IQ4 Lesson 11 of 23 45 min Practical Investigation

Combustion of Alcohols

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.

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Think First — Before You Read

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.

Key Equations — Combustion Calorimetry

Heat Absorbed by Water

q = mcΔT

m = mass of water (g) · c = 4.18 J g⁻¹ °C⁻¹ · ΔT = T_final − T_initial (°C) · q in joules

Moles of Fuel Burned

n = Δm / M

Δm = m₁ − m₂ (mass decrease of spirit burner, g) · M = molar mass of alcohol (g/mol)

Molar Enthalpy of Combustion

ΔHc = −q / n

convert q to kJ first (÷1000) · result in kJ/mol · always negative for combustion (exothermic)

Full Chain

q → n → ΔHc

Step 1: q = mcΔT (J) · Step 2: n = Δm/M · Step 3: ΔHc = −(q÷1000)/n (kJ/mol)

Learning Intentions

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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

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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

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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

Key Terms — scan these before reading

draught shield(b) Improvements: draught shield; repeat and average; cooling curve extrapolation; clean calorimeter; immediate reweighing.

waterThe mass of the spirit burner decreases by 0.48 g and the water temperature rises by 13.4°C.

HydrocarbonAn organic compound containing only carbon and hydrogen atoms.

Functional groupA specific atom arrangement determining characteristic chemical reactions.

Homologous seriesA family of compounds with the same functional group, differing by CH₂.

Addition polymerA polymer formed by monomers adding together without loss of atoms.

Core Concepts

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.

Must Do — Reliability Questions: For any reliability improvement answer, list at least THREE specific improvements, each linked to the specific source of error it addresses. "Repeat the experiment" earns one mark. "Use a draught shield to reduce heat loss from air currents, which is a systematic source of error that makes all experimental ΔHc values less negative than theoretical" earns marks for naming the source, describing the improvement, and linking them.

Common Error — Sign of Δm: Calculating Δm = m₂ − m₁ (final minus initial) gives a negative Δm — correct in sign but causes confusion. Standardise: always calculate Δm = m₁ − m₂ (initial minus final) to give a POSITIVE mass decrease. Then apply the negative sign explicitly in ΔHc = −q/n. That way, q > 0, n > 0, and the negative sign in the formula correctly gives ΔHc < 0.

Exam TipFor organic chemistry questions, draw full structural formulas showing all atoms and bonds — condensed or skeletal formulas alone may lose marks in HSC extended-response questions.

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.

Step 1

q = mcΔT

m = mass of water (NOT fuel)
c = 4.18 J g⁻¹ °C⁻¹
result in joules (J)

Step 2

n = Δm / M

Δm = m₁ − m₂ (fuel burned)
M = molar mass of alcohol
result in moles

Step 3

ΔHc = −q/1000/n

divide q by 1000 (J→kJ) first
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

Step 1 — Heat absorbed by water

ΔT = 34.6 − 19.2 = 15.4 °C

q = 200 g × 4.18 J g⁻¹ °C⁻¹ × 15.4 °C = 12 874.4 J

Step 2 — Moles of ethanol burned

Δm = 183.72 − 183.24 = 0.48 g

n = 0.48 / 46.07 = 0.01042 mol

Step 3 — Molar enthalpy of combustion

ΔHc = −(12 874.4 ÷ 1000) / 0.01042 = −12.874 / 0.01042 = −1235.5 kJ/mol

Step 4 — Compare to theoretical

% of theoretical = (1235.5 / 1367) × 100% = 90.4%

(90% is high — real experiments typically give 40–70%; good technique here)

Must Do — Units at Every Step: HSC markers specifically check units. Write "q = 200 g × 4.18 J g⁻¹ °C⁻¹ × 15.4 °C = 12 874 J" — not just "= 12874". Units serve as a self-check: if they don't cancel to give J (or kJ/mol), the formula has been applied incorrectly.

Common Error — Forgetting J→kJ Conversion: Students divide q (in J) by n directly, giving ΔHc in J/mol rather than kJ/mol. Their answer is 1000× too large in magnitude. If your ΔHc is in the hundreds of thousands of J/mol, check whether you forgot to divide by 1000. The correct unit is always kJ/mol.

Common Error — Using fuel mass in q = mcΔT: m in q = mcΔT is the mass of WATER, not the mass of alcohol. The mass of water heats up — that's what the thermometer measures. Using the mass of alcohol gives a completely wrong value for q.

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 ↓

Insight — Systematic vs. Random Error: All five sources produce the same direction of error (experimental |ΔHc| < theoretical). This is a SYSTEMATIC error, not random — it does not average out with repeat trials. Repeat trials improve precision (reduce scatter between results) but do not reduce systematic discrepancy from heat loss, incomplete combustion, or evaporation. These systematic errors can only be reduced by improving the experimental design.

