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To investigate the relationship between the structure and heat provided by combustion of a range of alcohols.

Extracts from this document...

Introduction

COURSE WORK PIECE 1:

INVESTIGATION OF THE COMBUSTION OF ALCOHOLS

HARRY HUDSON 10 U                                                                                                                     25/10/02

AIM: To investigate the relationship between the structure and heat provided by combustion of a range of alcohols.

INTRODUCTION:

In this investigation I will be burning a range of alcohol in a method known as calorimetry. This will allow me to see the amount of energy produced by each alcohol, and then look at the structure of the alcohol and investigate why an amount of energy is produced for each alcohol. Before I go on with the experiment there are several factors that must be first understood.

What is an alcohol?

The definition of an alcohol as taken from Richard Harwood’s Chemistry textbook is “a series of organic compounds containing the functional group – OH.” The – OH group, called a hydroxyl group is what defines the compound as an alcohol. The alcohol compounds are very similar to the alkanes, however alcohols contain one oxygen atom, creating the hydroxyl group, and making the alkane an alkanol (alcohol). As the hydroxyl group makes the compound different to an alkane, the hydroxyl group is seen to be “functional”. The formula for alcohol is:

In this investigation I will look at the first five alcohols. These are methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol. The classification of alcohols is similar to the classification of alkanes, where the name refers to the number of carbon atoms i.e. ‘meth’- one carbon atom, ‘eth-’=two carbon atoms, ‘prop-’=3 carbon atoms, ‘but-’=four carbon atoms, ‘pent-’=five carbon atoms and so on. The carbon atom that the hydroxyl group is bonded to may classify the alcohol further.

...read more.

Middle

1.640

2

19

39

20

250

141.030

139.180

1.850

3

19

39

20

250

139.180

137.340

1.840

1

Ethanol

19

39

20

250

103.430

101.940

1.490

2

19

39

20

250

105.020

103.430

1.590

3

19

39

20

250

106.520

105.020

1.500

1

Propan-1-ol

19

39

20

250

219.140

218.310

0.830

2

19

39

20

250

218.310

217.460

0.850

3

19

39

20

250

218.310

217.460

0.850

1

Butan-1-ol

18

38

20

250

234.420

233.450

0.970

2

18

38

20

250

235.190

234.420

0.770

3

18

38

20

250

236.060

235.190

0.870

1

Pentan-1-ol

20

40

20

250

130.450

129.800

0.650

2

20

40

20

250

129.800

129.130

0.670

3

20

40

20

250

129.130

128.450

0.680

Anomalies

At this stage it is quite hard to identify anomalies, as the results are not yet calculated in a way that allow me to compare them to my hypothesis well. We do see a fluctuation in the results, however this can be expected due to dependant variables not controlled properly.

DATA ANALYSIS:

Now that we have got the data required for the investigation it is important to make accurate and relevant calculations in order to represent it well. Below I have demonstrated the equation (as stated in the method) used to calculate this data as an example. For this example I am using the first experiment conducted, which was for methanol.

  • First I will find the heat transferred (∆ H (j)) during this reaction.

∆ H (j)  = mass of water (g) X specific heat capacity of water (J/g) (S.H.C) X temperature rise (°C)

mass of water = 250g

S.H.C of water = 4.2 j/g

Temperature rise = 20°C

These figures are put into the equation to give

250 X 4.2 X 20 = ∆ H (j)

=21000 J

(These figures are all constants, and so the energy released will be the same for each experiment)

  • Now I need to find the amount of energy per gram of alcohol burnt. This significance of this is that we are taking the amount of energy released overall, and looking at it in relation to the mass of the alcohol combusted.

∆ H per gram of alcohol burnt (j/g) = ∆ H (°C) / change in mass (g)

∆ H = 21000 kJ

Change in mass = 1.640 g

These figures are put into the equation to give

21000/1.640 = ∆ H per gram of alcohol burnt (j/g)

= 12.805j/g

  • This figure must now be calculated in relation to the actual structure of the atom, so as to support our investigation. The current figure is useless by itself to use, as the number of methanol atoms per gram is different to the number of propanol atoms per gram. Therefore in relation to one another these results mean very little. So, we must find the energy released per mole of alcohol. This is found using the RMM of the alcohol

∆ H per gram of alcohol burnt per mole (j/mole) = ∆ H per gram of alcohol burnt (g) X RMM of alcohol

