The Combustion of Alchohols
The Combustion of Alchohols
Chemistry Coursework
Introduction
In order to carry out this investigation, I first have to define the terms of the objectives so that an understanding is present and a clear process to acquire the answer can be devised.
What are alcohols?
Alcohols are a specific group of substances that are hydroxyl derivatives of hydrocarbons but more explicitly they are types of fuel. A fuel is a substance (e.g. methane) that readily undergoes combustion (the combustion of a substance involves its reaction with oxygen and the release of energy) and gives out large amounts of energy. The combustion of fuels is usually an exothermic reaction.
The distinction between alcohols and other fuels, such as methane, is that alcohols are a homologous series of compounds that contain the -OH group of atoms as the functional group. The functional group in a molecule is the group of atoms within the structure that determines the characteristic reactions of that substance. All the earlier, smaller alcohols have a neutral pH, and are colourless liquids that do not conduct electricity. The general chemical structure of an alcohol is as follows:
CnH(2n+1)OH
By far the most well known alcohol is ethanol, which often goes by the common name of alcohol itself.
Its chemical formula is C2H5OH. It is produced in industry by reacting ethene and steam together. They are first compressed to 60 atmospheres and passed over a catalyst (immobilised phosphoric(v) acid) at 300ºC.
300ºC, 60 atmospheres
ethene + steam -------› ethanol
phosphoric acid
C2H4(g) + H2O(g) --› C2H5OH(g)
It can also be produced via fermentation because ethanol and CO2 are the natural waste products of yeasts when they ferment sugar. However, ethanol is toxic to yeast so once the ethanol concentration has reached around 14% or the sugar runs out then the multiplying yeast die and the fermentation ends. The Babylonians and Egyptians discovered this fermentation and found the product of crushed grapes (also containing sugar) would be a drink with a kick.
yeast
glucose --› ethanol + carbon dioxide
enzymes
C6H12O6(aq) ---› 2C2H5OH(aq) + 2CO2(g)
This alcohol is not always used as a fuel for combustion but also has an array of different purposes in industry and everyday life. It is an important solvent and raw material in making many other organic chemicals. Many everyday items such as paints, glues, perfumes and aftershaves use ethanol as a solvent, and notoriously for alcoholic drinks. Ethanol can be safely consumed but must only be drunk in moderation. Mediocre amounts of alcohol seem to reduce coronary heart disease, however, heavy drinking can damage muscle tissue of the heart and eventually cause liver failure and death. Methanol is very toxic and even in small amounts can cause blindness and death.
Coming back to the idea of alcohols as a fuel; when fuels are combusted the reaction with oxygen produces two products - carbon dioxide and water as well as energy from the exothermic reaction. So:
fuel + oxygen --› carbon dioxide + water
and replacing the fuel with the alcohol gives:
alcohol(CnH2n+1OH) + oxygen(O2) --› carbon dioxide(CO2) + water(H2O)
What am I going to study?
The aim of the investigation is to work with the combustion of specific alcohols. However, just mindlessly burning fuels is not going to achieve anything extensive. I need to have something to study. Firstly I have to narrow down the investigation by stating which alcohols I am going to work with. Below are the alcohols I will use:
* Methanol (CH3OH) 1 carbon atom least carbon atoms
* Ethanol (C2H5OH) 2 carbon atoms
* Propanol (C3H7OH) 3 carbon atoms
* Butanol (C4H9OH) 4 carbon atoms
* Pentanol (C5H11OH) 5 carbon atoms most carbon atoms
These are the first five basic alcohols that share common properties such as being colourless, non-conductors and have a neutral pH. They are exclusively linked together by the pattern in the number of carbon atoms in each of the alcohols' molecules. As you can see from the chemical formulas of the alcohols they go in ascending order of number of carbon atoms - methanol starting with one up to pentanol, which has five carbon atoms in a molecule.
What I want to study and find out is that between the five alcohols, which alcohol is the best alcohol? But on what conditions and criteria can I determine which alcohol is the best?
