Why Energy per Gram and per Mole Give Different Rankings
Discover why two fuels can rank differently when comparing energy per gram and energy per mole. Learn how molar mass, calorimetry, and the amount of fuel reacted affect the calculations, with worked HSC Chemistry examples and practice questions.
Burn one mole of methane and one mole of hydrogen completely. Which fuel releases more energy?
Methane releases approximately 890 kJ per mole, while hydrogen releases approximately 286 kJ per mole. Methane wins comfortably.
Now compare one gram of each fuel. Hydrogen releases approximately 142 kJ per gram, while methane releases only about 55.5 kJ per gram. This time, hydrogen wins.
Both rankings are correct. The difference comes down to what you are dividing the energy by: a fixed number of particles or a fixed mass of fuel.
This is why energy per gram and energy per mole can give different rankings. A mole measures an amount of particles, while a gram measures mass. Different substances have different molar masses, so one gram of each substance does not contain the same number of molecules.
01Why the two rankings are different
The key is that energy per mole and energy per gram answer different questions.
- Energy per mole asks how much energy is released when one mole of fuel molecules reacts.
- Energy per gram asks how much energy is released when one gram of fuel reacts.
One mole always contains approximately \(6.022\times10^{23}\) particles. However, those particles do not all have the same mass.
For example, a mole of hydrogen molecules has a mass of approximately 2.016 g. A mole of methane molecules has a mass of approximately 16.04 g.
Consequently, one gram of hydrogen contains almost eight times as many molecules as one gram of methane.
That difference is large enough to reverse the energy rankings.
A numerical comparison: hydrogen versus methane
Consider the complete combustion reactions:
Hydrogen:
\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(l)} \]
Methane:
\[ \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \]
The following are rounded illustrative comparison values based on standard combustion enthalpies. Both reactions form liquid water, so the product states are consistent.
| Property | Hydrogen | Methane |
|---|---|---|
| Molar mass (g mol\(^{-1}\)) | 2.016 | 16.04 |
| Molar enthalpy of combustion (kJ mol\(^{-1}\)) | −285.8 | −890.3 |
| Energy released per mole (kJ) | 285.8 | 890.3 |
| Energy released per gram (kJ g\(^{-1}\)) | 141.8 | 55.5 |
Notice the reversal:
- Per mole, methane releases more energy.
- Per gram, hydrogen releases more energy.
The negative enthalpy values indicate that both reactions are exothermic. The positive energy-per-gram figures represent the magnitudes of the energy released.
The mathematical relationship
To convert energy per mole into energy per gram, divide by the molar mass:
\[ \boxed{ E_{\text{per gram}}= \frac{|\Delta H_c|}{M} } \]
where:
- \(E_{\text{per gram}}\) is the energy released per gram, in \(\mathrm{kJ\,g^{-1}}\).
- \(\Delta H_c\) is the molar enthalpy of combustion, in \(\mathrm{kJ\,mol^{-1}}\).
- \(M\) is the molar mass of the fuel, in \(\mathrm{g\,mol^{-1}}\).
- The absolute value indicates that we are comparing the amount of energy released, rather than its negative sign.
For hydrogen:
\[ E_{\text{per gram}}= \frac{285.8}{2.016} =141.8\ \mathrm{kJ\,g^{-1}} \]
For methane:
\[ E_{\text{per gram}}= \frac{890.3}{16.04} =55.5\ \mathrm{kJ\,g^{-1}} \]
Methane releases more energy per molecule, but each gram of hydrogen contains many more molecules.
That is the entire reason the rankings change.
What the comparison actually means
Imagine you have two identical containers, each holding 1.00 g of fuel.
The hydrogen container holds approximately:
\[ n=\frac{1.00}{2.016} =0.496\ \mathrm{mol} \]
The methane container holds:
\[ n=\frac{1.00}{16.04} =0.0623\ \mathrm{mol} \]
The hydrogen sample contains almost eight times as many moles of fuel.
Although each mole of methane releases more energy, the much larger number of hydrogen molecules in the equal-mass sample changes the overall result.
