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# To compare the given values of the molar heat of combustion

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Introduction

AIM: The aims of my experiment are to compare the given values of the molar heat of combustion with my experimental values, and to predict the theoretical value of heptanol. PREDICTION: I predict that my experimental values will be lower than the given values. This is because heat will be lost in unnecessarily obtaining the results. This will be by sound and light energy given off, incomplete combustion (complete combustion occurs when there are lots of oxygen atoms available when the fuel burns, and then you get carbon dioxide, as a carbon atom bonds with two oxygen atoms; a limited supply of oxygen results in carbon monoxide being produced), and also conduction, convection and radiation of heat through the air, and draughts speeding the process up. Water will be evaporated, meaning that there is a lower volume of water, causing the remaining water to absorb more heat energy, which will affect the results. To combat this, I will stir the water with the mercury thermometer immediately after the experiment, thus spreading the heat energy around the water. The beaker will also be heated, diverting some of the heat energy away from the water. ...read more.

Middle

ETHANOL PROPANOL BUTANOL PENTANOL HEXANOL START TEMP 19.4 18.7 19.5 19.5 19.3 END TEMP 102 101 85 78 71 TEMP. DIFFERENCE (*c) 82.6 81.3 65.5 58.5 50.7 WEIGHT LOSS (g) 4.2 3.65 3.32 2.98 2.67 My results seem to indicate that the temperature difference decreases as the number of carbons increase, and the weight loss increases as the number of carbons increase. To find the energy of the alcohol, you must multiply the mass of water by the heat capacity by the temperature change. Energy = mass of water x heat capacity x temp change Energy = 100 x 4.2 x 82.6 Energy = 420 x 82.6 Energy = 34962 To find the energy per gram of the alcohol, you must divide the energy by the weight loss in the experiment. 34962 = 8324.29 (2 d.p.) 4.2 Then, to finds the molar heat of combustion, you have to multiply the energy per gram of the alcohol by the relative molar mass (RMM) of the alcohol. This varies, as a carbon atom weighs 12 grams, a hydrogen atom weighs one gram, and an oxygen atom weighs 16 grams. ...read more.

Conclusion

Also the calorimeter is too far away to absorb heat energy, diverting some of the heat energy away from the water. CONCLUSION: Here are the given values, along with my experimental values: Alcohol Exp. values (KJ mol -1 ) Given values (KJ mol -1 ) Ethanol -382.92 -1367.3 Propanol -561.30 -2021.0 Butanol -613.17 -2675.6 Pentanol -725.56 -3328.7 Hexanol -829.43 -3983.8 As I have explained, the values rise at a constant rate, so for heptanol, I predict that if the experiment was done to the same standard as the experiment that produced the given results, then the molar heat of combustion of heptanol would be roughly 654 less than -3983.8 KJ mol -1, as that seems to be the difference each time between two consecutive alcohols, which would be -4628 KJ mol . However, if I had the option of experimenting with heptanol (time and equipment prevented m my method from being reliable enough), then I predict that the value would be on the line of best fit, which would make it roughly -940 KJ mol -1. Alcohol Exp. values (KJ mol ) Given values (KJ mol ) Ethanol -382.92 -1367.3 Propanol -561.30 -2021.0 Butanol -613.17 -2675.6 Pentanol -725.56 -3328.7 Hexanol -829.43 -3983.8 Heptanol -940 -4628 ...read more.

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