PROCEDURE
Making an Alcohol Lamp:
-
Take a 50 cm3 Conical Flask
-
Take cotton and roll it in your hands to make the wick.
- The wick needs to fit snug in the hole and reach to the bottom of the jar.
- Now thread your wick through the opening of the flask
- Now fill the flask with alcohol
- Make sure the cotton is fully dipped in alcohol.
- Take aluminum paper and make it into a cork shape.
- Make a small hole in the middle of the aluminum cork so as to fit the cotton through it.
Combustion of Alcohol
- Take a calorimeter and record its mass
-
Add 100 cm3 of water and record down the mass.
- Record down the mass of water
- Put the calorimeter on top of the tripod with the thermometer in it.
- Record the initial temperature of the water.
- Record down the mass of the methanol lamp before putting fire
- Use a retort stand and clamp the methanol lamp under the tripod and make it as close to the calorimeter as possible.
- Light the methanol lamp and immediately start timing
- Keep the thermometer in and make sure that the water does not boil.
- After exactly 1 minute 30 seconds, use hot hands to take off the beaker and quickly turn the fire off the methanol lamp.
- Record the final temperature of water
- Record the final mass of the methanol lamp
- Do 3 repeated trials using methanol.
- Repeat the same process with ethanol and butan-1-ol lamps.
- Record all data and process it.
HOW TO PROCESS DATA:
After we have collected all the data, we will use the formula
n = m/M
to find the moles of alcohol burnt
Then, we will use the formula:
Q = mC∆T
to calculate the heat energy given out by the alcohol.
Finally, we will divide the heat energy by the number of moles to calculate the enthalpy of combustion of the alcohol.
∆H = mC∆T
n
QUANTITATIVE DATA:
QUALITATIVE DATA:
DATA PROCESSING
METHANOL:
Step 1: Calculating the number of moles of Methanol combusted
n = m/M
CH3OH
M = (12) + (3 x 1) + (16) + (1)
M = 32.0 gmol-1
m = 1.53 g
n = 1.53g/32.0 gmol-1
n = 0.0480 mol
Step 2: Calculating heat energy given out
Q = mC∆T
m = 100g
C = 4.186 Jg-1k-1
∆T = 11.5k
Q = (100g) (4.186 Jg-1k-1) (11.5 k)
Q = 4813.9 J
Q = 4.81 kJ
Step 3: Calculating the enthalpy of combustion
ΔHC = Q/mol
ΔHC = 4.81kJ
0.0480mol
= 100 kJ/mol
However it is exothermic, that is why it has to be negative
= - 100 kJ/mol
Step 4: Calculating the uncertainties
Uncertainty for mass of alcohol combusted = (0.02g/1.53g) x 100
1.31%
Uncertainty for volume of water: (0.05cm3/100cm3) * 100
= 0.05 %
Therefore uncertainty for mass of water: 0.05%
Uncertainty for change in temperature = (0.02/11.5) x 100
= 0.17%
Total Uncertainty = 1.53 %
ENTHALPY OF COMBUSTION OF METHANOL = - 100 kJ/mol ± 1.53 %
ETHANOL:
Step 1: Calculating the number of moles of Ethanol combusted
n = m/M
C2H5OH
M = (2 x 12) + (5 x 1) + (16) + (1)
M = 46.0 gmol-1
m = 1.06 g
n = 1.06g/46.0 gmol-1
n = 0.0230 mol
Step 2: Calculating heat energy given out
Q = mC∆T
m = 100g
C = 4.186 Jg-1k-1
∆T = 11.0k
Q = (100g) (4.186 Jg-1k-1) (11.0k)
Q = 4604.6 J
Q = 4.60 kJ
Step 3: Calculating the enthalpy of combustion
ΔHC = Q/mol
ΔHC = 4.60kJ
0.0230mol
= 200 kJ/mol
However it is exothermic, that is why it has to be negative
= - 200 kJ/mol
Step 4: Calculating the uncertainties
Uncertainty for mass of alcohol combusted = (0.02g/1.06g) x 100
1.78%
Uncertainty for volume of water: (0.05cm3/100cm3) * 100
= 0.05 %
Therefore uncertainty for mass of water: 0.05%
Uncertainty for change in temperature = (0.02/11.0) x 100
= 0.18%
Total Uncertainty = 2.01 %
ENTHALPY OF COMBUSTION OF ETHANOL = - 200 kJ/mol ± 2.01 %
BUTAN-1-OL:
Step 1: Calculating the number of moles of Butanol combusted
n = m/M
C4H9OH
M = (4 x 12) + (9 x 1) + (16) + (1)
M = 74.0 gmol-1
m = 0.29 g
n = 0.29g/74.0 gmol-1
n = 0.00390 mol
Step 2: Calculating heat energy given out
Q = mC∆T
m = 100g
C = 4.186 Jg-1k-1
∆T = 11.0k
Q = (100g) (4.186 Jg-1k-1) (4.00k)
Q = 1674.4 J
Q = 1.67 kJ
Step 3: Calculating the enthalpy of combustion
ΔHC = Q/mol
ΔHC = 1.67kJ
0.00390mol
= 428 kJ/mol
However it is exothermic, that is why it has to be negative
= - 428 kJ/mol
Step 4: Calculating the uncertainties
Uncertainty for mass of alcohol combusted = (0.02g/0.29g) x 100
6.90%
Uncertainty for volume of water: (0.05cm3/100cm3) * 100
= 0.05 %
Therefore uncertainty for mass of water: 0.05%
Uncertainty for change in temperature = (0.02/4.00) x 100
= 0.50%
Total Uncertainty = 7.45 %
ENTHALPY OF COMBUSTION OF BUTANOL = - 428 kJ/mol ± 7.45 %
COMPARISON OF THE ENTHALPY OF COMBUSTION OF ALCOHOLS
PERCENTAGE ERRORS:
Literature Value – Experiment Value x 100
Literature Value
Methanol:
726 kJ/mol - 100 kJ/mol x 100
726 kJ/mol
= 86.2 %
Ethanol:
1367 kJ/mol - 200 kJ/mol x 100
1367 kJ/mol
= 85.4 %
Butan-1-ol:
2676 kJ/mol - 428 kJ/mol x 100
2676 kJ/mol
= 84.0 %
CONCLUSION