This distinction is useful whenever you compare fuels, reactants, or experimental enthalpy data. Before deciding which substance releases more energy, check what quantity is being held constant.
02Understanding the calorimetry experiment
The difference between the two reporting methods becomes especially important when you obtain energy values experimentally.
Suppose a student burns a fuel beneath a container of water. The water temperature increases, and the student uses that temperature change to estimate the energy released.
Before calculating anything, you need to understand what is being measured.
The system and surroundings
The system is the part of the universe being studied. In a combustion experiment, the system is the reacting fuel and oxygen.
The surroundings are everything outside that reacting system. In a simple calorimetry experiment, these include the water, the metal container, the thermometer, and the surrounding air.
When combustion occurs, chemical energy is transferred from the reacting system to the surroundings.
Consider what happens during an exothermic reaction.
The reacting chemicals release energy, which the water and calorimeter absorb. The water temperature rises.
The heat changes of the system and surroundings have opposite signs:
\[ q_{\text{system}}+q_{\text{surroundings}}=0 \]
For an ideal insulated calorimeter:
\[ q_{\text{system}}=-q_{\text{surroundings}} \]
The symbol \(q\) represents heat transferred, measured in joules or kilojoules.
If the water temperature increases:
- \(q_{\text{surroundings}}\) is positive because the surroundings gain heat.
- \(q_{\text{system}}\) is negative because the reaction loses heat.
Therefore, an exothermic reaction has a negative enthalpy change:
\[ \Delta H<0 \]
A common mistake is to see a temperature increase and report a positive enthalpy change. Remember that the thermometer measures the temperature change of the surroundings, not the reacting chemicals.
How the experiment measures energy
A simple flame calorimeter provides a practical way to estimate the energy released.
The apparatus contains a known mass of water in a metal container. A fuel burner sits underneath it, and a thermometer measures the water temperature.
The experimental sequence is:
- Measure the initial mass of the fuel burner.
- Record the initial water temperature.
- Burn the fuel beneath the container.
- Stir the water to distribute the absorbed heat.
- Record the highest water temperature.
- Reweigh the fuel burner to determine the mass of fuel consumed.
The temperature change is:
\[ \Delta T=T_{\text{final}}-T_{\text{initial}} \]
The heat absorbed by the water is calculated using:
\[ q=mc\Delta T \]
where:
- \(m\) is the mass of water in grams.
- \(c\) is the specific heat capacity of water, usually approximated as \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\).
- \(\Delta T\) is the temperature change in degrees Celsius.
The calculation gives the heat gained by the water, not automatically the total heat released by the reaction.
If the calorimeter itself absorbs heat, include its heat capacity:
\[ q_{\text{surroundings}} =mc\Delta T+C_{\text{cal}}\Delta T \]
Here, \(C_{\text{cal}}\) is the calorimeter heat capacity in \(\mathrm{J\,^{\circ}C^{-1}}\).
For an ideal experiment with no heat escaping:
\[ q_{\text{reaction}}=-q_{\text{surroundings}} \]
At constant atmospheric pressure, this reaction heat corresponds to the enthalpy change for the amount that reacted.
Why the original experimental design is imperfect
A simple metal container above an open flame is useful for understanding calorimetry, but it does not capture all the energy released.
Some heat escapes into the surrounding air. The flame may also be affected by draughts, and incomplete combustion can produce less energy than complete combustion.
There is another problem: the metal container absorbs some energy as well as the water.
If the student assumes that all energy enters the water, the calculated reaction enthalpy will be too small in magnitude.
A more controlled design uses:
- An insulating lid and draught shield.
- A fixed distance between the burner and container.
- A measured calorimeter heat capacity.
- Consistent water masses and starting temperatures.
- Sufficient oxygen for complete combustion.
- Accurate measurement of the fuel actually consumed.
These improvements reduce experimental uncertainty, although they do not guarantee that every source of heat loss has been eliminated.
This distinction matters when comparing fuels. A higher measured temperature rise does not automatically mean a fuel has a higher true enthalpy of combustion. The experimental conditions must also be considered.
The amount that reacted matters
There is one more quantity to establish before calculating molar enthalpy: the actual amount of fuel that reacted.
