Results
Methanol
Temperature of water before *c 18.00 18.00
Temperature of water after *c 38.00 38.00
Mass of burner before g 204.64 203.89
Temperature of water after *c 203.89 203.19
Ethanol
Temperature of water before *c 15.00 15.00
Temperature of water after *c 35.00 35.00
Mass of burner before g 160.00 159.10
Temperature of water after *c 159.10 158.10
Propan-1-ol
Temperature of water before *c 14.00 16.00
Temperature of water after *c 34.00 36.00
Mass of burner before g 156.48 155.38
Temperature of water after *c 155.38 154.30
Propan-2-ol
Temperature of water before *c 16.00 18.00
Temperature of water after *c 36.00 38.00
Mass of burner before g 115.82 114.71
Temperature of water after *c 114.71 113.58
Butan-1-ol
Temperature of water before *c 18.00 12.00
Temperature of water after *c 38.00 32.00
Mass of burner before g 167.66 166.43
Temperature of water after *c 166.43 165.23
Butan-2-ol
Temperature of water before *c 16.00 14.00
Temperature of water after *c 36.00 34.00
Mass of burner before g 180.12 178.93
Temperature of water after *c 178.93 177.78
Analysis/Conclusion
In this experiment, I recorded the mass of the burners and the temperature of the water before and after the experiment so that I can calculate the amount of heat absorbed for each fuel. To calculate the amount of heat absorbed, I can use the formula below:
Mass of water x temperature rise x 4.2
With all the fuels, I used 200grams of water and made the temperature rise 20*c. So the amount of heat absorbed will be:
200/1000 x 20 x 4.2 = 16.8KJ
In this experiment, I recorded two readings for all the fuels. I did this so that I could get more reliable and accurate results. Since I have got two readings for all my fuels, I am going to calculate the average of the two.
Methanol:
204.64 – 203.89 = 0.75grams
203.89 – 203.19 = 0.70grams
(0.75 + 0.70)/2 = 0.725grams
Ethanol:
160.00 – 159.10 = 0.90grams
159.10 – 158.10 = 1.00grams
(0.90 + 1.00)/2 = 0.95grams
Propan-1-ol:
156.48 – 155.38 =1.10grams
155.38 – 154.30 = 1.08grams
(1.10 + 1.08)/2 = 1.09grams
Propan-2-ol:
115.82 – 114.71 = 1.11grams
114.71 – 113.58 = 1.13grams
(1.11 + 1.13)/2 = 1.12grams
Butan-1-ol:
167.66 – 166.43 = 1.23grams
166.43 – 165.23 = 1.20grams
(1.23 + 1.20)/2 = 1.215grams
Butan-2-ol:
180.12 – 178.93 = 1.19grams
178.93 – 177.78 = 1.15grams
(1.19 + 1.15)/2 = 1.17grams
Now that I have worked out the averages for all the fuels, I can work out the enthalpy change of combustion of all the fuels. This is a measure of the energy transferred when one mole of the fuel burns completely.
To get the enthalpy change of combustion of a fuel, you have to find out the mass of the particular fuel then you divide the mass of the fuel you weighed from the mass of the fuel giving you a certain number of moles. Then you divide the amount of heat absorbed from the number of moles calculated previously.
Since heat is given off it is an exothermic reaction making your enthalpy change of combustion calculated negative.
Here are the enthalpy changes of combustion for all the fuels used.
Methanol:
Formula of the fuel = CH3OH
Mass of the fuel = (12 + 4 + 16) = 32grams
Mass of 1 mole of the fuel = 0.75grams
Number of moles of the fuel used = 0.75/32
= 0.0234moles
Energy transferred by 0.0243 moles of the fuel = 16.8J
Energy transferred by 1 mole of the fuel = 16.8/0.0234
= 716.80 KJ mol-1
So the enthalpy change of combustion is –716.80KJ mol-1.