Relating back to investigating combustion it would be logical to grade the alcohols by which one can produce the most energy. An experiment to find this, however, would have to be carefully planned because calculating energy output by means of a practical experiment can be endlessly flawed in a variety of ways. Also to find the best one would involve comparing one to another. This is not as straightforward as it sounds. It would not be viable to simply burn equal masses of each fuel and see which one release most energy. This is because the size and the mass of the molecule of each alcohol is different and so having the same mass of each alcohol would result in having a different number of moles of each, which are ultimately what are being reacted - the molecules. Therefore, each alcohol would have to be compared by the number of moles of each that are burnt.
The best alcohol can now be defined as follows:
The best alcohol is that which is the most efficient; that which releases the most energy for the same amount of moles combusted.
I can calculate the energy released by the reaction by setting up an experiment where a volume of water is heated by the energy from the reaction. Then using the specific heat capacity of the water as well as calculating how many moles of the alcohol is burnt, I can get an estimate of the total energy released by the reaction.
Hypothesis
...
This is a preview of the whole essay
The best alcohol can now be defined as follows:
The best alcohol is that which is the most efficient; that which releases the most energy for the same amount of moles combusted.
I can calculate the energy released by the reaction by setting up an experiment where a volume of water is heated by the energy from the reaction. Then using the specific heat capacity of the water as well as calculating how many moles of the alcohol is burnt, I can get an estimate of the total energy released by the reaction.
Hypothesis
This hypothesis is based on the knowledge that I have of fuels and alcohols and the knowledge I have about energy in reactions. This hypothesis will include my predictions as to which alcohol is the best one and why.
In order for a reaction to take place energy has to be taken in to break the bonds of the molecules being reacted. When the products of the reaction are created, they are being made by the re-forming of the broken bonds but in a new arrangement. When the new bonds are formed energy is released again. In combustion reactions the energy given out by the reaction exceeds the energy taken in to break the bonds. This leads me on to believe that the alcohol which will produce the most energy will be that which forms the most bonds.
I could also predict methanol to be the best alcohol because it has the fewest bonds to be broken, therefore it will take in the least energy. However, it will also make fewer products per molecule of the alcohol that is combusted, so less energy will be given out. That is why I predict the largest alcohol to be the best one because it will form the most bonds when the products are made per molecule of the alcohol combusted.
The bigger the alcohol, the more energy is released.
Preliminary Experiment
Aim:
To find the optimum conditions for undertaking the main experiment.
Apparatus:
Boss and clamp, carbonated drinks can, cardboard draft shield, heatproof mat, measuring tube, retort stand, ruler, stopwatch, thermometer, water, wax candle.
Diagram:
Method:
During this preliminary experiment I worked with several variables. Theses included the height of the can from the candle, the volume of water used in the can and use of the draught shield. I took the can and secured it to the retort stand on the heatproof mat using the boss and clamp. The can could be moved up or down depending on where it was required to be positioned. Then I placed the candle directly underneath the can. Now I could use the measuring tube to measure out different volumes of water for each test. I could also alter the height of the can above the candle to different values using a ruler for accuracy. Then I put the thermometer into the can and noted the temperature of the water. I lit the candle and, using the stopwatch, timed how long it took the candle to raise the temperature of the water by 20ºC. In some tests I also used the draught shield around the candle and can.
Results:
Below I have set out the results of the experiment with the variations in each test outlined:
Heigt of can
Mass of water
Time taken for
Draught
above candle (cm)
in can (ml)
temp change (s)
shield?
6
20
35
?
2
6
40
89
?
3
5
20
12
?
4
5
30
34
?
5
4
30
91
?
6
4
50
43
?
7
3
20
43
?
8
3
40
59
?
9
2
20
6
?
0
2
50
50
?
The preliminary experiment brought back a wide range of different times to heat the water by 20ºC. I was looking for conditions that would give back a time that would be suitable for accurately obtaining results. Therefore the heating of the water should not be so spontaneous that the time is too quick to record accurately, and on the other hand the heating should not be too long for the obvious reasons of wasting time.
