The room temperature in the classroom is usually 21 oc. To ensure that the water rises by 20 oc only, I will remove the draught shield and place the lid on the spirit burner 2 oc -3 oc before the water reaches 41 oc. The number of degrees in advance of the desired 10 oc rise that I will place the lid on the spirit burner depends on the amount how fast the temperature is rising. If it appears to be rising rapidly then the lid will be placed on more degrees in advance, if it is rising slower then of course the lid will be placed on fewer degrees in advance.
Even after the flame has been extinguished the thermometer should still be used to stir the water.
The final temperature is the maximum temperature that the water reaches. This is to be recorded in the table above.
The burner should be weighed with the lid on; the same procedure should be followed after heating, as before, the balance must settle before the tare button is pressed.
The weight as before should be recorded straight away.
The water should then be poured into a clean dry 100cm 3-measuring cylinder; the volume of the water should be recorded in to the table above. Assuming that 1cm 3 of water has a mass of 1g I will then know how much water was lost through evaporation.
This process should be repeated to more times with the same alcohol, so that average values can be obtained.
In between repeats new portions of cold water should be measured out and poured in to dry calorimeters to ensure that the temperature rises all take place in the same conditions. As water behaves differently as it approaches boiling temperature, it absorbs more energy with a lower change in temperature in order to break intermolecular attractive forces to become a gas. Also while the water is in the calorimeter there is no way of knowing the mass of water being heated, if it is not refreshed between repeats then the water definitely going to evaporate.
This whole process including the two repeats for each of the other alcohols.
I have chosen to compare the enthalpy change of combustion of consecutive straight chain alcohols. Firstly because I already understand that the physical properties of alcohols like hydrocarbons change as the relative molecular increases. By investigating consecutive straight chain alcohols I should be able to observe trends, which should make my analysis more interesting
I also know that the trends in properties that can be seen in alcohols are interfered with by the OH bond, because of hydroxyl groups polarising properties, it tends make the properties of the lower alcohols the exception from the trend. By observing trends in the lower alcohols I intend to determine whether the hydroxyl groups effect on physical properties extends to enthalpy change of combustion.
Analysis
Methanol
Average change in temperature recorded over the three attempts is worked out by subtracting the average start temperature from the average end temp.
ΔT= End
Start
(The line above end represents end bar, as in x the accepted mathematical symbol for mean average)
32.33 - 22.33 = 10 oc
ΔMass if alcohol used to obtain this 10 oc is
207.48-207.08= 0.4g
The average mass of water used is obtained by taking an average, of the average start and end volumes. To obtain a mass half way between the two values is the mass to use as the average mass used.
100 +99.33 =99.67g
Ethanol
ΔT 22.33 - 32.33 = 10 oc
ΔMass if alcohol used to obtain this 10 oc is
189.30-189.25=0.38 g
Estimated average mass of water used
100.00 + 99.67=99.84
2
Propan-1-ol
ΔT 22.67 - 32.67 = 10 oc
ΔMass if alcohol used to obtain this 10 oc is
204.87-204.53=0.34g
100 +100 = 100g
2
ΔT 22.33 - 32.33 = 10 oc
ΔMass if alcohol used to obtain this 10 oc is
192.17 –191.93=0.24 g
100 +100 = 100g
2
From the table containing the average changes that I observed in the combustion alcohols, I can clearly see a trend in the change in mass of fuel, burnt in order to raise the temperature of approximately 100cm3 of water by 10oc
As the length of the chain increases, (the table is ordered by number of carbons in a chain increasing by one each row) mass of alcohol used to give out the energy required to raise the temperature 10 oc decreases.
As the temperature change is kept constant the amount of energy that each alcohol is required to give out is constant. This suggests that energy is contained in a smaller volume. Which means that there are more covalent bonds being broken per cm 3 of liquid. Manifestly there most be an increase in the density of the alcohol as the length of the chain increases.
Calculation of the enthalpy of combustion
Assumptions made to make this calculation 1cm 3 of water has a mass of 1g
The average mass of water used is exactly half way between the start and end mass
The specific heat capacity of water is 4.2J/g/K
Methanol
q = cmΔT
q represents the quantity of energy used to obtain the rise in temperature of water
c represents the specific heat capacity, (the amount of energy needed to raise 1 gram of a substance’s temperature by 1 oK
m represents mass of the substance being heated
ΔT represents the temperature change observed in the substance being heated
q=4.2J/g/K x 99.67gx10 oK
Note the intervals in the Kelvin scale are the same as the intervals of the Celsius scale therefore a change of 10 oc is the same as change of 10 oK
q= 4186.14J
The average heat energy absorbed by water from the combustion of methanol was 4190J
To the enthalpy of combustion is energy given out (thermal) when one mole of a fuel is combusted
I have worked out how much thermal energy 0.4g of methanol has given out
There fore to work out the amount of energy one mole would give out, I need to work out the number of moles used to give out 4190J of thermal energy.