Ethanol:
Formula of the fuel = C2H5OH
Mass of the fuel = ((12 x 2) + 6+ 16) = 46grams
Mass of 1 mole of the fuel = 0.95grams
Number of moles of the fuel used = 0.95/46
= 0.02065 moles
Energy transferred by 0.0298 moles of the fuel = 16.8J
Energy transferred by 1 mole of the fuel = 16.8/0.02065
= 813.47 KJ mol-1
So the enthalpy change of combustion is –813.47 KJ mol-1.
Propan-1-ol:
Formula of the fuel = C3H7OH
Mass of the fuel = ((12 x 3) + 8 + 16) = 60grams
Mass of 1 mole of the fuel = 1.09grams
Number of moles of the fuel used = 1.09/60
= 0.0182 moles
Energy transferred by 0.0136 moles of the fuel = 16.8J
Energy transferred by 1 mole of the fuel = 16.8/0.0182
= 924.77 KJ mol-1
So the enthalpy change of combustion is –924.77 KJ mol-1.
Propan-2-ol:
Formula of the fuel = C3H7OH
Mass of the fuel = ((12 x 3) + 8 + 16) = 60grams
Mass of 1 mole of the fuel = 1.12grams
Number of moles of the fuel used = 1.12/60
= 0.0187 moles
Energy transferred by 0.0159 moles of the fuel = 16.8J
Energy transferred by 1 mole of the fuel = 16.8/0.0187
= 900 KJ mol-1
So the enthalpy change of combustion is -900 KJ mol-1.
Butan-1-ol:
Formula of the fuel = C4H9OH
Mass of the fuel = ((12 x 4) + 10 + 16) = 74grams
Mass of 1 mole of the fuel = 1.215grams
Number of moles of the fuel used = 1.215/74
= 0.0164 moles
Energy transferred by 0.0104 moles of the fuel = 16.8J
Energy transferred by 1 mole of the fuel = 16.8/0.0164
= 1023.21 KJ mol-1
So the enthalpy change of combustion is –1023.21 KJ mol-1.
Butan-2-ol:
Formula of the fuel = C4H9OH
Mass of the fuel = ((12 x 4) + 10 + 16) = 74grams
Mass of 1 mole of the fuel = 1.17grams
Number of moles of the fuel used = 1.17/74
= 0.0158 moles
Energy transferred by 0.0164 moles of the fuel = 16.8J
Energy transferred by 1 mole of the fuel = 16.8/0.0158
= 1062.56 KJ mol-1
So the enthalpy change of combustion is –1062.56 KJ mol-1.
The enthalpy change of combustion of alcohols are affected by their molecular structure because as the carbon chain becomes longer, the number of bonds in a molecule also increases; and since each bond has a specific bond enthalpy to it (example, each C-H bond has an average bond enthalpy of +413 KJ mol-1), therefore the greater the number of bonds, the higher the final enthalpy change of combustion value would be.
The main trend and pattern I noticed in my result is that when the alcohol group is in the body of the alcohol, the more fuel is burnt resulting to the enthalpy change of combustion being lower than the alcohols that the alcohol group is not in the body - the grams of fuel burnt for Propan-2-ol was 1.12 grams and Propan-1-ol was 1.09 resulting the enthalpy change of combustion of Propan-2-ol lower than the enthalpy change of combustion of Propan-1-ol but for Butan-1-ol, the enthalpy change of combustion was lower than Butan-2-ol. This is an anomaly.
Another trend noticed is when the number of carbon atoms increases the mass of the fuel burnt increases and so does the enthalpy change of combustion.
Evaluation
I had an anomaly result in my graph.
One of the trends I noticed was that when the alcohol is branched, the enthalpy change of combustion is lower than the un-branched alcohol but in this case I got the opposite for Butan-2-ol, which means it, is an anomaly.
At carbon 4, which is Butan-2-ol the enthalpy change of combustion was –1062.56 KJ mol-1 and Butan-1-ol was –1023.21 KJ mol-1.
