How to Calculate Energy Released by Burning Impure Fuel

A 10.0 g fuel sample is labelled 80.0% pure. You burn the sample and want to calculate how much energy it releases.

Should you calculate the energy using 10.0 g of fuel or 8.00 g?

The correct answer is 8.00 g, provided the remaining 20.0% is an inert impurity that does not burn. Only the actual fuel contributes its combustion energy.

This gives you the central method for calculating energy released by burning an impure fuel:

1. Calculate the mass of pure, combustible fuel using its percentage purity.
2. Convert that mass into moles.
3. Use the enthalpy of combustion to calculate the energy released.
4. If a thermochemical equation is given, make sure its enthalpy change corresponds to the correct number of moles of fuel.

There is an important second part, though. If you measure the temperature change during combustion, you are calculating the heat captured by the surroundings. This is an experimental result, not automatically the same as the theoretical energy released by the fuel.

Let’s work through both methods.

01First identify what is actually burning

An impure fuel contains the substance you want to burn, alongside other materials.

However, not every impurity behaves in the same way.

Consider three possible components:

For a straightforward purity calculation, we usually assume the impurity is inert.

Under this assumption, the percentage purity tells you what fraction of the sample’s mass is the actual fuel.

Calculating the mass of pure fuel

Use:

\[ m_{\text{fuel}}=m_{\text{sample}}\times p \]

Where:

• \(m_{\text{fuel}}\) is the mass of combustible fuel, in grams.
• \(m_{\text{sample}}\) is the total mass of the impure sample, in grams.
• \(p\) is the purity expressed as a decimal.

For example, a sample that is 75.0% pure has a purity fraction of:

\[ p=\frac{75.0}{100}=0.750 \]

If the sample has a total mass of 12.0 g:

\[ m_{\text{fuel}}=12.0\times0.750 \]

\[ m_{\text{fuel}}=9.00\text{ g} \]

The remaining 3.00 g is impurity.

Notice that we multiply by the purity fraction rather than dividing by it. We are finding the amount of fuel contained within the sample, not the amount of sample needed to obtain a particular mass of fuel.

02Convert the pure fuel mass into moles

Enthalpy of combustion is normally expressed in kilojoules per mole.

That means you cannot multiply the mass of fuel directly by its molar enthalpy of combustion.

You must first calculate the amount of fuel in moles.

Use:

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

Where:

• \(n\) is the amount of substance, in mol.
• \(m\) is the mass of pure fuel, in grams.
• \(M\) is the molar mass, in g mol\(^{-1}\).

Once you have the number of moles, you can calculate the energy released.

Understanding the enthalpy of combustion

The enthalpy of combustion, \(\Delta H_c\), is the enthalpy change when one mole of a substance undergoes complete combustion.

For ethanol, an illustrative thermochemical equation is:

\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]

\[ \Delta H_c=-1367\text{ kJ mol}^{-1} \]

The negative sign tells you that the reaction is exothermic. Energy leaves the reacting system and enters the surroundings.

If one mole of ethanol burns completely, 1367 kJ of energy is released.

If two moles burn, twice that amount is released.

Therefore, when the equation is written for one mole of fuel:

\[ q_{\text{released}}=n\times|\Delta H_c| \]

Here, \(q_{\text{released}}\) is the positive magnitude of energy released, measured in kJ.

We use the absolute value because the enthalpy change is negative, but the amount of energy released is conventionally reported as a positive quantity.

03Manipulating complete thermochemical equations

Sometimes a question gives you a complete thermochemical equation rather than the enthalpy of combustion per mole of fuel.

This creates another potential trap.

The enthalpy change belongs to the entire equation as written. It does not automatically represent the combustion of one mole of fuel.

Consider the following equation:

\[ \ce{2C3H8(g) + 10O2(g) -> 6CO2(g) + 8H2O(l)} \]

\[ \Delta H=-4440\text{ kJ} \]

This equation represents the complete combustion of two moles of propane.

Therefore, the release of 4440 kJ corresponds to two moles of propane, not one.

To find the energy released per mole:

\[ \Delta H_c=\frac{-4440}{2} \]

\[ \Delta H_c=-2220\text{ kJ mol}^{-1} \]

This is an example of manipulating a thermochemical equation.

Remember these rules:

The coefficients and enthalpy change must always be adjusted together.

04The difference between theoretical and experimental energy

So far, we have calculated energy using the composition of the fuel and its enthalpy of combustion.

These are theoretical calculations.

But what happens when you actually burn a sample in a calorimetry experiment?

You might place a container of water above a burning fuel and measure the temperature increase.

