 mol of oxygen, corresponding to 8.00 g, and produces 4.80 g of solid carbon.

Step 1: Calculate the amount of butane.

Use:

\[ n=\frac{m}{M} \]

where \(n\) is the amount in moles, \(m\) is the mass in grams, and \(M\) is the molar mass in grams per mole.

\[ n(\ce{C4H10})=\frac{5.80}{58.0} \]

\[ =0.100\ \mathrm{mol} \]

Step 2: Calculate the oxygen requirement.

From the balanced equation:

\[ 2\ \mathrm{mol\ butane}:5\ \mathrm{mol\ oxygen} \]

Therefore:

\[ n(\ce{O2})=0.100\times\frac{5}{2} \]

\[ \boxed{n(\ce{O2})=0.250\ \mathrm{mol}} \]

Step 3: Convert oxygen moles into mass.

Use:

\[ m=nM \]

\[ m(\ce{O2})=0.250\times32.0 \]

\[ \boxed{m(\ce{O2})=8.00\ \mathrm{g}} \]

Step 4: Calculate the amount of carbon produced.

The equation gives a ratio of two moles of butane to eight moles of carbon.

\[ n(\ce{C})=0.100\times\frac{8}{2} \]

\[ =0.400\ \mathrm{mol} \]

Step 5: Convert carbon moles into mass.

\[ m(\ce{C})=0.400\times12.0 \]

\[ \boxed{m(\ce{C})=4.80\ \mathrm{g}} \]

The reaction produces 4.80 g of solid carbon.

Notice that the solid carbon is a product of the incomplete combustion, not unreacted butane. The question explicitly states that all the carbon atoms end up as elemental carbon.

Question 4

A chemistry student compares two Bunsen burner conditions using the same fuel and the same gas-flow setting.

The following observations are recorded:

The student concludes:

“Burner A produces only solid carbon, while Burner B produces only carbon dioxide and water.”

Evaluate this conclusion. Explain what the observations establish, identify the limitation of the experiment, and propose an improved method for identifying the combustion products.

Solution 4

The student’s conclusion is not justified. The observations provide evidence of more incomplete combustion in Burner A and more complete combustion in Burner B, but they do not establish that either burner produces only one set of carbon-containing products.

Step 1: Interpret the yellow flame.

The yellow luminous flame indicates that hot carbon particles are present.

The black deposit on the porcelain provides direct evidence that some carbon has formed as a solid.

However, this does not mean that all carbon in the fuel becomes solid carbon. Carbon monoxide and carbon dioxide may also form.

Step 2: Interpret the blue flame.

Opening the air holes allows more air to mix with the fuel before combustion.

This supports more complete combustion, producing predominantly carbon dioxide and water.

However, the absence of visible soot does not prove that carbon monoxide is absent.

Step 3: Identify the experimental limitation.

The experiment uses flame colour and soot formation as qualitative observations.

Neither observation measures the actual amounts of carbon dioxide and carbon monoxide.

In particular, carbon monoxide is colourless and cannot be identified visually.

Step 4: Propose an improved experiment.

Use a controlled combustion apparatus with:

• A fixed fuel flow rate.
• A controlled and measured oxygen supply.
• An appropriate gas analyser to measure carbon dioxide, carbon monoxide, and residual oxygen.
• A suitable filter to collect any solid carbon.

Comparing the measured exhaust composition would provide much stronger evidence about the products formed.

Any investigation involving carbon monoxide must use appropriate laboratory safety controls and equipment.

The important distinction is between observing evidence of incomplete combustion and identifying every product of the reaction. Those are not the same task.

Question 5

This question combines combustion balancing, conservation of atoms, and the interpretation of experimental measurements.

A controlled combustion chamber contains exactly 1.00 mol of propane, \(\ce{C3H8}\), and 4.00 mol of oxygen, \(\ce{O2}\).

Assume the following:

• All the propane reacts.
• All 4.00 mol of oxygen is consumed.
• All hydrogen atoms form water.
• All carbon atoms end up in one or more of three products: carbon dioxide, carbon monoxide, and solid carbon.
• No other products or unreacted fuel remain.

A gas analyser establishes that the amount of carbon dioxide produced is exactly equal to the amount of carbon monoxide produced.

Determine:

(a) The amount, in moles, of carbon dioxide produced.

