- 100cm³ measuring cylinder
- Spirit burners (each containing 1 of the 5 fuels)
- Access to a balance
- Draught Shielding
Justification of apparatus
The Copper Calorimeter is what I will use to contain the water throughout the experiment. I chose a copper container because of its good conductivity of heat. I will ensure that I use the same copper calorimeter for each fuel, as changing the copper calorimeter each time could result in different thicknesses of copper, meaning a slight loss of energy through thicker containers.
The Thermometer is how I will ensure accuracy of recording temperature change. I will use a thermometer which has a maximum reading of no more than 60°C, as I do not intend to reach temperatures exceeding that amount.
Spirit Burners provide the safest means of burning the fuel for my experiment. Before I carry out my experiment, I will ensure that each wick of each spirit burner is the same length, which will ensure an equal rate of fuel is burned.
I will need Access to a balance to measure the fuel before and after the experiment. They will display the weight of fuel to an appropriate degree of accuracy.
The Draught Shield will minimise heat loss from the spirit burners. It will guide upwards the heat that is being radiated horizontally.
Method
Measure 100cm³ of water and pour it into the copper calorimeter. Record its temperature. Ensure that the start temperature for each experiment does not vary drastically. Keep the start temperatures within about 5oC of each other.
Support the calorimeter over the spirit burner. Arrange a draught shield to prevent heat loss.
Weigh the spirit burner.
Replace the spirit burner under the copper calorimeter. Ensure the wick length of the spirit burner is 1cm, then light the spirit burner.
Once the wick is alight, use the thermometer to stir the water all the time it is being heated. Keep heating until the temperature has risen by 15ºC. Ensure that the temperature rise for each experiment is as accurate as possible. Keep all temperature rises within 0.2oC of each other.
Immediately extinguish the spirit burner by replacing the cap over the wick. Keep stirring the water and record the highest temperature reached.
Weigh the burner to see what mass of fuel has been burned.
- Repeat the process twice for each alcohol. If the two recordings vary significantly, take a third recording.
- Once your results are reliable, take an average.
Molecular Mass of fuels
Methanol
CH4O
12.0107 + (1.00794 x4) +15.9994 = 32.04
Ethanol
C2H6O
(12.0107 x2) + (1.00794 x6) + 15.9994 = 46.07
Propan-1-ol
C3H80
(12.0107 x3) + (1.00794 x8) + 15.9994 = 60.1
Propan-2-ol
C3H80
(12.0107 x3) + (1.00794 x8) + 15.9994 = 60.1
Butan-1-ol
C4H100
(12.0107 x4) + (1.00794 x10) + 15.9994 = 74.12
Actual Enthalpy Changes
Here I will work out the actual enthalpy change of combustion of my alcohols, so that I have figures to compare my results to.
Methanol
CH3OH(l) + 1½O2(g)→ CO2(g) + 2H2O(l)
Ethanol
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Propan-1-ol
C3H7OH(l) + 4½O2(g) → 3CO2(g) + 4H2O(l)
Propan-2-ol
C3H7OH(l) + 4½O2(g) → 3CO2(g) + 4H2O(l)
Butan-1-ol
C4H9OH(l) + 6O2(g) → 4CO2(g) + 4H2O(l)
Analysis
Methanol Calculations
1st Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Methanol
Change in Weight = Initial Weight - Final Weight
217.07g – 215.11g = 1.96g
Now, I will need to find the change in temperature of the water from the combustion of Methanol.
Temperature Difference = Final Temperature – Initial Temperature
32°C – 16.8°C = 15.20°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100g x 4.2J x 15.2°C = 6,384.00J
Now I will get the amount of Methanol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.96g / 32.04 = 0.061M
Now I will find the amount of energy transferred to the water per mole of Methanol
Enthalpy per mole = Energy transferred / moles
6,384J / 0.061 = 104,655.73 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
104,655.73 J / 1000 = 104.66kJmol-1
2nd Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Methanol
Change in Weight = Initial Weight - Final Weight
200.3g – 198.4g = 1.90g
Now, I will need to find the change in temperature of the water from the combustion of Methanol.