The water gets warmer because it absorbs some of the energy released during combustion.

Calculating the heat absorbed by water

The equation is:

\[ q=mc\Delta T \]

Where:

• \(q\) is the heat absorbed, in joules.
• \(m\) is the mass of water, in grams.
• \(c\) is the specific heat capacity of water, \(4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1}\).
• \(\Delta T\) is the temperature change, in °C.

Calculate the temperature change using:

\[ \Delta T=T_{\text{final}}-T_{\text{initial}} \]

If the water temperature increases, \(q\) is positive because the water absorbs heat.

The combustion reaction, however, is exothermic, so its enthalpy change is negative.

If we assume all the heat released by combustion is transferred to the water:

\[ q_{\text{reaction}}=-q_{\text{water}} \]

This is an idealised model.

In a real experiment, some energy also heats the container and surrounding air. Some may escape through radiation or convection. Incomplete combustion can reduce the energy released as well.

Consequently, the energy absorbed by the water is usually less than the theoretical energy available from complete combustion.

What if the calorimeter itself absorbs heat?

If the apparatus has a known heat capacity, its energy absorption can also be calculated.

\[ q_{\text{cal}}=C_{\text{cal}}\Delta T \]

Where:

• \(q_{\text{cal}}\) is the heat absorbed by the calorimeter, in kJ.
• \(C_{\text{cal}}\) is its heat capacity, in kJ °C\(^{-1}\).
• \(\Delta T\) is the temperature increase, in °C.

The total measured heat captured is then:

\[ q_{\text{captured}}=q_{\text{water}}+q_{\text{cal}} \]

You should include this term whenever the question supplies the calorimeter’s heat capacity.

05A reliable method for any impure-fuel question

Before attempting a calculation, identify which information the question actually provides.

There are five mistakes worth watching for:

1. Using the total sample mass: The inert portion does not contribute combustion energy.
2. Using mass instead of moles: Enthalpy of combustion is usually expressed per mole.
3. Ignoring equation coefficients: The enthalpy change applies to the entire thermochemical equation as written.
4. Confusing signs: A negative enthalpy change represents energy released, while the magnitude of released energy is positive.
5. Treating experimental heat as total released energy: The water may capture only part of the energy produced.

One final distinction is particularly important.

If a sample contains two combustible substances, you cannot simply treat everything other than the main fuel as inert.

Instead, calculate the energy released by each combustible substance and add the contributions:

\[ q_{\text{total}}=\sum_i n_i|\Delta H_{c,i}| \]

Here, \(n_i\) is the amount of each combustible substance, and \(\Delta H_{c,i}\) is its enthalpy of combustion.

For a comparison of the energy content of different pure fuels, see Organic Fuels and Biofuels: HSC Chemistry Comparison Guide.

06Questions and solutions

The following questions progress from a standard purity calculation to more demanding problems involving calorimetry, multiple fuels, and the interpretation of experimental evidence.

Question 1

A 10.0 g sample of solid carbon fuel contains 75.0% graphite by mass. The remaining material is inert ash.

The complete combustion equation is:

\[ \ce{C(s) + O2(g) -> CO2(g)} \]

The enthalpy change is \(-394\text{ kJ mol}^{-1}\), and the molar mass of carbon is \(12.0\text{ g mol}^{-1}\).

Calculate the maximum energy released when the sample undergoes complete combustion.

Solution 1

The maximum theoretical energy released is 246 kJ.

Only the graphite reacts with oxygen. The inert ash must therefore be excluded from the mass used in the calculation.

Step 1: Calculate the mass of graphite.

\[ m=10.0\times0.750 \]

\[ m=7.50\text{ g} \]

Step 2: Calculate the amount of carbon.

\[ n=\frac{7.50}{12.0} \]

\[ n=0.625\text{ mol} \]

Step 3: Calculate the energy released.

The equation represents the combustion of one mole of carbon:

\[ q_{\text{released}}=0.625\times394 \]

\[ q_{\text{released}}=246.25\text{ kJ} \]

Therefore:

\[ \boxed{q_{\text{released}}=246\text{ kJ}} \]

The answer is positive because we are reporting the magnitude of energy released. The enthalpy change of the combustion reaction is approximately \(-246\text{ kJ}\).

Question 2

A 0.500 g sample of an impure propan-1-ol fuel is 80.0% pure. The remaining material is inert.

The fuel is burned beneath a container containing 350.0 g of water. The water temperature increases from 20.0°C to 26.5°C.

