The amount of energy needed to break the bonds of each of the alcohols in my experiment is shown in the table below:
- 2552, - 3788, - 5024, - 6260
1236, 1236, 1236
This shows that, if, for example, three times as much energy was given out making the bonds, compared to breaking the old ones, Pentanol would be the most efficient fuel, as it would have given out the most heat energy.
Here are the balanced symbol equations for each alcohol reactions:
Ethanol:
2C2H5OH + 6O2 → 4CO2 + 6H2O
Propanol:
2C3H7OH + 9O2 → 6CO2 + 8H2O
Butanol:
2C4H9OH + 12O2 → 8CO2 + 10H2O
Pentanol:
2C5H11OH + 15O2 → 10CO2 + 12H2O
To find the bond Energies above I worked out the bond energies needed to break each of the different bonds and calculated the totals for each side of the equation. Then I took the products form the reactants to find the delta H, for the alcohol. The workings for these calculations are on the following page:
Apparatus needed:
- 1 ethanol burner
- 1propanol burner
- 1 butanol burner
- 1 pentanol burner
- 1 steel can
- 100cm3 measuring cylinder
- Heat roof mat
- Clamp and clamp stand
- Thermometer
- Balance
Method
Using a 100cm3 measuring cylinder 100cm3 of water will be measured, to the accuracy of the nearest cm3. This water will be emptied into a tin can suspended by a clamp and clamp stand. There will be a 10cm gap between the can and the base of the stand. The alcohol burner will be placed directly under the can on a heatproof mat, with a flame that is big enough just to touch the bottom of the tin can when lit. The burner will be weighed with its lid on a balance that has the accuracy of a hundredth of a gram. Record the weight and then the burner is placed on the heatproof mat. The burner must remain with its lid on, so no alcohol evaporates. This would cause a safety hazard as it could cause a fire and would also affect the results if the alcohol escapes and is not burnt. The lid is removed and the flame is lit using a lighted split – not a Bunsen burner as this too could cause a fire. A thermometer will be used to measure the temperature and stir the water. It will have an accuracy of 1°C.
After the water’s temperature has risen by 40°C, the flame is blown out and the lid replaced to stop any alcohol evaporating. The burner will be weighed again and its weight will be recorded. From the two results, a difference in weight can be calculated, and his will also be recorded. The experiment will then be repeated with the other alcohols. All the alcohols will give out the same amount of energy in total, as the temperature of the water is being increased by 40°C, this means that the most efficient alcohol is the one that transfers this energy by burning the least amount of fuel.
The amount of energy given out by all of the alcohols will be:
Mass of Water x 4.2 x change in Temperature
100g x 4.2 x 40°C
= 16,800 Joules
I will repeat he experiment three times, as it will make my results more reliable, if I find any anomalous results, these will be repeated further.
As this is a dangerous experiment, there are a lot of safety precautions that must be attended with. Unless the flame is lit, the lid must always be on the burner, as it will minimize errors and if any alcohol evaporates, it could lead to an explosion. The burner must always be placed on a heatproof mat when alight, so the bench does not burn, and when using ethanol, a Bunsen burner must not be alight near to it as the ethanol could catch fire.
Fair Testing
Fair testing is very important, it will help minimize errors and make my results more reliable. To make my experiment a fair test, the same equipment, temperature increase, height from the can to the stand base, mass of water will be used, and most importantly the burner will always be weighed with its lid on. The variable that will be changed it’s the alcohol used.
Obtaining Evidence
The highlighted experiments gave odd results and so have not been included in the average.
Analysis
The results featured in the table above have made it possible to work out energy released for each fuel, per mole:
From my results I can see that in fact Butanol gives out the most energy when burnt. However, as the line of best fit on my graph shows, there is a slight anomaly. It also shows that Pentanol should have given out the most energy per mole. Also from my graph I can see that the relationship between the relative formula mass of the alcohol and the amount of energy given out is proportional. I believe that this is due to the fact that in alcohols, the bigger the formula mass, the more bonds the alcohol contains. As I referred to in my prediction of the experiment, the more bonds an exothermic reaction has, the more energy the reaction will give out. This tells me that if methanol were used, it would have given out less energy per mole than Ethanol.