Temperature Difference = Final Temperature – Initial Temperature
35.1°C -19.9°C = 15.20°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100g x 4.2J x 15.2°C =6,384.00J
Now I will get the amount of Methanol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.9g / 32.04 = 0.059
Now I will find the amount of energy transferred to the water per mole of Methanol
Enthalpy per mole = Energy transferred / moles
6,384J / 0.059 = 108,203.39J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
108,203.39J/mol-1 / 1000 = 108.20 kJmol-1
Average
(104.66 kJmol-1 + 108.20kJmol-1)/2 = -106.43 kJmol-1
Ethanol Calculations
1st Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Ethanol
Change in Weight = Initial Weight - Final Weight
181.1g –179.42g = 1.68g
Now, I will need to find the change in temperature of the water from the combustion of Ethanol.
Temperature Difference = Final Temperature – Initial Temperature
32.1°C – 16.9°C = 15.20°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100g x 4.2J x 15.2°C = 6,384.00J
Now I will get the amount of Ethanol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.68 / 46.07= 0.036M
Now I will find the amount of energy transferred to the water per mole of Ethanol
Enthalpy per mole = Energy transferred / moles
6,384J / 0.036 = 177,333.33 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
177,333.33 J / 1000 = 177.33kJmol-1
2nd Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Ethanol
Change in Weight = Initial Weight - Final Weight
182.46g – 180.86g = 1.60g
Now, I will need to find the change in temperature of the water from the combustion of Ethanol.
Temperature Difference = Final Temperature – Initial Temperature
33.1°C -17.9°C = 15.20°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100g x 4.2J x 15.2°C =6,384.00J
Now I will get the amount of Ethanol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.6g / 46.07 = 0.035
Now I will find the amount of energy transferred to the water per mole of Ethanol
Enthalpy per mole = Energy transferred / moles
6,384J / 0.035 = 182,400 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
182,400 J/mol-1 / 1000 = 182.40 kJmol-1
Average
(177.33 kJmol-1 + 182.40 kJmol-1)/2 = -179.87 kJmol-1
Propan-1-ol Calculations
1st Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Propan-1-ol
Change in Weight = Initial Weight - Final Weight
173.53g – 172.08 g = 1.45 g
Now, I will need to find the change in temperature of the water from the combustion of Propan-1-ol
Temperature Difference = Final Temperature – Initial Temperature
35°C -19.9°C = 15.1°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100g x 4.2 x 15.1°C =6,342J
Now I will get the amount of Propan-1-ol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.45g / 60.1 = 0.024
Now I will find the amount of energy transferred to the water per mole of Propan-1-ol
Enthalpy per mole = Energy transferred / moles
6,342J / 0.024 = 264250 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
264250 J/mol-1 / 1000 = 264.25 kJmol-1
2nd Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Propan-1-ol
Change in Weight = Initial Weight - Final Weight
155.88g - 154.48g = 1.4g
Now, I will need to find the change in temperature of the water from the combustion of Propan-1-ol
Temperature Difference = Final Temperature – Initial Temperature
33.1°C - 18°C = 15.1°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100 x 4.2 x 15.1 = 6342J
Now I will get the amount of Propan-1-ol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.4g / 60.1 = 0.023
Now I will find the amount of energy transferred to the water per mole of Propan-1-ol
Enthalpy per mole = Energy transferred / moles
6342J / 0.023 = 275,739.13 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
275,739.13 J/mol-1 / 1000 =275.74 kJmol-1
Average
(264.25 kJmol-1 + 275.74 kJmol-1)/2 = -267 kJmol-1
Propan-2-ol Calculations
1st Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Propan-2-ol
Change in Weight = Initial Weight - Final Weight
130.91g -129.27g = 1.64g
Now, I will need to find the change in temperature of the water from the combustion of Propan-2-ol
