Then in the actual procedure, the fact that there are many constants and repetitions of the experiment, it helps the results to be as precise as they can be. The constants that I have are: the water should start at the same temperature, use the same equipment and apparatus, place the spirit burner the same distance from the aluminium can each time (which is the purpose of the clamp), raise the temperature of the water the same amount of degrees each time, and use the same amount of water each time. The repetitions of the experiment with the same fuel help to make sure that the results received are correct, as well as confirm them, rather than an anonymous result. In the first step, a pipette and tweezers are used so that you don’t get other substances than the fuel on or in the spirit burner. Then, in step ten stirring the water with the thermometer helps to get more precise results because it is keeping the heat in the water circulating and not just sitting on the top. Also, the purpose of the spirit burner cap is to put the flame out immediately and when you use it to measure mass, it provides yet another constant
Sources
Salters Advanced Chemistry, Chemical Ideas, Second Edition, 2000
Salters Advanced Chemistry 200, Developing Fuels Activity 1.2 and 1.3
Averages
Data Analysis
Amount of Heat Absorbed by the Water
Equations for Combustion
Methanol:
CH3OH + O2 → CO2 + 2 H2O
OR
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O
Ethanol:
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
Propan-1-ol:
C3H7OH + O2 → 3 CO2 + 4 H2O
OR
2 C3H7OH + 9 O2 → 6 CO2 + 8 H2O
Butan-1-ol:
C4H9OH + 6 O2 → 4 CO2 + 5 H2O
Calculating Enthalpy Changes
Energy Transferred in Joules (J) = cm∆T
c = the specific heating capacity of water (4.2 Jg-1K-1)
m = the mass of water in grams (100 mL= 100 cm3 =100 grams)
∆T = the change or rise in the temperature of the water
∆Hc=enthalpy change of combustion
- Methanol
(4.2 Jg-1K-1) x (100 grams) x (20.0° C)
energy transferred= 8400 J
8400 J = 8.40 kJ
mass of fuel used = 0.82 grams
molar mass of methanol = 32.00 grams
0.82 grams/ 32.00 grams
= 0.025625 grams
= 0.025 grams
∆Hc = 8.40 kJ/0.025 grams = 336.000 kJ mol-1
Therefore, the enthalpy change for combustion in Methanol 1 is
-336.000 kJ mol-1.
- Methanol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.76 grams
molar mass of methanol = 32.00 grams
0.76 grams/ 32.00 grams
= 0.02375 grams
= 0.02 grams
∆Hc = 8.61 kJ/0.02 grams = 430.500 kJ mol-1
Therefore, the enthalpy change for combustion in Methanol 2 is
-430.500 kJ mol-1.
- Methanol
(4.2 Jg-1K-1) x (100 grams) x (20.0° C)
energy transferred= 8400 J
8400 J = 8.40 kJ
mass of fuel used = 0.83 grams
molar mass of methanol = 32.00 grams
0.83 grams/ 32.00 grams
= 0.0259375 grams
= 0.025 grams
∆Hc = 8.40 kJ/0.025 grams = 336.000 kJ mol-1
Therefore, the enthalpy change for combustion in Methanol 3 is
-336.000 kJ mol-1.
- Ethanol
(4.2 Jg-1K-1) x (100 grams) x (20.0° C)
energy transferred= 8400 J
8400 J = 8.40 kJ
mass of fuel used = 0.60 grams
molar mass of ethanol = 46.00 grams
0.60 grams/ 46.00 grams
= 0.0130434783 grams
= 0.013 grams
∆Hc = 8.40 kJ/0.013 grams = 646.1538462 kJ mol-1
= 646.154 kJ mol-1
Therefore the enthalpy change for combustion in Ethanol 1 is
-646.154 kJ mol-1.
- Ethanol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.60 grams
molar mass of ethanol = 46.00 grams
0.60 grams/ 46.00 grams
= 0.0130434783 grams
= 0.013 grams
∆Hc = 8.61 kJ/0.013 grams = 662.3076923 kJ mol-1
= 662.307 kJ mol-1
Therefore, the enthalpy change for combustion in Ethanol 2 is
-662.307 kJ mol-1.
