Variables
- The weight of the fuels being burnt
- The height of the test tube away from the flame
- The time of burning
- The amount of water in the test tube being heated
- Type of beaker, glass
- Same set of scales
The variable that has to be changed is…
· the type of alcohol used
The formulae of the alcohols that I will be using are…
· Methanol CH OH
· Ethanol C H OH
· Propanol C H OH
· Butanol C H OH
· Pentanol C H OH
Prediction
I predict that the more bonds there are holding the carbon, oxygen and hydrogen atoms together; more energy will be required to break them apart. For example Ethanol has the formula C H OH. In this formula you have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To separate these types of bonds you require a certain amount of energy.
TYPE OF BOND ENERGY REQUIRED TO BREAK THE BOND (j)
C-H = 412
C-O = 360
O-H = 463
O=O = 496
C=O = 743
C-C = 348
To separate C-H bond you need to apply 412 joules of energy. There are five such bonds in ethanol so you multiply 412 by five to get 2060 joules. You do these calculations for all the other types of bonds that make up ethanol and add them all together. All of the other alcohols can be broken up in this way.
Longer molecules take more energy to break its bonds, in this case Pentanol. Compared to a smaller molecule, Methanol that requires less energy to do so. So the longer the molecular structure in the alcohol the more energy it will take to remove the bonds. So I believe that pentanol will use more energy than methanol because it has more bonds to break.
Bonds
Analysis
First of all I will calculate the amount of energy evolved. I will achieve this by using this simple formula…
Energy evolved = temp rise °C x mass water g x 4.2 j/g °C
Energy evolved = …………. x 20cm cubed x 4.2
Type of alcohol Energy evolved in (j) Energy evolved in (j)
Methanol = 1 min = 134400. 2 min = 480480
Ethanol = 1 min = 144480. 2 min = 481488
Butanol = 1 min = 436800. 2min = 772800
To find out how much energy is produced per gram we use the formula…
Energy per gram of fuel = Energy evolved x Mass of fuel burnt
Energy per gram of fuel = ……… x ………
Below is a table showing how much energy is produced per gram when burning the alcohols in question…
Type of alcohol Energy per gram
Methanol = 1 min = 84672. 2 min = 4564560
Ethanol = 1 min = 621264. 2 min = 351486.24
Butanol = 1 min = 270816. 2 min = 1105104
As you can see the energy per gram decreases as the length of the molecule increases. This is because more fuel is burnt so there is more of it to be filled with the energy.
Below is a table showing the energy supposed to be produced per mole (as it suggests in the text book)…
Type of alcohol Energy per mole (kj/mol)
Methanol -726
Ethanol -1368
Butanol -2675
I predicted that the energy per gram would decrease as the molecular length increases. I think this because the alcohols with more carbon atoms in burnt more fuel so there would be less energy per gram because more fuel has been burnt. The reason why more fuel has been burnt is because of the large number of carbon atoms and large molecular length, therefore the surface area is large allowing more energy to be released.
Evaluation
I think my results were quite inaccurate.
I didn’t think about energy given off through sound and light, and the heat conducted away through the air had a part to play in my results. I need to do the experiment in a more controlled area, with no windows and an area that is maybe the same temperature all around.
I also didn’t think about the fact that at higher temperatures, heat is lost faster to the air and out of the beaker (diffusion), due to the bigger heat difference. There was some incomplete combustion which I should have thought about and maybe changed some things in the experiment because of them. The size of the wick may have been different in the different fuel pots. I did the butanol experiment under different conditions than the ethanol and methanol. It was a different room with wind coming from different directions and the water temperature was different at the start this may have made it harder for the water to heat up as quickly as the ethanol and methanol. The ethanol and methanol both started at 26°C and the butanol and 20°C.
I feel that this experiment could have been improved by using a wider range of alcohols. I only had time to record the results for 3 of the 5 fuels I was planning to do and maybe more than 5 alcohols would have been more accurate. If I did this experiment again reducing heat loss would be my main concern. I had a few miscellaneous results. Another error is that of incomplete combustion. Complete combustion occurs if there are lots of oxygen atoms available when the fuel burns, then you get carbon dioxide (carbons atoms bond with two oxygen atoms). If there is a limited supply of oxygen then you get carbon monoxide (each carbon atom can only bond with one oxygen atom). This is when incomplete combustion has occurred. This is so because the carbon monoxide could react some more to make carbon dioxide. If the oxygen supply is very limited then you get some atoms of carbon released before they can bond with any oxygen atoms. This is soot. Since heat is given out when bonds form, less energy is given out by incomplete combustion. So this is why it affects the outcome of the experiment. To overcome this problem, I would have to make sure a sufficient supply of oxygen was involved in the reaction.
BUTANOL: TIME 1 MINUTE
BUTANOL: TIME 2 MINUTES
AVERAGES:
WATER TEMPERATURE RISE:
(TIME 1 MINUTE) 13°C
(TIME 2 MINUTES) 23°C
FUEL LOSS:
(TIME 1 MINUTE) 0.62
(TIME 2MINUTES) 1.43