Therefore, from our experimental data and through data processing, we are able to determine the molar enthalpy of combustion of the three alcohols; methanol, ethanol and butan-1-ol. The molar enthalpies of combustion increase as the molecule becomes larger and this corroborates with my hypothesis and proves it right. It can be seen through the graph that the molar enthalpy doubles from methanol to ethanol and then doubles from ethanol to butan-1-ol. If we put this data into mathematical form, it almost becomes like
Molar Enthalpy of Alcohol =
100 kJ/mol + [100kJ/mol (No. of Carbons – 1)]
Through the experiment, we were able to determine the enthalpy change of combustion of methanol, ethanol and butan-1-ol. Although we had random errors involved, the systematic errors contributed far more in the final percentage error than the random errors. The difference between the given value of the enthalpy change of combustion of the three alcohols and the results of our experiment was huge as the percentage difference turned out to be huge. The percentage errors for methanol, ethanol and butan-1-ol were 86.2%, 85.4% and 84.0% respectively. This was due to the huge uncertainties and errors in the experiment.
First and foremost, the biggest problem in the lab which was the largest uncertainty was the wastage of heat to the air. We heated the water in the beaker using the spirit lamps; however, the issue with this was that a lot of heat was able to escape and therefore, there was less energy while a lot of energy was wasted. Thus, the enthalpy change we found was much lower than the actual value given in the Data Booklet as the enthalpy change of combustion is the absorption or releasing of energy and in this case, the energy was released. However, as a lot of energy was wasted, we were not able to find the exact energy release. This can be improved to a certain extent by putting a heat shield around the apparatus as the heat shield would prevent heat loss and this will improve the experiment by giving us a more accurate result. Also, the distance between the alcohol lamp flame and the beaker can be decreased so as to prevent as much energy loss as possible as less heat energy will escape to the surroundings and more will go into the beaker itself. Moreover, this lab can be further improved by using a calorimeter instead of a beaker. Although the lab design asked to use a calorimeter, due to limited time, we used a beaker instead. A calorimeter is a device used to measure the heat of reaction and as it is insulated, it gives prevents heat from escaping while beaker has no such capability to prevent heat escape into the atmosphere. However, a calorimeter would take much more time to heat the water inside. Therefore, by having a time extension, we would be able to use a calorimeter and therefore, it will give us a much more accurate value.
Another problem was that we assumed that complete combustions occurred and this became a problem as in fact, all the alcohols were being burned in the air. Also, the blackish grey soot on the flask indicated that carbon was being formed which reveals that actually, it was incomplete combustion as carbon is only formed when incomplete combustion occurs.
For the density of water, it could have been different than 1 gmcm-3 but we assumed it to be the density of pure water. There could have been impurities in the water or other factors could affect it to change its density. This assumptions might have also been one of the factors that affected our results. Furthermore, we made an assumption that none of the alcohols escaped through evaporation during the time the flame was extinguished and the alcohol lamp and the its contents were reweighed. This was an error as methanol and ethanol are volatile liquids and the fact that the alcohol lamps were hot after the combustion indicates that some of the alcohols would have escaped which increases the uncertainty in the labs.
In addition, one other reason that could affect the results of our lab is that the beaker might have absorbed some heat. The beaker was made of Pyrex glass and certainly absorbed some heat. This means that before the temperature of the water started to change, the beaker might have absorbed some heat and this affects the results as some energy is wasted in this process. Similar to this was the stirring rod and the thermometer which absorbed some heat as well further making the experiment less accurate. Furthermore, random errors certainly played a part in affecting the results as we had uncertainties for mass, temperature and volume. Even though random errors were comparatively small to affect the result, nevertheless, all these uncertainties add up to be large enough to affect the results of the lab.
For random errors, we can take certain measures to reduce the uncertainty and increase the accuracy of our lab. Also, we can measure the mass of water to try and reduce the uncertainty due to our assumption made of the density of water. For lowering the uncertainties even more, we can use a digital thermometer so that if the thermometer has a lower uncertainty, the percentage uncertainty will decrease.