For a fuel sample, calculate the mass consumed:
\[ m_{\text{consumed}} =m_{\text{initial}}-m_{\text{final}} \]
Then calculate the amount in moles:
\[ n_{\text{fuel}}= \frac{m_{\text{consumed}}}{M} \]
Use the mass of fuel consumed, not simply the mass initially placed in the burner.
For example, if a burner initially contains 2.00 g of fuel but only 1.20 g is consumed, the reacted amount must be calculated using 1.20 g.
The molar enthalpy of combustion is:
\[ \boxed{ \Delta H_c= \frac{q_{\text{reaction}}} {n_{\text{fuel reacted}}} } \]
For a general reaction, make sure the amount of reaction is consistent with the balanced equation and its coefficients. For combustion enthalpy reported per mole of fuel, the denominator is the moles of fuel consumed.
If you need a refresher on converting mass into moles, see Moles from Mass and Molar Mass: HSC Chemistry Guide.
For a more detailed explanation of determining the amount that actually reacted, see How to Find the Reacted Amount in Molar Enthalpy Calculations.
03Worked examples
Worked example: Comparing energy per gram and per mole
Hydrogen and methane have the following standard enthalpies of combustion:
- Hydrogen: \(-285.8\ \mathrm{kJ\,mol^{-1}}\), with a molar mass of \(2.016\ \mathrm{g\,mol^{-1}}\).
- Methane: \(-890.3\ \mathrm{kJ\,mol^{-1}}\), with a molar mass of \(16.04\ \mathrm{g\,mol^{-1}}\).
Both fuels undergo complete combustion, producing liquid water.
Calculate the energy released per gram for each fuel and identify which fuel releases more energy per mole and per gram.
Step 1
The molar enthalpy of combustion already gives the energy released per mole of fuel.
Therefore, methane releases more energy per mole because:
\[ 890.3>285.8 \]
The negative signs indicate that both reactions release energy.
Step 2
Divide the magnitude of its molar enthalpy by its molar mass:
\[ \begin{aligned} E_{\mathrm{H_2}}&=\frac{285.8}{2.016}\\ &=141.8\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Step 3
\[ \begin{aligned} E_{\mathrm{CH_4}}&=\frac{890.3}{16.04}\\ &=55.5\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Step 4
| Comparison | Higher energy release |
|---|---|
| Per mole | Methane |
| Per gram | Hydrogen |
Hydrogen releases approximately 2.56 times as much energy per gram as methane.
However, methane releases approximately 3.11 times as much energy per mole.
The result demonstrates that neither ranking is incorrect. Each uses a different basis for comparison.
Worked example: Calculating energy per gram and molar enthalpy from experimental data
A student burns an illustrative fuel X in a calorimeter.
The following data are recorded:
| Measurement | Value |
|---|---|
| Mass of fuel consumed | 0.0800 g |
| Mass of water | 150.0 g |
| Initial water temperature | 22.0 °C |
| Final water temperature | 25.0 °C |
| Calorimeter heat capacity | 42.0 J °C\(^{-1}\) |
| Molar mass of fuel X | 32.0 g mol\(^{-1}\) |
Assume complete combustion, constant atmospheric pressure, and no heat loss outside the measured surroundings.
Calculate:
a. The energy released per gram of fuel X.
b. The molar enthalpy of combustion of fuel X.
Step 1
Use final temperature minus initial temperature:
\[ \begin{aligned} \Delta T&=25.0-22.0\\ &=+3.0\ ^{\circ}\mathrm{C} \end{aligned} \]
The positive temperature change indicates that the surroundings have gained heat.
Step 2
Use:
\[ q_{\text{water}}=mc\Delta T \]
Substitute the values:
\[ \begin{aligned} q_{\text{water}} &=150.0\times4.18\times3.0\\ &=1881\ \mathrm{J}\\ &=1.881\ \mathrm{kJ} \end{aligned} \]
Step 3
The calorimeter also increases in temperature, so it absorbs additional energy.
