Results
From these results we can calculate the enthalpy change of combustion for methanol and hexane.
Hexane:
1g of water heated through 1ºC needs 4.2 Joules of heat
100g of water heated through 1ºC needs 4.2 x 100 joules of heat
100g of water heated through 22ºC needs 4.2 x 100 x 22 joules of heat.
200g of water heated through 22ºC needs 4.2 x 200 x 22 joules of heat.
=18.48 KJ
The 18.48 KJ must have come from the combustion of Hexane.
4.51g of Hexane gives 18.48 KJ of heat
1g of Hexane gives (18.48 / 4.51) KJ of heat
1 mole of Hexane gives (18.48 / 4.51 x 86) KJ of heat
=7167.7 KJ
So:
C6H14 + 91/2O2 →→→ 6CO2 + 7H2O
∆HcØ = - 7167.7 KJmol-1
Methanol:
1g of water heated through 1ºC needs 4.2 Joules of heat
100g of water heated through 1ºC needs 4.2 x 100 joules of heat
100g of water heated through 21ºC needs 4.2 x 100 x 21 joules of heat.
200g of water heated through 21ºC needs 4.2 x 200 x 21 joules of heat.
=17.64 KJ
The 18.48 KJ must have come from the combustion of Methanol.
9.19g of Methanol gives 17.64 KJ of heat
1g of Methanol gives (17.64 / 9.19) KJ of heat
1 mole of Methanol gives (17.64 / 9.19 x 32) KJ of heat
=5187.6 KJ
So:
CH3OH + 2/3O2 →→→ CO2 + 2H2O
∆HcØ = -5187.6 KJmol-1
In this experiment I calculated the percentage error so that I could see how accurate my results are. Here are my percentage errors.
Hexane
The accuracy of my experiment can be calculated by working out the percentage difference between the data book values and my values for the ∆HcØ of the two fuels.
Hexane: - 4163 – 1767.7 = -2395.3
- 2395.3 / - 4163 =0.575
0.575 x 100 = 57.5%
Methanol
Methanol: - 726 - - 5187.6 = 4461
4461 / - 726 =6.14
6.14 x 100 = 614%
From this I can see that my experiments were precise, using the equipment that I had available but not very accurate due to the data book values being a long way from my results.
Due to this I am going to adapt my method to make my results more accurate by attempting to prevent as much of the heat being lost to the surroundings, by setting up a more efficient draft exclusion system.
The apparatus and materials that I will use in my main experiment will be:
- A tripod
- Spirit burner containing one of four fuels
- Propan-1-ol
- Butan-1-ol
- Ethanol
- Methanol
- A copper calorimeter
-
A 100cm3 measuring cylinder
- A thermometer (0-100ºC)
- Clamp and stand
- Draft exclusion system (5 heat proof mats)
In this experiment my method will be slightly more accurate as I am going to make sure that at every possible stage I am reducing the error and therefore hopefully increasing the accuracy of my results, by making my ∆HcØ value closer to that of the data book.
Firstly I will set up the equipment as described in the original method. The main difference at this stage is the draft exclusion system. One of the heat proof mats needs to be positioned at the back behind the clamp stand, one on the left and one on the right hand side of the clamp stand to prevent the heat escaping from the sides and back.
Use the measuring cylinder to accurately measure out 200cm3 of cold water. Pour this into the copper calorimeter and take the temperature of the water.
Next, the mass of the first fuel needs to be recorded. In an attempt to reduce the error of the mass, three readings need to be taken and the average mass calculated.
Then the spirit burner will need to be positioned under the calorimeter at a height of 10cm above the top, and the remainder of the draft exclusion system needs to be arranged.
Balance a heat proof mat on the top of the two side ones ensuring that there is enough space left for the thermometer to move in the calorimeter.
Carefully light the spirit burner, and place the final heat proof mat in front of the spirit burner.
Monitor the temperature of the water, as it gradually increases. Once the temperature has increased by 17ºC snuff out the spirit burner and continue to stir the water in the calorimeter.
Record the highest temperature that the water reaches after heating. Once this has been recorded, reweigh the spirit burner (three times and calculate the average).
Empty the water from the calorimeter and rinse with cold water and make a note of any incomplete combustion (soot) around the base of the can.
Repeat the experiment for all four fuels and record all results.
Results
Mass tables
Final results table
Calculations
Ethanol:
Enthalpy
1g of water heated through 1ºC takes 4.2 Joules of heat.
100g of water heated through 1ºC takes 4.2 x 100 joules of heat.
200g of water heated through 1ºC takes 4.2 x 200 joules of heat.
200g of water heated through 19ºC takes 4.2 x 200 x 19 joules of heat.
=15.96 KJ
2.6g of ethanol give 15.96 KJ of heat
1g of ethanol gives (15.96 / 2.6) KJ of heat
1 mole of ethanol gives (15.96 / 2.6 x 46) KJ of heat.