The use of a draught shield was included in some of the tests in the preliminary experiment. Scientific basis for the use of a draught shield is for the apparent reason that it contains heat radiating out from the flame, which may not necessarily be focused on the can. Therefore more of the energy from the combustion is contributing to the heating of the water rather than being lost and introducing further inaccuracy into the results. The draught shield will also accelerate the rate of the increase in temperature so this will have to be taken into account.
I have chosen to use the conditions in the 8th test in the preliminary as you can see from the table above. With the can 3cm above the candle and with 40ml of water in the can, it takes just under a minute for the 20ºC rise, and this is test included the draught shield. So the time taken would be flexible and manageable to work with. 40ml of water is also a good value to work with. Obviously if any less is used then the time will be reduced; any more and the time increases. This mass also allows a good depth for the thermometer to be submerged into and to give a more accurate reading with little lag time (there will be an offset in the actual temperature and the reading on the thermometer). The height above the can is sufficient because even if the flame itself becomes quite tall then it will not be in full reach of the base and will not unwillingly distort results.
Between each test in the preliminary I noticed a blackening on the base of the can. This was the build-up of carbon on the can from the flame. I realised that the carbon would act as an insulator to the can and the water so I need to keep the can free from this build-up in the main experiment. The carbon could be easily washed of with water so this will not be too much of a problem.
In order to eliminate the effect of anomalous results of the final averages, I will repeat the test with each alcohol 3 times. Then the average for the three readings can be found to give the final values, and therefore dampening the effect that anomalous results play on the deviation of the recordings, if any.
Main Experiment
Aim:
To find the energy released by the combustion of the five alcohols as outlined previously.
Apparatus:
Boss and clamp, bunsen burner, cardboard draft shield, heatproof mat, measuring tube, retort stand, ruler, spirit burners: methanol; ethanol; propanol; butanol; propanol, steel carbonated drinks can, thermometer, water, wooden splints.
Diagram:
Method:
Firstly set up the retort stand, boss and clamp so that the clamp is positioned above the surface, joined to the stand using the boss. Place the heatproof mat under the stand. Record the mass of the empty steel can and then measure out 40ml of water using the measuring cylinder. Pour the water into the steel can. Fasten the can onto the clamp. Take the first spirit burner - methanol - and measure its mass using the electronic balance. Record the mass in grams then position the burner on the heatproof mat directly under the can. Using the ruler to measure exactly, raise the can and clamp till the base of the can is 3cm above the spirit burner. Tighten the boss to hold the can securely in place. Put the thermometer in the can and wait a minute, then record the initial temperature of the water.
Light the wooden splint from the bunsen burner and then after taking the lid off the burner, use the splint to ignite the spirit burner. From the moment you have ignited the burner, watch the temperature of the water with the thermometer. Make sure the thermometer always remains in the water. As soon as the temperature of the water has increased by 20ºC (from the initial recorded temperature) then extinguish the burner immediately using its lid. Re-record the mass of the spirit burner using the scales. Carry out the combustion of the same fuel another two times (the repetition of the test in order to dampen the effect of anomalous results). Then after recording the results carry out the whole test for the other four fuels - ethanol, propanol, butanol & pentanol.
Results:
Below are the results from the main experiment for all five alcohols.
Fuel
Mass at Start (g)
Mass at End (g)
Mass of Fuel Burnt (g)
(1)
73.99
73.09
0.90
Methanol
2
73.09
72.20
0.89
CH3OH
3
45.09
44.27
0.82
(2)
06.73
06.24
0.49
Ethanol
2
06.23
05.83
0.40
C2H5OH
3
05.80
05.40
0.40
(3)
46.06
45.61
0.45
Propanol
2
45.59
45.08
0.51
C3H7OH
3
45.08
44.56
0.52
(4)
250.73
250.46
0.27
Butanol
2
250.45
250.16
0.29
C4H9OH
3
250.18
249.85
0.33
(5)
88.74
88.46
0.28
Pentanol
2
88.46
88.17
0.29
C5H11OH
3
88.17
87.89
0.28
Now I have to find the average of these results because each alcohol has been tested thrice.
