METHOD
PRELIMINARY INVESTIGATION
To determine my final procedure, including apparatus, factors that will be changed and controlled and the range of results, I carried out a short preliminary investigation, mainly investigating volume of water and distance of flame from the bottom of the can.
The preliminary method was as follows:
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Measure out 100cm3 of water using a measuring cylinder
- Pour the water into an aluminium can
- Measure the temperature of the water with a thermometer
- Measure the mass of a candle using an electronic balance
- Place the candle 2cm away from the bottom of the can
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Light candle and heat the water until the water has risen 20oC in temperature
- Extinguish candle and weigh immediately
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Repeat this method for 200cm3 of water
- Repeat this method changing the distance between the candle and can to 3cm
The preliminary results were as follows:
In conclusion to the preliminary investigation we have firstly learnt to use a larger volume of water as it gives a larger reading and also it allows us to see the thermometer without taking it out of the water. However, the volume of water can’t be too big as it will take too long for the water to heat up. Also we should have a 2cm gap between the can and the candle because less heat is lost to the surroundings i.e. more is concentrated on the can. The preliminary experiment showed that a certain amount of soot accumulates at the bottom of the can after each experiment (a sign of incomplete combustion). This must be wiped off as it may affect other results. One problem encountered was excess wax on the candle. In the final experiment this will not matter, as spirit burners will be used.
Therefore we can give a list that determines the conditions (independent and fixed variables) of the final experiment:
- The can must be cleaned between experiments (in case soot has accumulated) but not changed
-
The volume of water must always be 250cm3 and the water must be changed for each experiment
- The wick must always remain 2cm away from the can
- Draught shields must be used at all times
-
The temperature rise must always be 20oC from the recorded temperature when the burner is lit – accuracy to half a degree
- The water must be stirred whilst it is heating
- There will be three replicates for each experiment
- The mass of the alcohol must be found to two decimal places of a gram
- The temperature of the water must be measured to half a degree
- I will use exactly the same can for each test so the material (aluminium), shape, surface area and volume (330ml) will always be the same
These variables will determine an accurate and fair test.
I will investigate the relationship between structure and energy release of alcohols by measuring the change in mass of the alcohol when it heats 250cm3 of water at an increase of 20oC.
Apparatus and Diagram:
Thermometer
Aluminium can
Spirit Burner
Draught shield
Heat proof mat
Water
Measuring cylinder
Ethanol
Methanol
Propan-1-ol
Butan-1-ol
Pentan-1-ol
Clamp Stand
Boss
Scales
Ruler
The method is as follows (see diagram to help):
- First set up the apparatus as in the diagram
- The wick of the alcohol must be 2cm from the base of the can, measured by a ruler
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Using a measuring cylinder, the can must be filled with 250cm3 of water
- Measure the mass of the alcohol on a balance and the current temperature of the water(60 seconds after being poured)
- Put the alcohol in place directly underneath the can and light the wick with a lit splint
- Stir the water continuously
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When the thermometer shows that the temperature has risen by 20oC, extinguish the wick and immediately measure the new mass of the alcohol
- Remove any soot at the bottom of the can
- Repeat three times to produce three results for each alcohol
- Do this experiment with methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol.
-
Calculate the energy transferred from the alcohol to the water per mole of alcohol using the formula ΔH = (MCΔT / ΔM2) x RMM
The following safety procedures will be taken out
- Safety goggles must be worn at all times
- Alcohols must not be let out of their relatives pots
- The flame must watched at all times
- Caution must be taken with the can and flame as they will be hot
RESULTS
The following table lists the results that were witnessed during the experiment.
