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Experiment to investigate the heat of combustion of alcohols Essay

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Introduction

The heat of combustion of alcohols is the change in kJ/mol when 1 mole of the alcohol is burnt in excess oxygen (O2). I will be investigating 6 alcohols, using predictions and a practical to guide me through this experiment and form an overall conclusion.

Using formulas and calculations, I can show how much energy is released from these substances, and work out their heat of combustion.

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From the homologous series of alcohols using the general formula of CnH2n+1OH, I have chosen 6 alcohols, giving the Mr for calculation purposes later on:

Mr

* Methanol – CH3OH 32

* Ethanol – C2H5OH 46

* Propanol – C3H7OH 60

* Butanol – C4H9OH 74

* Pentanol – C5H11OH 88

* Hexanol – C6H13OH 102

The reactants will produce the products of carbon dioxide and water (CO2):

Alcohol + Oxygen Carbon Dioxide + Water

I.E: C2 H5 OH + 3O2 2CO2 + 3H2O

Apparatus List/Safety Precautions.

* Tin can – For holding the water

* 200 cm3 water – being heated by burning the alcohol.

* Thermometer – measuring temperature accurately.

* Heat-Proof Mat – keeping heat loss to minimum and safety against fires.

* Spirit Burners – hold the alcohol that is going to be burnt.

* Clamp Stand – hold the tin can in place.

* Measuring Cylinders – Measure the amount of water accurately.

* Mass balance – Accurate weight measuring equipment.

N.B: As you are dealing with fire, it should be noted that you must wear goggles and use the heat-proof mat to put out any flames from using a splint or a match.

Heat of combustion

Burning: This is intense chemical oxidation reaction, which is accompanied by the release of heat and glow. Combustion occurs when there is a combustible substance, an oxidizing agent and a source of ignition. The nature of combustion can be described as a rapidly proceeding oxidation.

Alcohols: derivatives of hydrocarbons, in the molecules of which there is one or several hydroxyl groups OH. All alcohols are divided into monatomic and polyhydric alcohols. Monohydric alcohols are alcohols that have one hydroxyl group. Multi-atomic alcohols are having several OH hydroxyl groups.

The heat of combustion: the amount of heat released during the complete combustion of 1 kg of liquid (solid) or 1 m3 of gaseous fuel under normal conditions. In this case, the source fuel and combustion products must be at the same pressure and temperature.

Alcohol combustion: Like most organic substances, alcohols burn with the formation of carbon dioxide and water:

C2H5-OH + 3O2 -> 2CO2 + 3H2O

When they burn, a lot of heat is released, which is often used in laboratories (laboratory burners). The lower alcohols burn with an almost colorless flame; the flame has a yellowish color in the higher alcohols due to incomplete combustion of carbon.

?6H13OH: hexyls’ alcohol, which has a liquid state and an alcoholic odor.

Methanol: Methanol is a limiting alcohol that has a hydroxyl group in its molecule. CH3OH is colorless toxic flammable liquid, the simplest of Alcohols. It turns out to be artificial either from carbon monoxide and hydrogen, by oxidizing natural gas, or by dry distillation of wood,

Ethanol: CH3CH2OH, the second representative of the homologous series of monohydric alcohols, under standard conditions a volatile, flammable, colorless transparent liquid. Ethanol is one of the most important starting materials in the modern organic synthesis industry.

Fair Test

There are certain factors that could be considered to affect the results in any way, to keep a fair test:

* Room Temperature – A high temperature could heat up the beaker of water, thus pushing up the thermometer reading.

* Heat Loss – Heat is always lost during this reaction. Some of the heat will not be used to heat the water in the beaker, altering the overall result. Therefore heat proof mats are used to keep it in as much as possible.

* Equipment – Changing equipment may alter the result in some way due to flaws in the apparatus, which will alter the result.

* Beaker Material – The tin can must always be used to avoid using both the glass and the tin, changing results as they both conduct heat differently.

* Mass of Water – It must be kept constant to give accurate results.

* Temperature change – Keeping it constant will give more reliable results.

Hypothesis

Within all chemical reactions, bonds are broken (endothermic) and rebuilt, building new bonds (exothermic). There is an overall energy change between the two, bond breaking and bond making.

However in this experiment heat is given out, making the reaction exothermic. Increasing the amount of carbon atoms, the molecule gets larger and so it requires more energy to break apart, as well as making new bonds. Therefore I predict that:

* The longer the molecule of alcohol, the larger the amount kJ/mol is given out by burning the substance.

To back this prediction, I have calculated some theoretical values to show how the amount of energy released increases as the number of carbon atoms rises.