- Ethanol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.58 grams
molar mass of ethanol = 46.00 grams
0.58 grams/ 46.00 grams
= 0.0126086957 grams
= 0.013 grams
∆Hc = 8.61 kJ/0.02 grams = 662.3076923 kJ mol-1
= 662.307 kJ mol-1
Therefore, the enthalpy change for combustion in Ethanol 3 is
-662.307 kJ mol-1.
1. Propan-1-ol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.55 grams
molar mass of propan-1-ol = 60.00 grams
0.55 grams/ 60.00 grams
= 0.009166666 grams
= 0.009 grams
∆Hc = 8.61 kJ/0.009 grams = 956.66666 kJ mol-1
= 956.667 kJ mol-1
Therefore, the enthalpy change for combustion in Propan-1-ol 1 is
-956.667 kJ mol-1.
2. Propan-1-ol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.50 grams
molar mass of propan-1-ol = 60.00 grams
0.50 grams/ 60.00 grams
= 0.008333333 grams
= 0.008 grams
∆Hc = 8.61 kJ/0.008 grams = 1076.250 kJ mol-1
= 1076.250 kJ mol-1
Therefore, the enthalpy change for combustion in Propan-1-ol 2 is
-1076.250 kJ mol-1.
3. Propan-1-ol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.49 grams
molar mass of propan-1-ol = 60.00 grams
0.49 grams/ 60.00 grams
= 0.00816666 grams
= 0.008 grams
∆Hc = 8.61 kJ/0.008 grams = 1076.250 kJ mol-1
= 1076.250 kJ mol-1
Therefore, the enthalpy change for combustion in Propan-1-ol 3 is
-1076.250 kJ mol-1.
- Butan-1-ol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.47 grams
molar mass of propan-1-ol = 74.00 grams
0.47 grams/ 74.00 grams
= 0.00635135grams
= 0.006 grams
∆Hc = 8.61 kJ/0.006 grams = 1435.000 kJ mol-1
Therefore, the enthalpy change for combustion in Butan-1-ol 1 is
-1435.000 kJ mol-1.
- Butan-1-ol
(4.2 Jg-1K-1) x (100 grams) x (20.5° C)
energy transferred= 8610 J
8610 J = 8.61 kJ
mass of fuel used = 0.46 grams
molar mass of propan-1-ol = 74.00 grams
0.46 grams/ 74.00 grams
= 0.00621621 grams
= 0.006 grams
∆Hc = 8.61 kJ/0.006 grams = 1435.000 kJ mol-1
Therefore, the enthalpy change for combustion in Butan-1-ol 2 is
-1435.000 kJ mol-1.
- Butan-1-ol
(4.2 Jg-1K-1) x (100 grams) x (21.0° C)
energy transferred= 8820 J
8820 J = 8.82 kJ
mass of fuel used = 0.50 grams
molar mass of propan-1-ol = 74.00 grams
0.50 grams/ 74.00 grams
= 0.0067567 grams
= 0.007 grams
∆Hc = 8.82 kJ/0.007 grams = 1260.000 kJ mol-1
Therefore, the enthalpy change for combustion in Butan-1-ol 3 is
-1260.000 kJ mol-1.
Conclusion
In this particular experiment, I have used Methanol, Ethanol, Propan-1-ol, and Butan-1-ol. As these fuels are ordered in ascending order according to their molecular structure, I have noticed that as the molecular structures get larger, the enthalpy change of combustion has also gotten larger.
Through this experiment I have found that the enthalpy change of combustion of different fuels is affected by their molecular structure because it takes more energy to break and make bonds from butan-1-ol as opposed to methanol because butan-1-ol has more bonds to be broken and made than methanol. Therefore, as the molecular structures get larger, as well as the amount of bonds, it takes more energy to break and make the compounds because there are more bonds to be broken or made. So coincidently, the trend that I have found is that as the molecular structures become larger, so does the energy used to break and make bonds, and in the end result, the enthalpy change of combustion is larger too.
