| | O O
H H
C2H5OH + 3O2 → 3H2O + 2CO2
The next consecutive alcohol (in relation to the numbers of carbon atoms when put in order of least to most) is propanol:
H H H
| | | H
H – C – C – C – O – H + 4.5 O=O → 4 / \ + 3O=C=O
| | | O O
H H H
C3H7OH + 4.5O2 → 4H2O + 3CO2
Comparing these two sets of values, we see that the difference between ethanol and propanol is: 1 extra C-C bond broken, 2C-H extra bonds broken, 1.5 extra O=O bonds broken, 2 extra H-O bonds made, and 2 extra C-O bonds made.
ETHANOL: C2H5OH + 3O2 → 3H2O + 2CO2
PROPANOL: C3H7OH + 4.5O2 → 4H2O + 3CO2
BUTANOL: C4 H9OH + 6O2 → 5H2O + 4CO2
PENTANOL: C5H11OH + 7.5O2 → 6H2O + 5CO2
HEXANOL: C6H13OH + 9O2 → 7H2O + 6CO2
This difference is the same between any two consecutive alcohols.
Therefore, as this is the difference between any two consecutive alcohols, than the molar heat of combustion should increase by however much energy it takes to break and make these bonds each time, so my graph should be straight. I have included a prediction graph in my coursework.
For the molar heat of combustion for heptanol, I predict that this will follow on in the sequence, and be on my line of best fit. I will be able to make a more accurate prediction after obtaining my results.
METHOD: For this experiment, I need a stop clock, clamp stand, thermometer, tin can, 100cm3 of water, and a spirit burner of ethanol. Firstly, I weighed the ethanol spirit burner with the lid on. I put 100cm3 of water into the tin can, and set it up on a clamp stand. I placed the ethanol spirit burner under the tin can, and positioned the can so that the flame just touched the bottom of the can. As soon as I took the lid off the spirit burner, someone else lit the wick, and I started the stop clock. I timed the experiment for five minutes, before putting the lid back on the burner. I then weighed the burner with the lid on again. I also recorded the temperature of the water at the beginning and end of the experiment. I then repeated this method for propanol, butanol, pentanol and hexanol. These are the results that I obtained:
\ /
These results seem anomalous to me, so I have discarded them
To check that my first set of results was accurate, I have repeated the experiment.
I have taken the average of both sets of results, discarding my anomalous results.
My results seem to indicate that the temperature difference decreases as the number of carbons increase, and the weight loss increases as the number of carbons increase.
To find the energy of the alcohol, you must multiply the mass of water by the heat capacity by the temperature change.
Energy = mass of water x heat capacity x temp change
Energy = 100 x 4.2 x 82.6
Energy = 420 x 82.6
Energy = 34962
To find the energy per gram of the alcohol, you must divide the energy by the weight loss in the experiment.
34962 = 8324.29 (2 d.p.)
4.2
Then, to finds the molar heat of combustion, you have to multiply the energy per gram of the alcohol by the relative molar mass (RMM) of the alcohol. This varies, as a carbon atom weighs 12 grams, a hydrogen atom weighs one gram, and an oxygen atom weighs 16 grams.
So for ethanol, the equation is: C2H5OH
12 x 2 + 1 x 5 + 1 x 16 + 1 x 1
= 24 + 5 + 16 + 1
= 46
I have worked out the RMM of each of the alcohols I am experimenting with.
ETHANOL: C2H5OH →12 x 2 + 1 x 5 + 1 x 16 + 1 x 1 = 46
PROPANOL: C3H7OH → 12 x 3 + 1 x 7 + 1 x 16 + 1 x 1 = 60
BUTANOL: C4H9OH → 12 x 4 + 1 x 9 + 1 x 16 + 1 x 1 = 74
PENTANOL: C5H11OH → 12 x 5 + 1 x 11 + 1 x 16 + 1 x 1 = 88
HEXANOL: C6H13OH → 12 x 6 + 1 x 13 + 1 x 16 + 1 x 1 = 102
You must multiply the relative molar mass by the energy per gram to find the molar heat of combustion. So 46 is the RMM of Ethanol, and that is multiplied by 15981.82 to give the molar heat of combustion of Ethanol.
So 46 x 15981.82 = 382917.14 J
To find the molar heat of combustion of ethanol in KJ mol -1, we must divide the answer by 1000, and turn it into a negative number, which makes:
382917.14 – –1000 = -382.92
For the other alcohols, these are the results I obtained:
ANALYSIS: As I predicted, my experimental values were lower than the given values. This is because heat will was lost in unnecessarily obtaining the results. This was because by sound and light energy given off, incomplete combustion (complete combustion occurs when there are lots of oxygen atoms available when the fuel burns, and then you get carbon dioxide, as a carbon atom bonds with two oxygen atoms; a limited supply of oxygen results in carbon monoxide being produced), and also conduction, convection and radiation of heat through the air, and draughts speeding the process up. Water was evaporated, meaning that there was a lower volume of water, causing the remaining water to absorb more heat energy, which affected the results. The beaker was also heated in the process, diverting some of the heat energy away from the water. As I predicted, the heavier the one mole of the alcohol, the greater the molar heat of combustion was. This is because the heavier the mole, the more bonds there are, and the more heat energy is required to break them, resulting in a higher heat of combustion. As I predicted, the line on my graph of results were relatively straight, although propanol seemed slightly anomalous. This is because of the number of bonds broken and made between any two consecutive alcohols is constant, so the change in the results should also be constant.
As I mentioned in my prediction, there were things wrong in my experiment. Through research, I have found away to minimise the effect and presence of all the unintentional variables. For this experiment, you would need a copper calorimeter, a water pump, an electric heating coil, a joulemeter, and a low-voltage power supply. I would fill up the calorimeter with a set amount of water (100cm ) to near the top, and place it on a stand. I would attach a water pump to the calorimeter, and draw in a quite fast flow of air. I would then place an ethanol spirit burner and place it under the calorimeter, and then observe the behaviour of the flame: if the flame slowly goes out, then that is an indication that more air is required. If the flame seems unsteady, then the flow of air needs to be reduced. I would then weigh the spirit burner with the lid on, and record the temperature. I would time the experiment for five minutes, before putting the lid back on the burner. I would then weigh the burner with the lid on again, and record the temperature again after stirring the water thoroughly (so that the heat energy can be spread around the water). I would then repeat this method for propanol, butanol, pentanol and hexanol. This method should prevent conduction, convection and radiation of heat through the air, and draughts speeding the process up, and water evaporation. Also the calorimeter is too far away to absorb heat energy, diverting some of the heat energy away from the water.
CONCLUSION: Here are the given values, along with my experimental values:
As I have explained, the values rise at a constant rate, so for heptanol, I predict that if the experiment was done to the same standard as the experiment that produced the given results, then the molar heat of combustion of heptanol would be roughly 654 less than –3983.8 KJ mol -1, as that seems to be the difference each time between two consecutive alcohols, which would be –4628 KJ mol . However, if I had the option of experimenting with heptanol (time and equipment prevented m my method from being reliable enough), then I predict that the value would be on the line of best fit, which would make it roughly –940 KJ mol -1.