All results are measured to 2 decimal places
ANALYSIS
HEAT TRANSFERRED
The readings obtained and represented in the results table are the observations. They are simply what was directly found in the investigation. However to compare the results we must consider the actual heat transferred from the alkanol to the water. We must then find out how much energy was transferred per gram of alkanol. After that the heat energy transfer must be found per mole of alkanol to be able to compare results in a fair manner. To do this we must use the formula:
ΔH = (MCΔT / ΔM2) x RMM
(Where H= heat energy, M= mass of water, C= heat capacity of water (always 4.2kJ/g) and T= temperature, M2= mass of alkanol used, RMM = relative molecular mass of alkanol)
Worked example:
Take the first result for methanol. First we must use the formula MCΔT. This finds out how much heat energy was transferred from the methanol to the water in total:
First the numbers must be substituted into the formula
MCΔT
= 250 x 4.2 x 20
=21000J were transferred from the methanol to the water in total
Next we must find how many joules of energy were transferred from the methanol to the water per gram of methanol. This must be found so that the alkanols can be directly compared. The formula for the mass of methanol used is:
ΔM2
=1.49
Therefore to find the energy transferred per gram of methanol to the water we must finally work out the first answer divided by how much methanol was used:
ΔH = MCΔT / ΔM2
ΔH = 21000 / 1.49
ΔH =14093.96 (2 d.p.)
Therefore 14093.96J or 14.09kJ of energy was transferred per gram of methanol to the water
However, we still do not have a fair comparison between alkanols. To obtain a fair comparison we must multiply by the relative molecular mass of the alkanol:
ΔH = 14.09… x 32
ΔH = 451.01 (2 d.p.)
Therefore 451.01 kJ/mole of energy was transferred from the methanol to the water.
The following table lists the values for each part of the equation and shows the final energy transfers per gram of each alkanol.
All results are to 2 decimal places.
I have produced a graph that plots energy transfer per mole against the relative molecular mass (RMM) of the alkanol. This graph shows the trend that the results give. The graph shows that as the RMM increases, the energy transfer per mole of alkanol also increases. There is a line of best fit which gives evidence that the increase is a regular increase. This is shown by how the line is a straight line that is close to many of the points. Each result is not on the line but is very close, especially in relative terms i.e. the scale is quite big. This gives further evidence to suggest that the stated trend is accurate. There is one anomalous result or rather there is a set. Pentan-1-ol doesn’t fit into the line. As mentioned, the scale is quite large so the result for pentan-1-ol is not as bad as it seems, however, it doesn’t fit into the trend directly supported by the other alkanols. The anomaly may have been for many reasons but probably because the pentan-1ol may have been a lot fresher than the other alkanols. This is suggested by the fact that pentanol is anomalous because it has a greater energy transfer per mole than it should have according to the other results.
The results match my prediction, which stated that due to relative bond energies, there would be a regular increase in heat energy transfer. This happened because the energy absorbed in breaking the bonds was always less than the energy released when making bonds. Because the RMM increases (in a relative manner), in terms of moles, the products were greater compared to the reactant i.e. the energy transfer was greater. This can best be described by working out the actual bond energies for the alkanols.
Bond energies will find the exact number, which our results should be. They should therefore show us what we should have found. Therefore I will find the theoretical bond energy output for each alkanol to compare my results with what they should be.
BOND ENERGY VALUES
A bond energy is the energy absorbed to break a particular type of covalent bond or the energy released to make a certain bond. The difference between making and breaking bonds is that breaking bonds is an endothermic process as it absorbs energy and making bond is an exothermic process in that it releases energy. This works on the principle that ‘if a reaction causes on thing to happen, the opposite reaction will cause the reverse’.
The final energy output of a reaction is found by the following equation:
Final energy output = energy absorbed in breaking bonds - energy released in making bonds.
Therefore if the energy output were a positive number, the energy absorbed in breaking bonds would be greater than the energy released when making bonds, therefore the reaction would be endothermic. If the energy output is a negative number, then the energy released in making bonds would be greater than the energy absorbed when breaking bonds, and thus the reaction would be exothermic. As we know that combustion is an exothermic reaction, we expect to find that the energy released is greater than the energy absorbed.
