Why Oxygen Supply Changes Hydrocarbon Combustion Products

A Bunsen burner can produce a blue flame one moment and a yellow, smoky flame when its air hole is partly closed. The fuel hasn’t changed. So why have the products changed?

The answer is the amount of oxygen available to react with the fuel. With sufficient oxygen, a hydrocarbon undergoes complete combustion, producing carbon dioxide and water. When oxygen is limited, incomplete combustion occurs, and carbon monoxide and/or solid carbon (soot) can form instead.

The important detail is that oxygen supply determines how far the carbon is oxidised. It doesn’t simply switch combustion on or off. Real flames can produce a mixture of complete and incomplete combustion products.

01How does oxygen supply change combustion products?

Think about what happens to the atoms in a hydrocarbon when it burns.

A hydrocarbon contains carbon and hydrogen. During combustion, these atoms react with oxygen from the air.

If enough oxygen is available, the carbon forms carbon dioxide, while the hydrogen forms water.

If oxygen is restricted, some carbon may not be fully oxidised. It can form carbon monoxide or remain as solid carbon.

This gives us three important carbon-containing products:

• Carbon dioxide, \(\ce{CO2}\): the product of complete oxidation of carbon.
• Carbon monoxide, \(\ce{CO}\): a product of incomplete oxidation.
• Solid carbon, \(\ce{C}\): another possible product of incomplete combustion, often visible as soot.

Water, \(\ce{H2O}\), is also produced when the hydrogen in the fuel reacts with oxygen.

Complete combustion: sufficient oxygen

Consider propane, \(\ce{C3H8}\), burning with sufficient oxygen.

The balanced equation is:

\[
\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}
\]

Notice what happens to the atoms:

• Three carbon atoms produce three carbon dioxide molecules.
• Eight hydrogen atoms produce four water molecules.
• Five oxygen molecules provide the oxygen needed for both products.

Every carbon atom ends up as carbon dioxide.

The carbon has been oxidised to its highest oxidation state in this product, +4.

Incomplete combustion: limited oxygen

Now imagine the same amount of propane burning with less oxygen.

Instead of producing only carbon dioxide, the carbon may form carbon monoxide.

One possible balanced equation is:

\[
\ce{2C3H8 + 7O2 -> 6CO + 8H2O}
\]

The hydrogen still forms water, but the carbon forms carbon monoxide rather than carbon dioxide.

Carbon in carbon monoxide has an oxidation state of +2, compared with +4 in carbon dioxide.

With even more restricted oxygen, solid carbon can form:

\[
\ce{C3H8 + 2O2 -> 3C + 4H2O}
\]

Here, the carbon is present in its elemental form, with an oxidation state of zero.

These are idealised reaction pathways. An actual flame can produce carbon dioxide, carbon monoxide, soot, and unburnt hydrocarbons together.

Comparing the three pathways

For one mole of propane, the theoretical oxygen requirements are:

These values come from the balanced equations, assuming that all the propane follows the specified pathway and all hydrogen forms water.

Notice the pattern: producing carbon monoxide requires less oxygen than producing carbon dioxide. Producing solid carbon requires less again.

This is why the products depend on oxygen supply.

However, don’t treat these values as fixed boundaries for real flames. A flame can contain several reaction zones, each with different oxygen availability and temperature.

02What observations indicate incomplete combustion?

You can often recognise incomplete combustion from the appearance of a flame, but observations need to be interpreted carefully.

A useful example is a Bunsen burner.

When its air hole is open, more air mixes with the gas before ignition. The burner usually produces a blue, non-luminous flame.

When the air hole is closed, less oxygen enters the gas mixture before combustion. The flame becomes yellow and luminous, and it may produce black soot.

The yellow colour comes from tiny carbon particles heated to high temperatures. These particles glow, producing the luminous appearance.

Observations and what they tell you

One particularly important distinction is between carbon monoxide and carbon dioxide.

Carbon dioxide can be identified using limewater. It turns milky because calcium carbonate forms.

\[
\ce{CO2 + Ca(OH)2 -> CaCO3 + H2O}
\]

Carbon monoxide is colourless and odourless. You cannot identify it by looking at the flame, and it is highly toxic.

