The spirit lamps contain dangerous substances which may cause fatal reactions if not handled with properly. Thus we must make sure that the lamps are used when only appropriate and are not dropped or opened up.
The nature of the experiment means that we will be dealing with heat hence the use of a heat proof mat ensures that there isn’t any damage done by heat and that the area is clear.
The fact we are handling chemicals means that we need to wear goggles to make sure we are safe from the chemical.
Fair Testing/ Variables
To ensure that this is a fair test we must take into consideration the variables other than the one which we are investigating (i.e. the alcohol itself).
One of these variables is the height which the alcohol is burning the tin can. It will be unfair to vary the height of the tin can as it would result in causing the flames of the different alcohol’s burn the can at unfair advantages over others as the heat will travel at different distances.
The
Accuracy
To ensure accuracy we will be measuring the weights of our spirit lamps using an electronic scale. The accuracy of this test however is at a disadvantage as we are not able to repeat tests in each alcohol due to practicality and time restrictions.
Range
The range of this test is set at five as we are able to obtain these five alcohols easily and we are able to see a trend or pattern which occurs in the graph we will plot.
Theoretical Results
Energy needed to break bonds
Ethanol C2H5OH + 2O2 2CO2 + 3H2O
1 C-C 346 x 1 = 346
5 C-O 412 x 5 = 2060
1 C-O 358 x 1 = 358
1 O-H 464 x 1 = 464
3 O=O 498 x 3 = 1494
= +4722
Energy Given Out
4 C=O 743 x 4 = 2972
6 O-H 464 x 6 = 2784
= -5756
4722 + -5756 = -1034 kj/mol
Energy needed to break bonds
Propanol C3H7OH + 5O2 3CO2 + 4H2O
2 C-C 346 x 2 = 692
7 C-H 412 x 7 = 2884
1 C-O 358 x 1 = 358
1 O-H 464 x 1 = 464
5 O=O 498 x 5 = 2490
= +6888
Energy Given Out
6 C=O 743 x 6 = 4458
8 O-H 464 x 8 = 3712
= -8170
6888 + -8170 = -1282 kj/mol
Energy needed to break bonds
Butanol C4H9OH + 6O2 4CO2 + 5H2O
3 C-C 346 x 3 = 1038
9 C-H 412 x 9 = 3708
1 C-O 358 x 1 = 358
1 O-H 464 x 1 = 464
6 O=O 498 x 6 = 2988
= +8556
Energy Given Out
8 C=O 743 x 8 = 5944
10 O-H 464 x 10 = 4640
= -10584
8556 + -10584 = -2028 kj/mol
Energy needed to break bonds
Pentanol C5H11OH + O2 3CO2 + 4H2O
4 C-C 346 x 4 = 1384
11 C-H 412 x 11 = 4532
1 C-O 358 x 1 = 358
1 O-H 464 x 1 = 464
5 O=O 498 x 5 = 2490
= +6888
Energy Given Out
6 C=O 743 x 6 = 4458
8 O-H 464 x 8 = 3712
= -8170
6888 + -8170 = -1282 kJ/mol
Analysis of Theoretical Results
The results show that the most energy is given out in pentanol. This has the largest structure and the energy and the amount of atoms are directly proportional to each other. This supports my prediction.
Equipment
Spirit Lamps – Used to carry out experiment
Heat Proof Mat – Safety Measure
Thermometer – To measure the temperature of the water
Tin Can (calorimeter) – A conductor of heat to transfer to the water
Clamp – To place the tin can at a constant height
Obtaining Results
Method
Collect and set up apparatus shown in the diagram:
Measure out 100cm of water into the calorimeter. 100cm has the mass of 100g. Clamp the calorimeter into place so that it is just above the spirit burner. Take the initial temperature of the water. Measure and record the mass of the spirit burner with the lid on. Then remove the cap and light the spirit lamp. Take a keen eye on when the temperature has changed. In our case, this would be by 20 C. After the temperature has reached your desired point, place the lid on and measure the contents again. Record the change in mass. Repeat the test as necessary.
Results
Methanol – 190.95 189.99 = Change of 0.96g
Ethanol – 177.64 176.83 = Change of 0.81g
Propanol – 197.15 196.40 = Change of 0.75g
Butanol – 200.13 199.54 = Change of 0.59g
Pentanol – 206.24 205.70 = Change of 0.54g
Energy Given = mass of water x 4.2 (SHC of water) x Temperature Change
Out
= 100 x 4.2 x 20
= 8400J
Methanol CH3OH
0.96 of Methonal gives 8400J
1g 8400J
0.96 = 8750J
1 mol of methanol (32g) gives 8750J x 32
1000 = 280J
Ethanol C2H5OH
0.81 of Ethanol gives 8400J
1g 8400J
0.81 = 10370J
1 mol of Ethanol (46g) gives 10370J x 46
1000 = 477J
Propanol C3H7OH
0.75 of Propanol gives 8400J
1g 8400J
0.75 = 11200J
1 mol of Propanol (60g) gives 11200J x 60
- = 672J
Butanol C4H9OH
0.59g of Butanol gives 8400J
1g 8400J
0.59 = 14237J
1 mol of Butanol (74g) gives 14237J x 74
- = 1053J
Pentanol C5H11OH
0.54 of Pentanol gives 8400J
1g 8400J
0.54 = 1556J
1 mol of Pentanol (88g) gives 15556J x 88
- = 1369J
Analysis of Results
The results shows that there is a inverse proportion between the structure and the amount of time it takes to heat up the water. The graph shows this in the curve of best fit. The results are
Directly proportional to the structure it seems from the results shown here. They are similar to the theoretical results.
Supporting Prediction
The results were true my prediction. They followed the same pattern as I had mentioned and the reasons for this seems from the results were also true for the same reasons.
Conclusion
The structure of the alcohol seems to be important in the time it takes for water to heat up. Not only does it mean the alcohol itself have more structure to burn but as a result what is given off when burnt also has more molecules and structure.
Evaluation
The test was satisfactory. The results had come out as planned and the theory was correct. The factors which would effect the test were taken into account as much as possible. One variable which was not dealt with was the fact that the heat was lost into the atmosphere. This could have been dealt with using draught screens but this could not be done due to insufficient supplies. Also accuracy was lost in that we were not able to repeat the test again due to time and practicality.
I felt that everything else was accurate especially measuring the weight as we had an electronic scale to measure it for us. Overall the test was carried out sufficiently and the results were clear and of a high standard.
