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To investigate which fuel gives out the most energy when burnt. We are burning ethanol, methanol, propanol, butanol and pentanol

Extracts from this document...

Introduction

SHEHAN DE-SILVA

AMDG5th July

Chemistry Coursework

Aim:

To investigate which fuel gives out the most energy when burnt. We are burning ethanol, methanol, propanol, butanol and pentanol.

Prediction:

My initial prediction is that as the amount of carbon molecules in the fuel increases, the more energy it will give out when burnt. Therefore I am predicting that “pentanol” will give out more energy as “methanol” because it has more carbon molecules.When the fuel is burnt it reacts with the oxygen in the air to form the products water and carbon dioxide. This is true for all the fuels I will be using. This is because the amount reactant energy is more than the product energy; therefore some energy has been given out in the form of heat. The energy is given out when forming the bonds between the new water and carbon dioxide molecules.

There are two possible results of this experiment, one is endothermic, this is when energy is absorbed from the surroundings and is used to break the covalent bonds between the fuel molecules. This will cause temperatures of the fuel to fall.

The second reaction is exothermic,this is when energy is transferred to the surroundings and is used to form bonds between the fuel molecules. This will cause temperatures of the fuel to rise.

The following calculations will show the amount energy, I predict, will be transferred into the water. I will then compare these results to my actual results.

...read more.

Middle

165.77

164.77

28

39

Ethanol

168.97

168.23

29

39

Propanol

173.22

172.46

24

39

Butanol

172.26

171.76

24

36

Pentanol

175.22

174.68

24

36

 Repeat:

Experiment 2

Alcohol:

Starting Mass: (g)

Finishing Mass: (g)

Starting Temperature: Degrees Celsius

Finishing

Temperature: Degrees Celsius

Methanol

164.77

163.81

39

50

Ethanol

168.23

167.46

38

50

Propanol

172.46

171.69

39

50

Butanol

171.76

171.25

36

48

Pentanol

174.68

174.14

36

47

From these results I have calculated the following:

  • Methanol

This alcohol released 147.8KJ/MOL in the first experiment, and 127.4KJ/MOL in the second experiment.

The average of these results and my final result is: 137.6KJ/MOL

  • Ethanol

This alcohol released 287.13KJ/MOLin the first experiment, and 276.65KJ/MOLin the second experiment.

The average of these results and my final result is: 281.81KJ/MOL

  • Propanol

This alcohol released 497.24KJ/MOLin the first experiment, and 360.09KJ/MOLin the second experiment.

The average of these results and my final result is: 428.67KJ/MOL

  • Butanol

This alcohol released 745.6KJ/MOLin the first experiment, and 705.7KJ/MOL in the second experiment.

The average of these results and my final result is: 725.65KJ/MOL

  • Pentanol

This alcohol released 822.19KJ/MOLin the first experiment, and 781.86KJ/MOLin the second experiment.

The average of these results and my final result is: 801.88KJ/MOL

In order to get the above result I had to carry out the following calculations:

Energy per mole = Energy per gram x Formula mass

I have used the above formula for all the results in order to see how much energy is produced from each experiment. This can be found on the following pages, which explains my calculations as to how much energy is released per mole.

1. Methanol-weight                                   _        Temperature

Start-                172.22                                           Start- 24°C

End-                174.68                                        End-        36°C

Burnt-                0.54                                           Rise-   12°C

Methanol (RMM) = CH3OH= 32                   Energy released= 11 x 100 x 4.2

Moles = 1 ÷ 32 = 0.03125                                       = 4620

Energy/mole = energy/moles = 4620÷0.03125

                                         = 147.8KJ/MOL

2. Methanol (repeat)

...read more.

Conclusion

  • A wider range of alcohols, such has hexanol and heptanol, this will give me a wider range of results, and this will help further improve my experiment.
  • I could use alkenes or alkanes to see whether they mimic the alcohols.
  • I could see which metals released the most energy when burnt
  • I could investigate which alcohol evaporates the most of it contents in the fastest time.

Although overall my experiment went well, there were a few aspects I could not control:

  • We were not able to control the amount of energy lost to the surroundings, resulting in lots of energy being lost. Heat was lost to the air through conduction and convection.
  • In order to solve this problem, We could use a device called a bomb calorimeter. This piece of equipment is designed to avoid heat loss and prevent evaporation. The results will then be more precise and eradicate this problem. A diagram below describing the set up of how we would use the bomb calorimeter.

image00.jpg

  • Through out the experiment, incomplete combustion was taking place. Evidence of this was left on the bottom of the copper can as soot. This meant that there was not enough oxygen, which in turn produced carbon mononxide. This is a problem because some potential energy was lost because the carbon monoxide could still react to make more carbon dioxide. To solve this problem I would have to make sure and provide a sufficient amount of oxygen for the reaction to complete.

...read more.

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