Energy released- 10520 + 8536 = -1984KJ/MOL
Pentanol
Pentanol + oxygen → carbon dioxide + Water
C5H11OH + 7½02 → 5CO2 + 6H20
Bonds Broken
11 X C-H (11 X 410)
1 X C-O (1 X 360)
1 X O-H (1 X 460)
7½ X 0=0 (7½ X 496)
4 X C-C (4 X 350) = 10450KJ/MOL
Bonds Made
10 X O=C (10 X 740)
12 X O-H (12 X 460) = 12920KJ/MOL
Energy released- 12920 + 10450 = -2470KJ/MOL
Along with my initial prediction, these calculations will help me determine if the practical I am doing is been carried out correctly. They also show me what my results, roughly should be. These predictions have also told me the following:
Making energy > breaking energy
Preliminary Tests
In my preliminary tests, I used the alcohol, ethanol. In this preliminary test, I wanted to investigate the amount of energy given out when I was burned.
I first filled a copper can with 100cm3 of water using a measuring cylinder. I then clamped this 12cm above a spirit burner filled with ethanol. I then weighed the spirit burner, and recorded the temperature of the water before the burner is lit. I will then light the burner and wait for a temperature rise on 10°C. I will then weigh the spirit burner after the experiment and record the temperature after the burner is lit.
Apparatus:
- Ethanol
- Spirit Burner
- Measuring Cylinder
- 100ml of water (Tap Water)
- Metal Beaker (Copper Can)
- Clamp Stand
- Heat Proof Mat
- Safety Goggles
- Digital Weighing Scales
- Bunsen Burner/Matches
- Thermometer
Diagram
Results for preliminary
From the results I can gather the following bits of data:
- Temperature rise
- Mass of alcohol burnt
I will also use the relative molecular mass, specific heat capacity and the density of water to calculate the amount of moles, amount of energy released and finally the amount of energy/mole.
For example:
Start mass= 168.97
End mass = 168.23
Burnt = 0.74g
RMM= 46
Moles= 0.74/46 = 0.016
Temp at start = 28
Temp at end = 40
Rise = 12
Energy released = 12 x 100 x 4.2
Energy released = 5040
Energy/mole = energy/Moles
= 5040/0.016
= 315000J/mol 315KJ/MOL
From the preliminary test I have decided to make the following improvements:
In my preliminary I used measuring cylinders to measure the 100cm3 of water. In my actual I will be using a burette to ensure accuracy and to make sure that the test is fair.
- Heat mats (draft shields)
In my preliminary, a lot of heat was lost due to the flicker of the fuel’s flame. Energy was also lost to the surroundings. In my actual I will be placing heat mats on wither sides of the experiment, to try and reduce the amount of energy lost through the flicker of the fuel’s flame. This will also reduce the amount of energy lost to the surroundings slightly.
In order to make sure the temperature of the water in the copper can has reached the correct, accurate temperature, I will be stirring the water constantly using the thermometer in order to keep a constant temperature around the whole can
Taking into account all these factors I will be carrying out the actual experiment making the improvements above.
Method
-
First I used a burette and filled it with 50cm3; I will do this twice in order to fill my copper can with 100cm3 of water.
- I will be using the burette because in my preliminary tests, I discovered that it would be more accurate if I were to use a burette. I watched the water in the burette drip into the copper can, until the top of the viscous in burette reached the “0” mark.
-
I will then clamp copper can 12cm above a spirit burner containing the desired alcohol.
- Before I light the spirit burner I will weight it on weighing scales with its lid on. I will record this weight.
- At the same time I will be recording the temperature of the water.
- In order to prevent the alcohol from evaporating in the air, and loosing some of its weight, I will be keeping the lid on the spirit burner until I light it.
- Finally when everything is in place I will light the spirit burner.
- In order to maintain a fair test, I will be constantly stirring the water.
- When the temperature of the water has increased by 10°C, I will remove the spirit burner from the experiment and place its lid on.
- I will then weigh the spirit burner after the practical, with the lid on.
- At the same time, I will allow the water in the can to reach the highest temperature it can, and write this down
- I will repeat the test in order to maintain accuracy; I will then find the average of the two results and record them.
