Investigating the structure of a fuel and the amount of energy released during combustion
Investigating the structure of a fuel and the amount of energy released during combustion
Aim
We are trying to find out whether an increase in the number of carbon atoms and different structural formulae affect the amount of energy released during combustion.
We have chosen to investigate 4 different alcohols: Ethanol, Propanol, Butanol and Octanol.
Three of the alcohol's we are using have 2 different structures (these are called isomers):
• Propanol = Propan-1-ol and Propan-2-ol
• Butanol = Butan-1-ol and Butan-2-ol
• Octanol = Octan-1-ol and Octan-2-ol
We haven't chosen to use the different structures, because we only wish to investigate the number of carbon atoms in different alcohols, and see whether a different number of carbon atoms will affect the amount of energy released during combustion.
Background Information
Alcohol's
I know that the alcohols I have chosen all produce an exothermic reaction during combustion and the products of the combustion of alcohols are water and carbon dioxide.
The basic chemical formula for an alcohol is CnH(2n+1)OH.
Ethanol = C2H5OH
Propanol = C3H7OH
Butanol = C4H9OH
Octanol = C8H17OH
Propanol, Butanol and Octanol have two different structures, 1-ol and 2-ol. They still have the same number of carbon bonds as each other, but their structures are slightly different:
Ethanol =
Propan-1-ol =
Propan-2-ol =
Butan-1-ol =
Butan-2-ol =
Octan-1-ol =
Octan-2-ol =
Measuring the energy output
To measure the energy output it is not possible to measure the amount of heat directly, instead you have to use the standard method and look at the effect that energy has on the temperature of a known volume of water.
It takes 4.2J/kg, (this is the specific heat capacity of water) to raise 1g of water; therefore to calculate the energy released by each alcohol we have to use the sum:
ENERGY RELEASED = mass of water x temperature rise x 4.2J.
Heat of reaction
Chemical bonds are forces of attraction between the atoms or molecules in a substance. Energy is needed to break these bonds and energy is released when new bonds are made. In a chemical reaction bonds between atoms in the reactant molecules are broken and new bonds are made.
When the new bonds are made the energy given out is greater than the energy taken in to break the old bonds. This type of reaction is called an exothermic reaction.
When you burn alcohol/fuel it is an exothermic reaction, as heat is given out. This is because the total energy released in bond making is greater than the energy needed for bond breaking.
The greater the surface area, the greater the force of attraction between the molecules, making it less easy to vaporise. Therefore the more carbon molecules there are, the more bonds there are and the greater the surface area.
Energy diagrams
In a chemical reaction, the reactants and the products possess different chemical bonds. Therefore, they possess different amounts of energy.
This can be shown in an energy diagram, which displays the energy content of the reactants and the products:
Energy diagram for an exothermic reaction
The energy diagram above shows the energy content of the reactants and the products in a chemical reaction. The difference is called the heat of reaction. It is given the symbol ?H. So,
Heat of reaction (?H) = Energy of products - Energy of reactants
In an exothermic reaction the products of the reaction contain less energy than the reactants and therefore ?H is a negative quantity.
Heat of combustion
Heat is measured in Joules (J) and in kilojoules (KJ). 1KJ = 1000J. The heat of reaction is measured in kilojoules per mole of reactant (KJ/mol).
The heat of combustion is the heat absorbed when 1 mole of reactant is completely burnt in oxygen.
Bond energy values
To break a chemical bond you require a certain amount of energy. This energy is different for different chemical bonds.
The table below lists the bond energies of some bonds in KJ/mol:
Bond
Bond energy (KJ/mol)
H-H
436
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Heat of combustion
Heat is measured in Joules (J) and in kilojoules (KJ). 1KJ = 1000J. The heat of reaction is measured in kilojoules per mole of reactant (KJ/mol).
The heat of combustion is the heat absorbed when 1 mole of reactant is completely burnt in oxygen.
Bond energy values
To break a chemical bond you require a certain amount of energy. This energy is different for different chemical bonds.