You would expect the differences between the energies given out to be the same from one alcohol to the next, as only one CH2 is being added, or at least very similar. I have worked out the differences and here is what I found:
From this you can see that there are not similarities what so ever between the energies given out in my experiment. However, if the points had lay along the line of best fit, this may have been true. This can only conclude that my results are extremely inaccurate. Overall, my results have proven that my prediction that as the number of bonds increases, the amount of energy given out will also increase. However, they do not show very strong evidence and are not very reliable.
In conclusion to this experiment, I can see that as the molecules increase in size, the energy given out when the alcohol is burnt also increases.
Evaluation
It make my experiment as accurate as possible, I ensured that all the variables, excluding the one that was being changed were kept constant. However, a lot of heat energy was lost to the surroundings due to the flame not always being the same size. Also the tin can used up a lot of the heat energy, though heating itself up. These are the two main sources of error in the experiment. The draft from people walking past the workbench blew the flame, so it heated the air instead of the water. The reading of the thermometer, also could have added to the error. As, when the temperature had risen by 40°C and the burner been blown out, the temperature continued t rises up a couple of degrees. This too would have led to inaccurate results.
As you can see from my table of results, further back, there were four main anomalous results, and I have highlighted them on the table.
As the experiment with Butanol did not provide any reliable results, it was repeated twice more. This time, the results were more reliable, and were then used to calculate the average. I think these anomalies were due to sources of error, of which I have mentioned above. Smaller sources of error include using different burners in repeats, as they all had different wick sizes, which made the flame longer or shorter. The yellow flame was evidence that incomplete combustion occurred, this was due to the different bonds being made, and therefore the results would be affected. The soot that collected on the bottom of the can could have absorbed some of the heat energy.
I was able to draw a line of best fit, which means that my results must have been fairly reliable, although the line does not touch two points. Using four points, in my opinion is not sufficient enough to produce a graph that you can produce a firm conclusion from, however, due to safety reasons, I could not use methanol, as it is too explosive and could be dangerous.
I need to reduce some of the errors in my experiment to improve it. To do this I could use the same burner for the repeats so the length of the wick would be the same. By carrying out the experiment in a fume cupboard or ceiled container, I could reduce the heat escaping through rafts. If the soot were removed from the bottom of the can, by cleaning, the heat would get to the water quicker, which would also minimize errors.
Also I could measure he size of the wick to make sure that they are equal. I could further this experiment by investigating other homologous series, such as alkenes, or carboxylic acids. This would also me to see if there are any patterns or trends within the theory. To do this I would set up the same experiment, but instead of using alcohols, I would use the alkenes. To make the experiment more reliable I would use a wider range of fuels, for example, six. The first 6 alkenes are as follows: Ethane, Propene, Butene, Pentene, Hexane, and Heptene.
Below are the relative formula mass and the formulas of the first 6 alkenes:
Here are the Balanced Symbol Equations for the first six Alkenes:
Ethene:
2C2H6 + 7O2 → 4CO2 + 6H2O
Propene:
2C3H8 + 10O2 → 6CO2 + 8H2O
Butene:
2C4H10 + 13O2 → 8CO2 + 10H2O
Pentene:
2C5H12 + 16O2 → 10CO2 + 12H2O
Hexene:
2C6H14 + 19O2 → 12CO2 + 14H2O
Heptene:
2C7H16 + 21O2 → 14CO2 + 16H2O
This would be a starting point for the experiment. I think I would find that the pattern is the same and that the longer the alkene, the more bonds contains. From what I have discovered in this experiment I would predict that Heptene would give out the most energy as the more bonds in the exothermic reaction the more energy given out from the reaction.
