Combustion of Alcohols: Balancing HSC Chemistry Equations

A bottle is labelled propan-1-ol, \(\ce{C3H7OH}\). You burn it in plenty of oxygen. You know carbon dioxide and water should form, but what coefficient belongs in front of \(\ce{O2}\)?

Before calculating, make a prediction. Should you balance oxygen first because this is a combustion reaction? It sounds sensible, but it usually makes the equation harder. Oxygen is the one element arriving from two places: the alcohol already contains oxygen, and more oxygen enters as \(\ce{O2}\). The reliable move is to leave oxygen until last.

01What complete combustion is actually doing

When an alcohol undergoes complete combustion, there is enough oxygen for:

• every carbon atom to end up in carbon dioxide, \(\ce{CO2}\)
• every hydrogen atom to end up in water, \(\ce{H2O}\)

So the basic pattern is:

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

That tells you the products immediately. The balancing comes next.

Picture the atoms being reassigned. Carbon has one destination: \(\ce{CO2}\). Hydrogen has one destination: \(\ce{H2O}\). Oxygen is the awkward one because some is already hiding inside the alcohol.

A slightly silly way to picture it is a group arriving at a party. One oxygen atom is already with the alcohol, while the extra oxygen guests arrive in pairs as \(\ce{O2}\). You can’t work out how many pairs need to arrive until you know how many oxygen atoms the products need altogether.

The analogy stops there. Real atoms aren’t guests choosing where to stand. During combustion, bonds break, new bonds form, and energy is released.

02Predicting the products

For complete combustion of the alcohols studied in the HSC course, the product rule is:

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

Do not replace \(\ce{CO2}\) with \(\ce{CO}\) just because carbon monoxide is also a possible combustion product. Carbon monoxide is associated with incomplete combustion, where oxygen supply is insufficient.

Likewise, solid carbon, or soot, is not a product you should include in an equation explicitly described as complete combustion.

The word complete is doing important work.

What about the oxygen already in the alcohol?

Take ethanol, \(\ce{C2H5OH}\). It contains:

• 2 carbon atoms
• 6 hydrogen atoms
• 1 oxygen atom

The products therefore need 2 \(\ce{CO2}\) molecules and 3 \(\ce{H2O}\) molecules:

\[
\ce{C2H5OH + O2 -> 2CO2 + 3H2O}
\]

Count the oxygen atoms on the right:

• \(2\ce{CO2}\) contains \(4\) oxygen atoms
• \(3\ce{H2O}\) contains \(3\) oxygen atoms
• total \(=7\) oxygen atoms

One oxygen atom is already present in the ethanol. The remaining six must come from \(\ce{O2}\):

\[
6\text{ O atoms} = 3\ce{O2}
\]

So:

\[
\ce{C2H5OH + 3O2 -> 2CO2 + 3H2O}
\]

That is why simply counting all seven product oxygen atoms and dividing by two would fail. You would be pretending the oxygen atom inside ethanol did not exist.

03A reliable method for balancing alcohol combustion

Use this order:

1. Write \(\ce{CO2}\) and \(\ce{H2O}\) as the products.
2. Balance carbon.
3. Balance hydrogen.
4. Count the oxygen atoms required on the product side.
5. Subtract any oxygen already present in the alcohol, then balance \(\ce{O2}\).
6. If necessary, multiply every coefficient by the same number to remove fractions.
7. Check every element.

Carbon, then hydrogen, then oxygen is not a magic law. It is simply the most efficient order because oxygen appears in both reactants.

Never change a chemical formula to make an equation balance. You may change coefficients in front of formulas, but not subscripts inside them.

For example, changing \(\ce{CO2}\) into \(\ce{CO3}\) would not be balancing. It would be inventing a different substance.

04The general equation for saturated monohydric alcohols

The common saturated, acyclic alcohol homologous series can be written as:

\[
\ce{C_nH_{2n+2}O}
\]

Here, \(n\) is the number of carbon atoms.

For one mole of this alcohol:

• \(n\) carbon atoms produce \(n\) moles of \(\ce{CO2}\)
• \(2n+2\) hydrogen atoms produce \(n+1\) moles of \(\ce{H2O}\)

Start with:

\[
\ce{C_nH_{2n+2}O + O2 -> nCO2 + (n+1)H2O}
\]

The products contain:

\[
2n+(n+1)=3n+1
\]

oxygen atoms.

