Whether alcohol is suitable alternative for petrol, and Whether it matters which alcohol we use.
Introduction
Today, the prices of vital commodities derived from oil are increasing, and in particular, the cost of running a petrol driven car is becoming more expensive to do so, and this is all due to the fact that the amount of oil left to be extracted is diminishing. Thus, a suitable alternative is required. One such candidate is gasohol, which is about % alcohol.
This is much cleaner than normal petrol, and also makes the petrol last a lot longer.
Thus, my objectives, and the purpose of this piece of coursework is to find out:
) whether alcohol is suitable alternative for petrol, and
2) Whether it matters which alcohol we use.
Preparation
In order to complete either objective ("whether alcohol is a suitable alternative for petrol?" And; "Does it matter which alcohol we use?"), I need to make sure that I carry out a fair test, which means that I keep all the variables the same except one, which will be the alcohol. I will also need to repeat the experiment 3 times in order to check the accuracy of the experiment.
The variables are:
) the distance the calorimeter is above the flame
2) the amount of water used
3) the temperature change of the water
4) the size of the wick
5) the alcohol I use
From my preliminary work, I have decided to keep the variables as (respectively):
) 10cm
2) 50ml
3) +10°C
4) 3.5cm
5) This variable will change, in order to achieve a conclusion to objective two.
[To see how these variables relate to the experiment, please see the Method]
In order for this experiment to be not only fair, but safe as well, there are basic guidelines I will use, such as; I will wear goggles at all times, and I will keep the alcohols at 2 arms' length away from naked flames (except when I am burning them)
Theoretical results
I will now work out what I should, theoretically, get from one mole of ethanol, propanol, butanol, and petrol, assuming complete combustion. Complete combustion is where all of the alcohol is turned into carbon dioxide (CO2) and water (H2O). It is completely oxidised.
Ethanol
The molecular formula for ethanol is C2H5OH, and assuming complete combustion, it combines with 3O to give 2CO and 3HO. This can be represented in the formula:
C2H5OH + 3O2 › 2CO2 + 3H2O
It can also be represented in the following structural formula:
From this equation, and the bond energies in the appendix, I can work out how much energy is used to break all the bond in the reactant molecules, and how much energy is given out in the forming of the products' molecules.
Energy required breaking bonds in reactant molecules (kJ/mole):
bond type
No. of bonds
energy/bond
total energy
C-H
5
413
2065
C-C
347
34
C-O
358
358
O-H
464
464
O=O
3
498
494
TOTAL=
4728
Energy formed in forming bonds in products' molecules (kJ/mole):
bond type
No. of bonds
energy/bond
total energy
C=O
4
805
3220
O-H
6
464
2784
TOTAL=
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Energy required breaking bonds in reactant molecules (kJ/mole):
bond type
No. of bonds
energy/bond
total energy
C-H
5
413
2065
C-C
347
34
C-O
358
358
O-H
464
464
O=O
3
498
494
TOTAL=
4728
Energy formed in forming bonds in products' molecules (kJ/mole):
bond type
No. of bonds
energy/bond
total energy
C=O
4
805
3220
O-H
6
464
2784
TOTAL=
6004
To find the amount of energy released per mole we simply subtract the energy required to break bonds from the amount of energy released.
4728-6004=-1276
Thus, the net energy released per mole of ethanol completely oxidised is -1276kJ (that is to say that 1267kJ of energy is lost from the system, i.e. the molecules). If we divide this by the weight of one mole (46g), we get the amount of energy released per gram. This is handy for when I will compare the results of different alcohols.
276?46 = 27.7
Thus, for every gram of ethanol that is burnt, 27.7kJ of energy is given off to the surroundings.
This all shows that the reaction is exothermic. This means that it gives out energy. This can be demonstrated on the following diagram (see next page).
The activation energy [AE] is how much energy is needed to break the bonds in the products.