A yellow flame is evidence of soot formation, not a direct test for carbon monoxide.

Why isn’t the total amount of oxygen enough to predict the products?

Here’s a subtle point that often causes trouble in exam questions.

Imagine supplying a burner with enough oxygen, overall, to completely combust all the fuel. Does that guarantee that only carbon dioxide and water will form?

Not necessarily.

Oxygen must actually reach the fuel while the reaction conditions allow combustion to occur. Poor mixing can create fuel-rich regions where oxygen is locally insufficient.

Other factors, including temperature and the time available for reactions to occur, also influence the final products.

Think of oxygen supply as both a quantity and a distribution problem. Having enough oxygen somewhere in the system doesn’t guarantee that every fuel molecule encounters it.

This distinction becomes particularly important when interpreting combustion experiments and gas analysis.

03Don’t confuse combustion with addition and substitution

There’s another distinction worth getting right when studying hydrocarbon reactions.

Complete and incomplete combustion describe how much the carbon is oxidised. Addition and substitution describe different types of organic reactions.

They are not alternative names for combustion pathways.

Addition reactions

An addition reaction commonly occurs when an unsaturated hydrocarbon, such as an alkene, reacts with another substance.

Consider ethene reacting with bromine.

Ethene contains a carbon-carbon double bond, \(\ce{C=C}\). During addition, this double bond becomes a single bond, and one bromine atom attaches to each carbon atom.

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

The product is 1,2-dibromoethane.

The reaction is carried out under ordinary conditions, such as room temperature, without requiring ultraviolet light.

An important observation is that bromine water changes from orange-brown to colourless as bromine is consumed.

Substitution reactions

Substitution commonly occurs when an alkane reacts with a halogen in the presence of ultraviolet light.

Consider methane reacting with chlorine:

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

The product is chloromethane, along with hydrogen chloride.

One hydrogen atom in methane is replaced by a chlorine atom.

Unlike addition, the carbon framework does not gain two new atoms across a double bond. Instead, a carbon-hydrogen bond is replaced by a carbon-chlorine bond.

Ultraviolet light initiates the reaction by breaking chlorine molecules into reactive chlorine radicals.

An observable change is that the pale green chlorine colour fades as chlorine is consumed.

Addition versus substitution at a glance

Keep these reaction types separate from combustion.

In combustion, oxygen reacts with the fuel to produce oxides. In addition and substitution, the reactants and the bonds being changed are different.

For more practice identifying combustion products, see Combustion Reactions: Products and Equations for HSC Chemistry.

04Worked examples

05Questions and solutions

The following questions move from standard combustion calculations towards more demanding problems involving experimental observations, product analysis, and oxygen distribution.

Question 1

A 0.250 mol sample of ethane, \(\ce{C2H6}\), undergoes complete combustion.

Calculate:

a) The balanced chemical equation.

b) The amount, in moles, of oxygen required.

c) The amount, in moles, of carbon dioxide and water produced.

Solution 1

The balanced equation is:

\[
\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}
\]

The reaction produces carbon dioxide and water because the ethane undergoes complete combustion.

a) Balanced equation

\[
\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}
\]

b) Oxygen required

The mole ratio of ethane to oxygen is 2:7.

Therefore:

\[
\begin{aligned}
n(\ce{O2}) &=0.250\times\frac{7}{2}\\
&=0.875\text{ mol}
\end{aligned}
\]

c) Products formed

The mole ratio of ethane to carbon dioxide is 2:4.

\[
\begin{aligned}
n(\ce{CO2}) &=0.250\times\frac{4}{2}\\
&=0.500\text{ mol}
\end{aligned}
\]

The mole ratio of ethane to water is 2:6.

\[
\begin{aligned}
n(\ce{H2O}) &=0.250\times\frac{6}{2}\\
&=0.750\text{ mol}
\end{aligned}
\]

The final amounts are:

• Oxygen required: 0.875 mol.
• Carbon dioxide produced: 0.500 mol.
• Water produced: 0.750 mol.

The key is to use the coefficients from the balanced equation as mole ratios, rather than assuming that one mole of fuel always requires one mole of oxygen.