Each alcohol has certain bonds that need to be broken and require specific amounts of energy (joules) to do it:

Bond type Energy Required (J)

C – H 412

O = O 498

O – H 464

C = O 805

C – C 347

C – O 336

This indicates that you will need 412 joules to break a Carbon – Hydrogen bond. I have calculated the energy change for each alcohol, using the bond energies already given above:

Methanol

CH3 OH + 1.5O2 CO2 + 2H2O

Bond Breaking Bond Making

3 (C – H) = 3 x 412 2 (C = O) = 2 x 805

1 (C – O) = 336 4 (O – H) = 4 x 464

1 (O – H) = 464

1.5 (O = O) = 1.5 x 498

= + 2783 Joules = – 3466 Joules

Energy Change = 2783 – 3466 = – 683 kJ/mol.

Ethanol

C2H5 OH + 3O2 2CO2 + 3H2O

Bond Breaking Bond Making

5 (C – H) = 5 x 412 4 (C = O) = 4 x 805

1 (C – O) = 336 6 (O – H) = 6 x 464

1 (O – H) = 464

3 (O = O) = 3 x 498

1 (C – C) = 347

= + 4701 J = – 6004 J

Energy Change = 4701 – 6004 = – 1303 kJ/mol.

Propanol

C3H7 OH + 4.5O2 3CO2 + 4H2O

Bond Breaking Bond Making

7 (C – H) = 7 x 412 6 (C = O) = 6 x 805

1 (C – O) = 336 8 (O – H) = 8 x 464

1 (O – H) = 464

4.5 (O = O) = 4.5 x 498

2 (C – C) = 2 x 347

= + 6619 J = – 8542 J

Energy Change = 6619 – 8542 = – 1923 kJ/mol.

Butanol

C4H9 OH + 6O2 4CO2 + 5H2O

Bond Breaking Bond Making

9 (C – H) = 9 x 412 8 (C = O) = 8 x 805

1 (C – O) = 336 10 (O – H) = 10 x 464

1 (O – H) = 464

6 (O = O) = 6 x 498

3 (C – C) = 3 x 347

= + 8537 J = – 11080 J

Energy Change = 8537 – 11080 = – 2543 kJ/mol.

Pentanol

C5H11 OH + 7.5O2 5CO2 + 6H2O

Bond Breaking Bond Making

11 (C – H) = 11 x 412 10 (C = O) = 10 x 805

1 (C – O) = 336 12 (O – H) = 12 x 464

1 (O – H) = 464

7.5 (O = O) = 7.5 x 498

4 (C – C) = 4 x 347

= + 10455 J = – 13618 J

Energy Change = 10455 – 13618 = – 3163 kJ/mol.

Hexanol

C6H13 OH + 9O2 6CO2 + 7H2O

Bond Breaking Bond Making

13 (C – H) = 13 x 412 12 (C = O) = 12 x 805

1 (C – O) = 336 14 (O – H) = 14 x 464

1 (O – H) = 464

9 (O = O) = 9 x 498

5 (C – C) = 5 x 347

= + 12373 J = – 16156 J

Energy Change = 12373 – 16156 = – 3783 kJ/mol.

Therefore, this graph will back up my prediction. These values will clearly reflects that as the molecule becomes larger, with more bonds to make and break, and the amount of energy released (exothermic) as a pose to it being taken in (endothermic), is much larger as the Carbon atoms increase.

Preliminary Work

Also to back up my prediction further, I have produced results from a similar experiment, I had previously carried out, only with 5 alcohols instead of 6. These results are represented in a graph, shown on the next page. The values here are labelled as negative just to show that the reaction is exothermic. However on the graph, the values are positive to indicate how much energy is given off.

Methanol

Ethanol

Propanol

Butanol

Pentanol

Mass (g) of burner + lid (before)

195.48

214.11

194.80

171.41

146.98

Mass (g) of burner + lid (after)

192.03

212.25

193.25

170.20

146.02

Mass of alcohol burnt (g)

3.45

1.86

1.55

1.21

0.96

Temperature oC change

20

20

20

20

20

This graph indicates that the amount of energy released is proportional to the no. of carbon atoms present in the alcohol.

Obtaining Evidence

Method.

* Setup the apparatus as shown in the diagram shown previously.

* Using a 50 cm3 measuring cylinder, measure accurately 200 cm3 of water and place it into the tin can.

* Place the thermometer in the can and record the starting temperature.

* Weigh the spirit burner (plus the lid) and place it under the can.

* It is important that the lid of the wick is kept on unless it is being burned because of the possibility that some of the mass of alcohol may be lost.

* Light the spirit burner, heating directly under the tin can.

* Watch the thermometer carefully and replace the lid on the burner, stopping the fire from burning, 20o C more than the starting temperature.

* Reweigh the burner (+lid), recording the mass of alcohol burnt by subtracting this result from value recorded before the burning.

* Repeat these steps for each other alcohol and tabulate them as done in the following tables.