Evaluation
In the experiment that I have completed, there were only three anomalous results out of twelve possible results. The first one was that the second repetition of methanol used 0.76 grams, whereas the other two used 0.82 grams and 0.83 grams. The second one was that the first experiment with propan-1-ol used 0.55 grams, whereas the other two used 0.50 grams and 0.49 grams. The last anomalous result I had was for the third repetition for butan-1-ol, it used 0.50 grams, whereas the other two only used 0.46 grams and 0.47 grams.
Throughout this experiment there are many measurements that take place with the equipment, such as the measuring cylinder, the balance and the thermometer.
Uncertainty of apparatus x 100
Percentage uncertainty = Result of measurement
For the measuring cylinder, I measured 100 cm3, and the uncertainty value for this is 0.5 cm3. Therefore, the percentage uncertainty is 0.5 divided by 100, then multiplied by 100, and the percentage uncertainty 0.50%.
The temperature uncertainty for all measurements with the thermometer is ± 0.02 degrees C in each case, with the range being 0.04.
For example, the percentage uncertainty for the temperature rise of 21 degrees is 0.02/21 x 100, which equals 0.09% for percentage uncertainty, but you must double this because the uncertainty is ± 0.02 degrees, so the overall percentage uncertainty for this one measurement is 0.18%. All of the other percentage uncertainties for the thermometer would be similar since the temperature rises were relatively close.
The percentage uncertainty for the average mass of Methanol is 0.008 / 0.804 x 100, which equals a 1.00% uncertainty.
The percentage uncertainty for the average mass of Ethanol is 0.008 / 0.594 x 100, which equals a 1.35% uncertainty.
The percentage uncertainty for the average mass of Propan-1-ol is 0.008 / 0.514 x 100, which equals a 1.56% uncertainty.
The percentage uncertainty for the average mass of
Butan-1-ol is 0.008 / 0.477 x 100, which equals a 1.68% uncertainty.
Together, the percentage uncertainty for all of the average masses of the fuels, the measuring cylinder, and the thermometer readings (12 x 0.18 (not exact though)) is 8.25%. This would only affect the results in a minor way, but this in combination with other sources of error that can’t be measured add up to make my personal results less dependent.
The sources of error in this experiment lay in the measurements and the effectiveness of carrying out the correct methods. For instance, the percentage error affects your experiment by making results both larger and smaller, but also it is hard to put the flame out at the exact right amount of time, so obviously, more fuel is burned than what is exactly required in this experiment.
The main sources of error that would affect my results the most is heat loss and an incomplete combustion of the fuel. Heat loss is a main source of error because all the heat is not transferred into the water; it escapes out into the environment. This happens because the water in the aluminium can isn’t covered, so heat escapes from the top of this. Heat can also stay in the can itself and escape along the outside of the can into the environment. Incomplete combustion is also a main source of error because the flame was yellow (rather than blue showing complete combustion). The spirit burner wasn’t getting enough oxygen to achieve complete combustion and some of the fuel was not converted into the purest form of carbon dioxide and water (both in the form of gases). This affects my result because obviously not all the fuel that was used in the experiment was completely converted, which if they had been, it would produce more heat
Overall, I think my results were as precise as they could be in the conditions that the equipment allowed. My results seem to be pretty close together, and therefore I think they are precise, but they are not accurate because they are not spot-on due to the conditions that I was able to use.
Procedure Assessment
To make the procedure more accurate and to improve my results, the best solution would be to use alternative apparatus, such as using a balance that goes more than two decimal places,
you could also use a different form of the draught shielding and metal calorimeter, or you could use a more accurate thermometer with more increments on it. A different form of apparatus looks like this:
Using a different balance with more decimal places would improve my results by making them more accurate because it would give a more specific measurement. Using the particular different form of apparatus would improve my results because it would keep the draft away from the water because it is surrounded by glass. It also keeps heat from escaping the water as well as evaporation because there is a tube allowing the hot air or exhaust to escape through the top. Also, the coil through the water allows the heat to go throughout the whole of the water and spread throughout it. Lastly, using a different thermometer with more increment would be more specific in its measurement because it will be able to go further than one half increments.
