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Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
53%, Where Did the Other Half Go?
A student burns methanol in a spirit burner under a copper calorimeter containing 200 g of water. The mass of the spirit burner decreases by 0.48 g and the water temperature rises by 13.4°C. A textbook says the molar enthalpy of combustion of methanol is −726 kJ/mol. When the student calculates her experimental value she gets −383 kJ/mol, barely 53% of the theoretical value. She did not make a calculation error. The equipment was set up correctly.
Before you read on: Write down every reason you can think of for why a real spirit burner experiment might give a result so far below the theoretical value. List as many as you can, you will return to evaluate your list at the end of the lesson.
Know
- The three-step calculation: q = mcΔT → n = Δm/M → ΔHc = −q/n
- Five sources of discrepancy between experimental and theoretical ΔHc
- ΔHc increases in magnitude by ~650 kJ/mol per additional CH₂ unit
- Combustion equations for methanol through pentan-1-ol
Understand
- Why spirit burner experiments always underestimate ΔHc (systematic, not random error)
- Why ΔHc magnitude increases with chain length, bond energy basis
- Why alcohols have lower energy density than corresponding alkanes (-OH group)
- Why ethanol's carbon cycle advantage ≠ higher energy content
Can Do
- Execute the full three-step ΔHc calculation with correct units at every step
- Describe THREE sources of discrepancy, each with source → effect on q or n → effect on ΔHc
- Evaluate alcohols vs. fossil fuels as energy sources using energy density and carbon cycle arguments
The Spirit Burner Calorimeter, Method & Measurements
A spirit burner calorimeter is a deliberately simple piece of equipment, simple enough to introduce systematic errors, and understanding exactly what you are measuring (and what you are not) is the key to both accurate results and explaining your discrepancy.
The apparatus consists of: a spirit burner containing the alcohol fuel; a copper calorimeter (or tin can) clamped above the flame; a measured mass of water in the calorimeter; and a thermometer to record water temperature.
Spirit burner calorimeter, heat released by combustion heats the water (measured by q = mcΔT); heat lost to surroundings is not captured, causing experimental ΔHc < theoretical ΔHc
Measurement Procedure
- Record initial mass of spirit burner + alcohol (m₁) on a balance.
- Record initial water temperature (T₁).
- Light the spirit burner; burn for a set time or until a target temperature rise is achieved.
- Extinguish the flame; record the maximum water temperature (T₂).
- Record the final mass of spirit burner + alcohol (m₂), do this immediately to minimise evaporation loss.
- Calculate: ΔT = T₂ − T₁; Δm = m₁ − m₂; q = mcΔT; n = Δm/M; ΔHc = −q/n.
Three Key Assumptions (All Imperfect)
1. All combustion heat is transferred to the water, no loss to surroundings.
2. All mass decrease is fuel burned, no evaporation without combustion.
3. Combustion is complete, only CO₂ and H₂O are produced.
All three assumptions break down in a real experiment, which is why experimental ΔHc is always lower in magnitude than the theoretical value.
Reliability Improvements
- Use a draught shield (cardboard surround) to reduce heat loss from air currents.
- Polish/clean the base of the calorimeter to remove soot deposits (soot insulates, reducing heat transfer).
- Keep the flame close to the calorimeter base.
- Use the same apparatus, same water mass, same duration for each alcohol, fair comparison.
- Repeat trials and average results to reduce random error.
- Take temperature readings at 30-second intervals; plot a cooling curve and extrapolate back to the moment of flame extinction to find the true maximum ΔT.
Calculating Molar Enthalpy of Combustion, Step by Step
The calculation has four variables, q, m, ΔT, and n, and the most common errors come not from the formula itself but from unit conversions, sign conventions, and confusing the mass of water with the mass of fuel.
c = 4.18 J g⁻¹ °C⁻¹
result in joules (J)
M = molar mass of alcohol
result in moles
result in kJ/mol
must be negative
Worked Example, Ethanol
Given: m(water) = 200 g · T₁ = 19.2°C · T₂ = 34.6°C · m₁(burner) = 183.72 g · m₂(burner) = 183.24 g · Ethanol M = 46.07 g/mol · Theoretical ΔHc = −1367 kJ/mol
Explaining the Discrepancy, Why Experimental ΔHc < Theoretical
The discrepancy between experimental and theoretical enthalpy of combustion is not a mistake, it is chemically meaningful, and explaining it correctly with specific sources of error is one of the highest-value skills tested in this practical investigation.