- I will do the above for all five of the alcohols, and record them in the table below:
Results:
Experiment 1:
Repeat:
Experiment 2
From these results I have calculated the following:
This alcohol released 147.8KJ/MOL in the first experiment, and 127.4KJ/MOL in the second experiment.
The average of these results and my final result is: 137.6KJ/MOL
This alcohol released 287.13KJ/MOL in the first experiment, and 276.65KJ/MOL in the second experiment.
The average of these results and my final result is: 281.81KJ/MOL
This alcohol released 497.24KJ/MOL in the first experiment, and 360.09KJ/MOL in the second experiment.
The average of these results and my final result is: 428.67KJ/MOL
This alcohol released 745.6KJ/MOL in the first experiment, and 705.7KJ/MOL in the second experiment.
The average of these results and my final result is: 725.65KJ/MOL
This alcohol released 822.19KJ/MOL in the first experiment, and 781.86KJ/MOL in the second experiment.
The average of these results and my final result is: 801.88KJ/MOL
In order to get the above result I had to carry out the following calculations:
Energy per mole = Energy per gram x Formula mass
I have used the above formula for all the results in order to see how much energy is produced from each experiment. This can be found on the following pages, which explains my calculations as to how much energy is released per mole.
1. Methanol-weight _ Temperature
Start- 172.22 Start- 24°C
End- 174.68 End- 36°C
Burnt- 0.54 Rise- 12°C
Methanol (RMM) = CH3OH= 32 Energy released= 11 x 100 x 4.2
Moles = 1 ÷ 32 = 0.03125 = 4620
Energy/mole = energy/moles = 4620÷0.03125
= 147.8KJ/MOL
2. Methanol (repeat) -weight Temperature
Start- 164.77 Start- 39°C
End- 163.61 End- 50°C
Burnt- 1.16 Rise- 11°C
Methanol (RMM) = CH3OH= 32 Energy released= 11 x 100 x 4.2
Moles = 1.16 ÷ 32 = 0.03625 = 4620
Energy/mole = energy/moles = 4620÷0.03625
= 127.44KJ/MOL
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1. Ethanol-weight Temperature
Start- 168.97 Start- 28°C
End- 168.23 End- 39°C
Burnt- 0.74 Rise- 11°C
Ethanol (RMM) = C2H3OH= 46 Energy released= 11 x 100 x 4.2
Moles = 0.74 ÷ 46 = 0.01609 = 4620
Energy/mole = energy/moles = 4620÷0.01609
= 287.13KJ/MOL
2. Ethanol (repeat) -weight Temperature
Start- 168.23 Start- 39°C
End- 161.46 End- 50°C
Burnt- 0.77 Rise- 11°C
Ethanol (RMM) = C2H3OH= 46 Energy released= 11 x 100 x 4.2
Moles = 0.77 ÷ 46 = 0.0167 = 4620
Energy/mole = energy/moles = 4620÷0.0167
= 276.65KJ/MOL
1. Propanol-weight Temperature
Start- 173.22 Start- 24°C
End- 172.46 End- 39°C
Burnt- 0.76 Rise- 15°C
Propanol (RMM) = C3H7OH= 60 Energy released= 15 x 100 x 4.2
Moles = 0.76 ÷ 60 = 0.0.01267 = 6300
Energy/mole = energy/moles = 6300÷0.01267
= 497.24KJ/MOL
2. Propanol (repeat) -weight Temperature
Start- 172.96 Start- 39°C
End- 171.69 End- 50°C
Burnt- 0.77 Rise- 11°C
Propanol (RMM) = C3H7OH= 60 Energy released= 11 x 100 x 4.2
Moles = 0.77 ÷ 60 = 0.01283 = 4620
Energy/mole = energy/moles = 4620÷0.01283
= 360.09KJ/MOL
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1. Butanol-weight Temperature
Start- 172.26 Start- 24°C
End- 171.76 End- 36°C
Burnt- 0.5 Rise- 12°C
Butanol (RMM) = C4H9OH= 74 Energy released= 12 x 100 x 4.2
Moles = 0.5 ÷ 74 = 0.00676 = 5040
Energy/mole = energy/moles = 5040÷0.00676
= 745.6KJ/MOL
2. Butanol (repeat) -weight Temperature
Start- 171.76 Start- 36°C
End- 171.25 End- 48°C
Burnt- 0.56 Rise- 12°C
Propanol (RMM) = C4H9OH= 74 Energy released= 12 x 100 x 4.2
Moles = 0.56 ÷ 74 = 0.00757 = 5040
Energy/mole = energy/moles = 5040÷0.00757
= 705.7KJ/MOL
1. Pentanol-weight Temperature
Start- 175.22 Start- 24°C
End- 174.68 End- 36°C
Burnt- 0.54 Rise- 12°C
Butanol (RMM) = C5H11OH= 88 Energy released= 12 x 100 x 4.2
Moles = 0.54 ÷ 88 = 0.00613 = 5040
Energy/mole = energy/moles = 5040÷0.00613
= 822.19KJ/MOL
2. Pentanol (repeat) -weight Temperature
Start- 174.68 Start- 36°C
End- 174.14 End- 47°C
