C – O = 2 X 336 = 672 C – O = 24 X 336 = 8064
O – H = 2 X 464 = 928
C – C = 4 X 346 = 1384
O = O = 9 X 497 = 4473
Total Energy in = 13547 kJ Total Energy out = 15488 kJ
Energy Released = 13547 – 15488 = -1941 kJ
-1941 = -970.5 kJ
2
Butanol
C4H9OH = 74g
Butanol + Oxygen Water + Carbon Dioxide
C4H9OH + 6O2 5H2O + 4CO2
C – H = 9 X 435 = 3915 H – O = 10 X 464 = 4640
C – O = 1 X 336 = 336 C – O = 16 X 336 = 5376
O – H = 1 X 464 = 464
C – C = 3 X 346 = 1038
O = O = 6 X 497 = 2982
Total Energy in = 8735 kJ Total Energy out = 10016 kJ
Energy Released = 8735 – 10016 = -1281 kJ
Pentanol
C4H9OH = 88g
Pentanol + Oxygen Water + Carbon Dioxide
2C4H9OH + 15O2 12H2O + 10CO2
C – H = 22 X 435 = 9570 H – O = 24 X 464 = 11136
C – O = 2 X 336 = 672 C – O = 40 X 336 = 13440
O – H = 2 X 464 = 928
C – C = 8 X 346 = 2768
O = O = 15 X 497 = 7455
Total Energy in = 21393 kJ Total Energy out = 24576 kJ
Energy Released = 21393 – 24576 = -3183 kJ
-3183 = -1591.5 kJ
2
Hexanol
C6H13OH = 102g
Hexanol + Oxygen Water + Carbon Dioxide
C6H13OH + 9O2 7H2O + 6CO2
C – H = 13 X 435 = 5655 H – O = 14 X 464 = 6496
C – O = 1 X 336 = 336 C – O = 24 X 336 = 8064
O – H = 1 X 464 = 464
C – C = 5 X 346 = 1730
O = O = 9 X 497 = 4473
Total Energy in = 12658 kJ Total Energy out = 14560 kJ
Energy Released = 12658 – 14560 = -1902 kJ
These results are negative because there is more energy given off in the reaction than is taken in. Therefore energy is lost to the environment. This creates a negative amount of energy left in the substance at the end of the experiment. The results on the graph are positive because I changed it from the negative to a positive. This was possible because I assumed that theoretically all of the energy from the burning alcohol is being transferred to the water. This makes it a lot easier to compare the theoretical results with the practical results.
As preliminary work to this experiment I had already done the experiment once with Ethanol. I did, therefore, know how to do the experiment safely and properly.
Method
Equipment needed: 0.5g of Ethanol, Butanol, Propanol, Pentanol, Hexanol; Lighted Splint; Crucible; Clamp stand; Clasp; Thermometer; Metal Can; water.
- Set up the experiment as shown in the diagram below.
- Measure out 0.5g of Ethanol and place this into a crucible. To make sure that the results are accurate, I will use a set of scales, which can measure weights of up to two decimal places. This will make it much easier to be sure that I am using the same amount of alcohol for each experiment.
-
Measure 100cm3 of water and put this into a metal can.
- Suspend the metal can 5cm above the crucible using the stand and the clasp to keep it in the same position. I kept the distance of the can above the crucible the same for each of the experiments.
- Record the temperature of the water.
- Set the ethanol alight and leave until it is completely burnt.
- Record the highest temperature of the water.
- Repeat steps 1-7 using Propanol, Butanol, Pentanol and Hexanol. The amount of water and the amount of alcohol need to be kept the same for all of the experiments.
Results Table
Energy transferred in the experiment
I expect the total amount of energy for the experimental results to be less than that of the theoretical results. This is because during the experiment not all of the energy from the alcohol has been transferred to the water. Some of it will have been lost to the environment and some will have been lost to the metal can containing the water. To find out the total amount of energy transferred to the water I used the following equation.
Energy Transferred = Rise in Mass of Specific heat
temperature X water (kg) X capacity of
of water (oC) water (kJ/kg/oC)
This equation only gives the energy given off by 0.5g of the alcohol. To find the energy that would be given off by one mole then the answer first had to be divided by the mass of the alcohol used. It then had to be multiplied by the mass of one mole of the alcohol.
Ethanol = 10 x 0.100 x 4.2
= 4.2
= 4.2 x 46 = 386.4kJ/mole
0.5
Propanol = 10 x 0.100 x 4.2
= 4.2
= 4.2 x 60 = 504kJ/mole
0.5
Butanol = 10 x 0.100 x 4.2
= 4.2
= 4.2 x 74 = 621.6kJ/mole
0.5
Pentanol = 11 x 0.100 x 4.2
= 4.2
= 4.2 x 88 = 813.12kJ/mole
0.5
Hexanol = 11 x 0.100 x 4.2
= 4.2
= 4.2 x 102 = 942.48kJ/mole
0.5
Conclusion
The graph shows that for the theoretical results the amount of energy lost increases by a regular amount of 310.5kJ/mole as the chain increases in length by one. The line for the practical results has a different position and gradient because during the experiment heat was lost to the environment. Not all of the energy from the alcohol was transferred to the water so it didn’t heat up as much as it should have done. The reason the two lines are diverging is that as more energy is produced by the different alcohols, then more and more energy as lost to the environment.
Apart from this difference the ratio between the points for each of the alcohols should be the same. This is because although different amounts of heat were lost, it should be the same percentage for each alcohol:
Ethanol = 660 = 1.71
386.4
Propanol = 970.5 = 1.93
504
Butanol = 1281 = 2.06
621.6
Pentanol = 1591.5 = 1.96
813.12
Hexanol = 1902 = 2.02
942.48
Most of these results are around the same area. The only anomalous result was the ratio for the Ethanol. This may have happened if the water had been cooled by a draught during the experiment. This is unlikely though because any cooling of the water would have shown up on the graph as well as on the ratios for the experiment.
The table shows that there is a definite link between the predicted and actual amount of heat lost by the burning of the alcohols.
Evaluation
The experiment went very well, as there were very few anomalous results.
The experiment could be easily improved because during the experiment, a lot of heat was lost to the environment. This meant that the water didn’t heat up as much as it should have done. This is the reason why the results for the experiment are shown as giving off less energy than predicted.
This heat loss could be minimised through the use of a bomb calorimeter. This would trap all of the heat produced inside a container, which is inside a water bath. The heat is then passed through a copper coil, which is also suspended in the water. This creates the maximum amount of contact with the water possible so more heat is transferred. The apparatus also contains a stirrer so that the water heated up by the experiment doesn’t stay around the container and copper coil. This also helps to maximise the contact between the water and the energy from the experiment.
The evidence shown in the experiment shows that my prediction that the bond energies needed and given will increase as the length of the chain of carbons increases. The conclusions and the graph I made from the experiment also support my prediction.
Bibliography
- Nuffield Co-ordinated Sciences - Chemistry