* I will repeat this experiment to produce 2 sets of results for more accurate values with more variance.

* I then averaged the mass of alcohol burnt from both sets of results, producing a more general graph.

Results

Set 1

Methanol

Ethanol

Propanol

Butanol

Pentanol

Hexanol

Mass (g) of burner + lid (before)

168.68

160.17

157.00

198.15

192.79

245.35

Mass (g) of burner + lid (after)

166.34

158.61

155.44

197.14

191.05

244.35

Mass of alcohol burnt (g)

2.34

1.56

1.56

1.01

1.74

0.85

Initial Temperature oC

14

20

14

13

13.5

14

Final Temperature oC

34

40

34

33

33.5

34

Temperature oC change

20

20

20

20

20

20

Set 2

Methanol

Ethanol

Propanol

Butanol

Pentanol

Hexanol

Mass (g) of burner + lid (before)

174.91

241.18

188.80

222.48

200.42

266.53

Mass (g) of burner + lid (after)

171.23

238.84

186.93

221.23

198.83

265.55

Mass of alcohol burnt (g)

3.68

1.34

1.87

1.25

1.59

0.98

Initial Temperature oC

17.5

13.5

15

14

13

14

Final Temperature oC

37.5

33.5

35

34

33

34

Temperature oC change

20

20

20

20

20

20

Methanol

Ethanol

Propanol

Butanol

Pentanol

Hexanol

Average Value for alcohol burnt (g)

3.01

1.45

1.715

1.13

1.665

0.915

Calculation

To work out the heat of combustion of the 6 alcohols, I will have to work out the amount of energy involved; the heat change in joules, which is equal for all the alcohols because they all have the same temperature change of 20o C and an equal mass of water (200cm3).

Heat Change in Joules = Mass (g) x Specific Heat Capacity x Temperature Change

= 200 x 4.2 x 20o C

= 16800 Joules.

Heat of Combustion.

This same method is used for each alcohol, obtaining different results but the same principle applies, where the number, X and the Mr will vary with each alcohol:

X g of alcohol produces Y Joules

1 g of alcohol produces Y/X Joules

1 Mole of alcohol produces Y/X x Mr Joules

Finally, to convert the result into kilojoules, divide the answer by 1000, converting the result into kJ/mol.

Methanol

> 3.01g produces 16800 joules

> 1g produces 16800/3.01 joules = 5581.40 Joules.

> 1 mole produces (16800/3.01 joules) x 32(Mr) = 178604.65 Joules/mol

o = 178.60 kJ/mol

Ethanol

> 1.45g produces 16800 joules

> 1g produces 16800/1.45 joules = 11586.21 Joules.

> 1 mole produces (16800/1.45 joules) x 46 = 532965.52 Joules/mol

o = 532.97 kJ/mol

Propanol

> 1.715g produces 16800 joules

> 1g produces 16800/1.715 joules = 9795.92 Joules.

> 1 mole produces (16800/1.715 joules) x 60 = 587755.10 Joules/mol

o = 587.76 kJ/mol

Butanol

> 1.13g produces 16800 joules

> 1g produces 16800/1.13 joules = 14867.26 Joules.

> 1 mole produces (16800/1.13 joules) x 74 = 1100176.99 Joules/mol

o = 1100.18 kJ/mol

Pentanol

> 1.665g produces 16800 joules

> 1g produces 16800/1.665 joules = 100090.10 Joules.

> 1 mole produces (16800/1.665 joules) x 88 = 887927.93 Joules/mol

o = 887 kJ/mol

Hexanol

> 0.915g produces 16800 joules

> 1g produces 16800/0.915 joules = 18360.66 Joules.

> 1 mole produces (16800/0.915 joules) x 102 = 1872786.86 Joules/mol

o = 1872.79 kJ/mol

These averaged values were then plotted on a graph and another graph was plotted, from the calculated bond energies that were present in my planning before.

However there are 2 other graphs that were plotted from the original data before it was averaged. Here is the calculated energy change, which is a little simplified to save time and space.

Set 1

Methanol – CH3OH 32 – 229.74 kJ/mol

Ethanol – C2H5OH 46 – 495.38 kJ/mol

Propanol – C3H7OH 60 – 646.15 kJ/mol

Butanol – C4H9OH 74 – 1230.89 kJ/mol

Pentanol – C5H11OH 88 – 849.66 kJ/mol

Hexanol – C6H13OH 102 – 2016.00 kJ/mol

Set 2

Methanol – CH3OH 32 – 146.09 kJ/mol

Ethanol – C2H5OH 46 – 576.72 kJ/mol

Propanol – C3H7OH 60 – 539.04 kJ/mol

Butanol – C4H9OH 74 – 994.56 kJ/mol

Pentanol – C5H11OH 88 – 929.81 kJ/mol

Hexanol – C6H13OH 102 – 1748.57 kJ/mol

Conclusion

As my prediction had suggested, the relationship between the number of carbon atoms and the heat of combustion is linear, therefore producing a line of best fit, going through the origin for all the graphs. This means that the heat of combustion (kJ/mol) is proportional to the size of the alcohol molecule (number of carbon atoms increases as the alcohol molecule becomes longer).