Temperature Difference = Final Temperature – Initial Temperature
33.6°C -18.4°C = 15.20°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100 x 4.2 x 15.2 = 6384J
Now I will get the amount of Propan-2-ol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.64g / 60.1 = 0.027
Now I will find the amount of energy transferred to the water per mole of Propan-2-ol
Enthalpy per mole = Energy transferred / moles
6384J / 0.027 = 236,444.44 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
236,444.44 J/mol-1 / 1000 = 236.44 kJmol-1
3rd Recording (due to inaccurate 2nd recording)
Firstly, I will need to find the amount of fuel burned during the combustion of Propan-2-ol
Change in Weight = Initial Weight - Final Weight
180.57g – 178.94g = 1.63g
Now, I will need to find the change in temperature of the water from the combustion of Propan-2-ol
Temperature Difference = Final Temperature – Initial Temperature
35°C -19.8°C = 15.2°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100g x 4.2J x 15.2°C = 6384J
Now I will get the amount of Propan-2-ol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.63g / 60.1 = 0.027
Now I will find the amount of energy transferred to the water per mole of Propan-2-ol
Enthalpy per mole = Energy transferred / moles
6384J / 0.027 = 236,444.44 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
236,444.44 J/mol-1 / 1000 = 236.44kJmol-1
Average
(236.44kJmol-1 + 236.44kJmol-1) / 2 = -236.44kJmol-1
Butan-1-ol Calculations
1st Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Butan-1-ol
Change in Weight = Initial Weight - Final Weight
184.1g -182.93g = 1.17g
Now, I will need to find the change in temperature of the water from the combustion of Butan-1-ol
Temperature Difference = Final Temperature – Initial Temperature
33.5°C -18.4°C = 15.1°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100 x 4.2 x 15.1 = 6342J
Now I will get the amount of Butan-1-ol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.17g / 74.12 = 0.016
Now I will find the amount of energy transferred to the water per mole of Butan-1-ol.
Enthalpy per mole = Energy transferred / moles
6342J / 0.016 = 396,375 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
396,375 J/mol-1 / 1000 = 396.38 kJmol-1
2nd Recording
Firstly, I will need to find the amount of fuel burned during the combustion of Butan-1-ol
Change in Weight = Initial Weight - Final Weight
123.55g –122.3g = 1.25g
Now, I will need to find the change in temperature of the water from the combustion of Butan-1-ol
Temperature Difference = Final Temperature – Initial Temperature
33.1°C -17.9°C = 15.2°C
Now I will find the energy transferred from the fuel to the water by using the equation
Energy Transferred = Mass of water heated x 4.2 x temperature rise
100 x 4.2 x 15.2 = 6384J
Now I will get the amount of Butan-1-ol burned in terms of moles.
Moles = Mass of Fuel burned / Mr of fuel
1.25g / 74.12 = 0.017
Now I will find the amount of energy transferred to the water per mole of Butan-1-ol
Enthalpy per mole = Energy transferred / moles
6384J / 0.017 = 375,529.41 J/mol-1
To convert the answer into KJ, I divide this figure by 1000.
375,529.41 J/mol-1 / 1000 = 375.53 kJmol-1
Average
(396.38 kJmol-1 + 375.53 kJmol-1)/2 = -385.96 kJmol-1
From my results, I can see that there is a trend in the relation to the enthalpy change of a fuel and its structural formula. I see that as the number of Carbon atoms in an alcohol increases, its enthalpy change of combustion increases. This is due to the fact that with increasing carbon number, an extra CH2 group is added to the carbon chain. I will now work out how much energy this equates to. During combustion, each CH2 group will create 1 extra CO2 and 1 extra H2O molecule.
So, with an extra CH2 group in a chain, an alcohol will give out an extra
-1712kJmol-1. In my experiment this was not true due to the amount of heat lost to the environment.
There is, however, and exception to this rule. Propan-2-ol has a lower enthalpy change of combustion than that of Propan-1-ol. This is because Propan-2-ol is a more stable molecule.
