C-O x 1 = 358 x 1 = 358 = 4728.9
O-H x 1 = 464 x 1 = 464
O=O x 3 = 498.3 x 3 = 1494.9
(2) C=O x 2 = 805 x 4 = 3220
O = 6004
O-H x 6 = 464 x 6 = 2784
4728.9 – 6004 = -1275.1
Propanol
Propanol + oxygen → carbon dioxide + water
C3H8O + 4.5O2 → 3CO2 + 4H2O
(1) C-H x 7 = 413 x 7 = 2891
C-C x 2 = 347 x 2 = 694
C-O x 1 = 358 x 1 = 358 = 6649.35
O-H x 1 = 464 x 1 = 464
O=O x 4.5 = 498.3 x 4.5 = 2242.35
(2) C=O x 3 = 805 x 6 = 4830
O = 8542
O-H x 8 = 464 x 8 = 3712
6649.35 – 8452 = -1892.65
Butanol
Butanol + oxygen → carbon dioxide + water
C4H10O + 6O2 → 4CO2 + 5H2O
(1) C-H x 9 = 413 x 9 = 3717
C-C x 2 = 347 x 2 = 1041
C-O x 1 = 358 x 1 = 358 = 8571.6
O-H x 1 = 464 x 1 = 464
O=O x 6 = 498.3 x 6 = 2991.6
(2) C=O x 4 = 805 x 8 = 6440
O = 11080
O-H x 8 = 464 x 10 = 4640
8571.6 – 11080 = -2508.4
Pentanol
Pentanol + oxygen → carbon dioxide + water
C5H12O + 7.5O2 → 5CO2 + 6H2O
(1) C-H x 11 = 413 x 11 = 4543
C-C x 4 = 347 x 4 = 1388
C-O x 1 = 358 x 1 = 358 = 10490.25
O-H x 1 = 464 x 1 = 464
O=O x 7.5 = 498.3 x 7.5 = 3737.25
(2) C=O x 5 = 805 x 10 = 8050
O = 13618
O-H x 12 = 464 x 12 = 5568
10490.25 – 13618 = -3127.75
Hexanol
Hexanol + oxygen → carbon dioxide + water
C6H14O + 9O2 → 6CO2 + 7H2O
(1) C-H x 13 = 413 x 13 = 5369
C-C x 5 = 347 x 5 = 1735
C-O x 1 = 358 x 1 = 358 = 12410.7
O-H x 1 = 464 x 1 = 464
O=O x 9 = 498.3 x 9 = 4484.7
(2) C=O x 6 = 805 x 12 = 9660
O = 16516
O-H x 14 = 464 x 14 = 6496
12410.7 - 16516 = -3745.3
Heptanol
Heptanol + oxygen → carbon dioxide + water
C7H16O + 10.5O2 → 7CO2 + 8H2O
(1) C-H x 15 = 413 x 15 = 6195
C-C x 6 = 347 x 6 = 2082
C-O x 1 = 358 x 1 = 358 = 14331.15
O-H x 1 = 464 x 1 = 464
O=O x 10.5 = 498.3 x 10.5 = 5232.15
(2) C=O x 7 = 805 x 14 = 11270
O = 18694
O-H x 16 = 464 x 16 = 7424
14331.15 – 18694 = -4362.85
Method
The equipment will be set out as shown above.
I am going to use drinks can, a crucible, mineral wool, a thermometer, water, a clamp stand and different types of alcohol. I will put 90ml of water at room temperature in the can, which will be at a fixed height of 5.5cm from the tabletop. The crucible will have a small piece of mineral wool in with 0.5g of the alcohol. I will then set the alcohol alight and let it burn out. When it has burnt out, I will try to re-light it in case it has been blown out. I will then record the finishing temperature and work out the temperature rise, which will enable me to work out the amount of energy released.
Before I begin my experiment, I will run some preliminary tests to figure out how much fuel and water to use. I will also find a suitable height for the can where the temperature rise will be recordable. I am going to use the following alcohols, methanol, ethanol, propanol, butanol, pentanol, hexanol and heptanol.
Preliminary tests
I had to find out how much alcohol to use what amount of water and how high the can should be from the tabletop. We used 100ml and had the can at 6cm from the tabletop.
I found that 0.5g of methanol was better to use because we need to be able to measure hexanol, which should burn longer, and we will only have enough time if we use 0.5g of fuel.
Next, I had to find out how many millilitres of water to use because if we used too much the fuel wouldn’t last long enough to heat the water. We used 0.5g of fuel and had the can at 6cm from the tabletop.
I found that 90mls of water was enough to be able to record a reasonable temperature change.
Finally, I had to find a suitable height to put the can. I used 0.5g of fuel and 90ml of water.
I decided to put the can at 5.5cm from the tabletop because you lost too much energy I f you had it much higher.
I decided to use 0.5g of fuel, 90ml of water and the can at 5.5cm for my experiments.
Results
Analysis
To work out the energy released in kJ per mole I had find out how many moles of the alcohol there is to 1g using the periodic table. Then I divided that number by 0.5 because I had used 0.5g of fuel and multiplied the energy released in joules per 0.5g by it. To get the energy into kJ I had to divide that number by 100.
The graph shows that there is a positive correlation between the amount of carbons in the alcohol and the energy released.
The actual results are very similar to the predicted results because when you have another carbon to carbon bond in the alcohol more energy is needed to break it. This would result in more energy being used, which is why the more carbons that the alcohol has the higher the negative number becomes. Burning fuels is exothermic reaction, which will always result in energy being lost.
Less energy is needed to break the bonds of the alcohol than to make the bonds of the carbon dioxide and water
Evaluation
The actual results are close to the predicted results, which means that they are quite accurate. The actual results have lost more energy. This could have happened due to the heat loss between the burning fuel and the drinks can. My results follow the trend of more carbons more energy used which supports my conclusions very well.
I feel that my experiment was done fairly. The heat loss between the burning fuel and the can was kept the same and to a minimal. The average result for Heptanol wasn’t near to the line of best fit. This may have been because the heat loss was greater or not all of it was burnt in a few of the crucibles.
To make sure that my experiment is fairer, I will need to minimise the heat loss. Another way would be to weigh the fuel in moles and not grams. This would result in the calculations becoming more accurate and not to decimal places. I could also use the same amount of mineral wool so that alcohol in the very centre can be burnt easily.
BY STACEY ROBINSON.