The alcohol already supplies one oxygen atom, so \(\ce{O2}\) must supply \(3n\) oxygen atoms. Because each \(\ce{O2}\) molecule contains two oxygen atoms, the coefficient of oxygen is:

\[
\frac{3n}{2}
\]

Therefore:

\[
\ce{C_nH_{2n+2}O + \frac{3n}{2}O2 -> nCO2 + (n+1)H2O}
\]

If you want whole-number coefficients for every value of \(n\), multiply through by two:

\[
\ce{2C_nH_{2n+2}O + 3nO2 -> 2nCO2 + 2(n+1)H2O}
\]

This equation is a useful shortcut, but only when the alcohol actually fits the formula \(\ce{C_nH_{2n+2}O}\). If a different molecular formula is given, balance that formula directly rather than forcing it into the shortcut.

05The patterns worth noticing

For a saturated monohydric alcohol \(\ce{C_nH_{2n+2}O}\), one mole produces:

There is a useful consequence hiding in that table:

\[
n(\ce{H2O})-n(\ce{CO2})=1\text{ mol per mole of alcohol}
\]

For example, one mole of propanol produces three moles of \(\ce{CO2}\) and four moles of \(\ce{H2O}\). One mole of pentanol produces five and six.

That relationship becomes surprisingly useful when combustion data are used to work backwards to the original alcohol.

06The most tempting mistakes

The isomer point is easy to miss. Propan-1-ol and propan-2-ol are structurally different, but both have the molecular formula \(\ce{C3H8O}\). Their complete combustion equations therefore have the same stoichiometric coefficients.

Combustion data can tell you a great deal about elemental composition. It cannot, by itself, tell you where the \(\ce{-OH}\) group sits on a carbon chain.

07What about state symbols?

The balancing method does not depend on state symbols, but the states can matter in other contexts.

For example, in a standard enthalpy of combustion equation, substances are written in their standard states, so water is usually written as \(\ce{H2O(l)}\). Inside a hot flame, however, newly formed water may initially be gaseous.

If a question gives experimental conditions, use those conditions. Do not treat \(\ce{H2O(l)}\) and \(\ce{H2O(g)}\) as interchangeable when energy changes are being discussed.

That distinction becomes important when you move from balancing equations to enthalpy of combustion of alcohols.

08Questions and solutions

Question 1

Write the balanced equation, using the smallest whole-number coefficients, for the complete combustion of methanol, \(\ce{CH3OH}\).

Solution 1

The balanced equation is:

\[
\ce{2CH3OH + 3O2 -> 2CO2 + 4H2O}
\]

Methanol contains one carbon atom, so one methanol molecule produces one \(\ce{CO2}\). It contains four hydrogen atoms, so it produces two \(\ce{H2O}\):

\[
\ce{CH3OH + O2 -> CO2 + 2H2O}
\]

The products contain four oxygen atoms. Methanol supplies one, so another three must come from \(\ce{O2}\):

\[
\frac{3}{2}\ce{O2}
\]

This gives:

\[
\ce{CH3OH + \frac{3}{2}O2 -> CO2 + 2H2O}
\]

Multiplying every coefficient by two gives the smallest whole-number form.

The tempting mistake is to use two \(\ce{O2}\) because four oxygen atoms appear in the products. That ignores the oxygen atom already present in methanol.

Question 2

Pentan-1-ol has the molecular formula \(\ce{C5H12O}\).

A \(0.250\ \text{mol}\) sample undergoes complete combustion.

Determine:

a. the balanced combustion equation using whole-number coefficients
b. the amount of \(\ce{O2}\) consumed
c. the amounts of \(\ce{CO2}\) and \(\ce{H2O}\) formed

Solution 2

The balanced equation is \(\ce{2C5H12O + 15O2 -> 10CO2 + 12H2O}\), and \(0.250\ \text{mol}\) of pentan-1-ol consumes \(1.875\ \text{mol}\) of \(\ce{O2}\), producing \(1.25\ \text{mol}\) of \(\ce{CO2}\) and \(1.50\ \text{mol}\) of \(\ce{H2O}\).

For part a, balance carbon and hydrogen first:

\[
\ce{C5H12O + O2 -> 5CO2 + 6H2O}
\]

The products contain:

\[
5(2)+6=16
\]

oxygen atoms.

One oxygen atom comes from pentan-1-ol, so fifteen more must come from \(\ce{O2}\):

\[
\frac{15}{2}\ce{O2}
\]

Thus:

\[
\ce{C5H12O + \frac{15}{2}O2 -> 5CO2 + 6H2O}
\]

Multiplying through by two:

\[
\ce{2C5H12O + 15O2 -> 10CO2 + 12H2O}
\]

For part b, one mole of pentan-1-ol consumes \(7.5\ \text{mol}\) of oxygen:

\[
n(\ce{O2})=0.250\times7.5=1.875\ \text{mol}
\]

For part c:

\[
\begin{aligned}
n(\ce{CO2})&=0.250\times5=1.25\ \text{mol}\\
n(\ce{H2O})&=0.250\times6=1.50\ \text{mol}
\end{aligned}
\]

The trap is reading the whole-number equation as though \(0.250\ \text{mol}\) of alcohol must first be doubled. It does not. Stoichiometric ratios can be simplified to a per-mole basis: one mole of pentan-1-ol corresponds to \(7.5\) moles of oxygen, five moles of carbon dioxide, and six moles of water.