The electronegative Oxygen atom in the hydroxyl group (OH) creates more of a dipole nature when it is at the end of the molecule (called a primary alcohol), which slightly destabilises the molecule. In Propan-2-ol, the Hydroxyl group (OH) is situated on the central carbon atom, called a secondary alcohol. This creates a dipole charge in which the partially positive carbon atom joined to the Hydroxyl group is stabilised by the effect of two neighbouring methyl groups.
The increase in Enthalpy change of combustion for each fuel is due to the amount of bonds available for the atom to make with Oxygen in its combustion. Methanol is the shortest molecule in my experiment and therefore has fewer bonds available to make with oxygen. Butan-1-ol, the longest molecule in my experiment, has more bonds available to make with Oxygen, and therefore has the largest enthalpy change of combustion than Methanol.
ComparisonEvaluation of my experiment
Results
The measurements I took in my experiment were the initial and final weight of fuel burned (to calculate the amount of fuel burned) and the initial and final temperature of water (to calculate the temperature change).
The scales I used displayed the weight of fuel to 2 decimal places:
At this diagram shows, the scales have a possible uncertainty of 0.005g. This means that to find out the percentage error of a mass, I will use the formula (0.005/change in weight) x 100. Although very small, this was one of several inaccuracies within my experiment, which overall could affect the results which I collect.
When measuring the water, I used a test tube which was able to take readings with a 0.1cm³ accuracy. From this, I was able to take readings within 0.05cm³. This means that with measuring 100cm³, there was a percentage uncertainty of 0.05%. With measuring the amount of water in a measuring cylinder, there was also the factor of the meniscus. I attempted to take measurements from the bottom of the meniscus as much as possible, so therefore I do not think this will create any additional percentage error.
The thermometer I used was able to calculate a difference of 0.1 degree. This means that when taking the reading from a thermometer, it was possible to gain a recording within 0.05 of a degree. Therefore the percentage uncertainty will be found by the formula (0.05/temperature change) x 100.
Percentage ErrorsMethanol
1st Recording
Mass of fuel burned: 1.96g; Percentage uncertainty: (0.005/1.96) x 100 = 0.26%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.20°C Percentage uncertainty: (0.05/15.20) x 100 = 0.32%
Total error = 0.26 + 0.05 + 0.32 = 0.63%
2nd Recording
Mass of fuel used: 1.90g; Percentage uncertainty: (0.005/1.90) x 100 = 0.26%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature change: 15.20°C; Percentage uncertainty: (0.05/15.20) x 100 = 0.32%
Total error = 0.26 + 0.05 + 0.32 = 0.63%
Average error = 0.63%
Ethanol
1st Recording
Mass of fuel burned: 1.68g; Percentage uncertainty: (0.005/1.68) x 100 = 0.30%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.20°C Percentage uncertainty: (0.05/15.20) x 100 = 0.32%
Total error = 0.30 + 0.05 + 0.32 = 0.67%
2nd Recording
Mass of fuel used: 1.60g; Percentage uncertainty: (0.005/1.60) x 100 = 0.31%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.20°C Percentage uncertainty: (0.05/15.20) x 100 = 0.32%
Total error = 0.31 + 0.05 + 0.32 = 0.68%
Average error = (0.67 + 0.68)/2 = 0.68%
Propan-1-ol
1st Recording
Mass of fuel burned: 1.45g; Percentage uncertainty: (0.005/1.45) x 100 = 0.34%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.10°C Percentage uncertainty: (0.05/15.10) x 100 = 0.33%
Total error = 0.34 + 0.05 + 0.33 = 0.72%
2nd Recording
Mass of fuel burned: 1.40g; Percentage uncertainty: (0.005/1.40) x 100 = 0.36%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.10°C Percentage uncertainty: (0.05/15.10) x 100 = 0.33%
Total error = 0.34 + 0.05 + 0.33 = 0.72%
Average error = 0.72%
Propan-2-ol
1st Recording
Mass of fuel burned: 1.64g; Percentage uncertainty: (0.005/1.64) x 100 = 0.30%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.20°C Percentage uncertainty: (0.05/15.20) x 100 = 0.33%