Fuel
Average Mass of Fuel Burnt (g)
Methanol
0.87
2
Ethanol
0.43
3
Propanol
0.49
4
Butanol
0.30
5
Pentanol
0.28
On the next page is a graphical representation of these results in the form of a bar chart. I have used a bar chart instead of a scatter graph because on one axis will be the number of carbon atoms in the alcohol molecule. If this was displayed on a scatter graph then it would imply that there could be alcohol molecules with fractions of carbon atoms in them and this is not possible, therefore it is best shown on a bar chart.
Looking at the graph formed by the results, there doesn't seem to be any specific pattern in their position. However, as I explained in my hypothesis, comparing the alcohols merely by the mass of each burnt would not be viable. So in my analysis I will include the conversion of the results so that they can be compared more accurately.
Analysis
From the results of the main experiment there seems to be a varied selection of values. For each alcohol the quantities are within a small range, which is good and hopefully not drawing out any anomalous results. I cannot be sure as to whether they are "correct" or not because they are merely showing the mass of fuel burnt needed to release the same amount of energy and as I explained before in my hypothesis, the alcohols have to be compared by how many moles of each were burnt. I can be sure that the energy released in each experiment was very close to equal because on the best part all the conditions of the experiment for each alcohol were identical. Therefore I would have had a constant measure of the same amount of energy because the water was heated by the same amount each time.
To find out what the value of this energy released is I have to calculate the energy taken in by the water by using its specific heat capacity of
4.18 kj-1mol-1dm-3. However, I also have to take into account the presence of the can in the experiment. The can is also being heated up at the same time as the water. Therefore, energy being released from the combustion is also being put into heating up the can. So similarly with the water, I have to calculate the energy put into the can by using its specific heat capacity. The can was made out of steel; the SHC for steel is 4.5kj-1mol-1dm-3. I don't know what the temperature change of the can was in the experiment (this is required to calculate the energy) because I was monitoring the rise in the water temperature. However, I can take an educated guess and assume that the temperature rise of the can was uniform with the water, so rising by 20ºC each time.
Total Energy (assuming no energy was lost to the atmosphere) = Q
Mass = M
}
Standard Heat Capacity = C
}
of the can = c
of the water = w
Temperature Change = ?T
}
Q = (Mc x Cc x ?Tc) + (Mw x Cw x ?Tw)
Q = (31.93 x 4.50 x 20.0) + (40.0 x 4.18 x 20.0)
= 6217.7 joules
= 6.2177 kilojoules
I have calculated the value of energy released by the combustion of each alcohol as 6.2177 kilojoules.
Now I need to convert the results so that they show the number of moles of each alcohol burnt rather than the mass of each burnt. Then by integrating in the value I found for the amount of energy released by the combustion
(6.2177 KJ) I can continue to find the amount of energy released per mole of the fuel burnt.