Then divide the quantity of energy by this number
The formula of the fuel is CH3OH
To work out the number of moles used I need to work out the mass of one mole of the methanol
The relative molecular mass of the formula above is 12 + 3 + 16+1=32
One mole of CH3OH weighs 32g
To calculate the number of moles used I will need to divide the mass used by the mass of one mole
To work out the number of moles used to raise the water’s temperature by 10 oK
Mass used = no. mol
Mr
0.4g = 0.0125moles
32g
0.0125moles of methage my result is out is
335 x100
=46.14%
726
100-46.14 =53.86
My result is 53.86%
out
My total percentag
e uncertai
n
ty is 0.51%
Ethanol
The uncertainty in reading a thermomete
r
that me
a
sures to 0.1 oc is 0.05
The
refore the pe
r
centage uncertainty in reading the thermometer ianol is combusted is 4190J
The molecular formula of ethanol is C2H5OH
One mole of ethanol has a mass of 24+5+16+1=46g
The number of moles used is mass used
Molar mass
0.38 = 8.26 x 10 -3
46
Therefore the thermal energy that the combustion of one mole of ethanol would give out is
4190 J =5.07 x 103
8.26 x 10 –3moles
The enthalpy of combustion of ethanol is 507kJ/mol
Propan-1-ol
q =4.2J/g/K x 100.00g x10 oK
q=4200J
The thermal energy absorbed by water from the combustion of 0.34g of propan-1-ol is 4200J
The molecular formula of propan-1-ol is C3H7OH
One mole of ethanol has a mass of 36+7+16+1=60g
The number of moles used is mass used
Molar mass
0.34g = 5.67 x 10 –3
60
Therefore the energy that the combustion of one mole of propan-1-ol would give out is
4200 =7.41 x 10 6
5.67 x 10 –3
The enthalpy of combustion of propan-1-ol is 741kJ/mol
Butan-1-ol
q =4.2J/g/K x 100.00g x10 oK
q=4200J
The thermal energy absorbed by the water in the combustion of 0.24g of Butan-1ol is 4200J
The molecular formula of Butan-1-ol is C4H9OH
One mole of Butan-1-ol has a mass of 48+9+16+1=74g
The number of moles is used is mass used
Molar mass
0.24 =3.24 x10 –3
74
Therefore the thermal energy that one mole of butan-1-ol would give out is
4200 = 1.30 x 10 6
3.24x 10 –3
The enthalpy change of combustion of one mole of butan-1-ol is 1300kJ/mol
The enthalpy change of combustion of alcohols increases as the length of the chain increases.
The chain length increases with an addition of a CH2 group
When the alcohol is combusted what actually happens is the covalent bonds are broken, and new bond are formed.
More energy is released from the formation of bonds then is absorbed to break the bonds of the alcohol.
Subsequently the more atoms in the alcohol, more bonds are broken, and more bonds are formed.
The enthalpy change of this.
Because in the combustion of alcohols C-H and C-C bonds more energy is released then absorbed, the more carbons to from CO2 and the more hydrogen to form H2O the more energy is given out.
Hence the rise in enthalpy change as the chains length increases
Evaluation
The value that I worked out for the enthalpy change of combustion of methanol is 335kJ/mol.
The actual value is 726kJmol; I realised that my result would not be the same as the actual results as when I placed the lid on the fuel I could feel the air inside the draught shield had warmed significantly. Which meant that the heat energy had not gone straight to the water and that some may not have actually reached it at all. However I did not expect my register less then half the enthalpy change.
I repeated the procedure twice giving me three results to obtain an average, I observed no clear anomalous results, they appeared to support each other, which, suggests that any error was a procedural error and not just a random mistake.
Note the discrepancy between my worked value for the enthalpy of combustion of alcohol and the actual value was not isolated to Methanol
My worked Enthalpy change of combustion Actual value of enthalpy change of combustion
Ethanol 507kJ/mol 1367kJ/mol
Propan-1-ol 741kJ/mol 2021kJ/mol
Butan-1ol 1300kJ/mol 2676kJ/mol
I believe that most of the main source of error came from the heat loss to air and heat loss to apparatus. Furthermore my metal calorimeter was orange when I begun the heating process, however by the end its was black, it appeared to be covered in a black film of soot which is a clear sign of incomplete combustion.
To check how much of the error should be attributed to procedural error I shall work out how percentage uncertainty in reading instruments.
