Step 2:
R.M.M. of CH3OH = 12 + 1 + 1 + 1 + 16 + 1 = 32g per mole.
If 3.60g of methanol produce 16720J of energy, then 32g of methanol produces:
16720 J ÷ 3.60g = 4644.4
4644.4 x 32g = 148622.2 J or 148.6 kJ
Step 3:
CH3OH + 1½O2 = CO2 + 2H2O
3 x H – C
1 x O – H
1 x O – C
1½ x O = O
3 x 435 = 1305
1 x 464 = 464
1 x 358 = 358
1½ x 497 = 745.5
Total = 2872.5
2872.5 – 3462 = -589.5 kJ
Step 4:
148.6kJ ÷ 589.5kJ = 0.25
0.25 x 100 = 25%
This means that 75% of the energy produced from the exothermic reactions is lost to the atmosphere etc… and is not used to heat up the water inside the conical flask. To do this for the methanol and the other alcohols I will need to repeat steps 1 – 4 in my results to see if there is any difference in efficiency between different alcohols.
- I can work out the equations for each alcohol using the formula:
C (n) H (2n + 1) OH
Where n = number of carbons.
I.e.: -
Methanol = CH3OH
Ethanol = C2H5OH
Propan-1-ol = C3H7OH
Butan-1-ol = C4H9OH
Pentan-1-ol = C5H11OH
Prediction
I can predict with the research above that the longer/ bigger the hydrocarbon is, the more energy it will produce using a smaller amount of mass. This will also means that it will raise the temperature quicker than a smaller hydrocarbon. This is because when a bond is made, it releases out energy to the atmosphere. So if there are more bonds to make, then there will be more energy released.
I carried out a preliminary experiment to give me an idea of what I will happen and makes sure there are no major flaws in the investigation. I used methanol as the alcohol to be burnt, 100g of water to be heated and to raise the temperature of the water 40°C. Here were my results:
My results show it took 3.6g of methanol to raise the temperature of the water 40°C. The results from the experiment seem reliable with a good accuracy of measuring so now I can write out my plan and carry out the investigation using my different ranges (Different types of Alcohol).
Plan
Apparatus:-
- Conical Flask
- Spirit Burner with wick
- Alcohols (Methanol, Ethanol, Propan-1-ol, Butan-1-ol, Pentan-1-ol)
- Clamp
- Stand
- Thermometer
- Water (100cm³)
- Scales
- Measuring Cylinder
Method: -
Set up the apparatus as shown, and then get a spirit burner with a fuel in it, e.g.: - Methanol. Weigh the spirit burner with the fuel, wick and the cap of the spirit burner to 2 decimal places and record this in the result table. Fill a conical flask with 100cm³ of water and clamp to the stand. Measure the height from the bottom of the conical flask to the bench and adjust it so it is 10cm above the bench. Take the spirit burner and remove the cap. Then place it directly underneath the conical flask. Place a thermometer in the conical flask and measure the temperature of the water. Record this in the results table. Add 50°C to this temperature to work out how much the temperature of the water show up heated up to. Then light the wick in the spirit burner and gently stir the water in the conical flask to keep a constant temperature. When the water’s temperature has risen 50°C, put the cap on the spirit burner to put the flame out. Then weigh the spirit burner again with wick and cap to 2 decimal places again and record this in the results table. Clean the conical flask of any soot that might have built up on the bottom and let the water cool down back to room temperature. Then repeat the experiment for the other alcohols and the repeat the investigation to get an average result for each alcohol. If there are any obvious anomalous results, then repeat it for a third time. Once all the results have been recorded, work out the amount of mass of the alcohol that has been used. Then work out the average mass of the alcohol that has been used for the repeated results.
Fair Test: -
To make it a fair test you must carry out these precautions: -
- Make sure that the temperature is raised to exactly 50°C of the original temperature.
- The measuring of the amount of water, temperature of the water and the weights of the spirit burners are accurate.
- The height between the bottom of the flask and the bench is the same
- The same amount of wick is exposed outside the spirit burner.
- No soot is left on the bottom of the conical flask after each result.
Safety/Hazards: -
To make it a fair test you must carry out these precautions: -
- Wear Safety Goggles at all times.
- Bags and Stalls are put on the sides of the room, out of the way of any experiment.
- The normal Science Labs rules must be obeyed.
- Be careful with harmful substances and open flames, as alcohol is very flammable.
