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Alcohol Heat Combustion Investigation

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Introduction The investigation that is being performed is designed to find out which from a range of alcohols is giving out the most energy when it is burnt. I am using Methanol, Ethanol, Propenol and Butan. I will investigate this by burning the alcohols (separately) under a beaker full of water. In order to collect data I will weigh the alcohol being burned before and after the experiment. The experiment is over when the water in the beaker has been heated for a set time. Using the data collected and the bond energy values for the equation of the burning of each alcohol, I can calculate which contains most energy and precisely how much Ki/mol each gives out. Method The experiment was set up as below, a alcohol (and container) b wick c tripod d beaker e water f gauze g bench mat The experiment was done in 9 stages, 1. Set the equipment up as above. 2. Weigh and record the alcohol. 3. Place the alcohol 10cm underneath the beaker, which will be filled with 200ml of water. 4. Record temperature of water watching carefully until it reaches 40 C. 5. Extinguish the flame. 6. Weigh the alcohol again and recorded. 7. Replace water with cool water after the beaker has returned to room temperature 8. ...read more.


C3H7OH Melting = -126 Boiling = 97 Butanol(Butan-1-ol) C4H9OH Melting = -90 Boiling = 117(The names like propan-1-ol refer to the position of the -OH group on the carbon chain, the OH groups above are on the first Carbon atom, the "1" position)This table shows that I will need to investigate a series of different alcohols in my investigation. Ideally I would need at least 4 alcohols for a good range of results for comparison. The complete combustion of an alcohol involves reaction with Oxygen to produce Carbon Dioxide and Water. The general formula for this reaction is:Cn H 2n+1 OH + (n+n/2)O2 ? nH2O + nCO2Balanced equations for each of the available alcohols that I will use are as follows:Methanol 2CH3OH + 3O2 --- 2CO2 + 4H2OEthanol C2H5OH + 3O2 --- 2CO2 + 3H2OPropanol 2C3H7OH + 9O2 --- 6CO2 + 8H2OButanol C4H9OH + 6O2 --- 4CO2 + 5H2O Tables Results Burning of Methanol CH3OH + 11/2O2 - CO2 + 2H2O O -- H | H -- C -- H + O = O - O = C = O + H -- O -- H | H 2061 + 747.45 - 1610 + 1856 = 2808.45 - 3466 = -657.55Kj Burning of Ethanol C2H5OH + 3O2 - 2CO2 + 3H2O H O -- H | | H--C--C -- H + O = O - O = C = O + H -- O -- ...read more.


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. ?? ?? ?? ?? Combustion of Alcohols Sam Norton ...read more.

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