Experimental ΔHc values are ALWAYS lower in magnitude than theoretical values, typically 40–70% of the theoretical value. This systematic under-measurement has five specific causes:
| Source of Error | What happens physically | Effect on q or n | Effect on |ΔHc| |
|---|---|---|---|
| Heat loss to surroundings | Calorimeter walls, thermometer, bench, and air absorb heat, only water temperature rise is recorded | q too low | |ΔHc| too low ↓ |
| Incomplete combustion | Oxygen-limited flame produces CO and soot (C) instead of CO₂, evidence: black soot on calorimeter base | q too low (unreleased energy in CO, C) | |ΔHc| too low ↓ |
| Alcohol evaporation without combustion | Alcohol evaporates from wick/opening without burning, registers as Δm increase but no heat released | n too high (Δm inflated) | |ΔHc| too low ↓ |
| Calorimeter heat capacity ignored | Copper calorimeter itself heats up, this heat is not in q = mcΔT (water only) | q too low | |ΔHc| too low ↓ |
| Early temperature reading | Temperature read before reaching true maximum (calorimeter already cooling), ΔT underestimated | q too low (ΔT low) | |ΔHc| too low ↓ |
ΔHc Trend with Chain Length & Alcohols vs. Fossil Fuels
The trend in combustion enthalpy with chain length is completely predictable from bond chemistry, and the comparison between alcohols and fossil fuels as energy sources is one of the most directly HSC-relevant applications of this chemistry.
ΔHc Increases with Chain Length, Why?
Each additional CH₂ unit adds two C–H bonds and one C–C bond. When these combust, new C=O bonds (in CO₂) and O–H bonds (in H₂O) form. Energy released in forming the new bonds exceeds energy required to break the C–H and C–C bonds, net ~650 kJ/mol extra energy released per CH₂ unit.
Bars show relative ΔHc magnitude; values on right show energy density (kJ/g = |ΔHc|/M).
The increment per CH₂ is remarkably constant (~650 kJ/mol), confirming the bond energy analysis. Note that energy density (kJ/g) also increases with chain length, but all alcohols are lower than comparable alkanes because the –OH group adds mass (O = 16 g/mol) without contributing proportional combustion energy.
Alcohols vs. Fossil Fuels, Comparison Table
| Feature | Ethanol (C₂H₅OH) | Petrol (octane, C₈H₁₈) | Natural gas (methane) |
|---|---|---|---|
| ΔHc (kJ/mol) | −1367 | −5471 | −890 |
| Energy density (kJ/g) | 29.7 | 47.9 | 55.6 |
| Renewable feedstock | Yes (fermentation) | No (crude oil) | No (natural gas) |
| Complete combustion products | CO₂ + H₂O | CO₂ + H₂O | CO₂ + H₂O |
| Carbon cycle | Near-carbon-neutral (plant absorbs CO₂ during growth) | Net CO₂ addition (fossil carbon) | Net CO₂ addition (fossil carbon) |
| Oxygen in molecule | Yes (–OH), lowers energy density | No | No |
Common Misconceptions, Combustion of Alcohols
"The discrepancy is because the student made errors in the calculation." No, the discrepancy is systematic and appears in every spirit burner experiment, even when calculations are correct. The cause is physical: heat loss to surroundings, incomplete combustion, and alcohol evaporation without combustion.
"m in q = mcΔT is the mass of alcohol burned." m is the mass of WATER in the calorimeter. The water temperature rises, that's what you measure. Using the alcohol mass gives a completely wrong q value.
"Longer alcohols release more energy per gram because they are 'bigger'." True, but the correct explanation is bond-based: each additional CH₂ unit adds C–H and C–C bonds that, when combusted, form additional CO₂ and H₂O bonds releasing ~650 kJ/mol net. 'Bigger' is not a chemical explanation.
"Ethanol and petrol have different combustion products, ethanol is cleaner." Both produce only CO₂ + H₂O in complete combustion. The difference is the carbon cycle: ethanol CO₂ is near-carbon-neutral (recycled from atmosphere via plant growth); petrol CO₂ is a net addition from fossil reserves.
Full Spirit Burner Calculation, Propan-1-ol
ΔT = 32.1 − 18.5 = 13.6 °C
q = 150 g × 4.18 J g⁻¹ °C⁻¹ × 13.6 °C = 8 527.2 J
Δm = 212.48 − 211.93 = 0.55 g
n = 0.55 / 60.09 = 0.009153 mol
ΔHc = −(8527.2 ÷ 1000) / 0.009153 = −8.5272 / 0.009153 = −931.6 kJ/mol
% = (931.6 / 2021) × 100% = 46.1%
(1) Heat loss to surroundings, the copper calorimeter, thermometer, surrounding air, and bench all absorbed heat. Only the heat that raised the water temperature was captured in q = mcΔT; heat transferred elsewhere was not measured, making q_measured < q_actual and therefore |ΔHc_exp| < |ΔHc_theoretical|.
