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Combustion Of Alcohols Essay Research Paper Planning

Combustion Of Alcohols Essay, Research Paper Planning In this investigation I will be burning alcohols to heat up a beaker of water. I will be burning five alcohols, methanol, ethanol, propanol, butanol and pentanol. The aim is to find out how much energy is produced when burning these alcohols. ‘An alcohol is a series of organic homologous compounds, with the general formula Cn H n + 1OH´.

Combustion Of Alcohols Essay, Research Paper

Planning In this investigation I will be burning alcohols to heat up a beaker of water. I will be burning five alcohols, methanol, ethanol, propanol, butanol and pentanol. The aim is to find out how much energy is produced when burning these alcohols. ‘An alcohol is a series of organic homologous compounds, with the general formula Cn H n + 1OH´. Alcohols react with oxygen in the air to form water and carbon dioxide. The reaction that is involved in burning alcohols is exothermic because heat is given out. Form this reason the reactant energy is higher than that of the product.

The energy is given out when forming the bonds between the new water and carbon dioxide molecules. The amount of energy produced by such exothermic reactions can be calculated by using the formula Mass of the substance x rise in temp x SHC( specific heat capacity). The specific heat capacity is the number of joules required to heat one gram of water by 1ºC. I chose to use water because it is safe, easily found, and has a reliable specific heat capacity of 4.2.

The bond that are formed in an exothermic reaction can be of two types. The first could be ionic, where a metal is produced. Ionic bonding involves electrons transferring from one atom to the other consequently leaving an electrostatic force between them. The other form of bonding is covalent where atoms share electrons to complete their outer orbit. An example being Methane where four hydrogen atoms each share an electron with a carbon atom.

The method that I will use is as follows… ·Measure 100cm of water in to a glass beaker

·Place the beaker into the grasp of the clamp stand.

·Record the starting temperature of the water

·Weigh the spirit burner with the lid on.

·Put the chosen alcohol burner under the beaker allowing the flame to just touch the beaker.

·Leave to heat up until the temperature of the water is exactly 50ºC more than the original temperature.

·Weigh the spirit burner

·Record all results

The variables that must remain constant throughout the experiment are… ·Mass of the water 100cm

·Type of beaker, glass

·Temperature rise of 50ºC.

·Surrounding temperature of around 23ºC

·The height of the beaker from the wick

·Same set of scales

·Weigh the spirit burner with the lid on.

The variable that must be changed is…

·The type of alcohol used

The formulae of the alcohols that I will be using are…

·Methanol CH OH

·Ethanol C H OH

·Propanol C H OH

·Butanol C H OH

·Pentanol C H OHPrediction I predict that the more bonds there are holding the carbon, oxygen and hydrogen atoms together, more energy will be required to break them apart. For example Ethanol has the formula C H OH. In this formula you have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To separate these types of bonds you require a certain amount of energy which I will show in a table..

TYPE OF BONDENERGY REQUIRED TO BREAK THE BOND(j)

C-H410

C-O360

O-H510

O=O496

C=O740

C-C350

To separate C-H bond you need to apply 410 joules of energy. There are five such bonds in ethanol so you multiply 410 by five to get 2050 joules. You do these calculations for all the other types of bonds that make up ethanol, add them all together and you get 3270 joules. All of the other alcohols can be broken up in this way. Below is a table showing the energy required to break up the bonds in each alcohol. Type of alcoholEnergy required to break the bonds in the alcohol (j)

Methanol2100

Ethanol3270

Propanol3740

Butanol4560

Pentanol5380

As you can see a longer molecule takes more energy to break it´s bonds, in this case Pentanol. Compared to a smaller molecule, methanol which requires less energy to do so. I can come to predict that the longer the molecular structure in the alcohol the more energy it will take to remove the bonds. So when I come to predicting results I can safely say that pentanol will evolve more energy than methanol simply because it has more bonds to break. Type of alcoholBefore exp 1After exp 1Before exp 2After exp 2Before exp 3After exp 3Average temperature difference(ºC)Average increase in weight (g)

MethanolTemp(ºC)23732373237350

Weight(g)153.12152.12162.84159.56156.35154.09-2.18

EthanolTemp(ºC)26762373217150

Weight(g)169.78166.76169.21163.56176.00172.91-3.92

PropanolTemp(ºC)23732474237350

Weight(g)191.56189.21168.11157.00189.01187.32-5.05

ButanolTemp(ºC)23732171247450

Weight(g)159.76156.7198.1284.32187.56181.58-7.61

PentanolTemp(ºC)24742272237350

Weight(g)164.85157.23167.94155.38161.83153.19-9.61

Analysis

As you can see in the obtaining section I have recorded all the information from the results that I need to complete the calculations… ·Energy evolved

·Energy per gram

·Energy per mole First of all I will calculate the amount of energy evolved. I will achieve this by using this simple formula…

