Enthalpies of combustion.

This is what the table looked like:

We then did a table which showed the averages of all the alcohols. It looked like this:

Next we worked out the heat absorbed by the H2O. To do this we had to calculate it using this method:

(Mass of H2O*4.2*temp. rise)

= (50*4.2*temp. rise) joules

To make the joules into kilojoules we had to divide by 1,000. After this it needed to be worked out into KJ per mole. To do this we divided by the number of moles in the alcohol. The alcohol’s moles are as follows:

►Methanol (CH3OH) Mr=32 gm’s=0.545 32/0.545=58.7156

►Ethanol (C2H5OH) Mr=46 gm’s=0.39 46/0.39=117.949

►Propanol (C3H7OH) Mr=60 gm’s=0.21 60/0.21=285.714

►Butanol (C4H9OH) Mr=74 gm’s =0.23 74/0.23=321.739

This is the table for kilojoules per mole:

This graph shows the results from the table in graph form

Conclusion

I have found that as the number of carbon atoms goes up so does the enthalpy of combustion. This is because it takes more of the heat to break down the carbon atoms. We also found that the butanol left a sooty residue on the beaker. This happens because there is not enough oxygen in the air to react with the carbon and break it down. This could also explain why the graph stops sloping in a straight line after propanol, because there aren’t enough oxygen atoms to react with the carbon atoms therefore slowing the reaction down and causing the dip in the graph. If we had done it in a controlled environment the graph might have looked like this:

As you can see the graph has a constant slope now. This would be how it would look if the experiment was done in a controlled environment.