\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\ &=42.0\times3.0\\ &=126\ \mathrm{J}\\ &=0.126\ \mathrm{kJ} \end{aligned} \]
The total heat absorbed by the surroundings is:
\[ \begin{aligned} q_{\text{surroundings}} &=1.881+0.126\\ &=2.007\ \mathrm{kJ} \end{aligned} \]
Step 4
The reaction is exothermic, so its heat change has the opposite sign:
\[ q_{\text{reaction}}=-2.007\ \mathrm{kJ} \]
Step 5
Divide the positive magnitude of the reaction heat by the mass of fuel consumed:
\[ \begin{aligned} E_{\text{per gram}} &=\frac{2.007}{0.0800}\\ &=25.1\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Step 6
Use:
\[ n=\frac{m}{M} \]
Substitute:
\[ \begin{aligned} n_{\text{fuel}} &=\frac{0.0800}{32.0}\\ &=0.00250\ \mathrm{mol} \end{aligned} \]
Step 7
\[ \begin{aligned} \Delta H_c &=\frac{q_{\text{reaction}}}{n_{\text{fuel}}}\\ &=\frac{-2.007}{0.00250}\\ &=-802.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Therefore, the results are:
- Energy released per gram: \(25.1\ \mathrm{kJ\,g^{-1}}\).
- Molar enthalpy of combustion: \(-803\ \mathrm{kJ\,mol^{-1}}\).
Notice that both answers come from the same measured heat. The only difference is the denominator: mass for the first calculation and moles for the second.
04Questions and solutions
Question 1
A fuel sample releases 6.00 kJ of energy when 0.200 g is completely burned.
The fuel has a molar mass of \(40.0\ \mathrm{g\,mol^{-1}}\).
Assuming all the energy released is transferred to the surroundings, calculate:
a. The energy released per gram.
b. The molar enthalpy of combustion.
Solution 1
The energy released is \(30.0\ \mathrm{kJ\,g^{-1}}\), and the molar enthalpy of combustion is \(-1200\ \mathrm{kJ\,mol^{-1}}\).
The important distinction is that the energy-per-gram calculation uses the mass consumed, while the molar enthalpy calculation uses the moles consumed.
Part a: Energy per gram
\[ \begin{aligned} E_{\text{per gram}} &=\frac{6.00}{0.200}\\ &=30.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Part b: Molar enthalpy
First, calculate the amount of fuel:
\[ \begin{aligned} n&=\frac{m}{M}\\ &=\frac{0.200}{40.0}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]
The reaction releases energy, so the reaction heat is negative:
\[ q_{\text{reaction}}=-6.00\ \mathrm{kJ} \]
Calculate the molar enthalpy:
\[ \begin{aligned} \Delta H_c &=\frac{-6.00}{0.00500}\\ &=-1200\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Both results describe the same combustion event using different reporting bases.
Question 2
Two hypothetical fuels, A and B, are tested using an ideal calorimeter.
The following illustrative data are recorded:
| Measurement | Fuel A | Fuel B |
|---|---|---|
| Molar mass (g mol\(^{-1}\)) | 20.0 | 50.0 |
| Mass consumed (g) | 0.250 | 0.250 |
| Heat absorbed by surroundings (kJ) | 6.00 | 5.00 |
Assume complete combustion and negligible heat loss outside the measured surroundings.
a. Calculate the energy released per gram for each fuel.
b. Calculate the magnitude of the molar enthalpy of combustion for each fuel.
c. Identify which fuel ranks higher on each basis.
d. Explain why the rankings differ.
Solution 2
Fuel A releases more energy per gram, but fuel B releases more energy per mole.
This is possible because the fuels have different molar masses. The same mass of each fuel does not contain the same number of molecules.
Part a: Energy per gram
For fuel A:
\[ \begin{aligned} E_A&=\frac{6.00}{0.250}\\ &=24.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
For fuel B:
\[ \begin{aligned} E_B&=\frac{5.00}{0.250}\\ &=20.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Fuel A has the higher energy release per gram.
