2060 + 348 + 743 + 463
3231 + 1488 = 4719 5750 – 4719 = 1030 kJ
Propanol C3 H7 OH + 4.5O2 → 3CO2 + 4H2O
4458 + 3704= 8162
H H H O = O
⏐ ⏐ ⏐ O = O
H – C – C – C – O – H O = O
⏐ ⏐ ⏐ O = O = 2232
H H H O
( 412 x 7 ) + ( 348 x 2 ) + 360 + 463 = 4403 + 2232 = 6635
- 696
8162 – 6635 = 1527 kJ
Butanol C4 H9 OH + 6O2 → 4CO2 + 5H2O
5944 4625
H H H H O = O O = O
| | | | O = O O = O = 2976
H – C – C – C – C – O – H O = O O = O
| | | |
H H H H
( 412 x 9 ) + ( 348 x 3 ) + 360 + 463 = 5575 + 2976 =8551 10569 – 8551
= 2018 kJ
Therefore using these results I can predict the following.
Methanol will give out the least amount of energy then Ethanol then Propanol and Butanol will give out the most. I think this is because each time I react a alcohol that has an extra carbon bond (e.g. ethanol and methanol) the amount of energy given out is roughly 500j more than before.
Things I will need
In the experiments I am going to do I will need these following things:
Chemicals:
-
methanol (C H3 OH)
-
ethanol (C2 H5 OH)
-
propanol (C3 H7 OH)
-
butanol (C4 H9 OH)
Apparatus:
- copper calorimeter
- fuel burner
- lids for fuel burner
- thermometer
- balance
Fair testing
- same starting temp
- same finishing temp
- same amount of water
- height of copper calorimeter
- same flame size (second experiment)
If the above were not kept the same the experiment would not be a fair test and results may not be accurate.
Starting and finishing temp would be kept the same as the higher the temp is the more energy has to be used to heat it further
The same amount of water will be necessary because it takes less energy to heat less water again making the results inaccurate
The height of the copper calorimeter will have to be the same height each time so convection doesn’t occur which would change the amount of energy used
The same flame size will be used second time around to ensure the same amount of heat being lost and this will help the distance between the burner and calorimeter stay the same, convection will remain the same and heat loss into the calorimeter
How will I get my Results
To get my results I am going to light the burner and heat water in a copper calorimeter. I will be heating the water 10°, making sure of this by using a thermometer. I will then measure the amount of alcohol burnt using a balance straight before and after to get the most accurate results I can.
Obtaining the evidence
I used the following apparatus:
Chemicals:
Apparatus:
This is how I did the experiment:
I measured the temperature change (which was from a certain temperature to another certain temperature), and the amount of alcohol burnt off (by measuring the weight of the burner before and after and measure the different.), and the amount of water. I measured the temperature with a thermometer, the mass change with a balance and the water with a measuring cylinder.
To make shore my results are reliable repeated the experiment a three times, compared it to my predicted results and looked at other people’s results and compared them. I did calculations in the planning to work out the rough amount of energy given out by each alcohol (see plan for more)
Results
Experiment 1
Experiment 2
Experiment 3
Mass change in all experiments
Analysing the results and drawing up conclusions
Methanol Ethanol Propanol Butanol
Methanol Ethanol Propanol Butanol
From these conclusions and graphs the amount of alcohol burnt decreases though out the different experiments. The reason for this may be that there was a hotter flame so less fuel energy was needed to heat the water.
Results of energy produced
Conclusion
In my experiment I have discovered the amount of alcohol burnt to heat 50ml of water 10° and the energy output of the experiment. The most notable patterns in my results were that the amount of energy released per experiment decreased the same amount from Butanol to Methanol.
The reason for these results and patterns can be explained by the fact that butane has more bonds to break than propanol likewise propanol than ethanol releasing more energy when breaking the extra bonds. The predictions I made were 535kJ 1030kJ 1527kJ 2018kJ the predictions that I made were very different to the actual results obtained by the calculations of my results, which were 204kJ 345kJ 504kJ 740kJ.
From the graph it is most obvious to see that the predictions were not correct so the prediction method I used was not reliable.
Evaluation
The first experiment performed was not a fair test as the flame sizes were all different and the draught shelters were not used in the first experiment, but in the second and third I made sure all flame sizes were the same and I used draught shelters in each. This method could have been improved but not using classroom methods. To make the results as accurate as possible it would have been necessary to use a thermometer that measures to a degree of accuracy within 0.5(, and I would have to stop heating the water exactly on 30( so the change in water temp would have been the same. The results I obtained did make sense but they weren’t as accurate as I would have liked, they were accurate enough to obtain and support a firm conclusion apart from the predictions do not make much sense as to being so far off the actual results. In experiment 1 there were unexpected results that made the results slightly inaccurate ethanol burnt using more energy than methanol. This was resulting in an unfair test, as the flame sizes were different and there was no protection against convection and draught. The only limitations were that it was impossible to do the experiment in great detail. To support my conclusion I have the evidence of the bond energies and knowledge of the fact that there are more bonds to break in butanol. To further support my conclusion I could have experimented with different alcohols.
