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

Investigate and compare the amount of energy released during the combustion of alcohols with practical and theoretical combustions.

Extracts from this document...

Introduction

Chemistry Coursework Objective My aim is to investigate and compare the amount of energy released during the combustion of alcohols with practical and theoretical combustions. Combustion (burning) is the reaction of substance with oxygen. I aimed to use four different alcohols, these being: * Ethanol * Butanol * Proponol * Methanol Method I set up the apparatus as the diagram shows: To start the experiment I weighed the empty burner using electronic scales, and then added 75ml of the alcohol I was using, and then weighed it again. I then calculated the mass of alcohol in the burner. I then took the temperature of the water, and then lit the burner. I decided to keep the experiment running each time until the water had risen 200c. when this had happened I blew out the burner, and re-weighed it, so that I could calculate how much alcohol had been burned. At this point I used the formula: Energy = Mass of Water x Temperature Rise x Specific heat Capacity (g) ...read more.

Middle

+ (496 x6) (743 x8) (463 x10) O-C x1 360 ---------- ---------- ------------ O-H x1 463 2976 5944 4630 ------------ 5575 8551 10574 8551-10574 = -2023 Practical Start Temperature End Temperature Start Mass End Mass Mass Used 23 75 182.09 177.86 4.23 50g x 520 x 4.2 191035.461 ------------------- x 74 = --------------- = 191.035461 4.23 1000 Proponol C3H7OH + 4.5O2 3CO2 + 4H2O H H H O=O C=O H-O-H | | | O=O || H-O-H H - C - C - C - C - O - H + O=O C + H-O-H | | | O=O || H-O-H H H H (O=O%2) C=O || C || C=O || C Theoretical H - C x7 (412 x 7) O=O x (4.5) C=O x6 H - O x8 C - C x2 (348 x2) + (496 x 4.5 (743 x6) + (463 x8) O - C x2 360 --------------- ---------- ------------- O - H x1 463 2232 4458 3704 ---------------- 4403 6635 8162 6635 - 8162 = -1527 Practical Start Temperature End Temperature Start Mass End Mass Mass Used 25 45 184.9 184.0 0.9 21 41 ...read more.

Conclusion

The energy lost during the combustion has a)turned to water or steam and evaporated into the air. For example with Ethanol, when this is burned in air or oxygen the products are carbon dioxide and water. The water will evaporate into the air - this is where some of the energy goes as well as b)being taken up by breaking of bonds during the burning of the substance. When bonds are formed energy is released. If bonds that hold particles are to be broken or changed then energy has to be supplied to enable this to happen. The change is the difference between the energy liberated to form the bonds and that required to break the bonds. The experiment also proves that Exothermic process is followed because energy is liberated during the reaction proven by the end masses of the substances being lighter than the start masses. The quantities of energy produced depends upon the quantity of the chemical burnt as shown by the results. I.e. the higher the level of mass burnt or combusted the higher the level of energy released. ?? ?? ?? ?? Page: 1 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE 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 GCSE Organic Chemistry essays

  1. Marked by a teacher

    Experiment to investigate the heat of combustion of alcohols.

    4 star(s)

    C2H5 OH + 3O2 2CO2 + 3H2O Bond Breaking Bond Making 5 (C - H) = 5 x 412 4 (C = O) = 4 x 805 1 (C - O) = 336 6 (O - H) = 6 x 464 1 (O - H)

  2. Investigating the Combustion of Alcohols

    The theoretical value according to the Spreadsheet data and the Nuffield Data Book for the standard enthalpy change of combustion of methanol is -726 kJmol-1. Clearly, this value is much greater than that obtained in this experiment. This difference is most probably due to the large heat losses that occurred in this experiment.

  1. The energy produced of different alcohols.

    x 35.5 = 22365J Odd Result C4H9OH Butanol = 150 x 4.2 x 43.5 = 27405J C5H11OH Hexanol = 150 x 4.2 x 44.5 = 28035J Conclusion Looking at my results on investigating into the combustion of a range of alcohols, I noticed a number of patterns.

  2. An experiment to investigate the factors that determine the amount of energy released when ...

    Keeping any of the variables constant is relatively simple, although variables such as oxygen supply will change very slightly. Varying any of the variables is also relatively simple except again, the oxygen supply will pose a problem. I have chosen to vary the number of carbon atoms in an alcohol.

  1. Investigating the energy released from burning different alcohols.

    The alcohol burner is inside the heat shield, under the copper can. As hot air rises, this will heat the copper can through conduction, an in turn the copper will heat the water conduction. The copper will also absorb heat through radiation as the bottom of the can has been

  2. Comparing the enthalpy changes of combustion of different alcohols.

    the way I was carrying out this replicate was different to all the others (experimental technique). I cannot account for the anomalous results for pentan-1-ol. There is a chance that I mis-read the bottle and did another fuel e.g. picked up propan-1-ol instead of pentan-1-ol or used pentan-2-ol instead of Pentan-1-ol.

  1. Molar Heat of Combustion of Alcohols

    92.44 94.41 66.23 Final mass of Ethanol (g) 92.02 93.96 65.82 Difference in mass (g) 0.42 0.45 0.41 Mass of Water (g)* Specific Heat Capacity of Water (4.2)* Temp Change (OC) = Energy Supplied (j) Energy Supplied/ Amount of Alcohol Burnt= Energy Supplied for 1g Energy Supplied By One Gram* Weight of One Mole of Alcohol= Molar Heat

  2. GCSE Chemistry Revision Notes - everything!

    The outer electron is shielded from the full attraction of the nucleus by all the inner electrons. In every single atom in the element of Group 1, the outer electron will feel an overall attractive force 1+ from the nucleus, but the effect of the force falls very quickly as the distance increases.

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