Question 2

A burner receives 0.800 mol of butane, \(\ce{C4H10}\), and 3.00 mol of oxygen.

For a simplified calculation, assume that every mole of butane that reacts follows the same incomplete combustion pathway:

• All carbon forms carbon monoxide.
• All hydrogen forms water.
• No soot is formed.

Calculate the maximum amount of butane that can follow this pathway using the available oxygen.

How much of the original butane is not accounted for by this pathway? Explain why this does not necessarily represent the amount of unburnt butane in a real flame.

Solution 2

The maximum amount of butane that can follow the specified carbon monoxide pathway is 0.667 mol. The remaining 0.133 mol is not accounted for by this simplified pathway, but it cannot automatically be labelled unburnt fuel in a real flame.

Step 1: Write the balanced incomplete combustion equation.

\[
\ce{2C4H10 + 9O2 -> 8CO + 10H2O}
\]

Two moles of butane require nine moles of oxygen.

Therefore, one mole of butane requires 4.5 mol of oxygen.

Step 2: Calculate the maximum amount of butane that can react.

\[
\begin{aligned}
n(\ce{C4H10})&=3.00\times\frac{2}{9}\\
&=0.667\text{ mol}
\end{aligned}
\]

Step 3: Calculate the amount not accounted for by this pathway.

\[
\begin{aligned}
n_{\text{remaining}}&=0.800-0.667\\
&=0.133\text{ mol}
\end{aligned}
\]

Step 4: Interpret the result.

The calculation assumes that all reacting butane forms carbon monoxide.

However, a real flame could also produce solid carbon. That pathway requires less oxygen per mole of fuel because the carbon is not oxidised as far.

Unburnt hydrocarbons may also escape the reaction zone.

Therefore, 0.133 mol is the amount not accounted for by the specified model. It is not necessarily the amount of unburnt butane in an actual flame.

This is an important exam distinction: a balanced equation describes a particular reaction pathway, not every possible reaction occurring in a real combustion system.

Question 3

Two identical Bunsen burners receive the same methane flow rate at the same gas pressure.

• Burner A has its air hole fully open.
• Burner B has its air hole nearly closed.

Both burners are operating steadily.

A student briefly holds a cool porcelain surface above each flame. Burner A produces a blue, non-luminous flame and leaves almost no visible deposit. Burner B produces a yellow, luminous flame and leaves a black deposit.

Answer the following:

a) Explain why the two burners produce different flame appearances.

b) Identify the likely substance responsible for the black deposit.

c) Explain why the observations do not prove that all carbon in burner B forms this substance.

d) Explain why closing the air hole does not mean that the flame receives absolutely no oxygen.

Solution 3

Burner B undergoes more incomplete combustion because less oxygen mixes with the methane before ignition. Its black deposit is most likely solid carbon, but this does not mean that soot is the only carbon-containing product.

a) Different flame appearances

Burner A has an open air hole, allowing more air to mix with the methane before combustion.

This improves oxygen availability in the hot reaction zone, favouring more complete combustion.

Burner B receives less primary air because its air hole is nearly closed. This creates more fuel-rich regions where oxygen is insufficient for complete oxidation.

Tiny carbon particles can form and become heated until they glow yellow.

b) Black deposit

The black deposit is soot, which consists mainly of solid carbon particles.

These particles have escaped from the flame and collected on the cool porcelain surface.

c) Why soot is not the only possible product

Incomplete combustion can produce carbon monoxide as well as soot.

Some carbon may form carbon dioxide, some may form carbon monoxide, and some may remain as solid carbon.

The black deposit only demonstrates that solid carbon escaped from the flame. It does not reveal the complete composition of the exhaust gases.

d) Oxygen is still available

Closing the air hole restricts the amount of air entering the burner barrel before ignition.

It does not remove oxygen from the surrounding atmosphere.

The flame can still encounter oxygen from the surrounding air, particularly around its outer regions.

This explains why a Bunsen burner with a closed air hole can still burn rather than simply extinguishing immediately.

Question 4

A 0.500 mol sample of propane undergoes combustion.