Question 3

A \(0.120\ \text{mol}\) sample of a pure saturated monohydric alcohol has the formula \(\ce{C_nH_{2n+2}O}\). Complete combustion produces \(0.360\ \text{mol}\) of \(\ce{CO2}\).

Determine:

a. the value of \(n\)
b. the molecular formula of the alcohol
c. the balanced combustion equation
d. the amount of water formed
e. whether these combustion data are enough to determine the alcohol’s structural isomer

Solution 3

The alcohol has \(n=3\) and molecular formula \(\ce{C3H8O}\); its combustion produces \(0.480\ \text{mol}\) of water, but the data cannot distinguish propan-1-ol from propan-2-ol.

For one mole of \(\ce{C_nH_{2n+2}O}\), complete combustion produces \(n\) moles of \(\ce{CO2}\).

Therefore:

\[
0.120n=0.360
\]

so:

\[
n=\frac{0.360}{0.120}=3
\]

The molecular formula is therefore:

\[
\ce{C3H8O}
\]

Its complete combustion equation begins:

\[
\ce{C3H8O + O2 -> 3CO2 + 4H2O}
\]

The products contain ten oxygen atoms. One comes from the alcohol, so nine must come from oxygen gas:

\[
\ce{C3H8O + \frac{9}{2}O2 -> 3CO2 + 4H2O}
\]

Using whole-number coefficients:

\[
\ce{2C3H8O + 9O2 -> 6CO2 + 8H2O}
\]

The water produced is:

\[
n(\ce{H2O})=0.120\times4=0.480\ \text{mol}
\]

The tempting conclusion is that identifying \(\ce{C3H8O}\) has identified the substance. It has not. Propan-1-ol and propan-2-ol have the same molecular formula, so complete combustion gives the same mole ratios for both.

Combustion reveals composition here, not molecular structure.

Question 4

A mixture contains only methanol, \(\ce{CH3OH}\), and propan-1-ol, \(\ce{C3H7OH}\).

The mixture is completely combusted and produces \(0.180\ \text{mol}\) of \(\ce{CO2}\) and \(0.280\ \text{mol}\) of \(\ce{H2O}\).

Determine:

a. the amount of methanol originally present
b. the amount of propan-1-ol originally present
c. the total amount of \(\ce{O2}\) consumed

Solution 4

The mixture contained \(0.060\ \text{mol}\) of methanol and \(0.040\ \text{mol}\) of propan-1-ol, and it consumed \(0.270\ \text{mol}\) of \(\ce{O2}\).

The key is not to start by guessing a percentage composition. First use the different combustion patterns.

For methanol:

\[
\ce{CH3OH + \frac{3}{2}O2 -> CO2 + 2H2O}
\]

For propan-1-ol:

\[
\ce{C3H7OH + \frac{9}{2}O2 -> 3CO2 + 4H2O}
\]

Let \(x\) be the amount of methanol in moles and \(y\) the amount of propan-1-ol in moles.

Carbon dioxide gives:

\[
x+3y=0.180
\]

Water gives:

\[
2x+4y=0.280
\]

There is a neat shortcut hiding in these equations. Subtract the amount of carbon dioxide from the amount of water:

\[
0.280-0.180=0.100\ \text{mol}
\]

For every saturated monohydric alcohol, one mole of alcohol gives exactly one more mole of water than carbon dioxide. Therefore:

\[
x+y=0.100
\]

Now combine this with:

\[
x+3y=0.180
\]

Subtracting gives:

\[
2y=0.080
\]

so:

\[
y=0.040\ \text{mol}
\]

Then:

\[
x=0.100-0.040=0.060\ \text{mol}
\]

Now calculate the oxygen consumption.

Methanol consumes \(1.5\) moles of \(\ce{O2}\) per mole:

\[
n(\ce{O2,\ methanol})=0.060\times1.5=0.090\ \text{mol}
\]

Propan-1-ol consumes \(4.5\) moles of \(\ce{O2}\) per mole:

\[
n(\ce{O2,\ propanol})=0.040\times4.5=0.180\ \text{mol}
\]

Therefore:

\[
n(\ce{O2,\ total})=0.090+0.180=0.270\ \text{mol}
\]

A tempting route is to treat \(0.180\ \text{mol}\) of carbon dioxide as though it came from one pure alcohol. That throws away the most important condition in the question: two different alcohols contribute to the same measured products.

