I will use the following method to collect all the data that I will need to complete these calculations.
I will fill a beaker with 200g of water, and clamp it in the air with a retort stand, boss and clamp. Then fill a spirit burner, with the alcohol to be tested, then weigh it all, and record the results. Place this directly underneath the beaker of water, and light. Stir the water with a thermometer and heat until the temperature has risen by 40 C. When this has happened I will place the lid back over the burner straight away to stop any further combustion of the fuel. Reweigh the burner and its contents to measure the loss of fuel during combustion. I will repeat this process for all the alcohols: methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol. For safety reasons I will wear goggles throughout the experiment, to prevent any hot substances spitting into my eyes.
Diagram
List Of Equipment
Spirit Burner
Alcohols
Retort stand
Boss and clamp
200ml Metal Beaker
Water
-20 C to 150 C Mercury thermometer
Digital electronic scales
Preliminary Work
I conducted a preliminary experiment similar to the proposed method above, to check that everything in my experiment was feasible. I used the same setup and equipment, as in the diagram but used hexane instead of an alcohol.
Results of Preliminary work
Mass of fuel + burner + lid - Start: 235.02g
-Finish: 232.14g
Temperature of 200ml of water - Start: 24 degrees C
Calculating Heat Of Combustion
Energy transferred (joules) = Specific heat capacity x temperature rise x mass of water
= 4.18 x 40 x 200
= 3340 J
= 33.44 KJ
Moles = Mass
RMM
= 2.88 / 86
= 0.0334
1 Mole >> 1 x 33.4
0.0334
= 998.5 KJ mol energy was released
▲H = -998.5 KJ mol
This is the experimental value, the theoretical value can be worked out by bond energy calculations.
The reaction that took place was:
Hexane + Oxygen >> Carbon dioxide + Water
H14 C6 + 9.5 O2 >> 6 CO2 + 7 H2O
With the structural formula
The bonds broken: The bonds made:
5 x C-C = 1735 12 x C=O = 9660
14 x C-H = 5782 + 14 x O-H = 6496 +
9.5 x O=O = 4731 + = 16156
= 12248
12248 - 16156 = -3908 KJ mol
Conclusions From Preliminary Work
The total energy released should have been 3908 KJ for every mole that was burned. As our experiment showed that in practice only 998.5 KJ per mole released went to heating the water. It is clear that inaccuracies and wastage of energy were present in the experiment. Whilst it is clearly impossible to eradicate all of this, as it is inevitable that some energy will be lost as heat or light energy into the surroundings, we want to keep this to a minimum in our experiment. We could try and combat this by using some form of insulation, around the equipment so that the maximum amount of energy released possible goes to heating the water and is not dispersed in the surroundings.
As hexane is very volatile and will quickly lose mass when exposed to the air even when not burning, I have learnt to immediately replace the lid when I have finished burning. If I don’t do this more fuel will be lost that was burnt and therefore my readings will be less accurate.
The preliminary experiment was however generally successful. It showed some points I might wish to improve on but that overall this method can be used in my final experiment.
Number And Range Of Readings
I have chosen to use five different alcohols, as this will give me enough information to complete my calculations, and give a useable graph. I will repeat each reading once to achieve better accuracy in my results. I will measure the temperature to the nearest C as the thermometers are not accurate enough to show the temperature any more precisely. As I am using digital scales I can give accurate readings of mass change to 2 decimal places. I will also give the energy change in Kjmol to 2 decimal places
Number Of Variables
To make sure the experiment a fair test we will have only one variable. This will be the alcohol to be tested. All other equipment will stay the same. The same amount of water, in the same beaker, will be started at the same temperature each time, and the lab temperature will be kept constant. We will keep the distance from burner to beaker constant at 5cm. If this was varied, the percentage of energy used to heat the water would change, and this would therefore give an unfair test.
Obtaining
Results Table
Analysis
The two graphs are identical, only the energy ‘output’ graph has been reflected in the y-axis. This simply makes all the values positive. The energy ‘change’ will always be negative as it is subtracting the energy in the reactants from the energy in the products. As energy is being given off the energy left in the products will always be less than the reactants started with. The energy ‘output’ however is measuring the energy given off; as these reactions are exothermic it will always be positive.
These results show the prediction to be correct. As we were able to calculate the expected results in the form of bond energy diagrams then the prediction was relatively easy to form. As anticipated the graph of RMM against Energy Output per mole, shows linear positive correlation. With increased RMM, there are more bonds to be broken and then remade. As the combustion of alcohols is always exothermic this process always releases excess energy. The more of this process of bond breaking and reforming is done per mole, i.e. the higher the RMM, the more energy released per mole. It will be linear; as for every alcohol you move up the scale you add on the same each time, two extra H-C bonds.
The formulae for alcohols create a predictable pattern as the chain length increases. This allows us to say exactly how many of each bond type will be present in an alcohol of ‘n’ carbon atoms. We can also predict how many molecules of oxygen, carbon dioxide and water will be present in the combustion of a single alcohol molecule ‘n’ carbon atoms long. Therefore it is possible to create a simple formula to give us the exact amount of energy released for any alcohol.
n = Number of carbon atoms present in one alcohol molecule
The structure of an alcohol is (‘n’ carbon atoms long):
It will react with a certain amount of oxygen molecules (dependant on ‘n’)
To give a certain amount of Carbon dioxide molecules and a certain amount of water molecules (both dependant on ‘n’)
It is clear that from any length alcohol there will always be:
(Square bracketed is the energy used to break 1 bond or lost when 1 bond is made.)