These graphs had shown that as the chain of atoms became longer with each alcohol, the number of bonds that had to be broken rose and so more energy was required to break them. The same applies when constructing new bonds to form the carbon dioxide and water. With increasing molecule length, there is an increase in

I had also plotted a further graph to compare with my results that I had obtained from a data sheet, informing me of the energy values needed to break certain bonds. This graph was certainly accurate as most of the points were either on the line of bets fit or drifting ever so slightly from it. On the other hand, my line of best fit was overall, lower than the processed data graph. This seemed to happen because of energy loss is various places that I could not help as it could not all be absorbed by the tin can.

Evaluation

All three of my graphs had possessed anomalies of some sort, concentrated in the area of the fourth, fifth and sixth result (Butanol/Pentanol/Hexanol). This may be because I either did not take enough care in obtaining these results, in terms of accurately reading off the thermometer etc., or there was simply “too” much energy that could not be absorbed by the tin can, even though there were heat-proof mats to keep the energy loss to a minimal.

I believe that I had conducted the experiment accurately, and recording enough results to provide a sound conclusion. There weren’t many flaws to my technique as each experiment proved successful.

However there are some areas of inaccuracy during the experiment that would have altered the readings:

1. Energy being lost via sound and light and not heat, which lowers values as some energy has transferred into something else.

2. Conduction of heat from the tin can. The fact that the tin can has absorbed some of the heat already takes away some of the overall heat energy given off by the earlier alcohols.

3. The thermometer bulb was touching the bottom of the tin can, which may have inadvertently gained extra degrees on its reading, not giving the real temperature of the water. This may give rise to parts of the water is at a different temperature to the rest, altering my degree of accuracy at which I read off the thermometer.

4. Heat energy is lost via convection of air around the wick.

5. It can also be said that there was not enough oxygen supply for the alcohol to burn completely, in complete combustion, because of the heat-proof mats restricting the amount of oxygen that could be supplied. Therefore a case of incomplete combustion (carbon monoxide is formed instead of carbon dioxide) would occur and when new bonds are formed, this incompletion will result in the lessening of energy released from the reaction.

6. Each alcohol burns stronger with more carbon atoms, increasing the ferocity of the flame, which affects the height the wick was away from the tin can, providing more energy if the flame is higher or stronger. This would alter the reading by increasing the values.

Other improvements could be made for a more efficient and accurate experiment, such as better equipment to obtain more accurate results. Or by having a larger variety of alcohols, extending your range of results or repeating the experiment more than 4 times to secure extremely good readings, giving a more reliable conclusion.

Overall, this experiment has proven that it is able to record very reliable and accurate results but there an alternate method of measuring the heat of combustion, as the diagram shows below:

This diagram illustrates the use of a calorimeter, a container especially designed for this purpose to obtain very accurate results, restricting heat loss from its shape and size. The burner is filled two-thirds of the way with the alcohol and the starting temperature is recorded.

A small current of air is drawn through the spiral, to provide a steady supply of oxygen to the wick, with the water constantly being stirred by the stirrer. After a rise of a certain temperature (20oC), the maximum temperature is recorded and the burner is reweighed.

Calculating the heat capacity is difficult because there are different materials (copper/water and glass) that have different heat capacities. To get round this obstacle, an electric heater is placed in the water, and then connected to a joulemeter (measures the number of joules supplied to the calorimeter). It is kept on until the apparatus temperature reaches the equivalent of the temperature in the experiment with the alcohol. Here are some sample results:

Starting Temperature = 19.4oC

Final Temperature = 39.4 oC

Mass of Methanol burned = 1.02g

Starting reading on Joulemeter = 7030 J

Final reading on Joulemeter = 29020 J

Difference in Readings = 21990 J

Therefore it takes 21990 J, to increase the temperature of the calorimeter from 19.4oC to 39.4 oC, and that 1.02g of methanol releases 21990 J:

Energy released when 1 mole of Methanol is burned = 21990 x 32g/1.02g = 689, 882 J

Thus the heat of combustion is -689.88 kJ (negative for exothermic reacting), being a lot closer then my experimental result of 178 kJ. This just shows how effective this method is to getting closer to the extremely accurate result.

Cite this Experiment to investigate the heat of combustion of alcohols Essay

Experiment to investigate the heat of combustion of alcohols Essay. (2017, Jul 22). Retrieved from https://graduateway.com/experiment-investigate-heat-combustion-alcohols-121/

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