Must Do — Full Discrepancy Answer: For each source, provide all three components: (1) the SOURCE — what physical event causes the error; (2) the DIRECTION — does it make q too low, or n too high; (3) the CONSEQUENCE — makes |ΔHc| smaller (less negative) than theoretical. Three sources with full three-part explanations = full marks on a 4–6 mark discrepancy question.

Common Error — "Human Error": Writing "the experiment was inaccurate due to human error" earns zero marks. It is not specific, does not identify a source, and does not explain the direction of effect. "Human error" explains random scatter between trials — not the consistent and systematic 40–60% under-measurement seen across all spirit burner experiments, with all alcohols. The marker wants specific physical causes.

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.

Methanol (C1)

−726

22.7 kJ/g

Ethanol (C2)

−1367

29.7 kJ/g

Propan-1-ol (C3)

−2021

33.6 kJ/g

Butan-1-ol (C4)

−2676

36.1 kJ/g

Pentan-1-ol (C5)

−3329

37.8 kJ/g

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
Energy density (kJ/L liquid)	~23 400	~34 800	~0.04 (gas at STP)
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
Engine modification for pure fuel	Yes (E85 blend or flex-fuel engine)	No (standard engine)	Yes (CNG conversion)

Key Evaluation Points

Energy density: Petrol has significantly higher energy per gram (~48 kJ/g) and per litre (~35 kJ/L) than ethanol (~30 kJ/g, ~23 kJ/L). A car running on pure ethanol needs ~50% more fuel by volume to travel the same distance. This is partly because ethanol contains an oxygen atom (–OH) that contributes mass without contributing additional combustible energy.

Combustion products: Both ethanol and petrol produce only CO₂ + H₂O in complete combustion. Neither is "cleaner" in terms of combustion products alone. The advantage of ethanol is the carbon cycle — the CO₂ produced when ethanol burns was absorbed from the atmosphere when the sugar cane or corn grew, making it near-carbon-neutral. Fossil fuel CO₂ was sequestered millions of years ago and represents a net addition to the atmosphere.

Must Do — kJ/mol vs kJ/g: When comparing energy densities, always distinguish between kJ/mol and kJ/g. Larger molecules release more kJ/mol (more bonds to break/form) — but this does not mean they are better fuels. Energy per gram (kJ/g) is the relevant comparison for fuel applications because it accounts for the mass of fuel you need to carry. Octane (47.9 kJ/g) is a more energy-dense fuel than ethanol (29.7 kJ/g), even though ethanol releases more kJ/mol than methane.

Common Error: "Ethanol is a better fuel than petrol because it is renewable." Renewable ≠ better as a fuel. The energy content comparison must be addressed separately from the sustainability comparison. Ethanol has significantly lower energy density than petrol, requires engine modification for high-blend use, and its production (fermentation + distillation) requires energy inputs. A balanced evaluation acknowledges both the sustainability advantage AND the energy density disadvantage.

⚠️ Common Misconceptions — Combustion of Alcohols

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"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.

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"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.

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"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.

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"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.

Worked Examples

Example 1 Straightforward

Full Spirit Burner Calculation — Propan-1-ol

Problem: A student burns propan-1-ol (M = 60.09 g/mol) in a spirit burner to heat 150 g of water. The spirit burner mass decreases from 212.48 g to 211.93 g. Water temperature rises from 18.5°C to 32.1°C. Calculate the experimental molar enthalpy of combustion, the percentage of the theoretical value (−2021 kJ/mol) obtained, and suggest two reasons for the discrepancy.

Step 1 — Heat absorbed by water:
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

Step 2 — Moles of propan-1-ol burned:
Δm = 212.48 − 211.93 = 0.55 g
n = 0.55 / 60.09 = 0.009153 mol

Step 3 — Molar enthalpy of combustion:
ΔHc = −(8527.2 ÷ 1000) / 0.009153 = −8.5272 / 0.009153 = −931.6 kJ/mol

Step 4 — Percentage of theoretical:
% = (931.6 / 2021) × 100% = 46.1%

Step 5 — Two reasons for discrepancy:
(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).

Answer: ΔHc(exp) = −932 kJ/mol (3 s.f.) — 46.1% of the theoretical −2021 kJ/mol.
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.

Example 2 Intermediate

Trend in ΔHc and Chain Length — Including Energy Density

Problem: Molar enthalpies of combustion for primary alcohols: methanol −726, ethanol −1367, propan-1-ol −2021, butan-1-ol −2676 kJ/mol. (a) Describe and explain the trend. (b) Predict ΔHc of pentan-1-ol. (c) A student claims "longer alcohols release more energy per gram." Evaluate this claim.