Part b: Molar enthalpy
First, calculate the amount of fuel A:
\[ n_A=\frac{0.250}{20.0} =0.0125\ \mathrm{mol} \]
Its molar enthalpy is:
\[ \begin{aligned} \Delta H_{c,A} &=\frac{-6.00}{0.0125}\\ &=-480\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Now calculate the amount of fuel B:
\[ n_B=\frac{0.250}{50.0} =0.00500\ \mathrm{mol} \]
Its molar enthalpy is:
\[ \begin{aligned} \Delta H_{c,B} &=\frac{-5.00}{0.00500}\\ &=-1000\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Part c: Compare the rankings
| Reporting basis | Higher energy release |
|---|---|
| Per gram | Fuel A |
| Per mole | Fuel B |
Part d: Explain the difference
Fuel B has a much larger molar mass, meaning that 0.250 g contains fewer moles than the same mass of fuel A.
However, each mole of fuel B releases considerably more energy.
The two fuels are therefore being compared using different quantities:
- Energy per gram measures the energy released by a fixed mass.
- Energy per mole measures the energy released by a fixed amount of molecules.
There is no contradiction between the results.
Question 3
Propane undergoes complete combustion according to:
\[ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} \]
A student initially has 0.0500 mol of propane and 0.200 mol of oxygen.
After combustion, the calorimeter records that the surroundings have absorbed 88.0 kJ of energy.
Assume that:
- Oxygen is the limiting reactant.
- All the oxygen reacts.
- Propane undergoes complete combustion.
- No heat escapes the measured surroundings.
Use a molar mass of \(44.0\ \mathrm{g\,mol^{-1}}\) for propane.
a. Calculate the amount of propane that actually reacts.
b. Calculate the mass of propane consumed.
c. Calculate the energy released per gram of propane.
d. Calculate the molar enthalpy of combustion.
e. Explain why using the initial amount of propane would produce an incorrect result.
Solution 3
The propane that reacts is \(0.0400\ \mathrm{mol}\), giving an energy release of \(50.0\ \mathrm{kJ\,g^{-1}}\) and a molar enthalpy of combustion of \(-2200\ \mathrm{kJ\,mol^{-1}}\).
The key is that the initial propane sample does not react completely because oxygen runs out first.
Part a: Calculate the amount of propane that reacts
The balanced equation shows that:
\[ 1\ \mathrm{mol\ C_3H_8} : 5\ \mathrm{mol\ O_2} \]
Therefore, the amount of propane that can react with 0.200 mol of oxygen is:
\[ \begin{aligned} n(\ce{C3H8}) &=\frac{0.200}{5}\\ &=0.0400\ \mathrm{mol} \end{aligned} \]
Only 0.0400 mol of propane reacts.
The remaining 0.0100 mol is unreacted.
Part b: Calculate the mass consumed
\[ \begin{aligned} m&=nM\\ &=0.0400\times44.0\\ &=1.76\ \mathrm{g} \end{aligned} \]
Part c: Calculate energy released per gram
The surroundings gain 88.0 kJ, so the reaction releases 88.0 kJ.
\[ \begin{aligned} E_{\text{per gram}} &=\frac{88.0}{1.76}\\ &=50.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Part d: Calculate the molar enthalpy
The reaction is exothermic:
\[ q_{\text{reaction}}=-88.0\ \mathrm{kJ} \]
The amount of propane reacted is 0.0400 mol.
\[ \begin{aligned} \Delta H_c &=\frac{-88.0}{0.0400}\\ &=-2200\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Part e: Explain the incorrect denominator
Using the initial propane amount of 0.0500 mol would assume that all the propane reacted.
That would give:
\[ \frac{-88.0}{0.0500} =-1760\ \mathrm{kJ\,mol^{-1}} \]
This is incorrect because 0.0100 mol of propane remains unreacted.
The measured heat was produced by only 0.0400 mol of propane. Dividing by the larger initial amount incorrectly spreads the released energy across fuel that did not participate in the reaction.
This is why the amount reacted must be established before calculating molar enthalpy.
Question 4
Two students compare the energy released by fuels P and Q using simple flame calorimeters.
Both students calculate energy per gram by dividing the measured heat absorbed by the water and calorimeter by the mass of fuel consumed.
