I predict that the larger the molecule the more bonds there are to make and break. More energy has been used to break bonds; more energy would also be given out when making the bonds of the products of the reaction. It is possible to suggest that the amount of energy given off by an alcohol with a larger carbon chain (more carbon atoms) will be larger than an alcohol with a smaller carbon chain, for example, Butanol has twice as many carbon atoms as Ethanol, therefore there should be twice as much as energy given out by Butanol than Ethanol.
Method:
- A clamp-stand with a heat proof mat underneath it.
- 100cm
- ³ of water was measured out by using a measuring cylinder and then poured it into a copper calorimeter.
- The copper calorimeter was then placed into the clamp, which was on the clamp stand.
- The spirit burner of the chosen alcohol was weighed on a digital scale and was recorded in grams.
- The spirit burner was placed directly underneath the calorimeter (which was held by the clamp)
- The distance between the wick and calorimeter was 4.5cm
- A thermometer was placed into the water and the initial, final and difference between the two temperatures was recorded. (The rise in temperature will always be 15˚C)
- The wick was set alight by using a lighted splint.
- When the temperature on the thermometer reached the target temperature (15˚C rise above initial temperature) the flame was blown out and the spirit burner was reweighed again. Then the difference between the mass before and after the alcohol was burnt was calculated and recorded.
- The experiment was then repeated thee times for each of the five alcohols.
Diagram of apparatus:
When considering accuracy, measurements will be taken from a suitable level of accuracy. This will include appropriate apparatus:
- Volumes will be calculated to the nearest cm³ using the measuring cylinder.
- Masses will be recorded to 2 d.p.using the digital scales.
- Heights being measured to the nearest millimetre with a ruler.
- Temperatures being measured to the nearest degree Celsius using a thermometer.
- 3 Repeats for each alcohol, and an average to reduce error. Also any results that are anomalous will not be entered into the average.
Safety is important and should be considered. Goggles should be worn at all times, ties should be tucked into shirts, bags should be cleared underneath desks and as alcohol is flammable and toxic any spills must be immediately cleaned up safely.
After the results have been acquired I will calculate the amount of energy released per mole, to do this I will calculate the amount of energy released using this formula:
Q=mcΔt
Energy released (ΔH) = Mass x Specific Heat Capacity x Change in temperature
In this experiment for each alcohol the mass of the water is 100g, the specific heat capacity of water is 4.2 and the difference in temperature is 15˚C.
To calculate the amount of energy released per mole, the calculation of the number of moles in the reaction of each alcohol will be needed.
The formula to calculate the number of moles is:
n= m/RMM
Number of moles = mass of alcohol used / Relative atomic mass
To get the energy released in kilojoules per mole, I will have to divide the result by 1000.
Theoretical Calculations:
Methanol
Equation:
Structure:
Bonds Breaking (+)
3 × C-H: 3 × 412 = 1236
1 × C-O: 1 × 358 = 358
3/2 × O=O: 3/2 × 498 = 747
1 × O-H: 1 × 464 = 464
Total: + 4473 kJ/mole
Bonds Making (-)
2 × C=O: 2 × 805 = 1610
4 × O-H: 4 × 464 = 1856
Total: - 3466 kJ/mole
ΔH = + 3269 – 3466
= - 1007 kJ/mole
Ethanol
Equation:
Structure:
Bonds Breaking (+)
5 × C-H: 5 × 412 = 2060
1 × C-C: 1 × 347 = 347