The table below lists the bond energies of some bonds in KJ/mol:
Bond
Bond energy (KJ/mol)
H-H
436
O=O
496
C-C
348
C=C
612
C-H
412
C-O
360
C=O
743
H-O
463
We can then use the bond energy values to calculate a value for the heat of reaction.
Prediction
I predict that an increase in the number of carbon atoms will increase the amount of energy released during combustion. This is because in a chemical reaction energy is needed to break bonds and is supplied to create the new bonds. When the new bonds are made the energy given out is greater than the energy taken in to break the old bonds - exothermic reaction.
Also, the greater the surface area, the greater the force of attraction between the molecules, making it harder to vaporise. Therefore more energy will be released.
I can show this theoretically by using the following calculations:
Ethanol + Oxygen Carbon dioxide + Water
C2H5OH + 3O2 2CO2 + 3H2O
xC-C=348 3xO=O
5xC-H=5x412 + =3x496 4xC=O + 6xO-H
xC-O=360 =4x743 =6x463
xO-H=463
TOTAL= 4719 5750
4719 - 5750 = -1031KJ/mol = energy released when 1 mole of ethanol is burned
Propanol + Oxygen Carbon dioxide + Water
2C3H7OH + 9O2 6CO2 + 8H2O
4xC-C=4x348
4xC-H=14x412 9xO=O 12xC=O + 16xO-H
2xC-O=360 + = 9x496 = 12x743 =16x463
2xO-H=463
TOTAL= 17734 16324
(17734 - 16324) ÷ 2 = -1527KJ/mol = energy released when 1 mole of Propanol is burned
Butanol + Oxygen Carbon dioxide + Water
C4H9OH + 6O2 4CO2 + 5H2O
3xC-C=3x348
9xC-H=9x412 6xO=O 8xC=O + 10xO-H
xC-O=360 + =6x496 =8x743 =10x463
xO-H=463
TOTAL= 8551 10574
8551 - 10574 = -2023KJ/mol = energy released when 1 mole of Butanol is burned
Octanol + Oxygen Carbon dioxide + Water
C8H17OH + 12O2 8CO2 + 9H2O
7xC-C=7x348
7xC-H=17x412 6xO=O 16xC=O + 18xO-H
xC-O=360 + =6x496 =16x743 =18x463
xO-H=463
TOTAL= 13239 20222
3239 - 20222 = -6983KJ/mol = energy released when 1 mole of Octanol is burned
NB: In the 'nuffeild advanced science book of data', there is a list showing the combustion of alcohols. I decided to compare my results to the theoretical calculations I had worked out above, as I feel that they are more accurate. However, I will refer to the data book's calculations, as a secondary source.
However, if I wished to use the 1-ol and 2-ol isomers of the following alcohols, then I predict that a similar amount of energy will be released. This is because even though they have a slightly different structure the formula is still the same, CnH(2n+1)OH, because it still has the same number of carbon and hydrogen atoms.
Diagram
We did a preliminary experiment of ethanol to test the accuracy of our method. We decided to change the amount of water, the material of the can and the way of ending the experiment. We increased the amount of water, this meant the water was heated slower which meant the energy wouldn't be wasted and would give more accurate results. We changed the can material to copper because the copper can will conduct the energy well so it is transferred more efficiently and reduce radiation. We also decided to change the method of ending the experiment because the amount of time is not needed just the temperature rise, so we will make the experiment end when the temperature rise is 20 °c this will stop too much energy loss by the water evapourating.
Method
. Collect an oil burner filled with ethanol.
2. Weigh it, without glass lid, record weight and replace lid.
3. Set up a stand and clamp to hold a copper can.
4. Measure out 100cm³(change to 200 later) of water using a measuring tube, place in the tin can and record the temperature of the water using a standard thermometer.
5. Place the oil burner on a mat and then position the tin can so it is directly above the wick.
6. Surround the oil burner and tin can with tinfoil and light the wick.
7. Wait till the temperature rise is 20, then put out the lamp.
8. Weigh the ethanol, without the glass lid, and record weight.
9. Repeat this for propan-1-ol, butan-1-ol, and octan-1-ol.
0. Repeat the whole experiment 3 times; so as to get an average result - making it a fairer and more accurate result.
From this experiment we needed to know the mass of the water, the temperature rise and the weight change of the alcohol. We needed to know the mass of the water and the temperature rise, so that we could find out the total energy released (Energy released = mass of water x temperature rise x 4.2J). The weight change of the alcohol could then help us to transfer this amount into moles, so we could compare our results to the theoretical answers and find out how accurate our results are.