Burnt- 0.52 Rise- 11°C
Propanol (RMM) = C5H11OH= 88 Energy released= 11 x 100 x 4.2
Moles = 0.52 ÷ 88 = 0.005909 = 4620
Energy/mole = energy/moles = 4620÷0.005909
= 781.86KJ/MOL
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
ANALASIS
From my graph I can see that my initial prediction was correct. I was able to prove that pentanol would release the most energy when burnt in air. I done this by calculating the energy released by working out the energy required and given out when bonds are made and bonds are broken. The results of the experiment showed that Methanol released the least energy and as we the number of carbons it each fuel increased through each alcohol the amount released in turn increased. Although my actual result fail to match the exact predicted line, they still managed to prove my prediction and show clearly the increase in energy given out as the number of carbon in each fuel increases. However Butanol as plotted on the graph is an anomaly. This may have been due to factors such as small amount of wind flickering the flame, creating energy loss. It may have been due to the lid of the spirit burner not put on correctly leading to the evaporation of the alcohol or it may have just been an error in our calculations.
My predicted graph and the graph showing my results are also very similar. They both show a positive correlation once again supporting my prediction. The experiment was carried out correctly and successfully, I can say this because of the lack of anomalous points. Pentanol is the alcohol with the longest chain of molecules so therefore it has more bonds to break and more to be made. Both the energy required and the energy given out increase as the RMM increases, but the energy released increases more than the energy required. (Making energy > breaking energy)
Pentanol has the largest molecular mass of 88, so then the difference is greater than the rest. I have concluded that the energy given out per mole increase as the relative molecular mass of the alcohol increases. The higher the RMM, the more energy released per mole.
The table below shows the percentage difference between my predicted results and my actual results, of energy released:
EVALUATION
Overall, I think that the results obtained were accurate and successful and my graph was accurately drawn, and the line of best fit proved my prediction well. If I had the chance to do the experiment again I would change the following aspects:
- A wider range of alcohols, such has hexanol and heptanol, this will give me a wider range of results, and this will help further improve my experiment.
- I could use alkenes or alkanes to see whether they mimic the alcohols.
- I could see which metals released the most energy when burnt
- I could investigate which alcohol evaporates the most of it contents in the fastest time.
Although overall my experiment went well, there were a few aspects I could not control:
- We were not able to control the amount of energy lost to the surroundings, resulting in lots of energy being lost. Heat was lost to the air through conduction and convection.
- In order to solve this problem, We could use a device called a bomb calorimeter. This piece of equipment is designed to avoid heat loss and prevent evaporation. The results will then be more precise and eradicate this problem. A diagram below describing the set up of how we would use the bomb calorimeter.
- Through out the experiment, incomplete combustion was taking place. Evidence of this was left on the bottom of the copper can as soot. This meant that there was not enough oxygen, which in turn produced carbon mononxide. This is a problem because some potential energy was lost because the carbon monoxide could still react to make more carbon dioxide. To solve this problem I would have to make sure and provide a sufficient amount of oxygen for the reaction to complete.