?H is how much energy is released as new bonds form. In this case, ?H is negative, which means that more energy is released than is taken in.
a & b = the reactants E = energy in system
c & d = the products AE = activation energy
Propanol & Butanol
Propanol [C3H7OH + 4.5O2 › 3CO2 + 4H2O]
Energy required to break bonds in reactants
bond type
No. of bonds
energy/bond
total
C-H
7
413
2891
C-C
2
347
694
C-O
358
358
O-H
464
464
O=O
4.5
498
2241
TOTAL=
6648
Energy released as bonds form in products
bond type
No. of bonds
energy/bond
total
C=O
6
805
4830
O-H
8
464
3712
TOTAL=
8542
[6648-8542 = 1894kJ/g] ? 60g=
Net energy/gram: 31.6kJ
Butanol [C4H9OH + 6O2 › 4CO2 + 5H2O]
Energy required to break bonds in reactants
bond type
No. of bonds
energy/bond
total
C-H
9
413
3717
C-C
3
347
041
C-O
358
358
O-H
464
464
O=O
6
498
2988
TOTAL=
8568
Energy released as bonds form in products
bond type
No. of bonds
energy/bond
total
C=O
8
805
6440
O-H
0
464
4640
TOTAL=
1080
[8568-11080 = 2512kJ/g]? 74g=
Net energy/gram: 33.9kJ
Octane
C8H18 + 12.5O2 › 8CO2 + 9H2O
Energy required to break bonds in reactants
bond type
No. of bonds
energy/bond
total
C-H
8
413
7434
C-C
7
347
2429
O=O
2.5
498
6225
TOTAL=
6088
Energy released as bonds form in products
bond type
No. of bonds
energy/bond
total
C=O
6
805
2880
O-H
8
464
8352
TOTAL=
21232
[16088-21232 = 5144kJ/g]? 114g=
Net energy/gram: 45.1kJ
From these calculations, I can draw the conclusion that ethanol will not be an efficient alternative for petrol, as petrol produces much more energy per gram than ethanol. Also based on these calculations, I can conclude that the larger the alcohol molecule, the more energy is given out. However, in practice, a proportion of this energy will be lost to the surroundings.
In order to verify these theoretical results, I need to make some actual measurements. My experiment is set out in the method.
Prediction
I predict that as the size of the alcohol molecule increases, so will the amount of energy per mole released. I also predict that ethanol will not release as much energy per mole than petrol.
Method
Apparatus list:
Measuring cylinder
Copper calorimeter
Crucible & glass wool wick [spirit burner]
Heat proof mat
Clamp & stand
Goggles
Alcohols - ethanol, propanol and butanol.
I will set up the apparatus as shown, using the above equipment. From my preparation experiments, I know that to achieve the best results, I need to use 50cc's [50ml or 50g] of water, and set the calorimeter at 10cm above the top of the spirit burner, which used a 3.5cm high wick. Before I start the experiment, I will need to weigh the spirit burner with alcohol, so that after I have heated the water 10ºC, I can weigh it again, and find the mass of alcohol burnt. I can then put this information into the formula:
Mass of water x ? t of water x specific heat capacity of water = energy given out
This formula tells me how much energy (in joules) I get from X grams of fuel. I can already substitute in a few pieces of information, to give:
50 x 10ºC x 4.2 = energy given out
50 x 10ºC x 4.2 = 2100J
Thus, for every X grams of alcohol I burn, I should get 2100J of energy, although how much of this will be lost to the surroundings, I do not know. I will repeat this experiment three times for each alcohol, in order to ensure that I don't just get an anomalous result.
Results
Once I had collected my data, I put it into the following tables, and worked out how much energy one gram and how much one mole of the alcohol produced.
I already know that X grams produces 2100J. If I divide 2100 by X, I get the amount of energy one gram produces.
2100 ? X = Y Joules
If I then times Y by the mass of one mole of the alcohol, I get how much energy is produced by one mole of the alcohol.