- Watch out for glass objects, which might get broken.
- The spirit burner is placed in a metal dish.
Results
Experiments:
* = Anomalous Results.
Glancing at these results, I can tell that less mass is being used to raise the temperature of the water by an alcohol with a bigger hydrocarbon structure. This must mean that a bigger hydrocarbon structure gives out more energy per gram of mass, as these results look reliable with only one anomalous result throughout the whole investigation.
Efficiency:
Methanol
Step 1:
g x Jg-1K-1 x K = J
100 x 4.18 x 50 = 20900 Joules
Step 2:
R.M.M. of CH3OH = 12 + 1 + 1 + 1 + 16 + 1 = 32g per mole.
If 3.70g of methanol produce 20900 J of energy, then 32g of methanol produces:
20900 J ÷ 3.70g = 5648.65
5648.65 x 32g = 180756.76 J or 180.8 kJ
Step 3:
CH3OH + 1½O2 = CO2 + 2H2O
3 x H – C
1 x O – H
1 x O – C
1½ x O = O
3 x 435 = 1305
1 x 464 = 464
1 x 358 = 358
1½ x 497 = 745.5
Total = 2872.5
2872.5 – 3462 = -589.5 kJ
Step 4:
180.8 kJ ÷ 589.5 kJ = 0.31
0.31 x 100 = 31%
Ethanol:
Step 1:
g x Jg-1K-1 x K = J
100 x 4.18 x 50 = 20900 Joules
Step 2:
R.M.M. of C2H5OH = 12 +12 + 1 + 1 + 1 + 1 + 1 +16 + 1 = 46g per mole.
If 2.53g of methanol produce 20900 J of energy, then 46g of methanol produces:
20900 J ÷ 2.53g = 8260.87
8260.87 x 46g = 380000 J or 380.0 kJ
Step 3:
C2H5OH + 3O2 = 2CO2 + 3H2O
5 x H – C
1 x C - C
1 x O – H
1 x O – C
3 x O = O
5 x 435 = 2175
1 x 347 = 347
1 x 464 = 464
1 x 358 = 358
3 x 497 = 1491
Total = 4835
4835 – 5996 = -1161 kJ
Step 4:
380.0 kJ ÷ 1161 kJ = 0.33
0.33 x 100 = 33%
Propan-1-ol
Step 1:
g x Jg-1K-1 x K = J
100 x 4.18 x 50 = 20900 Joules
Step 2:
R.M.M. of C3H7OH = 12 +12 + 12 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +16 + 1 = 60g per mole.
If 2.53g of methanol produce 20900 J of energy, then 60g of methanol produces:
20900 J ÷ 2.33g = 8969.96
8969.96 x 60g = 538197.6 J or 538.2 kJ
Step 3:
C3H7OH + 4½O2 = 3CO2 + 4H2O
7 x H – C
2 x C - C
1 x O – H
1 x O – C
4½ x O = O
7 x 435 = 3045
2 x 347 = 694
1 x 464 = 464
1 x 358 = 358
4½ x 497 = 2236.5
Total = 6797.5
6797.5 – 8530 = -1732.5 kJ
Step 4:
538.2 kJ ÷ 1732.5 kJ = 0.31
0.31 x 100 = 31%
Butan-1-ol
Step 1:
g x Jg-1K-1 x K = J
100 x 4.18 x 50 = 20900 Joules
Step 2:
R.M.M. of C4H9OH = 12 +12 + 12 +12 + 1 + 1+ 1 + 1 + 1 + 1 + 1 + 1 + 1 +16 + 1 = 74g per mole.
If 2.53g of methanol produce 20900 J of energy, then 74g of methanol produces:
20900 J ÷ 2.18g = 9587.16
9587.16 x 74g = 709449.84 J or 709.4 kJ
Step 3:
C4H9OH + 6O2 = 4CO2 + 5H2O
9 x H – C
3 x C - C
1 x O – H
1 x O – C
6 x O = O
9 x 435 = 3915
3 x 347 = 1041
1 x 464 = 464
1 x 358 = 358
6 x 497 = 2982
Total = 8760
8760 – 11064 = -2304 kJ
Step 4:
709.4 kJ ÷ 2304 kJ = 0.31
0.31 x 100 = 31%
Pentan-1-ol
Step 1:
g x Jg-1K-1 x K = J
100 x 4.18 x 50 = 20900 Joules
Step 2:
R.M.M. of C5H11OH = 12 +12 + 12 + 12 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +16 + 1 = 88g per mole.