(2) Incomplete combustion, the spirit burner flame may have been oxygen-limited, producing CO and/or soot (C) rather than fully oxidising to CO₂. These products retain chemical energy not released as heat, reducing the measured q below the theoretical value (which assumes complete combustion).
Discrepancy: (1) significant heat loss to surroundings not captured by water thermometry; (2) incomplete combustion producing CO and soot rather than full oxidation to CO₂.
Trend in ΔHc and Chain Length, Including Energy Density
The magnitude of ΔHc increases with chain length, approximately 650 kJ/mol per additional CH₂ unit (641 → 654 → 655 kJ/mol in the series).
Explanation: Each additional CH₂ unit adds two C–H bonds and one C–C bond. During complete combustion these bonds break (requiring energy) and new C=O bonds in CO₂ and O–H bonds in H₂O form (releasing energy). The energy released in forming additional CO₂ and H₂O bonds exceeds the energy required to break the additional C–H and C–C bonds, approximately 650 kJ/mol net per CH₂. More CH₂ units = more bonds broken and formed = more total energy released.
Average increment = (641 + 654 + 655) / 3 ≈ 650 kJ/mol per CH₂
Predicted: −2676 − 650 = −3326 kJ/mol (actual: −3329 kJ/mol, excellent agreement)
Energy density (kJ/g) = |ΔHc| / M:
Methanol: 726/32.04 = 22.7 kJ/g · Ethanol: 1367/46.07 = 29.7 kJ/g · Propan-1-ol: 2021/60.09 = 33.6 kJ/g · Butan-1-ol: 2676/74.12 = 36.1 kJ/g
Energy density DOES increase with chain length, the claim is correct. ΔHc increases by ~650 kJ/mol per CH₂, while M increases by only 14 g/mol per CH₂, the ratio |ΔHc|/M increases with chain length.
However, all alcohols have lower energy density than the corresponding alkane, the –OH group adds 16 g of oxygen per mole without contributing proportional combustion energy. Octane: 47.9 kJ/g vs octan-1-ol: ~40.6 kJ/g.
Practical Investigation Design and Evaluation (7 marks)
(1) Mass of water in the calorimeter (200 g for all), ensures constant heat capacity of the water system across trials.
(2) Distance between flame and base of calorimeter, ensures consistent fraction of heat transferred to water across trials.
(3) Duration of burning / target temperature rise, comparable burn times mean heat loss rates are similar across trials.
(1) Draught shield around the apparatus, reduces heat loss from air currents (dominant error source).
(2) Repeat each trial ×3 and calculate a mean, reduces random error from variations in flame position and timing.
(3) Temperature readings every 30 s; plot a cooling curve and extrapolate back to flame extinction, determines true maximum ΔT, correcting for cooling during reading.
(4) Clean calorimeter base between trials, removes soot deposits that insulate and reduce heat transfer efficiency.
(5) Weigh spirit burner immediately before and after, minimises mass loss due to alcohol evaporation between readings.
Verification: q = 200 × 4.18 × 7.4 = 6186.4 J; n = 0.51/46.07 = 0.01107 mol;
ΔHc = −(6186.4/1000)/0.01107 = −6.186/0.01107 = −559 kJ/mol ✓
Ratio trend: Methanol 42.3% → Ethanol 40.8% → Propan-1-ol 37.0% → Butan-1-ol 36.7%, the ratio DECREASES with chain length (proportional discrepancy increases).
Explanation: Longer-chain alcohols have larger, more complex molecules that are harder to fully oxidise under spirit burner conditions. Soot formation (incomplete combustion) is more pronounced with longer-chain alcohols, the yellow smoky flame of butan-1-ol vs the cleaner blue flame of methanol illustrates this. Greater incomplete combustion for longer chains → larger fraction of theoretical energy not released → lower ratio of experimental to theoretical ΔHc.
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Octane releases more energy per gram (~47 kJ/g) than ethanol (~29 kJ/g). The key reason: ethanol already contains an oxygen atom bonded to carbon, it is partially oxidised. Carbon in ethanol is at a higher average oxidation state than in octane, so less oxidation (and thus less energy release) occurs when it burns. Octane's carbons start fully reduced (oxidation state ≈ −2 to −3), so complete combustion oxidises them all the way to CO₂, releasing more energy per gram.