Their reported results are:
| Measurement | Fuel P | Fuel Q |
|---|---|---|
| Reported energy per gram | 32.0 kJ g\(^{-1}\) | 28.0 kJ g\(^{-1}\) |
| Combustion appearance | Smoky flame | Clean flame |
| Experimental setup | Open container, no draught shield | Insulated container with draught shield |
Both students conclude that fuel P has a higher energy content per gram.
Evaluate this conclusion.
In your response:
a. Explain what the calorimeter actually measures.
b. Identify two reasons why the reported results may not represent the true energy released by complete combustion.
c. Describe improvements that would make the comparison more reliable.
Solution 4
The conclusion is not justified by the available data. Fuel P has the higher reported energy per gram, but the different experimental conditions mean that the two values may not represent comparable measurements of complete combustion.
Part a: What the calorimeter measures
The calorimeter measures the heat absorbed by its surroundings, particularly the water and the container.
The student estimates the reaction heat using:
\[ q_{\text{reaction}}=-q_{\text{surroundings}} \]
This assumes that the measured surroundings absorb all the energy released by the reaction.
In reality, some energy can escape into the surrounding air.
Part b: Identify experimental limitations
Two important problems are present.
First, fuel P burns with a smoky flame. This suggests incomplete combustion may be occurring.
Incomplete combustion means the fuel does not release the same amount of energy as complete combustion to the specified products. Some chemical energy remains in partially oxidised products.
Second, fuel P is tested using an open container without a draught shield.
This allows more heat to escape into the room, meaning that the measured temperature increase may underestimate the energy released.
Fuel Q is tested using a more insulated arrangement, so a larger proportion of its released energy may be captured.
The difference in experimental conditions makes the reported values difficult to compare fairly.
Part c: Improve the experiment
The students should use the same calibrated calorimeter and standardise the experimental conditions.
Useful improvements include:
- Use the same mass of water and the same starting temperature.
- Keep the burner-to-container distance constant.
- Use a draught shield and an insulating lid.
- Measure the calorimeter heat capacity.
- Ensure sufficient oxygen for complete combustion.
- Measure the actual mass of fuel consumed.
- Repeat each measurement and compare the results.
The important point is that the reported value of \(32.0\ \mathrm{kJ\,g^{-1}}\) is an experimental estimate, not automatically the true energy released per gram.
A fair comparison requires both fuels to undergo the same defined reaction under suitably controlled conditions.
Question 5
A chemist prepares two 100.0 g fuel mixtures using two hypothetical fuels, A and B.
Their properties are:
| Property | Fuel A | Fuel B |
|---|---|---|
| Molar mass (g mol\(^{-1}\)) | 10.0 | 40.0 |
| Energy released per mole (kJ mol\(^{-1}\)) | 500 | 1200 |
Both fuels undergo complete combustion to the same specified product states. Assume sufficient oxygen, no heat loss, and that the stated energy values are positive magnitudes of their combustion enthalpies.
The mixtures are prepared as follows:
| Mixture | Mass of A | Mass of B |
|---|---|---|
| R | 80.0 g | 20.0 g |
| S | 20.0 g | 80.0 g |
For each mixture, calculate:
a. The total energy released on complete combustion.
b. The energy released per gram of mixture.
c. The total amount of fuel molecules, in moles.
d. The average energy released per mole of fuel molecules.
e. Identify which mixture ranks higher per gram and which ranks higher per mole.
A student suggests that the average energy per mole of each mixture can be found by taking the mass-weighted average of the two fuels’ molar energy values.
Evaluate this suggestion.
Solution 5
Mixture R releases more energy per gram, while mixture S releases more energy per mole of fuel molecules.
The student’s suggested mass-weighted average is incorrect because the two fuels have different molar masses. A mass fraction is not the same as a mole fraction.
This question requires us to calculate total energy and total moles independently before calculating either average.
Part a: Calculate the total energy released by each mixture
Start by calculating the energy released per gram of each fuel.
For fuel A:
\[ \begin{aligned} E_A&=\frac{500}{10.0}\\ &=50.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
For fuel B:
\[ \begin{aligned} E_B&=\frac{1200}{40.0}\\ &=30.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Mixture R contains 80.0 g of A and 20.0 g of B.