We surrounded the oil burner and copper can with a barrier of tinfoil, as it helps limit energy loss, raises accuracy and we will make sure there is a hole so the oxygen can get to the lamp stopping incomplete combustion, this will make less soot meaning less energy loss.
Alcohol's are quite dangerous liquids to work with, as they are extremely flammable, so great care had to be taken that they weren't spilt and safety goggles were worn at all times.
Results
First results
Weight before (g)
Weight after (g)
Temperature before (°C)
Temperature after (°C)
Ethanol
94.80
92.70
6
36
Propan-1-ol
209.43
207.52
6
36
Butan-1-ol
99.66
97.64
4
34
Octan-1-ol
16.58
15.93
7
37
Second results
Weight before (g)
Weight after (g)
Temperature before (°C)
Temperature after (°C)
Ethanol
90.83
88.95
6
36
Propan-1-ol
81.52
78.72
7
37
Butan-1-ol
86.51
85.15
6
36
Octan-1-ol
22.74
21.23
7
37
Third results
Weight before (g)
Weight after (g)
Temperature before (°C)
Temperature after (°C)
Ethanol
75.36
73.34
6
36
Propan-1-ol
31.43
29.14
5
35
Butan-1-ol
205.73
204.54
7
37
Octan-1-ol
26.86
25.45
8
38
Average results
Weight before (g)
Weight after (g)
Temp before (°C)
Temp after (°C)
Weight change (g)
Temp change (°C)
Ethanol
87
85
6
36
2
20
Propan-1-ol
74.13
71.79
6
36
2.34
20
Butan-1-ol
97.3
95.78
6
36
.52
20
Octan-1-ol
22.06
20.87
7
37
.19
20
From these results we can then go on to work out the amount of energy released, so we can compare our results to the theoretical calculations:
Ethanol = mass of water x temperature rise x 4.2J
200 x 20 x 4.2J
= 16800J/2g of ethanol
per g of ethanol = 16800÷2
= 8400J/g
mass of 1 mole of ethanol = Mr
C2H5OH = (2x12) + (6x1) + (1x16)
= 46
mole has a mass of 46g
So, energy released per mole of ethanol = 8400 x 46
= 386400J/mol
= 386.4KJ/mol
= 386KJ/mol
Propan-1-ol = mass of water x temperature rise x 4.2J
200 x 20 x 4.2J
= 16800J/2.34g of propan-1-ol
per g of propan-1-ol = 16800÷2.34
= 7179J/g
mass of 1 mole of propan-1-ol = Mr
C3H7OH = (3x12) + (8x1) + (1x16)
= 60
mole has a mass of 60g
So, energy released per mole of propan-1-ol = 7179 x 60
= 430740J/mol
= 430.740KJ/mol
= 431KJ/mol
Butan-1-ol = mass of water x temperature rise x 4.2J
200 x 20 x 4.2J
= 16800J/1.52g of butan-1-ol
per g of butan-1-ol = 16800÷1.52
= 11053J/g
mass of 1 mole of butan-1-ol = Mr
C4H9OH = (4x12) + (10x1) + (1x16)
= 74
mole has a mass of 74g
So, energy released per mole of butan-1-ol = 11053 x 74
= 817922J/mol
= 817.922KJ/mol
= 818KJ/mol
Octan-1-ol = mass of water x temperature rise x 4.2J
200 x 20 x 4.2J
= 16800J/1.19g of octan-1-ol
per g of octan-1-ol = 16800÷1.19
14118 = J/g
mass of 1 mole of octan-1-ol = Mr
C8H17OH = (8x12) + (17x1) + (1x16)
= 130
mole has a mass of 130g
So, energy released per mole of octan-1-ol = 14118 x 130
= 1835340J/mol
= 1835.34KJ/mol
= 1835KJ/mol
Calculated heat of combustion
Theoretical heat of combustion
Data book combustion of alcohol
Ethanol
386 KJ/mol
031 KJ/mol
371 KJ/mol
Propan-1-ol
431 KJ/mol
527 KJ/mol
655.2 KJ/mol
Butan-1-ol
815 KJ/mol
2023 KJ/mol
2675.6 KJ/mol
Octan-1-ol
835KJ/mol
6983 KJ/mol
5293.6 KJ/mol
Graphs
Conclusion
I conclude that what I have predicted is partially correct.
