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Aninvestigationintothecombustionofalcohol

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Introduction

Name: Sarah Galal Sayed Ahmed Elrify An investigation into the combustion of alcohol Aim: I am going to compare the enthalpy change of combustion of alcohol The enthalpy of combustion is a measure of the energy transferred when one mole of the fuel is burnt completely. Example:- CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) Plan: You will first need the following requirements: * Thermometer 0-110oC to measure the temperature of the water and to stir the water with it while it is heating the division will be 1oC * The four different kind of alcohol which will be used as fuel: 1. methanol 2. ethanol 3. propan-1-ol 4. Butan-1-ol The alcohol will be used as the spirit burner that we have to light, where will be one for each alcohol. * 100cm3 measuring cylinder so that we can measure the amount of water needed (range 0-110 cm3 , divisions of 1.0 cm3 ) * 200cm3 of water the division will be 1.0 cm3 * electronic balance so that the spirit burners will be weighed to 2.d.p * goggles for eye protection * small copper can to act as a calorimeter so that we can heat the water in it * lab coat is for clothes protection * draught shielding to prevent my experiment from any draught interfering * spirit burners is where the alcohol would be ...read more.

Middle

alcohol repeat the experiment three times for reliability To find the enthalpy of combustion from the data: You have to find the moles of the alcohol used. This will be the mass of the alcohol divided by its RMM(grams). Find the amount energy released into the water from the burning alcohol: Energy released= SHC x mass of water x temperature change Results:- Alcohol Initial mass (g) Final mass (g) Change of mass (g) Average mass (g) Initial temp oC Final temp oC Change of temp oC Average temperature oC Methanol 168.98 166.38 1.16 2.60 15 35 20 18.5 Methanol 169.36 166.70 1.10 16 33 17 ethanol 161.76 160.60 2.60 1.13 15 31 16 16 ethanol 160.60 159.50 2.60 16 32 16 Propan-1-ol 169.33 168.33 0.93 0.87 16 32 16 16 propan-1-ol 168.48 167.75 0.59 15 31 16 Butan-1-ol 176.50 175.57 0.77 0.76 17 34 17 16 butan-1-ol 164.01 163.42 0.73 16 31 15 Analysis: Now I am going to calculate the amount of energy released in one mole for each alcohol by using the following formulas. The RMM of each alcohol= the number of carbons + the number of hydrogen's in an The moles used= change in mass/ RMM The amount of energy released= SHC x mass of water x change of temperature The amount of energy released in one mole= the amount of energy released / moles Methanol: The formula = CH3OH The structural formula= The RMM of the methanol= 12 + (3x1) ...read more.

Conclusion

Precision error: The error involved here is said to be half a division of the instrumental scale. A balance that measure to 0.01g The error here is +/- 0.005g thus a reading of 272.63g of methanol from my experiment could be 272.631g, 272.632g, 272.633g etc. The percentage error involved in this reading is 0.005 x 100/ 272.63= 1.83x 10-3 % A thermometer that reads to a 1oC reading. The error here is +/- o.5oC Thus a reading of 17oC could be 17.5oC, 17.6oC, 17.7oC etc The percentage error involved in this reading is o.5 x 100/17= 2.94% A measuring cylinder that reads to 1cm3 The error here is 0.5 cm3 Thus reading of 100 cm3 could be 100.5 cm3 , 100.6 cm3 , 100.7 cm3 etc The percentage error involved in this reading is 0.5 x 100/100=0.5% However when reading a change in mass or temperature, the error reading will double. Thus a change in temperature of 20oC the percentage error involved would be 0.5x 100/20= 5% And when measuring the change in mass the alcohol for example 1.00g of the percentage error involved would be 0.005 x 100/1.00= 0.5 x 2= 1% From the precision error you can see that the error of equipment was not a lot. There could be error involved in the experiment which would affect the experiment. ...read more.

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