Energy evolved = Mass x Rise in temperature x SHC Energy evolved = 100g x 50ºC x 4.2

As you can see all of the alcohols will have the same amount energy evolved because all the numbers that are filled in the formula remain constant for each alcohol and the same numbers are applied for each individual alcohol. Below is a table showing the amount of energy evolved in each case…

Type of alcoholEnergy evolved in (j)Energy evolved in (kj)

Methanol21 00021

Ethanol21 00021

Propanol21 00021

Butanol21 00021

Pentanol21 00021

To find out how much energy is produced per gram we use the formula…

Energy per gram of fuel = Energy evolved x Mass of fuel burnt Energy per gram of fuel = 21kj x ? Below is a table showing how much energy is produced per gram when burning the alcohols in question…

Type of alcoholEnergy per gram (kj)

Methanol9.63

Ethanol5.36

Propanol4.16

Butanol2.76

Pentanol2.19

As you can see the energy per gram decreases as the length of the molecule increases. This is because more fuel is burnt so there is more of it to be filled with the energy. This is shown in graph 1.

To find out how much energy is produced per mole you have to use this formula.. Energy per mole = Energy per gram x Formula mass

Here is a list comprising all the formula masses of the chosen alcohols…

·Methanol 32g

·Ethanol 46g

·Propanol 60g

·Butanol 74g

·Pentanol 88g

Below is a table showing the energy produced per mole…

Type of alcoholEnergy per mole(kj/mol)

Methanol308.16

Ethanol245.56

Propanol249.60

Butanol206.46

Pentanol192.72

Again the results decrease as the molecule length increases suggesting that during the experiment more energy was last in the longer alcohols, this is shown in graph 2.

As you can see, graph 1 has a nice smooth curve suggesting that the results are reliable. But graph 2 has an anomalous result (x), this does not allow the curve to follow a smooth pattern. I think this is because the energy per gram in ethanol and propanol was only slightly different and when multiplied with the formula mass, propanol comes out larger. My prediction was not clear enough because to vary the amount of heat evolved, time would have to come into it. But I did predict that the energy per gram would decrease as the molecular length increases. I think this because the alcohols with more carbon atoms in like pentanol burnt more fuel so there would be less energy per gram because more fuel has been burnt. The reason why more fuel has been burnt is because of the large number of carbon atoms and large molecular length, hence the surface area is large allowing more energy to be released.

EVALUATION

I believe my results were very inaccurate. Below is a list of reasons why it was an inaccurate experiment. 1.Energy given off through sound and light.

2.Heat conducted and convected away through the air.

3.Radiation of heat out into the atmosphere.

4.The fact that the beaker gets hot.

5.The rubber clamp transferred heat way.

6.Heat may be take away through guts of wind.

7.The fact that at higher temperatures, heat is lost faster to the air and out of the beaker, due to the bigger heat difference, making the higher temperatures more inaccurate, and making a shallower gradient on the graph.

8.By incomplete combustion

9.The amount of energy you give the alcohol originally.

10.The availability of alcohol for the wick to burn, if not enough then the wick would burn not the alcohol which would give an inaccurate result.

11.Evaporation of water so there will be less water to heat, making the water hotter.

12.The size of the wick.

13.Not all of the water was the same temperature.

14.The flame size changed due to the type of alcohol, hence it was a different distance away from the beaker each time. Numbers 1 – 10 would decrease the reading and Numbers 11 – 14 would increase the reading. The equipment that I used in this experiment was very inaccurate because heat is a bad way of transferring energy without any loss of it. Any molecule will conduct heat, radiation happens and can be reduced but not completely halts. I feel that the most limiting factor of the experiment is the convection of air and to a lesser extent, of water. Also during the experiment, some of the water will have evaporated, thus the water mass/temp reading will be altered. I feel that this experiment could have been improved by using a wider range of alcohols such as hexanol and heptanol. This would give a better graph reading and a wider range of results to support a firm conclusion. On the other hand, if I had started below room temperature, so that the amount of energy gained, from room temperature might equal the energy lost at temperatures higher than room temperature. Next time reducing heat lost would be my main priority. Improving insulation techniques would be a valuable asset in obtaining the most reliable data I could. Another error is that of incomplete combustion. Complete combustion occurs if there are lots of oxygen atoms available when the fuel burns, then you get carbon dioxide (carbons atoms bond with two oxygen atoms). If there is a limited supply of oxygen then you get carbon monoxide (each carbon atom can only bond with one oxygen atom). This is when incomplete combustion has occurred. This is so because the carbon monoxide could react some more to make carbon dioxide. If the oxygen supply is very limited then you get some atoms of carbon released before they can bond with any oxygen atoms. This is what we call soot. Since heat is given out when bonds form, less energy is given out by incomplete combustion. So this is why it affects the outcome of the experiment. To overcome this problem, I would have to make sure a sufficient supply of oxygen was involved in the reaction.

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