Total error = 0.30 + 0.05 + 0.33 = 0.68%
3rd Recording (due to inaccurate 2nd recording)
Mass of fuel burned: 1.63g; Percentage uncertainty: (0.005/1.63) x 100 = 0.31%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.20°C Percentage uncertainty: (0.05/15.20) x 100 = 0.33%
Total error = 0.31 + 0.05 + 0.33 = 0.69%
Average error = (0.68 + 0.69)/2 = 0.69%
Butan-1-ol
1st Recording
Mass of fuel burned: 1.17g; Percentage uncertainty: (0.005/1.17) x 100 = 0.43%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.10°C Percentage uncertainty: (0.05/15.10) x 100 = 0.33%
Total error = 0.43 + 0.05 + 0.33 = 0.81%
2nd Recording
Mass of fuel burned: 1.25g; Percentage uncertainty: (0.005/1.25) x 100 = 0.40%
Water used: 100g; Percentage uncertainty: (0.05/100) x 100 = 0.05%
Temperature difference: 15.20°C Percentage uncertainty: (0.05/15.20) x 100 = 0.33%
Total error = 0.40 + 0.05 + 0.33 = 0.78%
Average error = (0.81 + 0.78)/2 = 0.80%
The percentage errors I have calculated are very small. However, there is a huge difference between my enthalpy changes of combustion and the actual enthalpy change of combustion. For example, I calculated the enthalpy change of Methanol to be -658 kJ mol-1. The average enthalpy change of combustion of Methanol in my experiment was -106.43 kJmol-1. This means that in my experiment, the value of the enthalpy change of combustion of Methanol is 16.2% of what I have calculated it to be. Therefore the percentage errors that I have calculated for my experiment simply cannot account for this loss. The major cause for this inaccuracy was the amount of heat lost to the surroundings of my experiment.
Procedure
Overall, the practical method in which I used to calculate the enthalpy change of combustion in my experiment was not very effective. The main reasoning behind this was that a great majority of the heat made from the combustion of the fuel did not transfer to the sample of water, but instead was lost to the surroundings. This means that the enthalpy changes of combustion which I have worked out are significantly below the actual amount, but equally proportionate to each other.
As this diagram shows, a significant amount of heat was lost to the surroundings of my experiment. Heat was continually being lost. Heat was being radiated in all directions. Some heat was absorbed by the draught shield. Some heat was absorbed by the copper calorimeter. Only a small amount of the heat was actually transferred to the sample of water.
In this experiment, I was limited to the equipment and supplies of the laboratory. Had this not been an issue, I would have improved my experiment to increase accuracy. To begin with, I would have used a draught shield with a diameter exactly matching that of my copper calorimeter. This would reduce heat loss significantly, as less heat would be able to escape between the copper calorimeter and draught shield.
Other places in which I could improve my experiment are with using more advanced apparatus. In my experiment I could use an electronic thermometer, which displays temperature to greater accuracy than to which I could read. This would tell me exactly when to extinguish the flame, thus improving the overall accuracy of my results.
In this practical procedure, there were several particularly important aspects of ensuring that the data I collected was precise and reliable. One of which was ensuring that each recording of data that I included in my results had been carefully and competently measured. This was due to the many measurements I had to take for my experiment, and if one of the measurements I took was wrong, it would affect the rest of my results. Another aspect which was important to ensuring the data I collected was accurate was ensuring that the apparatus I used each time was the same. This is to avoid variation in the materials and thickness of some of the equipment, which has the potential to alter my results.
References
- [1] EHS Science Department – Hazcards
- [2] Salters Advanced Chemistry – Chemical Ideas
- [3] Salters Advanced Chemistry – Chemical Storylines