Relative Atomic Masses: Carbon(C) = 12, Hydrogen(H) = 1, Oxygen(O) = 16
Methanol
(CH3OH)
mole of methanol = 12 + 1 + 1 + 1 + 16 + 1
= 32 grams
Average fuel burnt = (0.9 + 0.89 + 0.82) / 3
= 0.87 grams
Average moles burnt = 0.87 / 32
= 0.027 moles
Energy released per mole = 6217.7 / 0.027
= 230285 joules/mol
(÷1000)
= 230.29 kilojoules/mol
Ethanol
(C2H5OH)
mole of ethanol = 24 + 5 + 16 + 1
= 46 grams
Average fuel burnt = (0.49 + 0.40 + 0.40) / 3
= 0.43 grams
Average moles burnt = 0.43 / 46
= 0.0093 moles
Energy released per mole = 6217.7 / 0.0093
= 668569.9 joules/mol
(÷1000)
= 668.57 kilojoules/mol
Propanol
(C3H7OH)
mole of propanol = 36 + 7 + 16 + 1
= 60 grams
Average fuel burnt = (0.45 + 0.51 + 0.52) / 3
= 0.49 grams
Average moles burnt = 0.49 / 60
= 0.0082 moles
Energy released per mole = 6217.7 / 0.0082
= 758256.1 joules/mol
(÷1000)
= 758.26 kilojoules/mol
Butanol
(C4H9OH)
mole of butanol = 48 + 9 + 16 + 1
= 74 grams
Average fuel burnt = (0.27 + 0.29 + 0.33) / 3
= 0.30 grams
Average moles burnt = 0.30 / 74
= 0.0041 moles
Energy released per mole = 6217.7 / 0.0041
= 1516512.2 joules/mol
(÷1000)
= 1516.51 kilojoules/mol
Pentanol
(C5H11OH)
mole of butanol = 60 + 11 + 16 + 1
= 88 grams
Average fuel burnt = (0.28 + 0.29 + 0.28) / 3
= 0.28 grams
Average moles burnt = 0.28 / 88
= 0.0032 moles
Energy released per mole = 6217.7 / 0.0032
= 1943031.3 joules/mol
(÷1000)
= 1943.03 kilojoules/mol
I have now got values for each alcohol which show the energy released per mole of the fuel burnt. I can now create a revised bar chart showing the results, this time elaborating on the previous graph because it plots the alcohols against the energy released per mole rather than just showing the mass burnt.
As you can see, now this new representation of the results varies considerably from the previous plot. You can now see a trend in the position of the results. In order to emphasise the pattern in the results I need to plot them again on a scatter graph with a line of best fit, however, be wary of the implication that the scatter graph gives when the points are connected, that it is possible to have fractions of carbon, but obviously this us not true. I am using the scatter graph purely to show more clearly the pattern between alcohols.
The points do not stray far from the line of best fit with the exception of propanol which may just be an anomalous result. Since they tend to follow a straight line this shows that as the number of carbon atoms in an alcohol molecule increases, so does the energy released per mole of alcohol burnt.
Regarding my hypothesis, the point I made was that the bigger the alcohol molecule, the more energy is released. My results seem to be in accordance with this statement as the graph shows. There is a positive correlation, which means that as the number of carbon atoms in the alcohol molecule increases, so more energy is released with the combustion per mole of alcohol. I know that the values I got back from my experiment are not going to be accurate because of several factors affecting it. Loss of energy is the main concern. There is obviously going to be a certain amount of energy lost from the combustion of the alcohols because in reality, not all of the energy released is going to go into heating the can and the water. A lot of the energy would be lost to the surroundings via conduction or convection. The draught shield was put in place to try and reduce loss through convection but it would still not be an immaculate countermeasure. Energy is also lost through light and radiation from the flame as well as the heat expelled. However, those values would be minute in comparison. I wanted to find out the exact amount of energy that was expelled by each reaction.
Through my research I came upon a website compiled by a science university situated in Chicago. Here they had published a list of values for the energy of combustion for alcohols. They were as follows:
Alcohol
Density
Formula
Energy of Combustion (kj mol-1)
Methanol
0.793
CH3OH
-726
2
Ethanol
0.789
C2H5OH
-1367.3
3
Propanol
0.804
C3H7OH
2021.0
4
Butanol
0.810
C4H9OH
-2675.6
5
Pentanol
0.815
C5H11OH
-3328.7
The energy of combustion is probably written as a negative value because it is showing the decrease in enthalpy for the combustion reaction. Enthalpy is thermodynamic property and a measure of the internal energy of a substance usually measured in kj mol-1. In an exothermic reaction (e.g. combustion of alcohols), there is a decrease in enthalpy for the reaction mixture; in other words heat is given out. So the values can be interpreted as the positive energy released by the combustion. These values plotted on a graph looks like this:
However, these values would probably also have been obtained through practical experiment and there could always be the possibility of flaws in its design. This means that the values may not be the exact amounts of energy released. In order to find the exact amount you would have to calculate the values manually by using theory rather than experiment. Calculations using bond energies would be the answer to this problem. The formula for the combustion of a fuel is fuel + oxygen --› carbon dioxide + water. Adding up the energy needed to break the bonds in the fuel and oxygen and then deducting the energy needed to reconstruct the bonds in CO2 and water can seek the theoretical values.
