Spirit-burner calorimetry: why the fuel value falls short
In a school laboratory, a student burns propan-1-ol (Mr = 60.0) in a spirit burner beneath a copper can. The can contains 200 g of water and has a mass of 85 g; the specific heat capacities are 4.18 J g⁻¹ K⁻¹ for water and 0.385 J g⁻¹ K⁻¹ for copper. The thermometer reads 21.5 °C before and 54.0 °C after the experiment. The burner loses 1.50 g in mass. The draught shield is in place and the flame just touches the base of the can, but the room is draughty and the flame is visibly yellow at times. The data book gives ΔcH°(propan-1-ol) = –2021 kJ mol⁻¹. A classmate says the result is low only because some fuel evaporated and that covering the burner will make the experimental value match the data book.
1. Calculate the experimental enthalpy of combustion of propan-1-ol in kJ mol⁻¹, including the heat absorbed by the copper can. Give the sign and the percentage difference from the data-book value.
Read the worked answer
Heat gained by water: q(water) = m c ΔT = 200 × 4.18 × (54.0 – 21.5) = 27 170 J.
Heat gained by copper can: q(can) = 85 × 0.385 × 32.5 = 1 064 J.
Total heat released by combustion ≈ 27 170 + 1 064 = 28 234 J = 28.23 kJ.
Amount burned: n = 1.50 / 60.0 = 0.0250 mol.
Experimental enthalpy change: ΔcH = –28.23 / 0.0250 = –1.13 × 10³ kJ mol⁻¹ (about –1129 kJ mol⁻¹).
The sign is negative because combustion releases heat. The magnitude is smaller than the data-book value: percentage difference = (2021 – 1129) / 2021 × 100% ≈ 44%.
The can correction is small: without it the result would be about –1087 kJ mol⁻¹, only about 3.7% different. Including the can is correct, but it does not explain the large shortfall.
2. Evaluate the classmate's claim that evaporation alone explains the low result and that covering the burner will make the value match the data book. Use an estimate to support your judgement, and identify the most important improvement.
Read the worked answer
Evaporation does cause an error in the same direction as the observed one: fuel lost by evaporation is counted in the 1.50 g mass loss but has not released combustion energy, so the amount burned is overestimated and the calculated |ΔH| is too small. However, the size of the effect is far too small here.
The missing energy is about (0.0250 × 2021) – 28.23 ≈ 22.3 kJ. If evaporation were responsible, roughly 22.3 kJ would have to be supplied by vaporising fuel. With ΔvapH ≈ 47 kJ mol⁻¹, that corresponds to about 0.47 mol, which is nearly twenty times the 0.0250 mol actually burned. Even evaporating the whole 0.0250 mol would remove only about 1.2 kJ, around 5% of the missing energy.
The dominant losses are heat transfer to the surroundings and incomplete combustion, consistent with the yellow flame (CO and soot rather than only CO2). Covering the burner may improve precision slightly, but it cannot remove heat loss. The most important improvement is to use a bomb calorimeter with excess oxygen and good insulation, or at least lag the can, control draughts and use a more precise thermometer. The classmate is right about the direction of the evaporation error but wrong about its magnitude and about the conclusion.