Burning Alcohol
Aim
To find out how much energy in kJ is given off when 1 mole of different alcohol burns, and to devise a way of comparing these results and predicting values for other alcohols.
Prediction
I think that as we move down the group of alcohols, the energy given off by the burning of the alcohol will increase. As we move from Methanol (CH4O) down to Pentanol (C5H12O), there are more atoms in the molecule, and therefore more bonds between the atoms, as shown:
Methanol
Ethanol
(Propanol*)
Butanol
Pentanol
As you can see, there are more bonds in Pentanol than in Methanol. When alcohol is burnt, energy is put in to break down the bonds between the atoms in the molecule of alcohol. This alcohol reacts with the air and burns, giving off water and carbon dioxide. Taking Methanol as an example, this is what the word and chemical equation looks like:
Methanol + Oxygen --> Carbon Dioxide + Water
2CH4O + 3O2 --> 2CO2 + 4H2O
This is an exothermic reaction, meaning that when energy is put in, more energy is given off:
EOUT > EIN
This can be shown in another way, showing the breaking down of the bonds:
E = Energy H = Hydrogen O = Oxygen C = Carbon
This proves that more energy will be given off when more bonds are formed.
Set Variable
Each time I do the experiment, I will change the alcohol. The different alcohols I will use each time are Methanol, Ethanol, Butanol, and Pentanol.
Measured Variable
Each experiment, I will measure the difference in weight of the alcohol before and after the experiment, to calculate the energy given off. I will do three replicates and take an average to make sure my results are accurate.
Controlled Variables
Each experiment, I will keep the same:
* The mass of water (100g)
* The tin can
* The distance ...
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Set Variable
Each time I do the experiment, I will change the alcohol. The different alcohols I will use each time are Methanol, Ethanol, Butanol, and Pentanol.
Measured Variable
Each experiment, I will measure the difference in weight of the alcohol before and after the experiment, to calculate the energy given off. I will do three replicates and take an average to make sure my results are accurate.
Controlled Variables
Each experiment, I will keep the same:
* The mass of water (100g)
* The tin can
* The distance between the wick of the alcohol bottle and the tin can (20mm)
* The temperature by which the water is raised (10oC)
Apparatus
Method
. Set up apparatus as shown
2. Weigh the bottle of alcohol
3. Measure temperature of water with thermometer
4. Light the wick
5. Wait for water to raise in temperature by 10oC
6. Put out the flame
7. Measure weight of the bottle of alcohol and record the weight difference
8. Repeat 1-7 for each alcohol three times.
Safety Aspects
* Be careful not to knock over or drop the bottles of alcohol
* Wear goggles before and during when the wick is lit.
Results
Conclusion
As the number of carbons (and atoms) per molecule of alcohol increases, the energy released by one mole of the alcohol increases proportionally. This supports my original prediction, that the more atoms per molecule of alcohol increases, the energy released will increase.
Propanol
I left Propanol out of my original experiment. From my graph, I can extrapolate the approximate value that an experiment with Propanol would give. Reading it off my line, it comes to 1050kJ of energy. I did the experiment with Propanol, and it gave these results:
This result is not very close to the result I extrapolated from my graph, but it is quite close to the mid point to the results for Ethanol and Butanol, which is 890.41. This result fits nicely into the trend that my graph shows.
Using the same method on my graph, I could predict the value of Hexanol, which
Explanation
As I predicted, the energy released per mole of alcohol increased with the number of carbon atoms in each molecule of alcohol. This is because there are more bonds in the alcohols with more carbons per molecule than in the alcohols with fewer carbons per molecule (see Fig. 1, page 1).
When energy is put in, the bonds are broken. For the bonds to form again, energy is given out, and the amount of energy given out in the reaction is greater than that put in, meaning the reaction is exothermic. (Fig. 3 and 4, page 1). This means that to start the reaction, you need energy. Once the reaction is started, you get more energy back than you put in. This means that if you put more energy in, you get more energy out. You need more energy to break the bonds of Pentanol than Methanol, so you will get more energy out of burning Pentanol per mole, that you will with burning Methanol per mole.
Evaluation
My method is good, as it shows a definite trend, as shown in the graph and the conclusion, and they support my original prediction. The method I used, although is gave consistent results, was fundamentally flawed, as shown here:
There is one factor that the discrepancy in my results and the actual figures can be put down to - heat loss. While I kept the tin can as close to the flame as I could, there was still heat loss - to the air surrounding, and into the can itself, not just the water. While the trend is still shown, the results aren't as accurate as they could be. The problem comes when you try to make the results more accurate, because preventing heat loss it very difficult.
One approach to preventing heat loss to the air would be to do the experiment in a vacuum, but that in itself is a problem, because creating a large enough vacuum and stabilising would be difficult without very expensive equipment.
Another possibility is to contain the experiment in an environment which reflects all the heat back and focuses the heat on the can - perhaps using foil, or a more expensive reflective material if budget is not a problem. However, this can still result in loss of heat.
If we take a look at a thermos flask, the liquid inside is surrounded by two sheets of a reflective material. In between these two sheets is a vacuum, meaning it's very hard for heat to pass between the two layers. As further insulation, the two sheets are reflective, so any heat trying to get out, just gets reflected back in, and any heat trying to get out, just gets reflected off the side and doesn't get through. If this experiment could be contained within a larger 'thermos flask', it would prevent a lot of the heat loss.
Another way to increase the accuracy of results would be to measure the weight loss to more decimal places. The scales I used only went up to one decimal place. Another two decimal places would increase the accuracy of my results.
All in all, the experiment could be improved for accuracy, but at the very least, it does dictate a trend.
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Ross Beavis