(a) Trend and explanation:
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.

(b) Predict ΔHc for pentan-1-ol:
Average increment = (641 + 654 + 655) / 3 ≈ 650 kJ/mol per CH₂
Predicted: −2676 − 650 = −3326 kJ/mol (actual: −3329 kJ/mol — excellent agreement)

(c) Evaluate energy density claim:
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.

Answer: (a) |ΔHc| increases ~650 kJ/mol per CH₂ — bond energy basis: additional bonds combusted release ~650 kJ/mol more than they require to break. (b) −3326 kJ/mol (actual −3329 kJ/mol). (c) Claim correct — energy density (kJ/g) increases with chain length. But all alcohols have lower kJ/g than comparable alkanes due to the –OH group adding mass without proportional combustion energy.

Example 3 Extended Response

Practical Investigation Design and Evaluation (7 marks)

Problem: A student compares the molar enthalpies of combustion of four alcohols (m = 200 g water for all). Results are shown below. (a) Identify THREE controlled variables. (b) Describe reliability improvements. (c) Verify the ethanol calculation and explain the trend in the ratio of experimental to theoretical ΔHc.

Methanol

Δm (g): 0.62

ΔT (°C): 7.1

Exp. ΔHc (kJ/mol): −307

Theoretical (kJ/mol): −726

Ethanol

Δm (g): 0.51

ΔT (°C): 7.4

Exp. ΔHc (kJ/mol): −559

Theoretical (kJ/mol): −1367

Propan-1-ol

Δm (g): 0.44

ΔT (°C): 6.6

Exp. ΔHc (kJ/mol): −748

Theoretical (kJ/mol): −2021

Butan-1-ol

Δm (g): 0.39

ΔT (°C): 6.2

Exp. ΔHc (kJ/mol): −983

Theoretical (kJ/mol): −2676

(a) Three controlled variables:
(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.

(b) Reliability improvements:
(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.

(c) Verify ethanol and explain ratio trend:
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.

Answer: (a) Controlled: water mass, flame-calorimeter distance, burn duration. (b) Improvements: draught shield; repeat and average; cooling curve extrapolation; clean calorimeter; immediate reweighing. (c) Ethanol verified at −559 kJ/mol. Ratio decreases with chain length — longer molecules undergo more incomplete combustion, releasing less heat relative to their theoretical value.

SPIRIT BURNER CALORIMETRY:
Step 1: q = mcΔT (m = mass of water, c = 4.18 J g⁻¹ °C⁻¹)
Step 2: n = Δm/M (Δm = m₁ − m₂, fuel burned)
Step 3: ΔHc = −(q÷1000)/n in kJ/mol
ΔHc is always negative (exothermic)

DISCREPANCY SOURCES (all make |ΔHc| too low):
1. Heat loss to surroundings → q too low
2. Incomplete combustion (CO, soot) → q too low
3. Alcohol evaporation without burning → n too high
4. Calorimeter heat capacity ignored → q too low
5. Early temperature reading → q too low

ΔHc TREND:
|ΔHc| increases ~650 kJ/mol per extra CH₂ unit
More bonds broken/formed → more energy released
Methanol −726 → Ethanol −1367 → Propanol −2021 →
Butanol −2676 → Pentanol −3329 kJ/mol

ENERGY DENSITY:
kJ/g = |ΔHc|/M
Ethanol: 29.7 kJ/g · Petrol (~octane): 47.9 kJ/g
Alcohols lower than alkanes due to –OH oxygen
Ethanol advantage: carbon cycle (near-carbon-neutral)

Activities

Activity A

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.

Activity B

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

Check Your Understanding

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?

A−898 kJ/mol

B−898 J/mol

C+898 kJ/mol

D−8.98 kJ/mol

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?

AThe student made calculation errors in all trials

BThe alcohol used was impure, reducing its energy content

CSignificant heat was lost to the surroundings (calorimeter walls, air, bench) and was not measured by the water temperature rise; incomplete combustion also reduced the heat released

DThe theoretical values are incorrect and the experimental values are more accurate

Question 3. Which correctly explains why butan-1-ol has a larger magnitude molar enthalpy of combustion than ethanol?

AButan-1-ol has a higher boiling point, so it releases more energy when burned

BButan-1-ol has more C–H and C–C bonds than ethanol; complete combustion of these additional bonds releases more energy in forming additional CO₂ and H₂O

CButan-1-ol has stronger intermolecular forces that release energy on combustion

DButan-1-ol contains more oxygen than ethanol, allowing more complete combustion

Question 4. Which statement correctly compares ethanol and petrol (octane) as fuels?