(b) The amount, in moles, of carbon monoxide produced.

(c) The amount, in moles, of solid carbon produced.

(d) The balanced chemical equation using the smallest whole-number coefficients.

Explain why an equation that produces only carbon monoxide and water cannot satisfy all the information.

Solution 5

The products per mole of propane are:

• \(1.333\) mol of \(\ce{CO2}\)
• \(1.333\) mol of \(\ce{CO}\)
• \(0.333\) mol of solid carbon
• \(4.00\) mol of \(\ce{H2O}\)

The balanced equation using whole-number coefficients is:

\[ \boxed{ \ce{3C3H8 + 12O2 -> 4CO2 + 4CO + C + 12H2O} } \]

The key is to use both the experimental ratio and conservation of atoms. We cannot assume that all carbon forms the same product.

Step 1: Determine the amount of water produced.

One mole of propane contains eight moles of hydrogen atoms.

Since every water molecule contains two hydrogen atoms:

\[ n(\ce{H2O})=\frac{8}{2} \]

\[ \boxed{n(\ce{H2O})=4.00\ \mathrm{mol}} \]

This is fixed by the hydrogen balance.

Step 2: Calculate the oxygen available for carbon-containing products.

The 4.00 mol of oxygen molecules supply:

\[ 4.00\times2=8.00\ \mathrm{mol} \]

of oxygen atoms.

The four moles of water already use four moles of oxygen atoms.

Therefore, the remaining oxygen atoms available for carbon-containing products are:

\[ 8.00-4.00=4.00\ \mathrm{mol} \]

Step 3: Use the gas analyser result.

Let the amount of carbon dioxide be \(x\) mol.

The question states that the amount of carbon monoxide is equal to the amount of carbon dioxide.

Therefore:

\[ n(\ce{CO2})=x \]

\[ n(\ce{CO})=x \]

Carbon dioxide contains two oxygen atoms per molecule, while carbon monoxide contains one.

The total oxygen atoms used by these two gases are therefore:

\[ 2x+x=3x \]

All four remaining moles of oxygen atoms must be accounted for:

\[ 3x=4.00 \]

\[ x=\frac{4.00}{3} \]

Therefore:

\[ \boxed{n(\ce{CO2})=1.333\ \mathrm{mol}} \]

and

\[ \boxed{n(\ce{CO})=1.333\ \mathrm{mol}} \]

Step 4: Calculate the amount of solid carbon.

Initially, there is one mole of propane, containing three moles of carbon atoms.

The carbon in the two gaseous products is:

\[ n_{\text{carbon in gases}} = 1.333+1.333 \]

\[ =2.666\ \mathrm{mol} \]

The remaining carbon must be present as solid carbon.

\[ n(\ce{C})=3.00-2.666 \]

\[ \boxed{n(\ce{C})=0.333\ \mathrm{mol}} \]

All the carbon atoms are now accounted for.

Step 5: Construct the balanced equation.

For one mole of propane, the product amounts are:

The equation with fractional coefficients is:

\[ \ce{ C3H8 + 4O2 -> 4/3CO2 + 4/3CO + 1/3C + 4H2O } \]

Multiply every coefficient by three:

\[ \boxed{ \ce{3C3H8 + 12O2 -> 4CO2 + 4CO + C + 12H2O} } \]

Check the atoms:

The equation is balanced.

Why doesn’t the all-carbon-monoxide equation work?

A tempting approach is to write:

\[ \ce{C3H8 + 7/2O2 -> 3CO + 4H2O} \]

This is a perfectly valid balanced equation for a different incomplete combustion pathway.

However, it contradicts the experimental information because it produces no carbon dioxide.

The question specifically states that carbon dioxide and carbon monoxide are produced in equal amounts.

The oxygen supply, gas analyser result, and carbon balance together force some carbon to remain as solid carbon.

This is the important conceptual point: incomplete combustion does not mean that every carbon atom must follow the same reaction pathway.

06Where to go next

Once you can balance the combustion equation, you can use its coefficients as mole ratios to calculate fuel requirements, oxygen consumption, and product quantities.

The next useful step is connecting those mole ratios to the energy released during combustion.

For that, see Heat of Combustion: Calculations for HSC Chemistry, which develops the relationship between balanced combustion equations, calorimetry, and molar enthalpy of combustion.