· Weigh the crucible, then weigh the alcohol to as near as possible 0.5 grams.
· Put the chosen alcohol under the beaker allowing the flame to just touch the beaker.
· Leave to heat up until the alcohol evaporates.
· Take the temperature when the alcohol has evaporated.
· Record the temperature of the water now.
The variables that must remain constant throughout the experiment are…
· Mass of the water 150ml
· Type of can, metal, aluminium.
· Surrounding temperature of around 21°C.
· The height of the can from the crucible.
· Same set of scales.
· I will stir the water with the thermometer throughout heating to keep the water at an even temperature.
The variable that must be changed is…
· The type of alcohol used.
Prediction
Chemical formula for fuels:
Ethanol C2H5OH
Propanol C3H7OH
Butanol C4H9OH
Pentanol C5H11OH
Hexanol C6H13OH
Heptanol C7H15OH
Octanol C8H17OH
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 C2H5 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 which I will show in a table.
TYPE OF BOND ENERGY REQUIRED TO BREAK THE BOND (kj)
C-H 410
C-O 360
O-H 510
O=O 496
C=O 740
C-C 350
To separate C-H bond you need to apply 410 joules of energy. There are five such bonds in ethanol so you multiply 410 by five to get 2050 joules. You do these calculations for all the other types of bonds that make up ethanol, add them all together and you get 3270 joules.
All of the other alcohols can be broken up in this way. Below is a table showing the energy required to break up the bonds in each alcohol.
Results
Table of averages
Amount of energy produced from each alcohol
Energy = Mass of water x 4.2 x Rise in Temperature
Difference in weight
Or:
E = Mc x T
M
Ethanol
E = 150 x 4.2 x 8
= 5040 kj
per gram = 5040/1.1= 4581.8kj
Propanol
E = 150 x 4.2 x 9
= 5670 kj
per gram = 5670/4 = 1417.5kj
Butanol
E = 150 x 4.2 x 10
= 6300 kj
per gram = 6300/4.3 = 1465.1kj
pentanol
E = 150 x 4.2 x 11
= 6930kj
per gram = 6930/5.8 = 1194.8kj
Hexanol
E = 150 x 4.2 x 12
=7560kj
per gram = 7560/6.4 = 1181.2kj
Heptanol
E = 150 x 4.2 x 13
= 8190kj
per gram = 8190/1.5 = 5460kj
Ocatanol
E = 150 x 4.2 x 14
= 8820kj
per gram = 8820/1.4 = 6300kj
Analysing and drawing conclusions
I think my results and tables clearly show the pattern that I have found in this experiment. That is that heat combustion does increase when the amount of carbon atoms increases. I believe that my results do show a positive correlation and do show that the more carbon atoms there are the heat of combustion goes up. Another reason for these results is that the molecular length becomes longer in the bigger molecules increasing the surface area hence allowing more energy to be released. These results do support my initial prediction. After this I can conclude that my initial prediction was actually right but I didn’t allow for all of the experimental errors. I conclude that carbon atoms in alcohols do have an effect on the heat of combustion. As the amount of carbon atoms go up the heat of combustion does. This is because everytime you add another carbon atom you are also adding 15 onto the relative atomic mass that plays a big part in calculating the end results.
Evaluation
Sound and light energy could have been lost into the room. I could have placed heatproof mats around my experimental area, they could not have kept all of the heat in and much of this would have been taken away in the convection currents through the air. The tin that the water was being held in would have used up some of the heat energy to heat itself up. The alcohol containers had varying amounts of alcohol in them to start with along with varying sizes of wicks. This all contributed to the fact that the flame coming from the alcohol was varying in size so was sometimes not even touching the tin can. The room temperature would also have acted as a cooling agent. One of the less important factors could have been if there was a lacking of oxygen leading to incomplete combustion. Then the oxygen molecules would only form with one carbon molecule producing carbon monoxide but I doubt this actually happened.
This was a very difficult experiment to conduct in a classroom because there are lots of potential ways of losing heat because everything likes to gain heat energy. I think the thing that hindered our results the most was the fact that gusts of air and convection currents were taking the heat away from the experimental area and there was no way to stop this. Perhaps if I started the experiment below room temperature, so that the amount of gained energy, from room temperature, might equal the energy lost at temperatures higher than room temperature, then the experiment could produce better results. 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 what we call 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.