The bond energies we will be looking at our as follows:
BOND BOND ENERGY (kJ)
H – C 412
C – O 360
O – H 463
O = O 496
C = O 743
C – C 348
To work out bond energies we must first write out balanced equations for the combustion of all the alkanols:
Methanol
2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g)
Ethanol
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (g)
Propan-1-ol
2C3H7OH (l) + 9O2 (g) → 6CO2 (g) + 8H2O (g)
Butan-1-ol
C4H9OH (l) + 6O2 (g) → 4CO2 (g) + 5H2O (g)
Pentan-1-ol
2C5H11OH(l) + 15O2(g) → 10CO2(g) + 12H2O(g)
Worked example:
Take the combustion of methanol
Using the chemical equation, we must first find which bonds are broken and which bonds are formed:
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
BONDS BROKEN BONDS MADE
6 x C – H 4 x C = O
2 x C – O 8 x H – O
2 x H – O
3 x O = O
(There are no C – C bonds in methanol)
Next we must work out the energy absorbed when the bonds are broken using the bond energy values:
6 x 412 =2472
2 x 360 = 720
2 x 463 = 926
3 x 496 = 1488
Total energy absorbed when the bonds are broken =
2472 + 720 + 926 + 1488
=5606 kJ
Next we must work out the energy released when the bonds are formed using bond energy values:
4 x 743 = 2972
8 x 463 = 3704
Total energy released when the bonds are formed =
2972 + 3704
= 6676 kJ
Finally we must find the final energy output:
Final energy output = energy absorbed in breaking bonds - energy released in making bonds
Final energy output = 5606 – 6676
= -1070 kJ
This means that the reaction is exothermic because 1070 kJ are released.
However, we must work out how much energy is released per mole of methanol:
2 moles of methanol are used, therefore
Final energy output = - 1070 / 2
= -535 kJ per mole of methanol
The following table shows the quantity of each bond that is broken and made and thus the final calculation for the total energy output per mole of alkanol.
The fact that the numbers are negative denotes the reaction would be exothermic but this is not important, as we already know that the results in our experiment were exothermic, after all it was combustion. The important thing to consider is how the figures compare to the figures obtained in the experiment.
On the graph I have plotted the points for the actual bond energies that should have been shown. The graph of my investigation, and theoretical bond energies are on the same axis. This allows them to be compared much more easily. The line for bond energy is straight and goes through all the points showing how RMM is exactly proportional to the energy transfer per mole of alkanol. As we can see, the lines do not match exactly. The line for actual bond energies is overall of a higher value than that of the ones produced in the investigation. Also, the bond energy line has a steeper gradient, which means each alkanol is much more efficient than the previous one than the investigation results would suggest. This can all be explained due to heat loss. The more energy transferred per mole of alkanol, the more efficient the alkanol is as a fuel. Therefore, the investigation would have produced smaller results than what should have been because the bond energies do not account for heat loss, which was in abundance in the investigation. However, because the investigation gradient was different we can see that the heat loss would have been greater as the energy transfer increased. This makes sense in that heat loss would have been relative to the alkanol. This means that heat loss is not a constant, instead, it is a percentage of the energy released. Therefore, due to how percentages work, the actual heat loss would have been greater if the heat transfer is greater. The investigation heat transfer is not exactly the same percentage of what it should be for every alkanol. In fact, the relative heat loss increases as the RMM increases. This also makes sense as the can was the same size for all experiments so as the heat transfer increased, the area it had to heat up decreased, therefore more heat energy was lossed.
Overall, the results can support a firm conclusion that as RMM increases, heat energy transfer per mole also increases regularly.
EVALUATION
The experiment quite clearly showed us that as that as the relative molecular masses (RMM) of the alkanol increase; the energy transfer per mole of alkanol also increases. This also means that as RMM increases, the efficiency of the alkanol increases. Therefore we now know that apart from anything else, in practical terms, as the RMM of an alkanol increases, it becomes a better fuel.
The big question is whether we can trust the results. This can be answered in two sections, considering the aspect of how fair a test the experiment was, and also by accuracy, in how accurate the results are.