1 x O-H bond [464 KJ]
1 x C-O bond [358 KJ] Bonds in
(2n + 1) x C-H bonds [413 KJ] alcohol molecule
( n –1 ) x C-C bond(s) [347 KJ]
1.5 n x O=O bonds [498 KJ] Oxygen molecules
Add all these terms together to get the total energy used in breaking all bonds in the reaction of any alcohol in oxygen.
- + 358 + (2n + 1) x 413 + (n – 1) x 347 + 747n
Energy lost in the breaking of original bonds (KJ) = 1920n + 888
The energy created when the different bonds reform needs to be subtracted from this to give overall energy transfer. There will always be the following bonds created in the combustion of an alcohol in air.
2n x C=O bonds [805 KJ] Carbon dioxide
Molecule
(2n + 2) x H-O bonds [464 KJ] Water molecule
1610n + 928n + 928
energy created by the bond formations = 2538n + 928
subtract this from energy used to break the original bond to give the overall energy change:
(888 + 1920n) - (2538n + 928)
Therefore in the burning of any alcohol in air, the total energy change will be equal to:
-618 n – 40
Testing the formula:
Alcohol n Bond Energy Calculation Outcome - 618 n – 40
Methanol 1 -658 Kj / mol -658 Kj / mol
Ethanol 2 -1276 Kj / mol -1276 Kj / mol
Propan-1-ol 3 -1894 Kj / mol -1894 Kj / mol
Butan-1-ol 4 - 2512 Kj / mol - 2512 Kj / mol
Pentan-1-ol 5 - 3130 Kj / mol - 3130 Kj / mol
The formula –618n – 40, gives us the theoretical energy change for any alcohol in combustion. To make this give the energy output, simply reverse all the signs, i.e.
618n +40. It proves both the prediction and the collected results to be correct. The graph of y = 618n +40, will always be linear as there is no squared term in the equation. Our results show this, as the graphs shown above have straight lines of best fit. The exothermic reaction, that was predicted and then found, is also shown by the way this formula will always give a positive energy output, provided n is positive. (This will always be true as no alcohol has negative amounts of carbon atoms.)
Evaluation
This graph compares the theoretical energy output (pink) with the experimental (blue).
As we can see there is substantially less energy output in the experimental data than in the theoretical results. The percentage difference between the two lines however stays fairly similar, and therefore the two lines follow roughly the same pattern. Although the data that was collected was not very accurate, it does give us a valuable graph. The inaccuracy in our results is accounted for in some fundamental flaws in the method.
The biggest inaccuracy by far was the wastage of energy. With this experiment it is almost impossible to prevent large amounts of energy not being accounted for in our calculations. Our theoretical results calculate the total amount of energy given out, in any form of energy. In our experiment we only took into account the energy that went into heating the water. Much of the energy lost will have gone to light energy in the flame, or the sound energy given off as the alcohol burns. There is no way of irradiating this, as a flame and a small noise will always be present during combustion. We could try and account for this by conducting the experiment in darkness and measuring light levels. These results could then be added this to the heat energy output in an attempt to total all energy out. This however would be very dangerous and so inaccurate that it is not even worth considering. Not only was there wastage of energy in light and sound, but also in the heat energy that did not go towards heating the water. A lot of the heat energy that was produced will have been lost by being conducted or convected away in the air or radiated into the atmosphere. We could have set up insulation to prevent some of these losses although this would only have reduced the wastage slightly. Energy was also lost to heating some of the apparatus such as the beaker or the clamp used to hold the beaker in place. This is impossible to compensate for, and is inevitable in this particular experiment. These wastages account for the fact that the experimental data is always considerably less than the theoretical results.
Although there are no major anomalies in the graph, it is far from the theoretical, perfect straight line. This is down to further inaccuracies in our method. Whilst energy wastage was common between all of the readings, which had similar in the percentage losses, it primarily affects the relationship between the theoretical outcome and the experimental data. Some inaccuracies however, are likely to distort the relationship between the different pieces of experimental data themselves.
For example, although the distance from wick to beaker was kept constant, some alcohols seemingly burned with a larger flame, hence the heat getting closer to the beaker. I could have tested the height of each flame before conducting the experiment and then taken this into account when measuring distance from wick to beaker. This would be an improvement to make if I was to conduct the experiment again. A larger flame also means more energy wasted as light compared with other readings.
Despite these flaws in the plan we did get enough good results to identify the trend we were expecting. It is far from conclusive experimental evidence for the prediction, however, as there are so many clear inaccuracies. There is seemingly no other way however, to test the energy released in the combustion of alcohols. So in future I would have to try and cut out as many of the inaccuracies as I could without completely changing the method. If I was to repeat this experiment I think it would be beneficial to test more alcohols, possibly hexanol, or heptanol. This would improve my graph, as it would give me a clearer idea of experimental trends that are occurring. It would make spotting anomalous readings easier if I was to do two repeats, and not just one. This is also likely to make my results more accurate.