Use the following data:

• Molar mass of propan-1-ol: \(60.10\text{ g mol}^{-1}\).
• Enthalpy of combustion: \(-2021\text{ kJ mol}^{-1}\).
• Specific heat capacity of water: \(4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1}\).

Assume complete combustion. Ignore the heat absorbed by the container.

(a) Calculate the theoretical energy released by the sample.

(b) Calculate the heat absorbed by the water.

(c) Calculate the percentage of theoretical energy captured by the water.

(d) Explain why this percentage does not represent the purity of the fuel.

Solution 2

The theoretical energy released is 13.45 kJ, and the water absorbs 9.51 kJ. The percentage captured is approximately 70.7%.

Step 1: Calculate the mass of pure propan-1-ol.

\[ m=0.500\times0.800 \]

\[ m=0.400\text{ g} \]

Step 2: Calculate the number of moles.

\[ n=\frac{0.400}{60.10} \]

\[ n=0.00666\text{ mol} \]

Step 3: Calculate the theoretical energy released.

\[ q_{\text{theoretical}}=0.00666\times2021 \]

\[ q_{\text{theoretical}}=13.45\text{ kJ} \]

Step 4: Calculate the heat absorbed by the water.

The temperature change is:

\[ \Delta T=26.5-20.0=6.50^\circ\text{C} \]

Using:

\[ q=mc\Delta T \]

\[ q=\frac{350.0\times4.18\times6.50}{1000} \]

\[ q=9.51\text{ kJ} \]

Step 5: Calculate the percentage of theoretical energy captured.

\[ \text{Percentage captured} =\frac{9.51}{13.45}\times100 \]

\[ =70.7\% \]

Step 6: Explain the difference.

The 70.7% value represents the proportion of theoretical combustion energy absorbed by the water.

It does not represent fuel purity.

The sample is already known to be 80.0% pure. The difference between theoretical energy and measured heat can result from heat transfer to the surroundings, energy absorbed by the apparatus, or incomplete combustion.

It would be incorrect to conclude that the fuel was only 70.7% pure.

Question 3

An impure fuel mixture has a total mass of 8.00 g and contains:

• 50.0% propane by mass.
• 30.0% butane by mass.
• 20.0% inert material by mass.

The following thermochemical equations are provided:

Propane:

\[ \ce{2C3H8(g) + 10O2(g) -> 6CO2(g) + 8H2O(l)} \]

\[ \Delta H=-4440\text{ kJ} \]

Butane:

\[ \ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)} \]

\[ \Delta H=-5754\text{ kJ} \]

Calculate:

(a) The energy released by the complete combustion of the propane.

(b) The energy released by the complete combustion of the butane.

(c) The total theoretical energy released by the sample.

(d) The energy released per gram of the original impure mixture.

Explain why directly averaging the two enthalpy changes is not a valid method.

Solution 3

The mixture releases approximately 320 kJ in total, equivalent to 40.0 kJ per gram of the original impure sample.

Each combustible component must be treated separately because propane and butane have different molar masses and enthalpies of combustion.

Step 1: Calculate the mass of each component.

For propane:

\[ m_{\text{propane}}=8.00\times0.500 \]

\[ m_{\text{propane}}=4.00\text{ g} \]

For butane:

\[ m_{\text{butane}}=8.00\times0.300 \]

\[ m_{\text{butane}}=2.40\text{ g} \]

The remaining 1.60 g is inert and contributes no combustion energy.

Step 2: Calculate the energy released by propane.

The molar mass of propane is \(44.10\text{ g mol}^{-1}\).

\[ n_{\text{propane}}=\frac{4.00}{44.10} \]

\[ n_{\text{propane}}=0.0907\text{ mol} \]

The thermochemical equation represents the combustion of two moles of propane.

Therefore:

\[ \Delta H_c=\frac{-4440}{2} \]

\[ \Delta H_c=-2220\text{ kJ mol}^{-1} \]

The energy released is:

\[ q_{\text{propane}}=0.0907\times2220 \]

\[ q_{\text{propane}}=201.4\text{ kJ} \]

Step 3: Calculate the energy released by butane.

The molar mass of butane is \(58.12\text{ g mol}^{-1}\).

\[ n_{\text{butane}}=\frac{2.40}{58.12} \]

\[ n_{\text{butane}}=0.0413\text{ mol} \]

The supplied equation represents the combustion of two moles of butane.

Therefore:

\[ \Delta H_c=\frac{-5754}{2} \]

\[ \Delta H_c=-2877\text{ kJ mol}^{-1} \]

The energy released is:

\[ q_{\text{butane}}=0.0413\times2877 \]

\[ q_{\text{butane}}=118.8\text{ kJ} \]

Step 4: Calculate the total energy released.

