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Observation - Alcohol No. Of Carbon atoms per Alcohol Initial Mass Final Mass?  

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Observation Alcohol No. Of Carbon atoms per Alcohol Initial Mass Final Mass? Mass (g) Initial Temperature Final Temperature? Temperature ?H/g (kJ/g) ?H/mol (kJ/mol) Methanol 1) 2) 3) Average) 1 243.34g 238.59g 236.46g 238.59g 236.46g 233.94g 4.75g 2.13g 2.52g 17ºC 16ºC 18ºC 52ºC 54ºC 50ºC 34ºC 35ºC 38ºC 3.006 6.901 6.333 96.2 220.8 202.7 211.8 Ethanol 1) 2) 3) Average) 2 234.96g 217.36g 205.17g 232.2g 213.79g 201.02g 2.76g 3.57g 4.15g 16ºC 17ºC 19ºC 51ºC 60ºC 52ºC 35ºC 43ºC 33ºC 5.326 5.059 3.340 245.0 232.7 153.6 238.9 Propanol 1) 2) 3) Average) 3 203.61g 201.35g 242.21g 201.35g 199.54g 240.52g 2.26g 1.81g 1.69g 19ºC 18ºC 23ºC 50ºC 50ºC 54ºC 31ºC 32ºC 31ºC 5.761 7.425 7.704 345.7 445.5 462.2 453.85 Butanol 1) 2) 3) Average) 4 197.7g 196.26g 194.63g 196.26g 194.63g 192.31g 1.44g 1.63g 2.32g 15ºC 19ºC 18ºC 47ºC 51ºC 50ºC 32ºC 32ºC 32ºC 9.333 8.245 5.793 690.6 610.1 428.7 650.35 So as you can see by my results Butanol has the highest enthalpy of combustion.? To calculate the energy produced or lost during these experiments I used the formula: DH= mass of water x 4.2 x Dt = J/g DM This is the formula m x x t where m is the mass of the water and c is the energy needed to heat up 1kg of the substance 1°C. We then divide this by 1000 to get the answer in kJ/mol but this requires me to multiply the answer by the r.m.m of the different alcohols these are: (R.M.M is the relative molecular mass which is the values of all the atoms added up together) Methanol = 32 Ethanol = 46 Propanol = 60 Butanol = 74 ? So as you can see by the two...

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