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Comparing the enthalpy changes of combustion of different alcohols

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Introduction

AS Chemistry Coursework Comparing the enthalpy changes of combustion of different alcohols Apparatus: * Spirit burners containing: o Methanol o Ethanol o Propan-1-ol o Butan-1-ol * Copper can * 0-110?C thermometer * 100cm3 measuring cylinder * Balance * 5 heatproof mats * Clamp and clamp stand Method * Set up apparatus as shown, clamping the copper can 10cm above the spirit burner, all on top of a heatproof mat. * Carefully measure 200cm3 of water and fill copper can. * Record mass of water (1g=1cm3, ? 200g=200cm3). * Use thermometer to measure water temperature, record temperature in a results table. * Weight spirit burner with lid on, record mass in grams. * Arrange 3 heatproof mats around burner and stand, covering back and both sides. * Put goggles on. * Take lid off spirit burner, light burner, place 5th heatproof mat in front off burner and begin stirring water. * When water temperature rises to near 20?C put lid on burner, extinguishing flame. * Keep stirring water, record the highest temperature. * Weight spirit burner, record weight in grams and calculate mass of fuel used. * Calculate rise in temperature. * Rinse out copper can with cold water and clean can if soot appears on bottom before carrying out experiment with another fuel. ...read more.

Middle

200g of water heated through 18.5 �C needs 840 x 18.5 = 15,540 = 15.54KJ If 2.41g of Ethanol was burnt: 2.41g of Ethanol gives 15.54KJ of heat. ? 1g of Ethanol gives (15.54/2.41) KJ of heat O=16 H=1 C=12 Ethanol=CH3CH2OH 1 Mole=46 ? 1 mole of Ethanol gives (15.54/2.41 x 46) KJ of heat =296.60 KJ of heat So CH3 CH2OH + 3O2 ==> 2CO2 + 3H2O (Hc = -296.60 KJ mol-1 Propan-1-ol 1 gram of water heated through 1�C needs 4.2 Joules of heat energy. ? 200g of water heated through 1�C needs 4.2 x 200 = 840J ? 200g of water heated through 18.0 �C needs 840 x 18.0 = 15,120 = 15.12KJ If 2.36g of Propan-1-ol was burnt: 2.36g of Propan-1-ol gives 15.12KJ of heat. ? 1g of Propan-1-ol gives (15.12/2.36) KJ of heat O=16 H=1 C=12 Propan-1-ol =CH3CH2CH2OH 1 Mole=60 ? 1 mole of Propan-1-ol gives (15.12/2.36 x 60) KJ of heat =384.41KJ of heat So CH3CH2CH2OH + 5O2 ==> 3CO2 + 4H2O (Hc = -384.41 KJ mol-1 Butan-1-ol 1 gram of water heated through 1�C needs 4.2 Joules of heat energy. ? 200g of water heated through 1�C needs 4.2 x 200 = 840J ? 200g of water heated through 18.0 �C needs 840 x 18.0 = 15,120 = 15.12KJ If 1.70g of Butan-1-ol was burnt: 1.70g of Butan-1-ol gives 15.12KJ of heat. ...read more.

Conclusion

Because of this my results will be very precise, but compared to data book results, inaccurate. Evaluation of procedures Heat loss to air and heat loss to apparatus are huge sources of procedural error because if any heat is lost, it means fuel was burnt, but I have no way of measuring energy, I had to ensure as much heat as possible went to the water because that was the only thing I was measuring. Incomplete combustion and use of different burners are also main sources of error because they are hard to control and can easily give unfair tests. I reduced these errors by effective shielding using insulating heatproof mats underneath and all around experiment, these helped contain the heat and ensure as much heat as possible went to the water. I washed the container out with clean cold water in between experiments to help keep it a fair test. When the container got a layer of soot on the bottom, due to incomplete combustion I cleaned it off before the next experiment. I used identical spirit burners every time. I would of used an enriched oxygen atmosphere if the facilities were available; I used the best methods of heat loss prevention available. The least precise pieces of equipment should be more precise, and then measurements and results will be more accurate, for example using a burette for measuring water volume accurately and a more accurate thermometer. Then my percentage error will decrease given more precise results. Alex Graff ...read more.

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