Practice
Apply what you've learned. Complete Activities, then answer the MC and Short Answer questions.
Calculate ΔHc, Methanol Revisited
The Think First hook described a student who burned methanol (M = 32.04 g/mol) with 200 g of water. The spirit burner decreased by 0.48 g and the water temperature rose by 13.4°C. Reproduce the full three-step calculation and show that the experimental ΔHc = −383 kJ/mol as stated. Then calculate what percentage of the theoretical value (−726 kJ/mol) this represents.
Error Analysis, Match Source to Improvement
For each source of discrepancy listed below, write: (a) its effect on q or n (which one, and too high or too low); and (b) the specific experimental improvement that would reduce this error.
- Heat loss to surroundings
- Incomplete combustion producing soot
- Alcohol evaporating from wick without burning
- Copper calorimeter absorbing heat not measured by thermometer
Question 1. A student burns 0.36 g of ethanol (M = 46.07 g/mol) and heats 100 g of water. The temperature rises by 16.8°C. What is the experimental molar enthalpy of combustion?
Question 2. A spirit burner experiment consistently gives experimental ΔHc values that are about 50% of the theoretical values. Which is the BEST explanation for this systematic discrepancy?
Question 3. Which correctly explains why butan-1-ol has a larger magnitude molar enthalpy of combustion than ethanol?
Question 4. Which statement correctly compares ethanol and petrol (octane) as fuels?
Question 5. In a spirit burner experiment, some alcohol evaporates from the wick without being burned. What effect does this have on the calculation of ΔHc?
Question 6 4 marks
A student burns butan-1-ol (M = 74.12 g/mol) in a spirit burner under a calorimeter containing 200 g of water. The spirit burner mass decreases from 198.43 g to 197.96 g. The water temperature rises from 20.1°C to 30.7°C. Calculate the experimental molar enthalpy of combustion of butan-1-ol, showing all working with units.
Question 7 5 marks
The experimental molar enthalpy of combustion of alcohols measured using a spirit burner calorimeter is always lower in magnitude than the theoretical value. Identify and explain THREE specific sources of this systematic discrepancy. For each, state: (a) the source; (b) its effect on the measured value of q or n; (c) the resulting effect on the calculated |ΔHc|.
Question 8 6 marks
A student claims: "Ethanol should replace petrol as a fuel because it is renewable and produces the same combustion products as petrol." Evaluate this claim with reference to: (i) the accuracy of the statement about combustion products; (ii) a comparison of energy density; (iii) the carbon cycle advantage of ethanol; and (iv) at least one limitation of ethanol as a fuel.
Show All Answers
Multiple Choice Answers
Short Answer Sample Answers
ΔT = 30.7 − 20.1 = 10.6 °C [½ mark]
q = 200 g × 4.18 J g⁻¹ °C⁻¹ × 10.6 °C = 8861.6 J [1 mark]
Δm = 198.43 − 197.96 = 0.47 g; n = 0.47 / 74.12 = 0.006341 mol [1 mark]
ΔHc = −(8861.6 ÷ 1000) / 0.006341 = −8.8616 / 0.006341 = −1397 kJ/mol [1 mark for answer, ½ mark for correct sign and units, ½ mark for showing full working]
Source 1, Heat loss to surroundings: Combustion heat is distributed to the copper calorimeter walls, thermometer, air above the flame, and surrounding bench, not only the water. q = mcΔT only captures heat absorbed by the water; heat lost to other objects is not measured. ∴ q_measured < q_released → |ΔHc_exp| < |ΔHc_theoretical|.
Source 2, Incomplete combustion: The spirit burner flame may be oxygen-limited, producing CO and/or soot (C) rather than fully oxidising to CO₂. These products retain chemical energy that is not released as heat in the experiment. ∴ q too low → |ΔHc_exp| < |ΔHc_theoretical|.
Source 3, Alcohol evaporation without combustion: Some alcohol evaporates from the wick/burner opening without burning. This mass decrease (evaporation) is recorded as Δm, inflating n = Δm/M, but contributes no heat (q unchanged). With n too high in the denominator, |ΔHc| = q/n is reduced. ∴ n too high → |ΔHc_exp| < |ΔHc_theoretical|.
