Alcohol investigation

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Alcohol investigation

Background

        In this country, we use petrol as a source of our fuel for cars. However, Formula 1 cars have been using methanol, a type of alcohol, for years. Alcohols burn more efficiently releasing less harmful products, like carbon monoxide. Also, fossil fuels are running out, and an alternative fuel is needed, since we rely on this so much in every day life. However, there are problems with methanol to overcome before we all consider changing, as long term exposure causes blindness and brain damage.

Aim

        

        In this investigation, I am going to find out which alcohol releases the most energy when it is burnt. The alcohols I will use are: methanol, ethanol, propanol, and butanol.

Apparatus

  • Spirit burners (methanol, ethanol, propanol, butanol)
  • Clamp stand
  • Clamp
  • Measuring cylinder (to measure 50cm3 of water)
  • Copper beaker
  • Thermometer
  • Weighing scales
  • timer

Diagram

Fair testing

        I am going to test four alcohols, as mentioned above. Each alcohol should be tested three times, so an average can be found. Preferably, each alcohol should be repeated until consistent results are found, however, this may take a longer time than that which is available to do the experiment.

        I shall carry out a preliminary test to determine flame size, time to keep the spirit burners lit etc.

Control variables

  • Flame size
  • Distance of beaker from/ above flame
  • Volume of water
  • Quick extinguishments of flame
  • Same atmosphere/ room temperature

If these factors are not kept the same, it would make the experiment unfair, as the results would be different. The experiment would be unreliable and discredited. For example, a larger or small flame size will heat the water more or less, in the same time period, or the volume of water can change the temperature increase, as more or less molecules will need to be heated.

Method

  • Weigh spirit burner
  • Measure 50cm3  of water in the measuring cylinder
  • Pour water into a copper beaker
  • Take water temperature
  • Place beaker into clamp, 15.5cm above the surface
  • Place spirit burner under the beaker and light the wick
  • (NOTE-The flame size should be about the same in each experiment. If it is not, adjust the wick size and go back to step 1)
  • keep the wick burning for three minutes
  • put out the flame immediately
  • take the temperature of the water again
  • weigh the spirit burner again, to determine how much alcohol has been used
  • record results in a table (see page 5)
  • repeat for all different spirits until results are consistent

Theoretical workings

Methanol        +        oxygen        =        carbon dioxide        +        Water

CH3OH        +        1.5 O2                =        CO2                        +        2H2O

    H                +        O=O                =        O=C=O                +        H-O-H

H-C-O-H        +        0.5(O=O)        =                                +        H-O-H

    H

Reactants- Bonds present                        Products- Bonds present        

3XC-H=3X413=1239KJ                        2XC-O=2X740=1480KJ

1XC-O=1X360=360KJ                        4XO-H=4X463=1852KJ

1XO-H=1X463=463KJ

1.5XO=O=1.5X747=747KJ

TOTAL ENERGY IN=2809KJ        TOTAL ENERGY OUT=3332KJ

Energy change for = total bond energy – total bond energy

The reaction                      of reactants            of products

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∆Hcombustion = 2809 – 3332 = -523kJ per mole

Ethanol        +        oxygen        =        carbon dioxide        +        water

C2H5OH        +        3O2                =        2CO2                +        3H2O

    H H                +        O=O                =        O=C=O        +        H-O-H

H-C-C-O-H        +        O=O                =        O=C=O        +        H-O-H

    H H                +        O=O                =                        +        H-O-H

Reactants-bonds present                        Products-bonds present

5XC-H=5X413=2065kJ                        4XO=C=4X740=2960kJ

1XC-C=1X346=346kJ                        6XO-H=6X463=2778kJ

1XC-O=1X360=360kJ        

1XO-H=1X463=463kJ

3XO=O=3X498=1494

TOTAL ENERGY IN=4728kJ                total energy out=5738kJ

∆Hcombustion = 4728 – 5738 = -1010kJ per mole

Propanol        +        oxygen        =        carbon dioxide        +        water

C3H7OH        +        4.5 O2                =        3CO2                        +        4H2O

    H H H                 O=O                        O=C=O                        H-O-H

H-C-C-C-O-H +         O=O                =        O=C=O                +        H-O-H

    H H H                 O=O                        O=C=O                        H-O-H

                        0.5(O=O)                                                H-O-H

Reactants-bonds present                        Products-bonds present

7XC-H=7X413=2891kJ                        6XO=C=6X740=4440kJ

2XC-C=2X346=692kJ                        8XO-H=8X463=3704kJ

1XC-O=1X360=360kJ                                                

1XO-H=1X463=463kJ

4.5xO=O=4.5x498=2241

TOTAL ENERGY IN=6647kJ                TOTAL ENERGY OUT=8144kJ

 ∆Hcombustion = 6647 – 8144 = -1497kJ per mole

Butanol        +        oxygen        =        carbon dioxide        +        water

C4H9OH        +        6O2                =        4CO2                        +        5H2O

    H H ...

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