1 × C-O: 1 × 358 = 358
1 × O-H: 1 × 464 = 464
3 × O=O: 3 × 498 = 1494
Total: + 4723 kJ/mole
Bonds Making (-)
4 × O=C: 4 × 805 = 3220
6 × O-H: 6 × 464 = 2784
Total: - 6004 kJ/mole
ΔH = + 4723 - 6004
= - 1281 kJ/mole
Propanol
Equation:
Structure:
Bonds Breaking (+)
7 × C-H: 7 × 412 = 2884
2 × C-C: 2 × 347 = 694
1 × C-O: 1 × 358 = 358
1 × O-H: 1 × 464 = 464
4.5 × O=O: 4.5 × 498 = 2241
Total: + 6641 kJ/mole
Bonds Making (-)
6 × O=C: 6 × 805 = 4830
8 × O-H: 8 × 464 = 3712
Total: - 8542 kJ/mole
ΔH = + 6641 - 8542
= - 1901 kJ/mole
Butanol
Equation:
Structure:
Bonds Breaking (+)
9 × C-H: 9 × 412 = 3708
3 × C-C: 3 × 347 = 1041
1 × C-O: 1 × 358 = 358
1 × O-H: 1 × 464 = 464
6 × O=O: 6 × 498 = 2988
Total: + 8559 kJ/mole
Bonds Making (-)
8 × O- -C: 8 × 805 = 6440
10 × O-H: 10 × 464 = 4640
Total: - 11080 kJ/mole
ΔH = + 8559 - 11080
= - 2521 kJ/mole
Pentanol
Equation:
Structure:
Bonds Breaking (+)
11 × C-H: 11 × 412 = 4532
4 × C-C: 4 × 347 = 1388
1 × C-O: 1 × 358 = 358
1 × O-H: 1 × 464 = 464
7.5 × O=O: 7.5 × 498 = 3735
Total: + 10477 kJ/mole
Bonds Making (-)
10 × O=C: 10 × 805 = 8050
12 × O-H: 12 × 464 = 5568
Total: - 13618 kJ/mole
ΔH = + 10477 - 13618
= - 3141 kJ/mole
CH4O (l) + 1.5O2 (g) CO2 (g) + 2H2O (l)
C2H6O (l) + 3O2 (g) 2CO2 (g) + 3H2O (l)
C3H8O (l) + 4.5O2 (g) 3CO2 (g) + 4H2O (l)
C4H10O (l) + 6O2 (g) 4CO2 (g) + 5H2O (l)
C5H12O (l) + 7.5O2 (g) 5CO2 (g) + 6H2O (l)
When I group all the equations together a pattern emerges: Every time n increases by 1.
In the alcohol molecule, H increases by 2n. the number of; O2 molecules increases by 1.5, CO2 increases by 1 and H2O molecules increase by 1. therefore the general formula of this equation is :
CnH2n+1OH + 1.5nO2 nCO2 + (n+1)H2O
I have already discussed what I will do after the experiment, and now I will be doing that. I will work out the energy released per mole, also known as enthalpy change (ΔH). I have just worked out the enthalpy change for each alcohol I will be using, by scientific methods. Therefore I predict the energies released per mole for the different alcohols to be similar to the results shown in the table below. Pentanol will give off more heat than methanol and I predict that less pentanol will need to react to raise the temperature of the water than methanol. Therefore the higher the number of carbons, the lower the amount of alcohol used up, which increases the level of enthalpy change per mole. The enthalpy changes by the same amount every time the number of carbon atoms goes up by 1. Therefore the number of carbon atoms should be proportional to the energy released per mole.
The larger the molecule the more energy needed to break more bonds, but more energy is released when more new bonds are formed. Overall the theoretical calculations shows that the enthalpy change in KJ/mol increases with the increase to the number of carbon atoms present in the alcohol.
Obtaining evidence:
Methanol
Q= ΔH = 100 × 4.2 × 15
ΔH1= 6300 for 1.28g of methanol burned.
= 6300 ÷ 1.28
= 4921.87 J/g
ΔH2= 4.921 kJ/g
n= 4.921 × 32
ΔH3 = -157.72 kJ/mol of heat released
Ethanol
Q= ΔH = 100 × 4.2 × 15
ΔH1= 6300 for 1.21g of ethanol burned.
= 6300 ÷ 1.21
= 5206.61 J/g
ΔH2= 5.206 kJ/g
n= 5.206 × 46
ΔH3 = -239.50 kJ/mol of heat released
Propanol
Q= ΔH = 100 × 4.2 × 15
ΔH1= 6300 for 0.82g of propanol burned.
= 6300 ÷ 0.82
= 7682.92 J/g
ΔH2= 7.682 kJ/g
n= 7.682 × 60
ΔH3 = -460.91 kJ/mol of heat released
Butanol
Q= ΔH = 100 × 4.2 × 15
ΔH1= 6300 for 0.76g of butanol burned.
= 6300 ÷ 0.76
= 8289.47 J/g
ΔH2= 8.289 kJ/g
n= 8.289 × 74
ΔH3 = -613.38 kJ/mol of heat released
Pentanol
Q= ΔH = 100 × 4.2 × 15
ΔH1= 6300 for 0.70g of pentanol burned.