The bond energies required for these calculations are:
C - C 347 kj mol-1
C - H 413 kj mol-1
C - O 358 kj mol-1
C = O 805 kj mol-1
O - H 464 kj mol-1
O = O 498.3 kj mol-1
In order to know how many of each bond there are, I need to write out balanced equations for each complete combustion reaction. This will tell me how many moles of oxygen are needed to fully react a mole of alcohol and how many moles of CO2 and water are produced.
methanol + oxygen --› carbon dioxide + water
2CH3OH + 3O2 --› 2CO2 + 4H2O
ethanol + oxygen --› carbon dioxide + water
C2H5OH + 3O2 --› 2CO2 + 3H2O
propanol + oxygen --› carbon dioxide + water
2C3H7OH + 9O2 --› 6CO2 + 8H2O
butanol + oxygen --› carbon dioxide + water
C4H9OH + 6O2 --› 4CO2 + 5H2O
pentanol + oxygen --› carbon dioxide + water
2C5H11OH + 15O2 --› 10CO2 + 12H2O
I will explain in detail how to calculate the values but only for methanol. The rest I will calculate without the method shown because it is the same.
Methanol
(2CH3OH + 3O2 --› 2CO2 + 4H2O)
Add up all the bonds so:
(6 x 413) + (2 x 358) + (2 x 464) + 3 x 498.3) --› (4 x 805) + (8 x 464)
= 5616.9 kj mol-1 --› 6932 kj mol-1
If the total energy taken in is 5616.9kj mol-1 and the total needed to re-form
bonds was 6932 kj mol-1 then the difference would be the total energy released.
6932 - 5616.9 = 1315.1 kj mol-1
However, this is the energy released by two moles of methanol as you can see
from the equation so:
315.1 ? 2 = 657.55 kj mol-1
Ethanol
(5 x 413) + 347 + 358 + 464 + (3 x 498.3) --› (4 x 805) + (6 x 464)
= 4728.9 kj mol-1 --› 6004 kj mol-1
6004 - 4728.9 = 1275.1 kj mol-1
Propanol
(14 x 413) + (4 x 347) + (2 x 358) + (2 x 464) + (9 x 498.3)
--› (12 x 805) + (16 x 464)
= 13298.7 kj mol-1 --› 17084 kj mol-1
7084 - 13298.7 = 3785.3 kj mol-1
3785.3 ? 2 = 1892.65 kj mol-1
Butanol
(9 x 413) + (3 x 347) + n358 + 464 + (6 x 498.3) --› (8 x 805) + (10 x 464)
= 8569.8 kj mol-1 --› 11080 kj mol-1
1080 - 8569.8 = 2510.2 kj mol-1
Pentanol
(22 x 413) + (8 x 347) + (2 x 358) + (2 x 464) + (15 x 498.3)
--› (20 x 805) + (24 x 464)
= 20980.5 kj mol-1 --› 27236 kj mol-1
27236 - 20980.5 = 6255.5 kj mol-1
6255.5 ? 2 = 3127.75 kj mol-1
These values can also be plotted onto a graph so that I can see the trend between each.
These results also form a straight line like the values obtained from the website. However, I want to compare these results with the ones researched and with my own ones to see how different they are. Combining all three plots onto one scatter graph can easily do this. I have done this on the following page. As you can see, and as I expected, my results are plotted well below the other two lines. This shows that the energies I calculated to have been released by the combustion were lower than the theoretical/actual values. This is obviously because of the loss of energy in the experiment.
Investigation into the Combustion of Specific Alcohols
- Page 21 -