Y x Mr = Z Joules
Thus, for my first experiment for ethanol, I get:
.27g gives 2100J
2100 ? 1.27 = 1653
g gives 1653J
653 x 46 = 76038
mole gives 76038J, or 76kJ
I repeated this for each experiment for each alcohol, and produced the following results:
ethanol
expt. 1
mass of fuel burnt
gives
J of energy
.27g
2100
g
653
46g
76038
mole
76kJ
expt. 2
mass of fuel burnt
gives
J of energy
.2g
2100
g
750
46g
80500
mole
80.5kJ
expt. 3
mass of fuel burnt
gives
J of energy
.36g
2100
g
544
46g
71024
mole
71kJ
average
mass of fuel burnt
gives
J of energy
649
46
75854
mole
75kJ
propanol
mass of fuel burnt
gives
J of energy
expt. 1
0.65
2100
3230
60
93800
mole
93.8kJ
mass of fuel burnt
gives
J of energy
expt. 2
0.94
2100
2234
60
34040
mole
34kJ
mass of fuel burnt
gives
J of energy
expt. 3
0.64
2100
3281
60
96860
mole
96.8kJ
average
mass of fuel burnt
gives
J of energy
2915
60
74900
mole
74.9kJ
butanol
mass of fuel burnt
gives
J of energy
expt. 1
0.85
2100
2470
74
82780
mole
82.7kJ
mass of fuel burnt
gives
J of energy
expt. 2
0.86
2100
2441
74
80634
mole
80.6kJ
mass of fuel burnt
gives
J of energy
expt. 3
0.67
2100
3134
74
231916
mole
231.9kJ
average
mass of fuel burnt
gives
J of energy
2681
74
98394
mole
98.4kJ
Analysis or results
As can be seen from this graph, there is a discrepancy between the theoretical results, and what I actually recorded. This is because not all the energy went into the water, a lot was dissipated into the surroundings, i.e. the calorimeter, the heatproof mat, the air, etc. However, we can still see a definite positive correlation between the size of the molecule and the amount of energy released as it is burnt. As the size increases, so does the amount of energy released. This is exactly what I predicted.
Knowing that there is a positive correlation between the size of the molecule and the energy released, I can calculate the efficiency of my practical results. This is done by putting them into the following formula:
Efficiency = Practical x100
Theoretical
Expt.
Ethanol
Propanol
Butanol
5.96%
0.23%
7.27%
2
6.31%
7.07%
7.19%
3
5.56%
0.39%
9.23%
Avg.
5.88%
9.23%
7.90%
Total average=7.67%
As the size of the molecules increases, efficiency should decrease, as there will be more carbon to be oxidised, and therefore less energy as not all the carbon is turned into carbon dioxide (some will be turned into carbon monoxide; some into soot [just carbon]). However, this does not appear to be the case, as the efficiency was greatest for propanol, and least for ethanol.
Evaluation
I am satisfied with my results, I received no anomalous results, and I repeated my experiment three times, ensuring a good level of accuracy. I achieved a reasonable level of efficiency, although a higher level would have been much better. The results back each other up; there is a definite positive correlation between the size of the alcohol molecule and the energy released per mole, suggesting a very firm conclusion. This supports my original prediction. Also, even theoretically, none of the alcohols produced as much energy as petrol. So I don't think ethanol would be a viable alternative for petrol, although, some larger alcohols maybe. But, these molecules require more energy to get them started.
There is also a trend in the balanced formulas for the complete combustion of the alcohols.
C2H5OH + 3O2 › 2CO2 + 3H2O
C3H7OH + 4.5O2 › 3CO2 + 4H2O
C4H9OH + 6O2 › 4CO2 + 5H2O
With each extra carbon atom in alcohol molecule, there are two extra hydrogen atoms, and 1.5 extra oxygen molecules are needed to completely combust the alcohol. An extra carbon dioxide and an extra water molecule are also produced. This can be turned into a mathematical formula, thus:
CnH2n+1OH + 1.5nO2 › nCO2 + n+1H2O
This sort of formula is called an homologous series. Alcohols are a homologous series, and other such series include alkanes, and alkenes.
What could I do to make this experiment better?
To make this experiment work better, and achieve a higher efficiency, I could do several things. First, I could conduct the experiment inside a sealed container, as this would stop air heated by the combustion of the alcohol escaping. The container could also be silvered (this would stop heat being radiated away), as well as being filled with oxygen, which would ensure that as much of the alcohol was oxidised as possible. I could also use a much cleaner copper calorimeter, as this would ensure a good transfer of energy to the water.
Peter Hill 11z GCSE Chemistry Coursework