∆ H per gram of alcohol burnt = 12.805j/g

RMM of alcohol = 32

These figures are put into the equation to give

12.805 X 32 = ∆ H per gram of alcohol burnt per mole (j/mole)

                     = 409.756 j/mole

These calculations are made for each experiment using the equation functions of Microsoft Excel for speed and accuracy reasons. These calculations may be seen on the following table (over page):

Some features of this table should be noted:

  • Results are given to three decimal places, as this gives us enough accuracy without using figures unnecessarily long.
  • Averages are also given, not to better accuracy, eliminate anomalies.
  • The results generally fluctuate, however the differences that are large enough to be classed as anomalous are highlighted in black.

DATA ANALYSIS:

To look at the accuracy of our data, we compare our own results to the ‘theoretical results’, which we may calculate using bond energy values. This is a system, looking at the structure of an alcohol molecule and the number and different types of bonds it has. We know the energy values of these bonds. Therefore we may calculate the energy released for each reaction. The process of doing this is as follows:

  1. Write out a balanced formula equation for the combustion of the alcohol tested.
  2. Using a table with bond types and energy values, look at the balanced equation and the structure of the molecule to record how many of each bond there is. Do this for both bonds broken and bonds made.
  3. Calculate the energy for the different bonds and add these up to give total energy used to make bonds and total energy released breaking bonds.
  4. Take away the total energy used to make bonds from the total energy released breaking bonds, to give the total energy for the reaction.
  5. To allow comparison to our own results calculate this figure to one mole of the alcohol.

The bond energy calculations for each alcohol may be seen below. These values were calculated using the equation function on Microsoft Excel.

Methanol

Balanced chemical equation: 2CH3OH + 302 → 2CO2 + 4H2O

Bond Type

Bond Energy Value

Bonds Broken

Energy In (J)

Bonds Made

Energy Out (J)

O = O

498

3

1494

0

0

C - H

413

6

2478

0

0

C - C

347

0

0

0

0

H - O

464

2

928

8

3712

C = 0

805

0

0

4

3220

C - 0

358

2

716

0

0

Total Energy In

5616

Total Energy Out

6932

Energy total (kJ): 5616 – 6932 = - 1316

Energy total per mole of alcohol (kJ/mole): -1316/2= - 658

Ethanol

Balanced chemical equation: C2H5OH 2CO2 + 4H2O

Bond Type

Bond Energy Value

Bonds Broken

Energy In (J)

Bonds Made

Energy Out (J)

O = O

498

3

1494

0

0

C - H

413

5

2065

0

0

C - C

347

1

347

0

0

H - O

464

1

464

6

2784

C = 0

805

0

0

4

3220

C - 0

358

1

358

0

0

Total Energy In

4728

Total Energy Out

6004

...read more.

Conclusion

These were controlled quite well, as we see only minor fluctuations in the results. However the difference between the theoretical results and these results suggests that our method was not greatly efficient. The main areas that were ineffective in providing accurate results were:

  • Although draft excluders minimized cold drafts entering the experiment area, it did not stop heat energy being transferred to the surroundings.
  • For the water to be heated it was necessary for the can to be heated first, thus energy is “wasted” in heating the can.
  • The methods of measuring may not always have been totally accurate. For example it was noticed on a few occasions, that although the burner had been blown out, the temperature continued to rise. Therefore it has taken time for the energy transferred to the water to be measured on the thermometer, and thus a smaller temperature recording of 20˚C has been used in a calculation with a mass loss that is relative to a greater temperature.

If able to repeat this investigation with better equipment it would be possible to improve the method so as to provide more accurate results. Some improvements might be:

  • Electronic measuring equipment used to increase accuracy by eliminating human error.
  • The energy taken to heat the can accounted for accurately in the calculations.
  • The same type of burners used each time, so that the type of flame produced by each alcohol is similar.
  • Somehow controlling the heat conditions around the experiment so that the surrounding temperature is higher, and thus less heat is transferred to the surroundings.
  • Then use of a bomb calorimeter measures the energy transfer to a greater accuracy.
  • Supply oxygen directly to prevent incomplete combustion.

Further investigation could include continuing to test different alcohols in the series to see if this trend continues. Also to look at the combustion of other fuels could aid investigating the relationship between complexity and energy produced.

EVALUATE….

Incomplete combus

Taking off blackness

Wax on base

...read more.

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