To see if my results are precise and reliable, I have calculated some percentage errors for all the fuels used.
Methanol:
Mass of fuel on balance = 0.01/204.64 x 100 = 0.00489%
0.01/203.89 x 100 = 0.00490%
0.00489 + 0.00490= 0.00979% - 1st repetition
Mass of fuel on balance = 0.01/203.89 x 100 = 0.00489%
0.01/203.19 x 100 = 0.00492%
0.00489+ 0.00492 = 0.00981% - 2nd repetition
(0.00979 + 0.00981)/2 = 0.0098% - Average
Temperature change = 0.1/18 x 100 = 0.556%
0.1/38 x 100 = 0.263%
0.556 + 0.263 = 0.82%
Temperature change = 0.1/18 x 100 = 0.556%
0.1/38 x 100 = 0.263%
0.556 + 0.263 = 0.82%
(0.82 + 0.82)/2 = 0.82% -Average
Volume of measuring cylinder = 0.1/100 x 100 = 0.1%
0.1/100 x 100 = 0.1%
0.1 + 0.1 = 0.2%
Ethanol:
Mass of fuel on balance = 0.01/160.00 x 100 = 0.00625%
0.01/159.10 x 100 = 0.00629%
0.00625 + 0.00629 = 0.0125% - 1st repetition
Mass of fuel on balance = 0.01/159.10 x 100 = 0.00629%
0.01/158.10 x 100 = 0.00633%
0.00629 + 0.00633 = 0.0126% - 2nd repetition
(0.0125 + 0.0126)/2 = 0.0126% - Average
Temperature change = 0.1/15 x 100 = 0.667%
0.1/35 x 100 = 0.286%
0.667 + 0.286 = 0.953%
Temperature change = 0.1/15 x 100 = 0.667%
0.1/35 x 100 = 0.286%
0.667 + 0.286 = 0.953%
(0.953 + 0.953)/2 = 0.953% -Average
Volume of measuring cylinder = 0.1/100 x 100 = 0.1%
0.1/100 x 100 = 0.1%
0.1 + 0.1 = 0.2%
Propan-1-ol:
Mass of fuel on balance = 0.01/156.48 x 100 = 0.00639%
0.01/155.38 x 100 = 0.00644%
0.00639 + 0.00644 = 0.0128% - 1st repetition
Mass of fuel on balance = 0.01/155.38 x 100 = 0.00644%
0.01/154.30 x 100 = 0.00648%
0.00644 + 0.00648 = 0.0129% - 2nd repetition
(0.0128 + 0.0129)/2 = 0.01285% - Average
Temperature change = 0.1/14 x 100 = 0.714%
0.1/34 x 100 = 0.294%
0.714 + 0.294 = 1.01%
Temperature change = 0.1/16 x 100 = 0.625%
0.1/36 x 100 = 0.278%
0.625 + 0.278 = 0.903%
(1.01 + 0.903)/2 = 0.96% -Average
Volume of measuring cylinder = 0.1/100 x 100 = 0.1%
0.1/100 x 100 = 0.1%
0.1 + 0.1 = 0.2%
Propan-2-ol:
Mass of fuel on balance = 0.01/115.82 x 100 = 0.00863%
0.01/114.71 x 100 = 0.00872%
0.00863 + 0.00872 = 0.0174% - 1st repetition
Mass of fuel on balance = 0.01/114.71 x 100 = 0.00872%
0.01/113.58 x 100 = 0.00880%
0.00872 + 0.00880 = 0.0175% - 2nd repetition
(0.0174 + 0.0175)/2 = 0.01745% - Average
Temperature change = 0.1/16 x 100 = 0.625%
0.1/36 x 100 = 0.278%
0.625 + 0.278 = 0.903%
Temperature change = 0.1/18 x 100 = 0.556%
0.1/38 x 100 = 0.263%
0.556 + 0.263 = 0.819%
(0.903 + 0.819)/2 = 0.86% - Average
Volume of measuring cylinder = 0.1/100 x 100 = 0.1%
0.1/100 x 100 = 0.1%
0.1 + 0.1 = 0.2%
Butan-1-ol:
Mass of fuel on balance = 0.01/167.66 x 100 = 0.00596%
0.01/166.43 x 100 = 0.00601%
0.00596 + 0.00601 = 0.01197% - 1st repetition
Mass of fuel on balance = 0.01/166.43 x 100 = 0.00596%
0.01/165.23 x 100 = 0.00605%
0.00596 + 0.00605 = 0.0118% - 2nd repetition
(0.01197 + 0.0120)/2 = 0.012% - Average
Temperature change = 0.1/18 x 100 = 0.556%
0.1/38 x 100 = 0.263%
0.556 + 0.263 = 0.82%
Temperature change = 0.1/12 x 100 = 0.833%
0.1/32 x 100 = 0.313%
