Measuring the enthalpy change of combustion for an alcohol

Raw Results:

Alcohol Initial Mass of spirit burner (g)

(+/-0.01g)

Final Mass of spirit burner (g)

(+/-0.01g)

Initial Temperature of water (0C) (+/-

0.50C)

Final Temperature of water (0C) (+/-

0.50C)Methanol 118.00 116.94 24.0 46.0Ethanol 169.76 169.08 24.0 45.0

Propanol 119.13 117.80 23.0 45.0Butanol 136.65 136.12 22.0 43.0

Pentanol 160.84 160.01 24.0 45.0

Qualitative Results: There was no colour change during this experiment. Each alcohol burnt with a different sized flame with propanol being the tallest and thickest

yellow flame, and butanol having the smallest flame.

Analysis of results:Alcohol Change in weight of spirit

burner (g) (+/-0.01g)Change in temperature of water

(0C) (+/-0.50C)Methanol 1.06 22.0Ethanol 0.68 21.0

Propanol 1.33 22.0Butanol 0.53 21.0

Pentanol 0.83 21.0

Q=mc∆T: m=100cm3 c=4.18J/kg/0C ∆T= change in temperature of water Alcohol Amount of energy

Methanol Q=mc∆TQ= 100 × 4.18 × 22Q=9196J9196/1000=9.196kJ

Ethanol Q=mc∆TQ= 100 × 4.18 × 21Q=8778J8778/1000=8.778kJ

Propanol Q=mc∆TQ= 100 × 4.18 × 22Q=9196J9196/1000=9.196kJ

Butanol Q=mc∆TQ= 100 × 4.18 × 22Q=8778J8778/1000=8.778kJ

Pentanol Q=mc∆TQ= 100 × 4.18 × 22Q=8778J8778/1000=8.778kJ

Number of moles burnt: n=m/MrAlcohol Number of moles (mol)

Methanol Mr= 32.05m=1.06n=1.06/32.05=0.033mol

Ethanol Mr=46.08m=0.68n=0.68/46.08=0.015mol

Propanol Mr=60.11m=1.33n=1.33/60.11=0.022mol

Butanol Mr=74.14m=0.53n=0.53/74.14=0.0071mol

Pentanol Mr=88.17m=0.83n=0.83/88.17=0.0094mol

Therefore: 0.033mol of methanol releases 9.169kJ. 1 mole releases 9.196/0.033 =-278.67kJ mol-1

0.015mol of ethanol releases 8.778kJ. 1 mole releases 8.778/0.015 = -585.20kJ mol-1

0.022mol of propanol releases 9.169kJ. 1 mole releases 9.196/0.022 = -418.00kJ mol-1

0.0071mol of butanol releases 8.778kJ. 1 mole releases 8.778/0.0071 = -1236.34kJ mol-1

0.0094mol of pentanol releases 8.778kJ. 1 mole releases 8.778/0.0094 =- 933.83kJ mol-1

Uncertainties:Alcohol UncertaintiesMethanol Balance: (0.01/1.06)×2×100= +/-1.89%

Thermometer: (0.5/22.0) ×2×100= +/-4.55%Measuring cylinder: (0.5/100) ×1×100=+/-0.5%Total= +/-6.94%

Ethanol Balance: (0.01/0.68)×2×100= +/-2.94%Thermometer: (0.5/21.0) ×2×100= +/-4.76%Measuring cylinder: (0.5/100) ×1×100=+/-0.5%Total= +/-8.20%

Propanol Balance: (0.01/1.33)×2×100= +/-1.50%Thermometer: (0.5/22.0) ×2×100= +/-4.55%Measuring cylinder: (0.5/100) ×1×100=+/-0.5%Total= +/-6.55%

Butanol Balance: (0.01/0.53)×2×100= +/-3.77%Thermometer: (0.5/21.0) ×2×100= +/-4.76%Measuring cylinder: (0.5/100) ×1×100=+/-0.5%Total= +/-9.03%

Pentanol Balance: (0.01/0.83)×2×100= +/-2.41%Thermometer: (0.5/21.0) ×2×100= +/-4.76%Measuring cylinder: (0.5/100) ×1×100=+/-0.5%Total= +/-7.67%

Methanol: -278.67kJ mol-1+/-6.94% -278.67kJ mol-1+/- 19.34 kJ mol-1

Ethanol: -585.20kJ mol-1 +/-8.20% -585.20kJ mol-1 +/-47.99kJ mol-1

Propanol: -418.00kJ mol-1+/-6.55% -418.00kJ mol-1+/- 27.38kJ mol-1

Butanol: -1236.34kJ mol-1 +/- 9.03% -1236.34kJ mol-1 +/- 111.64kJ mol-1

Pentanol: - 933.83kJ mol-1 +/- 7.67% - 933.83kJ mol-1 +/- 71.62kJ mol-1

Literature values:Methanol: -725.70kJ mol-1

Ethanol: -1367.60kJ mol-1

Propanol: -2021.31kJ mol-1

Butanol: -2670.00kJ mol-1

Pentanol: -3330.91kJ mol-1

https://uk.answers.yahoo.com/question/index?qid=20130822173953AAsReD8

Conclusion:From my results there isn’t a definite pattern. I expected the enthalpy change of combustion to increase as the length of the hydrocarbon chain increased. The results do show an increase with methanol releases 278.67kJmol-1 and pentanol releasing 933.83kJmol-1. There is a definite increase here but with fluctuations with the other alcohols. Butanol had the largest enthalpy change of combustion (1236.34kJmol-1) even though it does not have the longest hydrocarbon chain. My results are very far away from the literature value, for example my value for pentanol was -933.83kJ mol-1 and the literature value is -3330.91kJ mol-1. The percentage difference is -71.96% ((-933.83—

3330.91/-3330.91)×100). There is a significant difference between these values and this is the case with all of my results.

Evaluation of results:I felt that the experiment was carried out correctly and we made minimal errors but our prediction did not match our results so something must have been carried out incorrectly. Were didn’t take any repeats due to time restraints so we were unable to calculate the mean which means our results are likely to not be entirely accurate. My results were very different from the literature values, this could be because we didn’t carry out he experiment correctly or our methods of collection were not accurate.

Evaluation of method:

Limitation How/why it affects the results ImprovementsNo repeats were taken so the mean could not be calculated or the standard deviation.

The raw data may therefore be unreliable because these results may be anomalies.

Take at least 5 repeats so that the mean can be calculated to ensure reliability in results.

The can was not at the same height above the flame for each experiment.

The water may have been heated up quicker if the can was nearer the flame so the results may be unreliable.

Using a ruler, measure a certain distance from the flame and make sure the bottom of the can is at this height for all of the experiments.

The same can was used for each experiment.

Not enough was left in between each experiment so the can didn’t have time to cool completely. This could have affected the temperature of the water and the time taken for the temperature to rise 200C.

Use a new can for each experiment to make the results more reliable.