(i) Combustion products: The student is correct that both ethanol and petrol produce only CO₂ and H₂O in complete combustion. Neither fuel has "cleaner" combustion products. [1 mark]
(ii) Energy density: Ethanol: 29.7 kJ/g; petrol (~octane): 47.9 kJ/g. Petrol has ~61% higher energy density per gram than ethanol. A car running on pure ethanol would need approximately 50% more fuel by volume. This is because ethanol contains an oxygen atom (–OH) that adds molecular mass without proportional combustion energy. [2 marks]
(iii) Carbon cycle advantage: Ethanol produced by fermentation of plant biomass is near-carbon-neutral, the CO₂ released when ethanol burns is offset by the CO₂ absorbed during photosynthesis when the sugar cane or corn grew. Fossil fuel CO₂ was sequestered millions of years ago and represents a net addition to the atmospheric carbon pool. [2 marks]
(iv) Limitation: Lower energy density means larger volumes required; many engines require modification; distillation of fermented ethanol is energy-intensive, partially eroding the carbon cycle advantage; land and water resources required for biomass crops may compete with food production. [1 mark, any valid limitation]
Conclusion: The claim is partially correct but oversimplified. Ethanol's combustion products claim is accurate; the sustainability benefit is real but applies to the carbon cycle. However, significantly lower energy density means ethanol is not a straightforward replacement for petrol.
53%, Now You Know Where the Other Half Went
Return to your list of reasons for the discrepancy. How many did you identify correctly?
The five sources are: (1) heat loss to surroundings, the dominant source; (2) incomplete combustion producing CO and soot; (3) alcohol evaporation from the wick without combustion (inflates n); (4) the copper calorimeter absorbing heat not captured by the thermometer; (5) reading the temperature before it reaches its true maximum. All five push the experimental |ΔHc| below the theoretical value, and none of them are "mistakes" by the student. They are fundamental limitations of the simple open-flame calorimeter design.
The 53% result for methanol is actually slightly better than average for a spirit burner experiment. A well-set-up experiment with a draught shield, clean calorimeter, and cooling curve extrapolation can get to 70–80%. A bomb calorimeter (sealed, oxygen-pressurised, no heat loss) achieves 99%+ of theoretical, at a cost of around $50,000.
Connections: This lesson builds on L10's alcohol production pathways, the fermentation and hydration routes both yield ethanol, which is what you have been burning here. In L12, the same alcohols undergo different reactions: dehydration, substitution, and oxidation.
Combustion of Alcohols & Comparison with Fossil Fuels
Every time you light a spirit burner in the lab you are running the same fundamental experiment that engineers use to evaluate fuels, and the gap between your experimental result and the theoretical value tells you something real and important about the limits of simple calorimetry.
Review
Quick-fire drill, read the prompt, recall the answer, then reveal to check.
Reveal
Step 1: q = mcΔT, m is mass of WATER (g), c = 4.18 J g⁻¹ °C⁻¹, result in joules (J)
Step 2: n = Δm/M, Δm = m₁ − m₂ (fuel burned, g), M = molar mass of alcohol
Step 3: ΔHc = −(q÷1000)/n, divide q by 1000 to convert J→kJ; result in kJ/mol; must be negative (exothermic)
Reveal
(1) Heat loss to surroundings, q too low (heat goes to calorimeter walls, air, bench, not measured by thermometer)
(2) Incomplete combustion (CO and soot), q too low (energy remains in CO and C products, not released as heat)
(3) Alcohol evaporation without combustion, n too high (evaporated mass is included in Δm but no heat is released)
Reveal
Methanol −726 · Ethanol −1367 · Propan-1-ol −2021 · Butan-1-ol −2676 · Pentan-1-ol −3329 kJ/mol
Increment per CH₂: ~650 kJ/mol (constant across the series)
Reason: each CH₂ adds C–H and C–C bonds; combustion forms additional CO₂ and H₂O releasing ~650 kJ/mol net per CH₂
Reveal
Ethanol: 29.7 kJ/g. Petrol (octane): 47.9 kJ/g. Petrol is higher (~61% more energy per gram).
Reason: Ethanol contains an –OH group, the oxygen atom adds 16 g/mol of mass to the molecule without contributing proportional combustion energy. Oxygen is already partially "oxidised" in the alcohol, so less energy is released per gram when it burns. Hydrocarbons contain only C and H bonds, which fully oxidise to CO₂ and H₂O releasing more energy per gram.
Reveal
Advantage, carbon cycle: Ethanol (from fermentation) is near-carbon-neutral, the CO₂ released when it burns was recently absorbed during sugar cane/corn photosynthesis. Fossil fuel CO₂ was sequestered millions of years ago and represents a net addition to the atmosphere.
Main limitation: Much lower energy density (29.7 kJ/g vs 47.9 kJ/g for petrol), a car needs ~50% more ethanol by volume for the same range. Also: engine modification needed for high-blend use; distillation of fermented ethanol is energy-intensive.