Both fuels burn completely, so their energy contributions are added.

\[ q_{\text{total}}=201.4+118.8 \]

\[ q_{\text{total}}=320.2\text{ kJ} \]

Step 5: Calculate the energy released per gram of the original mixture.

\[ \frac{320.2}{8.00} =40.0\text{ kJ g}^{-1} \]

Final answers:

• Propane releases 201.4 kJ.
• Butane releases 118.8 kJ.
• Total energy released is approximately 320 kJ.
• Energy released per gram of mixture is approximately \(40.0\text{ kJ g}^{-1}\).

Directly averaging the two enthalpy changes would be incorrect because they describe different amounts of different substances.

The actual amounts of propane and butane are not equal in moles, even though both are present in the same sample. You must convert each component’s mass into moles before calculating its energy contribution.

Question 4

An impure methane sample has a mass of 4.00 g and is 90.0% pure. The remaining material is inert.

During combustion, the water produced remains in the gaseous state.

The relevant thermochemical equation is:

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

\[ \Delta H=-802\text{ kJ mol}^{-1} \]

Another data source gives an enthalpy of combustion of \(-890\text{ kJ mol}^{-1}\) for methane when liquid water is produced.

Use a methane molar mass of \(16.04\text{ g mol}^{-1}\).

(a) Calculate the energy released under the stated conditions.

(b) Calculate the energy that would be predicted if the liquid-water enthalpy value were used instead.

(c) Calculate the amount by which the second calculation overestimates the energy released.

(d) Explain why the two enthalpy values differ.

Solution 4

The correct energy released is approximately 180 kJ. Using the liquid-water enthalpy value would predict approximately 200 kJ, an overestimate of about 19.7 kJ.

The difference arises because the two enthalpy values describe different physical states of the water produced.

Step 1: Calculate the mass of pure methane.

\[ m=4.00\times0.900 \]

\[ m=3.60\text{ g} \]

Step 2: Calculate the amount of methane.

\[ n=\frac{3.60}{16.04} \]

\[ n=0.2244\text{ mol} \]

Step 3: Calculate the energy released using the correct gaseous-water value.

\[ q_{\text{released}}=0.2244\times802 \]

\[ q_{\text{released}}=180.0\text{ kJ} \]

This is the correct theoretical energy for the stated reaction.

Step 4: Calculate the alternative prediction using liquid water.

\[ q_{\text{liquid}}=0.2244\times890 \]

\[ q_{\text{liquid}}=199.7\text{ kJ} \]

Step 5: Calculate the overestimate.

\[ \text{Overestimate}=199.7-180.0 \]

\[ =19.7\text{ kJ} \]

Step 6: Explain the difference.

The liquid-water equation releases additional energy because water vapour must condense into liquid water.

Condensation is an exothermic physical change, so energy is released as the water changes state.

The tempting mistake is to assume that the two enthalpy values are interchangeable because the same fuel is burning. They are not. The physical states of the products are part of the thermochemical equation and affect its enthalpy change.

This is why state symbols matter when manipulating thermochemical equations.

Question 5

A 1.000 g fuel sample contains three components:

• Ethanol.
• Methanol.
• Inert ash.

An independent chemical analysis establishes that the sample contains exactly 0.600 g of ethanol. The masses of methanol and inert ash are unknown.

The sample is burned completely in an idealised, calibrated calorimeter. The calorimeter contains 500.0 g of water and has a heat capacity of \(0.500\text{ kJ }^\circ\text{C}^{-1}\).

The temperature of the water and calorimeter increases by 9.27°C.

Use:

\[ c_{\text{water}}=4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1} \]

The following thermochemical equations and data are supplied.

Ethanol:

\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]

\[ \Delta H_c=-1367\text{ kJ mol}^{-1} \]

Methanol:

\[ \ce{CH3OH(l) + 3/2O2(g) -> CO2(g) + 2H2O(l)} \]

\[ \Delta H_c=-726\text{ kJ mol}^{-1} \]

Molar masses:

• Ethanol: \(46.07\text{ g mol}^{-1}\).
• Methanol: \(32.04\text{ g mol}^{-1}\).

Assume:

• Both alcohols undergo complete combustion.
• The ash is completely inert.
• No heat escapes the calorimeter.
• All corrected heat gained by the water and apparatus corresponds to the energy released by combustion under this idealised model.
• There are no other substances in the sample.

Calculate the mass percentage of methanol and inert ash in the original sample.

Explain why treating all non-ethanol material as inert would give an incorrect result.