FAIR TEST
The following is a list of the things that were kept constant, in our control:
- The can must be cleaned between experiments (in case soot as accumulated) but not changed
-
The volume of water must always be 250cm3 and the water must be changed for each experiment
- The wick must always remain 2cm away from the can
- Draught shields must be used at all times
-
The temperature rise must always be 20oC from the recorded temperature when the burner is lit – accuracy to half a degree
- The water must be stirred whilst it is heating
- There will be three replicates for each experiment
- The mass of the alcohol must be found to two decimal places of a gram
- The temperature of the water must be measured to half a degree
- I will use exactly the same can for each test so the material (aluminium), shape, surface area and volume (330ml) will always be the same
First was cleaning the can. This is a constant due to soot accumulating due to incomplete combustion occurring. Soot would affect other results because less heat could be transferred to the water, as the soot would absorb some. This was kept constant however all the soot could not always be cleaned off. Another factor was that the soot accumulating during each experiment might have affected each experiment. The only way this could have been improved would have been to add oxygen to the environment in some form to avoid incomplete combustion. Another factor involving the can was the use of the same can. This was obviously easy to keep however it wasn’t completely fair. There is a chance that the can may have been slightly distorted during an experiment due to the heat. An improvement would be to use different cans but make sure they were the same shape and materials.
The volume of the water was always accurately kept to 250cm3. One improvement may have been to use a measuring cylinder that had smaller increments than the one used (which had 1cm3 increments).
Keeping the can 2cm from the wick was not a constant that was easy to keep. One problem was that all the wicks were different at the tip, therefore it was difficult to judge where to measure from. The very tip was used but obviously this wasn’t really fair. The same type of spirit burner and the same length of wick should have been used for each experiment.
Draught shields were obviously an easy constant to keep and these worked successfully. Stirring is another factor that was successfully kept to.
The issue of temperature is a difficult to keep constant. The reading to half a degree was successfully kept constant however the problem lied mainly in where to begin and end the experiment. The experiment involved starting 60 seconds after the water had been poured. This was a good idea however it would have been better to start everything at the same temperature to make the test fair. The room should also have tried to be kept at a constant temperature although this would be difficult to achieve. The climax of each experiment was hard to judge. A 20degree change was hard to see exactly as it was difficult to stop the experiment at the exact moment that the required temperature was reached. This was partly due to it being difficult to judge this exact point and also because it was difficult to quickly extinguish the flame. The only way this could have been improved would be to use a better method of extinguishing the flame.
All of the constants relating to accuracy were easy to keep to and worked successfully
ACCURACY
As I have discussed in the fair test section, the constants were not kept to perfectly, and they weren’t the best constants to use either. This would have affected the accuracy of my results. However, it can be said that the difference that these problems would have made when the final results of energy/mole were found under perfect circumstances would be minimal. Everything that could be controlled by the hand or eye was done. In this way, the accuracy of my results is very high.
One thing to support this claim would be the similarity between my results and the actual bond energy calculations. The specific results themselves aren’t the same however this may be due to a number of factors, particularly heat loss. Heat would have been lossed in a number of ways. Firstly, simply to the surroundings. This may have been lossed by the water or by the flame. Surrounding temperature is the problem here. One improvement would be to have tried to keep the surrounding temperature constant. Secondly, heat would have been lossed to the can itself. To reduce heat loss more efficiently, a better system than a can could have been used such as a container which covers the flame (still allowing oxygen in), but also reducing heat loss from the water.
If we again consider the similarity between the results of the investigation, and the bond energies, we see that the pattern of the lines is the same in that it is a straight line with a positive gradient. This is the most important thing as it shows the results are accurate, as the relationship between the results is the same. There is one clear anomaly in pentanol. This may have been due to the difference in spirit burner, or the certain volume of methanol used.
GENERAL IMPROVEMENTS AND EXPANDING
There is one broad improvement that could be made to increase the accuracy, and the aspect of as fair test. This lies in the spirit burner and alcohol used. Different spirit burners and different mass of alcohols were used. To make the test much fairer, the same spirit burner should have been used for each experiment. This would mean cleaning the spirit burner out after each experiment, and also putting in a new wick which would be exactly the same for each experiment. Also the mass of alcohol should always be the same. This may have made a small difference in the result as a larger mass of alcohol may have meant an increased rate of reaction (combustion), as there would be more alcohol to combust.
There are many ways in which the experiment could be expanded. Firstly would be to increase the number of replicates to increase the accuracy, after all, the more replicates, the higher the accuracy. Another method would be to test more different alkanols to create more results to secure a firm conclusion including hexanol, heptanol and octanol. My results, however, definitely support a firm conclusion as they are.
