• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Salters AS Chemistry Assessed Practical - Enthalpy of Combustion

Extracts from this document...

Introduction

Salters AS Chemistry Assessed Practical Activity DF 1.3: Enthalpy of Combustion For my assessed practical I will be comparing the enthalpy changes of different alcohols. The enthalpy change of combustion of a fuel is the measure of energy transferred when one mole of fuel burns completely. To find this change I will be using the fuel to heat up some water and I know that to heat up 1g by 1�C it takes 4.2 J of energy. Requirements * A small copper can or other metal container to act as a calorimeter (base approximately 10cm in diameter). * 0-110�C thermometer * 100cm3 measuring cylinder * Spirit burners containing: -Methanol -CH3OH -Ethanol -CH3CH2OH -Propan-1-ol- CH3CH(OH)CH3 -Butan-1-ol -CH3CH2CH2CH2OH * A measuring balance * Bunsen burner * Draught shielding * Clamp Method 1. Set up the apparatus as shown in the diagram above; the calorimeter should be placed into the clamp and the draught shielding should be placed either side of this so that when the spirit burner is alight the flame isn't interfered with. Heatproof mats may be used as draught shielding. 2. Using the measuring cylinder, pour 200cm3 of cold water into the calorimeter and use the thermometer to record its temperature. 3. Place the spirit burner with the selected alcohol contained within underneath the calorimeter that is being held by the clamp. Make sure that the draught shield is suitable as energy loss has to be reduced as much as possible. 4. Weigh the spirit burner on the weighing balance with the lid on it so that the fuel doesn't evaporate and an inaccurate reading is recorded. ...read more.

Middle

12600J Enthalpy change of combustion = 12600J � 0.015g/mol-1= 840000J Conversion into KJ= -840 KJ Butan-1-ol Formula of alcohol= CH3(CH2)2CH2OH Mass of 1 mole of alcohol= 74g Number of moles of alcohol used= 0.91g � 74g = 0.012g/mol-1 Energy transferred by 1 mole of alcohol= 12600J � 0.012g/mol-1= 1050000J Conversion into KJ= -1050 KJ When looking at my results, I can see that when the molecular formula mass of the alcohol is higher, the larger the enthalpy change of combustion is. This is because more energy is needed to break the bonds in the alcohol and the enthalpy change of combustion is always negative as a combustion reaction is always exothermic. When looking at the molecular formula mass of methanol, there is only one carbon atom present and the enthalpy change of combustion is -213KJ mol-1. However if we compare this to Butan-1-ol, in its molecular structure there are 4 carbon atoms which means that there are more bonds to break (the bonds that join the carbon atoms to the hydrogen atoms), therefore its enthalpy change of combustion is -1050KJ mol-1 as more energy is needed. You can see from the diagram below that as the number of carbon atoms increase, the number of hydrogen also increases. When looking at my results, as the number of carbon atoms present in the alcohol increases, the higher the enthalpy change of combustion is. Methanol has the lowest enthalpy change as it has a two carbon structure- it doesn't need as much energy to break down these bonds. Ethanol's enthalpy change is higher than Methanol's as there is an extra carbon atom in Ethanol's molecular structure, meaning that more energy is needed to break down this extra bond. ...read more.

Conclusion

Care had to be taken when recording measurements and readings from the thermometer and from the measuring cylinder also. Relative significance of aspects of practical procedures and uncertainties associated with measurements. When completing my experiment there were various limitations however I have produced results that are as accurate as possible and they follow a trend: the more carbon atoms there are, the more atoms of hydrogen are joined to them meaning that it takes more energy for these bonds to be broken- the enthalpy change of combustion is greater. When looking at the percentage errors of the experiment all of them add up to make the total experiment error. Any experiment is not going to be perfect as there are always some uncertainty associated with measurement etc however my percentage errors are so small (around 0.002%- 0.0026%) that it doesn't contribute that greatly to my results. The greatest error was the heat loss from the experiment to the surroundings as the experiment was exothermic and therefore produced negative enthalpy changes of combustion. This is where most of the energy was lost. To prevent this heat loss we could have used a bomb calorimeter that is specifically designed to reduce heat loss to the environment but the equipment that was available to use was an ordinary calorimeter and we insulated it with heatproof mats. Overall, I feel that my experiment was a success as I obtained a set of results that follow the same trend and I have no anomalous results. A risk assessment was carried out before completing the experiment to ensure that people's safety was a priority and the experiment was carried out in the most accurate way as possible. Laura Bailey 12JG Chemistry 9th October 2005 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Organic Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Organic Chemistry essays

  1. Marked by a teacher

    Preparation of propanone from propan-2-ol

    5 star(s)

    There was a loss of product during the extractions. First, when we distilled out the propanone from the reaction mixture, some propanone might be vapoured and condense and stuck to the wall of the pear-shaped flask, the thermometer and other apparatus it passed through. So, the product obtained was slightly reduced.

  2. Marked by a teacher

    Comparing the enthalpy changes of combustion of different alcohols

    3 star(s)

    While analysing the results, all the units of the experiment were close and the average mass and temperature were almost right and accurate except one or two of them. There are two types of error which can be occurred in this experiment: 1.

  1. Find the enthalpy change of combustion of a number of alcohol's' so that you ...

    Final weight of spirit burner + lid + alcohol (g) Overall mass loss of fuel (g) = initial- final weight of spirit burner + lid + alcohol Initial temperature of the water (oC) Final temperature of the water (oC) Overall temperature change (oC)

  2. Determination of the formula of Hydrated Iron (II) Sulphate crystals

    and would neither significantly affect the accuracy of the results. Therefore, when deciding on which method is more accurate, just because method one has a lower measurement percentage error does not mean that it is more accurate as it's procedural uncertainties may be much more significant than method two's.

  1. The Relationship Between The Number of Carbon Atoms In An Alcohol And Its Standard ...

    0.77 16.38 0.006637931 -2467.63636364 Octan-1-ol C8H17OH 130 300 ml 19 35 16 198.18 197.42 0.76 20.16 0.005846154 -3448.42105263 *Note: Due to the fact that there are two results for some of the experiments as shown above. Having two different results in a graph for the same alcohol would prove difficult.

  2. The aim of this experiment is to investigate the enthalpy change of combustion for ...

    There was also no stirring device to make sure that the heat was equally spread throughout the water easily. The alcohol also spread out which increased its surface area, which may have also had an effect on the results.

  1. Experiment Hypothesis: The energy released by an alcohol increases as the number of carbon ...

    = 775272.73J = - 775Kjmol-1 Butanol Q = mcât = 50 * 4.18 * 22.4 = 4681.6J 0.35g = 4681.6J 74g = 74 * 4681.6 0.35 = 989824 J = 989.824 = - 990Kjmol-1 Pentanol Q =mcât = 50 * 4.18 * 20.2 = 4221.8J but 0.28g = 4221.8J 88g

  2. Evaluate the most suitable vegetable oil for conversion into a biodiesel fuel.

    Additionally it requires significantly less energy to break the double bonds, which will result in a higher chance of error in the conversion process. Other options such as Linseed and Sunflower oil would also be unsuitable as they too contain a large percentage of polyunsaturated acids.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work