0.833 + 0.313 = 1.15%
(0.82 + 1.15)/2 = 0.99% -Average
Volume of measuring cylinder = 0.1/100 x 100 = 0.1%
0.1/100 x 100 = 0.1%
0.1 + 0.1 = 0.2%
Butan-2-ol:
Mass of fuel on balance = 0.01/180.12 x 100 = 0.00555%
0.01/178.93 x 100 = 0.00559%
0.00555 + 0.00559 = 0.0111% - 1st repetition
Mass of fuel on balance = 0.01/178.93 x 100 = 0.00559%
0.01/177.78 x 100 = 0.00562%
0.00559 + 0.00562 = 0.0112% - 2nd repetition
(0.0111 + 0.0112)/2 = 0.0112% - Average
Temperature change = 0.1/16 x 100 = 0.625%
0.1/36 x 100 = 0.278%
0.625 + 0.278 = 0.903%
Temperature change = 0.1/14 x 100 = 0.714%
0.1/34 x 100 = 0.294%
0.714 + 0.294 = 1.01%
(0.903 + 1.01)/2 = 0.96% -Average
Volume of measuring cylinder = 0.1/100 x 100 = 0.1%
0.1/100 x 100 = 0.1%
0.1 + 0.1 = 0.2%
Overall, I think that my results are quite precise and reliable because I repeated the experiment twice resulting me to do an average and the two readings obtain weren’t too far apart.
Also my percentage errors were very little it did not even reach up to 1%.
I think my experiment went well but could have been better because since I carried out the experiment as an individual it was quite difficult to ensure that no heat is lost. Holding the heatproof mats in place and reading off the thermometer was quite distracting because the heatproof mat could drop and your eyes are off the thermometer.
The main significant error, which could have interfered with my results, was that the copper calorimeter that I had was covered in soot before and after heating. This meant that incomplete combustion took place.
I could have improved this experiment by using a clean calorimeter each time I am burning the fuels.
This could make my results more accurate and reliable because that is when I will get the correct enthalpy change of combustion meaning that complete combustion has taken place.
Another error was the distance between the calorimeter and the fuel. It varied.
This meant that the fuel did not heat the water inside the copper calorimeter directly.
I could have improved this experiment by keeping the distance between the fuel and calorimeter constant.
This could make my results more accurate and reliable because the fuel will not just burn and little heat would touch the calorimeter but the full heat.
This will allow me to have the accurate mass of fuel burnt changing the enthalpy change of combustion.
Another error was that when moving o to a next fuel the calorimeter has absorbed the heat and if you don’t cool it down the fuel will not fully burn.
I could have improved this experiment by cooling the calorimeter with water so it can return to the original temperature.
This could make my results more accurate and reliable because again the correct amount of fuel was not burnt and this will change my enthalpy change of combustion.
Acknowledgements
Chemical Ideas – George Burton
John Holman
John Lazonby
Gwen Pilling
David Waddington
Chemical Storylines – George Burton
John Holman
John Lazonby
Gwen Pilling
David Waddington