Solution 5

The sample contains approximately 27.4% methanol and 12.6% inert ash by mass.

The important insight is that the non-ethanol portion cannot automatically be classified as inert. Methanol is also a fuel, so it contributes additional combustion energy.

We can use the measured heat to determine how much methanol must have been present.

Step 1: Calculate the total heat absorbed by the water.

The water temperature increases by 9.27°C.

\[ q_{\text{water}}=mc\Delta T \]

\[ q_{\text{water}} =\frac{500.0\times4.18\times9.27}{1000} \]

\[ q_{\text{water}}=19.37\text{ kJ} \]

Step 2: Calculate the heat absorbed by the calorimeter.

\[ q_{\text{cal}}=C_{\text{cal}}\Delta T \]

\[ q_{\text{cal}}=0.500\times9.27 \]

\[ q_{\text{cal}}=4.635\text{ kJ} \]

Step 3: Calculate the total energy released by combustion.

Under the idealised assumptions, all released energy is captured by the water and calorimeter.

\[ q_{\text{total}}=19.37+4.635 \]

\[ q_{\text{total}}=24.01\text{ kJ} \]

This is the total energy contribution from both combustible substances.

Step 4: Calculate the energy released by the ethanol.

We know the ethanol mass is 0.600 g.

Convert this mass into moles:

\[ n_{\text{ethanol}}=\frac{0.600}{46.07} \]

\[ n_{\text{ethanol}}=0.01302\text{ mol} \]

Calculate its combustion energy:

\[ q_{\text{ethanol}}=0.01302\times1367 \]

\[ q_{\text{ethanol}}=17.80\text{ kJ} \]

Step 5: Determine the energy released by the methanol.

The total combustion energy is 24.01 kJ, of which 17.80 kJ comes from ethanol.

Therefore:

\[ q_{\text{methanol}}=24.01-17.80 \]

\[ q_{\text{methanol}}=6.21\text{ kJ} \]

This energy must come from the methanol because the ash is inert and the question states that there are no other substances.

Step 6: Calculate the mass of methanol.

First, calculate the number of moles of methanol.

\[ n_{\text{methanol}} =\frac{6.21}{726} \]

\[ n_{\text{methanol}}=0.00855\text{ mol} \]

Now convert moles into mass:

\[ m=nM \]

\[ m_{\text{methanol}}=0.00855\times32.04 \]

\[ m_{\text{methanol}}=0.274\text{ g} \]

Step 7: Calculate the mass of inert ash.

The total sample mass is 1.000 g.

\[ m_{\text{ash}}=1.000-0.600-0.274 \]

\[ m_{\text{ash}}=0.126\text{ g} \]

Step 8: Calculate the mass percentages.

For methanol:

\[ \%\text{ methanol} =\frac{0.274}{1.000}\times100 \]

\[ =27.4\% \]

For inert ash:

\[ \%\text{ ash} =\frac{0.126}{1.000}\times100 \]

\[ =12.6\% \]

Final answer:

The tempting approach is to subtract the ethanol mass from the total sample mass and assume that the remaining 0.400 g is inert.

That would ignore the energy contributed by methanol.

The measured energy is greater than the energy that ethanol alone could release. Under the stated assumptions, the additional energy must come from another combustible substance.

This is why the distinction between an impurity and an inert impurity matters. A substance can be an impurity relative to the main fuel while still contributing its own combustion energy.

The solution also depends on the supplied assumptions. If heat were lost, if combustion were incomplete, or if other combustible substances were present, the measured energy would not uniquely determine the methanol mass.

07What to remember for your HSC Chemistry calculations

When calculating energy released by burning an impure fuel, keep two different questions separate:

• How much energy should the fuel release according to its composition and enthalpy of combustion?
• How much energy was actually captured and measured during the experiment?

For an inert impurity, the theoretical calculation follows:

\[ m_{\text{fuel}}=m_{\text{sample}}\times p \]

\[ n_{\text{fuel}}=\frac{m_{\text{fuel}}}{M} \]

\[ q_{\text{released}}=n_{\text{fuel}}|\Delta H_c| \]

If the thermochemical equation represents more than one mole of fuel, scale its enthalpy change before using it.

For experimental calorimetry, calculate:

\[ q_{\text{water}}=mc\Delta T \]

Then include the calorimeter’s heat capacity if it is supplied.

Finally, check whether every component of the sample is genuinely inert or whether another combustible substance needs its own calculation.

That last decision is what separates a routine purity calculation from a more demanding question involving mixtures, thermochemical equations, and experimental evidence.