After the exhaust is cooled and the water is collected separately, gas analysis identifies:

Assume that:

• No other carbon-containing products are formed.
• All hydrogen from the propane forms water.
• All carbon from the propane is accounted for in the two measured products.

Calculate:

a) The amount of water produced.

b) The amount of oxygen consumed.

c) The fraction of carbon atoms that formed carbon dioxide.

d) Compare the oxygen consumed with the amount required for complete combustion of the original propane.

Solution 4

The sample produces 2.00 mol of water and consumes 2.25 mol of oxygen. Two-thirds of its carbon forms carbon dioxide, while one-third forms carbon monoxide.

a) Calculate the water produced

Each propane molecule contains eight hydrogen atoms.

The balanced complete combustion equation is:

\[
\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}
\]

The mole ratio of propane to water is 1:4.

Therefore:

\[
\begin{aligned}
n(\ce{H2O})&=0.500\times4\\
&=2.00\text{ mol}
\end{aligned}
\]

b) Calculate the oxygen consumed

Use the products to determine the total number of oxygen atoms.

Carbon dioxide contains two oxygen atoms per molecule.

Carbon monoxide contains one oxygen atom per molecule.

Water also contains one oxygen atom per molecule.

The total amount of oxygen atoms is:

\[
\begin{aligned}
n(\text{O atoms})&=(1.00\times2)\\
&\quad +(0.500\times1)\\
&\quad +(2.00\times1)\\
&=4.50\text{ mol}
\end{aligned}
\]

Each oxygen molecule contains two oxygen atoms.

Therefore:

\[
\begin{aligned}
n(\ce{O2})&=\frac{4.50}{2}\\
&=2.25\text{ mol}
\end{aligned}
\]

c) Calculate the fraction of carbon forming carbon dioxide

The original propane contains three carbon atoms per molecule.

Therefore, 0.500 mol of propane contains:

\[
0.500\times3=1.50\text{ mol of carbon atoms}
\]

Of these, 1.00 mol ends up in carbon dioxide.

The fraction is:

\[
\begin{aligned}
\text{Fraction}&=\frac{1.00}{1.50}\\
&=\frac{2}{3}\\
&=0.667
\end{aligned}
\]

Therefore, approximately 66.7% of the carbon atoms form carbon dioxide.

The remaining 33.3% form carbon monoxide.

d) Compare with complete combustion

Complete combustion requires five moles of oxygen per mole of propane.

For 0.500 mol of propane:

\[
\begin{aligned}
n(\ce{O2})_{\text{complete}}&=0.500\times5\\
&=2.50\text{ mol}
\end{aligned}
\]

The actual combustion consumes 2.25 mol.

The difference is:

\[
2.50-2.25=0.250\text{ mol}
\]

The sample consumes 0.250 mol less oxygen than would be needed for complete combustion.

This is consistent with the formation of carbon monoxide, which requires less oxygen per carbon atom than carbon dioxide.

The important technique here is to use conservation of atoms. Even when a reaction produces several different products, you can work backwards from the measured composition to determine how much oxygen was consumed.

Question 5

A simplified combustion system receives:

• 1.00 mol of methane, \(\ce{CH4}\).
• 2.50 mol of oxygen, \(\ce{O2}\).

The exhaust is sampled immediately after two separate gas streams meet, before they have sufficient time to react further.

The measured products are:

Assume there are no other products.

A student argues:

“The methane requires only 2.00 mol of oxygen for complete combustion. Since 2.50 mol was supplied, all the carbon must form carbon dioxide. The measured carbon monoxide proves that the data are incorrect.”

Evaluate the student’s argument.

Your response must:

a) Check whether the measured composition obeys conservation of atoms.

b) Explain how carbon monoxide and unreacted oxygen can coexist in the exhaust.

c) Calculate the final amounts of carbon dioxide and oxygen if the exhaust is subsequently mixed thoroughly while hot and given sufficient time for all carbon monoxide to oxidise.

Solution 5

The student’s argument is incorrect. The measured composition is consistent with conservation of atoms, even though the total oxygen supply exceeds the amount required for complete combustion.

The key is that oxygen must be available in the same reaction zone as the fuel, with sufficient temperature and time for the reactions to occur.

a) Check conservation of atoms

Start with carbon.