= 6300 ÷ 0.70
= 9000 J/g
ΔH2= 9.000 kJ/g
n= 9.000 × 74
ΔH3 = -792 kJ/mol of heat released
Analysis:
I have already discussed that I will plot a graph containing my results with the number of carbons on the x-axis and the enthalpy change on the y-axis. After the results have been plotted I will add a trendline with the correlation coefficient and equation. The correlation coefficient (C on graph) measure the depth of the correlation of my results. The strength of the correlation depends how close it is to one.
There are three graphs;
- Predicted graph of results
- graph of actual results
- graph with predicted and actual results on one graph for comparison
The separate graphs (1 & 2) will allow observation with greater detail. Graph 3 will allow a comparison of the two gradients of the best fit line of the predicted and actual results.
The anomalies found were not included in the averages to reduce error.
In my prediction I concluded that the enthalpy change increased by the same amount each time the carbon number increased by one. Therefore the enthalpy change is proportional to the carbon number. My actual results in terms of heat energy released per mole was far less than the predicted results. Although the results do appear to fit the prediction. The number of carbon atoms are proportional to the enthalpy change, it is directly proportional as the best fit line goes through the origin. The correlation is strong as it has a coefficient of 1 and the best fit line is straight.
In the prediction I stated that the fewer the number of carbon atoms in an alcohol, the less energy required to break the bonds of the reactants and then the less energy given off when forming the products. This is backed up by my results and shown in the graph. The results show that the larger the molecule the more energy needed to break more bonds, but more energy is released when more new bonds are formed. Overall this means and is shown in the theoretical calculations, that the enthalpy change(KJ/mol) increases as the number of carbon atoms present in the alcohol increases.
All of these combustion reactions have been exothermic; energy in the form of heat has been given off.
The final results of the calculations are negative that indicates the reaction is exothermic just as predicted in the planning section.
Evaluation:
Overall the experiment went very well producing good results and a strong correlation. This was caused by the accuracy of the apparatus, and the repeats that were carried out for each alcohol. Any anomalies were excluded from the average to reduce error and increase reliability of results. Although I have two anomalies in two separate alcohols which means I only have two valid results for the alcohols Ethanol (1.71g change) and Methanol (0.78g change). These may have occurred as there were many possible chances for error in the equipment, such as evaporation of the alcohols before weighing them. A draught also kept blowing the flame and disrupted the burning of alcohols,(the strength of the draught was different for each alcohol) the size of the wick on the spirit burner varied, and the distance between the calorimeter and the flame varied for each alcohol and when the calorimeter had black soot on the bottom, limits the water being heated. The thermometer was stirred but not constantly so different places in the water may have been hotter than others. These small inaccuracies are caused by slight differences in the values of the fixed variables, like the mass of water not being exactly 100g, due human error.
The main source of error is the heat loss from the calorimeters and all the equipment used. There was also incomplete combustion as the flame was orange and soot was formed. The alcohols did not burn efficiently as of incomplete combustion and my results may have been affected. If these errors had been excluded my results would have been far more accurate. The distance between the spirit burner and the calorimeter did not change as I made sure the distance was always approximately 4.5cm. However, the length of the wick did vary sometimes, a centimetre at most; this did affect the results though.
To improve the experiment the wicks should all be the same length to make sure of a fair test. The distance between the calorimeter and the flame should be constant at 2cm for example. Draught shields (excluders) could be put up to reduce draught and insulate the heat loss escaping into the environment. The experiment could take place in a fume cupboard to minimize the heat loss. But the only realistic method could be the draught excluders. Further work could include using more alcohols and more repeats of them this would give clear evidence and increase the reliability of the results.
I would say that the evidence is strong enough to draw conclusions from, because three repeats were just enough to maintain a strong average. My graph shows that the numbers of carbon atoms are directly proportional to the enthalpy change. This shows that the more carbon atoms the higher the level of enthalpy change. Overall my experiment was quite reliable considering all the sources of error. My prediction was not close to the results, the graph displayed the line of best fit was close to all the points so there was not a wider scatter of results. Except for my anomalies all the repeats were very close. This indicates that the results are reliable. Although the anomalies suggest they were not entirely reliable.
If this experiment was to be done again, then all the possible sources of error mentioned would have to be counteracted and controlled, as well as using a wider range alcohols. I would also take more repeats so that a much more accurate average could be taken and the results would be more accurate and reliable. I suggest 6 or 7 repeats for each alcohol. More results would make a far conclusive and reliable and accurate. More repeats would show that the trend line stays the same and that the enthalpy change is directly proportional to the number of carbons atoms present.