If 2.53g of methanol produce 20900 J of energy, then 88g of methanol produces:
20900 J ÷ 2.01g = 10398.01
10398.01 x 88g = 915024.88 J or 915.0 kJ
Step 3:
C5H11OH + 7½O2 = 5CO2 + 6H2O
11 x H – C
4 x C - C
1 x O – H
1 x O – C
7½ x O = O
11 x 435 = 4785
4 x 347 = 1388
1 x 464 = 464
1 x 358 = 358
7½ x 497 = 3727.5
Total = 10722.5
10722.5 – 13598 = -2875.5 kJ
Step 4:
915.0 kJ ÷ 2875.5 kJ = 0.32
0.32 x 100 = 32%
Looking at these results, the efficiency seems to stay the same so this is obviously not affected by changing the alcohol. Maybe if I was to repeat the investigation using a different independent factor then this might be different.
Conclusion
The graph shows a weak inverse proportionality between the Mass used and RMM. This means that as the RMM increases, the mass used decreases. The line of best fit shows that as the hydrocarbon structures increased in size, the smaller amount of alcohol that is used to produce same amount of energy (heat energy). This means that Pentan-1-ol used the smallest amount of mass to produce the same amount of heat energy to raise the temperature of the water 50°C, so as shown in Step 3, the bigger/ longer the hydrocarbon structure is the more energy that will be produced per mole. This is because as the hydrocarbon chains increase, the more covalent bonds there are to be broken during the combustion of alcohol. This means more energy is release from the combustion than a smaller hydrocarbon would give out with the same amount of mass used. This supports that my prediction is correct as I predicted, with the research and results from the preliminary experiment, that the longer/ bigger the hydrocarbon, the more energy is produced using less mass than a smaller hydrocarbon chain. However, I cannot completely backup my prediction as when I worked out the efficiency results, about 70% of the energy was not used to raise the temperature of the water 50°C. The energy was lost to the surrounding atmosphere, heating the air so this could have affected my results, but I need to think of a new experiment in my evaluation to stop this from happening and completely backup my prediction. I can draw a theoretical graph against the results, using the RMM of each alcohol and the amount of energy produced to heat 100cm³ of water 50°C with 100% efficiency for the theoretical values, to show what really should happen without lose of energy to the atmosphere.
The theoretical line of best fit shows a strong positive correlation. This means that when the mass is increased by 14g, an extra 571.5kJ. But the Results line of best fit has a weaker positive correlation, which is much lower than the theoretical line. This means the results produce a lot less heat per g to heat up the water 50°C, than the 100% efficient transfer of energy with the theoretical line. This is also shown with the steepness of the lines, as the theoretical line is much steeper than the results line, so the results line produces less energy to heat the water per gram.
Evaluation
In this investigation I believe that I have gone deep enough to get some justified results with good accuracy (2 decimal places for mass) and with this I have produced evidence to support my conclusion. We used 6 different alcohols to experiment with and repeated the results twice. We then noticed that there was an anomalous result so we repeated that experiment for the third time, so we could disregarder this result. This result might have occurred as another independent factor might have changed during the experiment. I.e.: - there might have been a bit more or less than 100g of water in the conical flask or we raised the temperature of the water just over or below 50°C for human error. None of the alcohols were changed in the experiment so they were the same concentration throughout the experiment. We also used the same measuring apparatus, so the accuracy stayed the same throughout the investigation. I don’t believe that this anomalous result is part of a new trend either in this investigation but if I was to change a different independent variable that might not be the case.
I believe our results were accurate enough for this investigation to provide reliable, usable results for my conclusion and analysis. All my results seem to follow the pattern shown on the graph and this has helped me to produce a good conclusion. However, as I said in the conclusion, there is a lot of energy being produced that is escaping from the experiment and is instead heating the air. To stop this from happening, if I was to carry out this investigation again, I could use calorimeter. This traps the heat energy so it has to heat up the water and not the surrounding air and it would make the results more accurate. By doing this as well I can expand my investigation and investigate bigger hydrocarbon structures like hexan-1-ol and see if there is any difference or something else to notice from the expanded results.
I got my research from the Chemistry GCSE textbook, revision books, past preliminary work and class work notes in my exercise book.