I, firstly, predicted that the more carbon atoms there were, the more energy would be released. This is proved to some extent, as, if you look at the graph you can see that the amount of energy released increases as the number of carbon atoms increases. This is because when the alcohol burns it produces exothermic energy and forms carbon dioxide and water. When the bonds in the alcohol break it requires energy and new bonds are formed to create the carbon dioxide and water - giving out energy. Due to this being an exothermic reaction the total amount of energy that is released is more than that is required to break the bonds of the reactants. This energy released from the exothermic reaction then increases in size when there are more carbon atoms, i.e. In Octanol. This is because there are more bonds to break, so more energy is needed and therefore more energy is given out.
This proves the first part of my prediction, as if you look at the graph it is obvious that Ethanol has the smallest amount of energy given off during combustion, as it has the smallest number of carbon atoms (C2H5OH) and that Octan-1-ol has the largest amount of energy given off during combustion, as it has the most number of carbon atoms (C8H17OH).
By looking at the results it seems as though the result for the butan-1-ol and is too low or the result for the octan-1-ol is too high. However, this is due to the fact that we didn't use the alcohol's in-between Butanol and Octanol.
Butanol has 4 carbon atoms and Octanol has 8 carbon atoms and the alcohol's containing 5 carbon atoms, 6 carbon atoms and 7 carbon atoms were missed out, causing the gradient of the line of best fit to be much steeper.
If you look at the results showing the weight change, you can see that we gained a couple of anomalous results. This could be due to many reasons, such as human error or a change in the set-up of the equipment, meaning there was a greater energy loss into the surroundings.
Overall, I feel that I have produced results that have helped me to prove my prediction and ones that have followed my aim. I obtained 3 results from each alcohol, allowing me to produce a more accurate, average result.
Evaluation
In our experiments we tried to make sure that our results were obtained using the greatest accuracy. However, we did gain a few anomalous results, which could've been due to many reasons:
Throughout the investigation we weighed the alcohol's to 3decimal places, carried out the experiment on each alcohol 3 times, so that I could then find an average, making my results more precise and accurate and also made sure that our equipment was set up as alike as possible. However, this was not always entirely conceivable, as this investigation was done over 3 lessons and therefore we weren't always able to set-up our equipment identically.
Also, due to lack of tinfoil-covered boards, we were unable to always use the most accurate method of insulating our oil burner and tin can. Consequently there was a risk of different amounts of energy loss into the surroundings, also there would be a degree of incomplete combustion which would lead to energy loss.
If I was to re-do this experiment I would use distilled water instead of the tap water we were using, as distilled water is pure and the tap water could have contained all sorts of things that could cause a change to our results.
I would also use a buirette instead of the measuring tube we used, as the measuring tube is only accurate to half a millimetre.
We measured the temperature of the water using a thermometer, which is also only accurate to half a degree, and if I did re-do this experiment I would measure the temperature using a data logger, as it is a lot more precise and accurate.
The part of the experiment, which I thought, was the least accurate element was the insulating device, as even though it did reduce energy loss, it didn't work to a very high standard. If I was to improve this investigation I would devise some sort of sealed insulated unit, which stopped as much heat loss as possible and had a stirring device in it to continuously stir the water to make sure it was heated evenly.
In my findings I worked out the accuracy of my results by comparing them to the theoretical equations of heat combustion, as well as the ones in the data book and then finding an average accuracy:
If I was to further my investigation I would like to also investigate the alcohols between Butanol and Octanol, which have the structures: C5H11OH, C6H13OH and C7H15OH. This would enable me to prove my theory I explained in my conclusion.