Question 5

A \(100\ \text{mL}\) sample of the vapour of a saturated monohydric alcohol, \(\ce{C_nH_{2n+2}O}\), is mixed with \(600\ \text{mL}\) of oxygen. Assume all gas volumes being compared are measured at the same temperature and pressure unless stated otherwise, so gas-volume ratios equal mole ratios.

The alcohol undergoes complete combustion.

While the products are still hot enough for all the water to remain as vapour, the total gas volume is \(850\ \text{mL}\).

The gases are then cooled so that the water condenses. When the remaining dry gases are measured at the original temperature and pressure, their total volume is \(450\ \text{mL}\).

Determine:

a. \(n\) and the molecular formula of the alcohol
b. the volume of oxygen consumed
c. the volumes of \(\ce{CO2}\), water vapour, and excess \(\ce{O2}\) immediately after combustion
d. why the \(450\ \text{mL}\) dry-gas reading cannot simply be treated as \(450\ \text{mL}\) of carbon dioxide
e. whether the combustion results identify the alcohol’s structural isomer

Solution 5

The alcohol has \(n=3\) and molecular formula \(\ce{C3H8O}\); \(450\ \text{mL}\) of oxygen is consumed, producing \(300\ \text{mL}\) of \(\ce{CO2}\) and \(400\ \text{mL}\) of water vapour, with \(150\ \text{mL}\) of oxygen left over. The dry \(450\ \text{mL}\) is therefore a mixture of carbon dioxide and excess oxygen, not pure carbon dioxide.

For a saturated monohydric alcohol:

\[
\ce{C_nH_{2n+2}O + \frac{3n}{2}O2 -> nCO2 + (n+1)H2O}
\]

A \(100\ \text{mL}\) sample therefore requires:

\[
100\left(\frac{3n}{2}\right)=150n\ \text{mL}
\]

of oxygen.

It produces:

\[
100n\ \text{mL}
\]

of carbon dioxide and:

\[
100(n+1)\ \text{mL}
\]

of water vapour.

Because \(600\ \text{mL}\) of oxygen was supplied, the excess oxygen volume is:

\[
600-150n
\]

The total hot product-gas volume is therefore:

\[
100n+100(n+1)+(600-150n)
\]

Simplifying:

\[
850=100n+100n+100+600-150n
\]

\[
850=700+50n
\]

so:

\[
50n=150
\]

and:

\[
n=3
\]

The alcohol therefore has molecular formula:

\[
\ce{C3H8O}
\]

Its balanced complete combustion equation is:

\[
\ce{2C3H8O + 9O2 -> 6CO2 + 8H2O}
\]

For \(100\ \text{mL}\) of alcohol vapour, the oxygen consumed is:

\[
100\times\frac{9}{2}=450\ \text{mL}
\]

The carbon dioxide volume is:

\[
100\times3=300\ \text{mL}
\]

The water-vapour volume is:

\[
100\times4=400\ \text{mL}
\]

The oxygen remaining is:

\[
600-450=150\ \text{mL}
\]

Immediately after combustion, while water remains gaseous:

\[
300+400+150=850\ \text{mL}
\]

which agrees with the observation.

After cooling, the \(400\ \text{mL}\) contribution from water vapour disappears from the gas phase because the water condenses. The remaining dry gases are:

\[
300\ \text{mL }\ce{CO2}+150\ \text{mL }\ce{O2}=450\ \text{mL}
\]

This is the conceptual hinge. A student might see \(450\ \text{mL}\) of dry gas and assume it is all carbon dioxide. That would only be justified if all the supplied oxygen had been consumed. Here it was deliberately supplied in excess.

The combustion data also cannot distinguish propan-1-ol from propan-2-ol. Both have the formula \(\ce{C3H8O}\), so both have the same combustion stoichiometry.

The extra gas measurement gives more information about amounts, but it still does not reveal how the atoms were connected in the original molecule.

09What this lets you do next

Once you can predict the products and balance the equation without guessing, combustion calculations become much more mechanical: masses, moles, gas volumes, and limiting-reactant questions all come from the same coefficients.

The next important step is recognising that a balanced equation tells you how much reacts, but not how much energy is released. That requires enthalpy data and careful attention to states, especially the difference between \(\ce{H2O(l)}\) and \(\ce{H2O(g)}\). That is where the enthalpy of combustion of alcohols builds directly on the equation-balancing method used here.