AEthanol has higher energy density per gram than petrol because it is a renewable fuel

BBoth produce only CO₂ and H₂O in complete combustion; ethanol's sustainability advantage lies in its near-carbon-neutral life cycle, not in different combustion products

CEthanol has a higher molar enthalpy of combustion than petrol, making it more energy-dense

DPetrol produces CO₂ while ethanol produces only water vapour in complete combustion

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?

AIt increases q because evaporation releases heat into the calorimeter

BIt decreases n because less fuel mass is recorded as burned

CIt increases n (Δm is inflated by evaporation) without changing q, making the calculated |ΔHc| smaller than the true value

DIt has no effect because evaporation is accounted for in the q = mcΔT formula

Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Short Answer Questions

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.

Multiple Choice Answers

Q1 — A: −898 kJ/mol. q = 100 × 4.18 × 16.8 = 7022.4 J. n = 0.36/46.07 = 0.007814 mol. ΔHc = −(7022.4/1000)/0.007814 = −7.022/0.007814 = −898.6 ≈ −898 kJ/mol. B has wrong units (J/mol — forgot ÷1000). C has wrong sign (combustion is exothermic; must be negative). D is 100× too small (divided by 1000 twice, or forgot to convert g to the correct unit).

Q2 — C: Heat loss and incomplete combustion. The discrepancy is systematic (consistent across all trials in one direction — always less than theoretical), not random. A (calculation errors) would produce random, non-systematic variation. B (impure alcohol) would change the fuel's actual ΔHc but does not explain the reproducible 50% under-measurement across all alcohols. D is wrong — theoretical values from bomb calorimetry under controlled conditions are the reference standard.

Q3 — B: More C–H and C–C bonds in butan-1-ol. Butan-1-ol (C4) has two more CH₂ units than ethanol (C2), adding extra C–H and C–C bonds. Combustion of these additional bonds forms additional CO₂ and H₂O, releasing ~650 kJ/mol per extra CH₂. A confuses boiling point (IMF property) with combustion enthalpy (bond energy property). C confuses IMFs with bond energies — IMFs do not release energy on combustion. D is wrong — both alcohols contain one oxygen atom (–OH); butan-1-ol does not contain more oxygen than ethanol.

Q4 — B: Both produce CO₂ + H₂O; ethanol advantage is carbon cycle. Complete combustion of both ethanol and octane produces only CO₂ + H₂O — there is no difference in combustion products. Ethanol's sustainability advantage is its near-carbon-neutral life cycle (CO₂ released on combustion is offset by CO₂ absorbed during plant growth), not different products. A is wrong — ethanol has lower energy density (~29.7 kJ/g) than petrol (~47.9 kJ/g). C is wrong — octane has much higher molar enthalpy (−5471 kJ/mol vs −1367 kJ/mol for ethanol). D is wrong — petrol also produces CO₂ in complete combustion.

Q5 — C: n inflated by evaporation, q unchanged, |ΔHc| too low. Alcohol evaporating without combustion registers as a mass decrease (Δm increases → n calculated is too high) but contributes no heat to the water (q is unchanged). ΔHc = −q/n: since n is too high in the denominator, the magnitude of ΔHc is reduced — the calculated |ΔHc| is smaller than the true value. A is wrong — evaporation is endothermic (absorbs heat) so if anything it slightly cools the wick area. B is wrong — evaporation inflates Δm, which increases n (not decreases). D is wrong — q = mcΔT measures only water temperature, with no correction for evaporation.

Short Answer Sample Answers

Q6 — Butan-1-ol calculation (4 marks):
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]

Q7 — Three sources of discrepancy (5 marks — ~1.5 marks each, rounded):
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|.

Q8 — Evaluate ethanol vs petrol claim (6 marks):
(i) Combustion products: The student is correct that both ethanol and petrol produce only CO₂ and H₂O in complete combustion — ethanol: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O; petrol (octane): C₈H₁₈ + 12.5O₂ → 8CO₂ + 9H₂O. Neither fuel has "cleaner" combustion products on a per-molecule basis. [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 to travel the same distance as on petrol. This is because ethanol contains an oxygen atom (–OH) that adds molecular mass without proportional combustion energy — the –OH oxygen is already partially oxidised. [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. This is ethanol's primary environmental advantage over petrol. [2 marks]
(iv) Limitation: Lower energy density means larger volumes required; many engines require modification (flex-fuel engine or blend <E10 for standard engines). 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, not combustion chemistry. However, significantly lower energy density means ethanol is not a straightforward replacement for petrol without considering engine efficiency, volume, and supply chain energy costs.

Revisit — Think First

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.

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Combustion of Alcohols & Fossil Fuels

Climb platforms, hit checkpoints, and answer questions on Combustion of Alcohols & Fossil Fuels. Quick recall from lessons 1–11.

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