There is 1.00 mol of methane, containing 1.00 mol of carbon atoms.

The products contain:

\[
0.750+0.250=1.00\text{ mol of carbon}
\]

Carbon is conserved.

Now check hydrogen.

One mole of methane contains four moles of hydrogen atoms.

The water contains:

\[
2.00\times2=4.00\text{ mol of hydrogen atoms}
\]

Hydrogen is conserved.

Finally, check oxygen.

The products contain:

• Carbon dioxide: \(0.750\times2=1.50\) mol of oxygen atoms.
• Carbon monoxide: \(0.250\times1=0.250\) mol of oxygen atoms.
• Water: \(2.00\times1=2.00\) mol of oxygen atoms.
• Unreacted oxygen: \(0.625\times2=1.25\) mol of oxygen atoms.

Total oxygen atoms:

\[
1.50+0.250+2.00+1.25=5.00
\]

The incoming oxygen is:

\[
2.50\times2=5.00\text{ mol of oxygen atoms}
\]

Therefore, oxygen is also conserved.

All three elements balance, so the measured composition is chemically consistent.

b) Explain how carbon monoxide and oxygen can coexist

This is the main conceptual challenge.

The student’s calculation is correct:

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

One mole of methane requires two moles of oxygen for complete combustion.

The system supplies 2.50 mol, which is more than enough overall.

However, this calculation assumes that all the oxygen reaches all the methane and reacts with it under suitable conditions.

That assumption is not necessarily true.

In the model system, the central methane-rich stream can contain regions with insufficient oxygen. Carbon monoxide forms in these regions.

Meanwhile, some oxygen travels through the separate air-rich stream without reacting with the methane.

When the streams meet at the sampling point, the exhaust contains both carbon monoxide and unused oxygen.

The distinction is between total oxygen supplied and oxygen that actually reaches the fuel in the hot reaction zone.

The tempting prediction of complete combustion ignores how the gases are distributed and how much time they have to react.

c) Calculate the products after further oxidation

Once the exhaust is thoroughly mixed while hot, carbon monoxide can react with the remaining oxygen.

The balanced equation is:

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

The mole ratio of carbon monoxide to oxygen is 2:1.

The exhaust contains 0.250 mol of carbon monoxide.

Therefore, the oxygen required to oxidise it completely is:

\[
\begin{aligned}
n(\ce{O2})&=0.250\times\frac{1}{2}\\
&=0.125\text{ mol}
\end{aligned}
\]

The carbon dioxide produced in this further reaction is:

\[
\begin{aligned}
n(\ce{CO2})&=0.250\times\frac{2}{2}\\
&=0.250\text{ mol}
\end{aligned}
\]

Calculate the final carbon dioxide amount:

\[
\begin{aligned}
n(\ce{CO2})_{\text{final}}&=0.750+0.250\\
&=1.00\text{ mol}
\end{aligned}
\]

Calculate the final oxygen amount:

\[
\begin{aligned}
n(\ce{O2})_{\text{final}}&=0.625-0.125\\
&=0.500\text{ mol}
\end{aligned}
\]

The final products are therefore:

• 1.00 mol of carbon dioxide.
• 2.00 mol of water.
• 0.500 mol of excess oxygen.
• No remaining carbon monoxide.

This final result is exactly what we would expect from complete combustion of one mole of methane supplied with 2.50 mol of oxygen: one mole of carbon dioxide, two moles of water, and half a mole of excess oxygen.

The important distinction is that the initial exhaust composition depends on how well the fuel and oxygen mix and react before sampling. The final composition can change if the remaining gases are mixed and allowed to react further while conditions are suitable.

06Where to go next

When you encounter a combustion question in an HSC Chemistry exam, don’t automatically assume that limited oxygen produces only carbon monoxide.

First identify the stated combustion conditions. Then determine which carbon-containing products are possible, write the appropriate balanced equation, and check that every atom is conserved.

Once you can identify the products, the next skill is calculating how much oxygen a fuel requires and using mole ratios to determine the limiting reactant.

Practise that separately with How to Calculate Oxygen Needed for Hydrocarbon Combustion.