Therefore:
\[ \begin{aligned} E_R&=(80.0\times50.0)+(20.0\times30.0)\\ &=4000+600\\ &=4600\ \mathrm{kJ} \end{aligned} \]
Mixture S contains 20.0 g of A and 80.0 g of B.
\[ \begin{aligned} E_S&=(20.0\times50.0)+(80.0\times30.0)\\ &=1000+2400\\ &=3400\ \mathrm{kJ} \end{aligned} \]
Part b: Calculate energy released per gram
Both mixtures have a total mass of 100.0 g.
For mixture R:
\[ \begin{aligned} E_{R,\text{per gram}} &=\frac{4600}{100.0}\\ &=46.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
For mixture S:
\[ \begin{aligned} E_{S,\text{per gram}} &=\frac{3400}{100.0}\\ &=34.0\ \mathrm{kJ\,g^{-1}} \end{aligned} \]
Mixture R releases more energy per gram.
Part c: Calculate the total amount of fuel molecules
Calculate the moles of each fuel separately.
For mixture R:
\[ \begin{aligned} n_A&=\frac{80.0}{10.0}=8.00\ \mathrm{mol}\\ n_B&=\frac{20.0}{40.0}=0.500\ \mathrm{mol} \end{aligned} \]
Therefore:
\[ n_R=8.00+0.500=8.50\ \mathrm{mol} \]
For mixture S:
\[ \begin{aligned} n_A&=\frac{20.0}{10.0}=2.00\ \mathrm{mol}\\ n_B&=\frac{80.0}{40.0}=2.00\ \mathrm{mol} \end{aligned} \]
Therefore:
\[ n_S=2.00+2.00=4.00\ \mathrm{mol} \]
Part d: Calculate average energy released per mole
For mixture R:
\[ \begin{aligned} E_{R,\text{per mole}} &=\frac{4600}{8.50}\\ &=541\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
For mixture S:
\[ \begin{aligned} E_{S,\text{per mole}} &=\frac{3400}{4.00}\\ &=850\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Part e: Compare the rankings
| Reporting basis | Mixture R | Mixture S | Higher energy release |
|---|---|---|---|
| Per gram | 46.0 kJ g\(^{-1}\) | 34.0 kJ g\(^{-1}\) | R |
| Per mole | 541 kJ mol\(^{-1}\) | 850 kJ mol\(^{-1}\) | S |
The rankings reverse again.
Mixture R contains more of fuel A, which has the higher energy release per gram.
Mixture S contains a larger proportion of fuel B by mass, but its mole composition is different. In fact, mixture S contains equal amounts of A and B in moles.
This is why the student’s proposed mass-weighted average does not work for molar energy.
To calculate energy per mole, divide the total energy by the total number of moles. You cannot simply average the individual molar values using mass percentages.
The deeper lesson is that a reporting basis determines how different components contribute to an overall result. Mass-based comparisons use mass, while mole-based comparisons use amounts of particles.
05Choosing the correct reporting basis
When a chemistry question asks you to compare energy values, identify the required basis before beginning the calculation.
| If the question asks for… | Divide by… | Correct unit |
|---|---|---|
| Energy released per gram | Mass of fuel consumed | \(\mathrm{kJ\,g^{-1}}\) |
| Molar enthalpy of combustion | Moles of fuel consumed | \(\mathrm{kJ\,mol^{-1}}\) |
| Molar enthalpy per mole of reaction | Amount of reaction, accounting for coefficients | \(\mathrm{kJ\,mol^{-1}}\) of reaction |
Keep the sign convention separate from the ranking:
- Exothermic reaction: negative enthalpy change.
- Endothermic reaction: positive enthalpy change.
- Energy released: usually reported as a positive magnitude.
Also remember that energy per gram is not energy per litre. A volumetric comparison requires density as well as energy per unit mass.
The next useful step is learning how experimental heat measurements become reliable enthalpy values. In particular, understanding heat losses, calorimeter corrections, and the amount of reactant consumed will help you decide whether two reported energy values are genuinely comparable.