=- 282.36 KJ mol-1
Error
The precision of this experiment is quite high as all of the percentage error figures are quite low. The stage that the highest error could occur is the temperature stages as these are the two with the highest percentage errors.
- 1367 – - 282.36= - 1084.64
- 1084.64 / -1367 = 0.79
- 0.79 x 100 = 79%
The accuracy of this experiment is low because there are lots of ways in which our results could be affected.
Methanol
Enthalpy
1g of water heated through 1ºC needs 4.2 Joules of heat
100g of water heated through 1ºC needs 4.2 x 100 joules of heat
100g of water heated through 22ºC needs 4.2 x 100 x 22 joules of heat.
200g of water heated through 22ºC needs 4.2 x 200 x 22 joules of heat.
=18.48 KJ
4.51g of Methanol gives 18.48 KJ of heat.
1g of Methanol gives (18.48 / 2.15) KJ of heat
1 mole of Methanol gives (18.48 / 2.15 x 32) KJ of heat
=- 1271.42 KJmol-1
Error
The precision of this experiment is quite high as all of the percentage error figures are quite low. The stage that the highest error could occur is the temperature stages as these are the two with the highest percentage errors.
- 726 - - 1271.42 = 545.42
545.42 / - 726 = -0.75
- 0.75 x 100 = 75%
The accuracy of this experiment is low because there are lots of ways in which our results could be affected.
Propan-1-ol
Enthalpy
1g of water heated through 1ºC needs 4.2 Joules of heat
100g of water heated through 1ºC needs 4.2 x 100 joules of heat
100g of water heated through 22ºC needs 4.2 x 100 x 21 joules of heat.
200g of water heated through 22ºC needs 4.2 x 200 x 21 joules of heat.
=17.64 KJ
2.23g of Propan-1-ol gives 17.64 KJ of heat
1g of Propan-1-ol gives (17.64 / 2.23) KJ of heat
1 mole of Propan-1-ol gives (17.64 / 2.23 x 70) KJ of heat
=- 553.72 KJ mol-1
Error
The precision of this experiment is quite high as all of the percentage error figures are quite low. The stage that the highest error could occur is the temperature stages as these are the two with the highest percentage errors.
- 2021- - 553.72 = - 1467.28
- 1467.28 / - 2021 = 0.73
0.73 x 100 = 73%
The accuracy of this experiment is low because there are lots of ways in which our results could be affected.
Butan-1-ol
Enthalpy
1g of water heated through 1ºC needs 4.2 Joules of heat
100g of water heated through 1ºC needs 4.2 x 100 joules of heat
100g of water heated through 22ºC needs 4.2 x 100 x 20 joules of heat.
200g of water heated through 22ºC needs 4.2 x 200 x 20 joules of heat.
=16.8 KJ
1.65g of butan-1-ol gives 16.8 KJ of heat
1g of Hexane gives (16.8 / 1.65) KJ of heat
1 mole of Hexane gives (16.8 / 1.65 x 74) KJ of heat
=753.45 KJ mol-1
Error
The precision of this experiment is quite high as all of the percentage error figures are quite low. The stage that the highest error could occur is the temperature stages as these are the two with the highest percentage errors.
- 2676 - -753.45 = - 1922.55
- 1922.55 / - 2676 = 0.34
0.34 x 100 = 34%
The accuracy of this experiment is low because there are lots of ways in which our results could be affected.
The molecular structures of the compounds are as follows:
Ethanol:
H H
| |
H-C-C-O-H
| |
H H
Methanol:
H
|
H-C-O-H
|
H
Propan-1-ol:
H H H
| | |
H - C – C – C – O - H
| | |
H H H
Butan-1-ol:
H H H H
| | | |
H –C –C –C –C –O –H
| | | |
H H H H
The enthalpy of combustion of alcohols, within the same homologous series is affected by the molecular structure of the compound. The numbers of CH2 groups have an effect on the enthalpy of the molecule, because it increases the number of bonds that will need to be broken and reformed within the compound during combustion.
For example, in Methanol, there is a CH3 group and an OH group. These bonds, when formed are what determines the enthalpy change of combustion of the alcohol. When Methanol is burned in oxygen the thermo chemical equation is:
CH3OH + 2/3 O2 →→→ Co2 + 2H2O
This means that when Methanol is burned the following bonds need to be broken.
C –H C –H C –H C –O O –H
These bonds all absorb energy when they are broken and therefore endothermic giving a positive value.
The enthalpy of the bond breaking al ads up to:
413 x 3 = 1239 KJ
358 x 1 = 358 KJ
464 x 1 = 464 KJ Total: + 2061 KJ
When bonds are formed in a compound the enthalpies will be negative because they release energy into the surroundings.
The bonds that will be formed in the CO2 and H2O are:
C – O C – O O – H O – H O – H O – H
358 x 2 = - 716
464 x 4 = - 1856 Total: - 1140
Therefore the total bond enthalpy in the combustion of methanol = 921