Methanol
The uncertainty in reading a thermometer that measures to 0.1 oc is 0.05
Therefore the percentage uncertainty in reading the thermometer is the uncertainty divided by average temperature change of water being heated by the combustion of methanol
0.05 x100 = 0.5%
10
The uncertainty in reading the 100cm–3 measuring cylinder that measures to 1cm–3 is 0.005
The average value of water recorded was 99.67cm3
0.005 x 100 = 5.02x 10–3%
99.67
The uncertainty in reading a balance that meat measure to 2 decimal places it 0.005 the average mass recorded is the average (start mass + average end mass)/2
207.48 +207.08 =207.28g
2
0.005 x100 = 2.41 x 10–3%
207.28
The percentage my result is out is
335 x100=46.14%
726
100-46.14 =53.86
My result is 53.86% out
My total percentage uncertainty is 0.51%
Ethanol
The uncertainty in reading a thermometer that measures to 0.1 oc is 0.05
Therefore the percentage uncertainty in reading the thermometer is the uncertainty divided by average temperature change of water being heated by the combustion of ethanol
0.05 x100 = 0.5%
10
The uncertainty in reading the 100cm–3 measuring cylinder that measures to 1cm–3 is 0.005
The average value of water recorded was 99.84cm3
0.005 x 100 = 5.01x 10–3%
99.84
The uncertainty in reading a balance that meat measure to 2 decimal places it 0.005 the average mass recorded is the average (start mass + average end mass)/2
189.63 + 189.25 = 189.43g
2
0.005 x 100 = 0.0264%
189.43
The percentage out my result is
- x100=37.1%
1367
My result is
37.1-100 = -62.9% of the actual result
My total percentage uncertainty in working out the enthalpy change of combustion of ethanol is 0.53%
Propan-1-ol
The uncertainty in reading a thermometer that measures to 0.1 oc is 0.05
Therefore the percentage uncertainty in reading the thermometer is the uncertainty divided by average temperature change of water being heated by the combustion of propan-1-ol
0.05 x100 = 0.5%
10
The uncertainty in reading the 100cm–3 measuring cylinder that measures to 1cm–3 is 0.005
The average value of water recorded was 100.00cm3
0.005 x 100 = 5.00x 10–3%
100
The uncertainty in reading a balance that meat measure to 2 decimal places it 0.005 the average mass recorded is the average (start mass + average end mass)/2
204.87+204.53 =204.7
2
0.005 x 100=2.44 x 10–3%
204.7
The percentage my result is out is
741 x 100=36.7%
2021
It is unsettling to see that my result is just over 1/3 of the actual result
36.7-100= -63.3% from the actual enthalpy change of combustion of propan-1-ol
My total percentage uncertainty is 0.51%
Butan-1-ol
The uncertainty in reading a thermometer that measures to 0.1 oc is 0.05
Therefore the percentage uncertainty in reading the thermometer is the uncertainty divided by average temperature change of water being heated by the combustion of butan-1-ol
0.05 x100 = 0.5%
10
The uncertainty in reading the 100cm–3 measuring cylinder that measures to 1cm–3 is 0.005
The average value of water recorded was 100.00cm3
0.005 x 100 = 5.00x 10–3%
100
The uncertainty in reading a balance that meat measure to 2 decimal places it 0.005 the average mass recorded is the average (start mass + average end mass)/2
192.17+191.93 =192.05
2
0.005 x 100=2.60 x 10–3%
1920.05
The percentage my result is out is
1300 x 100=48.6%
2676
48.6-100= -51.4% from the actual enthalpy change of combustion of butan-1-ol
My total percentage uncertainty is 0.51/%
As I have established that the percentage uncertainty hasn’t had a considerable effect on my experiment the only way I can think of improving my procedure is eradicate some of the procedural errors.
The easiest error to remove, is the incomplete combustion of the alcohol, The experiment simply has to take place in an oxygen rich atmosphere to ensure that there is oxygen way in excess of quantity needed for the complete combustion of the alcohols.
Also if I could redesign the procedure to remove the space between the burner and the calorimeter, and introduced oxygen feed to ensure enough oxygen is supplied I believe that it is quite possible to cut my percentage out from 2/3 to 1/3.
The reason why incomplete combustion could have a drastic effect on the worked enthalpy of combustion is the fact that the products release less energy then the products of complete combustion.
My investigation is based on the premises that complete combustion has occurred, and even if I did manage to collect all of the soot weigh it and work out how many moles of carbon is present, there is no way of getting it all. Therefore there is no way of working out the ratio of complete combustion to incomplete. Hence unless I get rid of the incomplete combustion there is no way gaining an accurate result.
Reducing the distance between the burner, to a minimal will maximise the efficiency of the heat transfer from flame to the water
