Risk Assessment
There are many hazards encounted in this experiment because it involves flames and highly flammable fuels, the spirit burners must be kept away from the fuels, a naked flame could easily ignite a bottle of ethanol. Ethanol, Methanol and Propan-1-ol are all highly flammable, their vapour will catch fire at temperatures above 13°C, 12°C and 21°C respectively. Methanol, Propan-1-ol and Butan-1-ol can all cause irreversible effects and seriously harm skin, eyes and can cause drowsiness, dizziness and is very harmful if swallowed. This is why I will be wearing goggles and taking extra precautions when handling these harmful alcohols. To ensure safety I will keep stock bottles of alcohols well away from any flames.
Results:
ANALYSIS
Calculations
Methanol
1 gram of water heated through 1°C needs 4.2 Joules of heat energy.
∴ 200g of water heated through 1°C needs 4.2 x 200 = 840J
∴ 200g of water heated through 20.0 °C needs 840 x 20.0
= 16,800J
= 16.8KJ
If 3.29g of Methanol was burnt:
3.29g of Methanol gives 16.8KJ of heat.
∴ 1g of Methanol gives (16.8/3.29) KJ of heat
O=16 H=1 C=12 Methanol=CH3OH 1 Mole=32
∴ 1 mole of Methanol gives (16.8/3.29 x 32) KJ of heat
=163.40 KJ of heat.
So CH3OH + 3/2O2 ⇒ CO2 + 2H2O
Hc = -163.40 KJ mol-1
Ethanol
1 gram of water heated through 1°C needs 4.2 Joules of heat energy.
∴ 200g of water heated through 1°C needs 4.2 x 200 = 840J
∴ 200g of water heated through 18.5 °C needs 840 x 18.5
= 15,540
= 15.54KJ
If 2.41g of Ethanol was burnt:
2.41g of Ethanol gives 15.54KJ of heat.
∴ 1g of Ethanol gives (15.54/2.41) KJ of heat
O=16 H=1 C=12 Ethanol=CH3CH2OH 1 Mole=46
∴ 1 mole of Ethanol gives (15.54/2.41 x 46) KJ of heat
=296.60 KJ of heat
So CH3 CH2OH + 3O2 ⇒ 2CO2 + 3H2O
Hc = -296.60 KJ mol-1
Propan-1-ol
1 gram of water heated through 1°C needs 4.2 Joules of heat energy.
∴ 200g of water heated through 1°C needs 4.2 x 200 = 840J
∴ 200g of water heated through 18.0 °C needs 840 x 18.0
= 15,120
= 15.12KJ
If 2.36g of Propan-1-ol was burnt:
2.36g of Propan-1-ol gives 15.12KJ of heat.
∴ 1g of Propan-1-ol gives (15.12/2.36) KJ of heat
O=16 H=1 C=12 Propan-1-ol =CH3CH2CH2OH 1 Mole=60
∴ 1 mole of Propan-1-ol gives (15.12/2.36 x 60) KJ of heat
=384.41KJ of heat
So CH3CH2CH2OH + 5O2 ⇒ 3CO2 + 4H2O
Hc = -384.41 KJ mol-1
Butan-1-ol
1 gram of water heated through 1°C needs 4.2 Joules of heat energy.
∴ 200g of water heated through 1°C needs 4.2 x 200 = 840J
∴ 200g of water heated through 18.0 °C needs 840 x 18.0
= 15,120
= 15.12KJ
If 1.70g of Butan-1-ol was burnt:
1.70g of Butan-1-ol gives 15.12KJ of heat.
∴ 1g of Butan-1-ol gives (15.12/1.70) KJ of heat
O=16 H=1 C=12 Butan-1-ol =CH3CH2CH2CH2OH 1 Mole=74
∴ 1 mole of Butan-1-ol gives (15.12/1.70 x 74) KJ of heat
=658.16 KJ of heat.
So CH3CH2CH2CH2OH + O2 ⇒ 4CO2 + 5H2O
Hc = -658.16 KJ mol-1
ANALYSIS
The enthalpy change of combustion of a fuel is a measure of the energy transferred when one mole of the fuel burns completely. I used the fact that 4.2J of energy is required to raise the temperature of 1g of water by 1°C to find the enthalpy change of combustion for 4 alcohols, Methanol, Ethanol, Propan-1-ol and Butan-1-ol.
I can see from these results that as the alcohols increase in size, they get more exothermic; this is related to their molecular structure:
Methanol=CH3OH
In this reaction, 7 bonds are broken (3 C-H bonds, 1 C-O bond, 1 O-H bond and 2 O-O bonds) and 6 bonds are made (4 O-H bonds and 2 O-C bonds). Oxygen breaks 4 bonds and makes 6 bonds, because this is a combustion reaction, the oxygen needs to be fully combusted, and it is the oxygen bonds being made that give out the most energy when bonds are being formed.
The energy taken in when the bonds break is less than the energy given out when the new bonds are made, so the resultant energy will be negative, and the overall reaction is exothermic.
Alcohol General Formula: CnH2n+1OH
These four alcohols are part of the alcohol homologous series, possessing similar properties and structures and only differing structurally by CH2. So as you go down the group, they increase by CH2 each time.
There is an upward trend, but one anomalous result, so no accurate analysis can be made of the similarity of difference between enthalpy change of the four alcohols. All I can deduct is that as alcohols get bigger, as more CH2 molecules join, the reaction becomes more exothermic because more energy is given out. I cannot however identify any pattern of differences between the four alcohols, my experiment and results are not accurate enough to make those conclusions.
EVALUATION
Evaluation of evidence
The method I used was not very accurate but very precise, taking into account the equipment used. I did not get any anomalous results; I averaged all readings to ensure maximum precision.
I can only be as precise as my least precise piece of equipment, e.g. thermometer and measuring cylinder. For the thermometer and measuring cylinder I could only read to one decimal place, so: temperature = 20.0°C ± 0.5°C
Percentage error = Difference x 100
Measurement
= 0.5 x 100
20.0
= 2.5%
Percentage error of thermometer for Methanol = 20.0°C ± 2.5%
Water volume % error = 0.5 x 100
200.0
= 0.25%
Percentage error of 200cm3 measuring cylinder = 200cm3 ± 0.25%
These percentage errors are very small and will not really affect my end results and conclusions.
Because of this my results will be very precise, but compared to data book results, inaccurate.
Evaluation of procedures
Heat loss to air and heat loss to apparatus are huge sources of procedural error because if any heat is lost, it means fuel was burnt, but I have no way of measuring energy, I had to ensure as much heat as possible went to the water because that was the only thing I was measuring. Incomplete combustion and use of different burners are also main sources of error because they are hard to control and can easily give unfair tests.
I reduced these errors by effective shielding using insulating heatproof mats underneath and all around experiment, these helped contain the heat and ensure as much heat as possible went to the water. I washed the container out with clean cold water in between experiments to help keep it a fair test. When the container got a layer of soot on the bottom, due to incomplete combustion I cleaned it off before the next experiment. I used identical spirit burners every time.
I would of used an enriched oxygen atmosphere if the facilities were available; I used the best methods of heat loss prevention available.
The least precise pieces of equipment should be more precise, and then measurements and results will be more accurate, for example using a burette for measuring water volume accurately and a more accurate thermometer. Then my percentage error will decrease given more precise results.
