Wednesday 11 March 2015

Hydrocarbons (2) Combustion of fuels with equations

The major use of hydrocarbons is still as fuels mainly for transport: petrol/gasoline, diesel and kerosine and fuel oil.

Few homes are heated these days(2015) using fuel oil but ships burn fuel oil, jet aircraft burn kerosine, lorries and cars burn petrol and diesel.

Hydrocarbons are useful fuels because they burn, reacting exothermically with oxygen to release energy.

Their combustion can experimentally be used to confirm their identity as hydrocarbons compounds.

Hydrocarbons are formed from just the elements hydrogen and carbon.

I guess if you have seen this experiment you will most probably have seen a demonstration of it at a distance where you wouldn't be able to see much.

Here are a couple of diagrams of typical apparatus that is often used:












There are mistakes in both diagrams so you have to be careful getting stuff off the net.

In the first diagram anhydrous copper sulphate is white it only ever turns blue when water contacts it.

In the second limewater is never blue it's always clear and colourless until carbon dioxide turns it chalky white!!

You'll also see that cobalt chloride paper is being used in the second experiment but it should be labelled anhydrous i.e. without water or the use of it is pointless!!

Anhydrous cobalt chloride paper is pale blue and it turns purple with addition of water.

So the experiment is set up you can't see much because its so far away for you and after about 10mins your teacher rejoices that there have been changes in the funnel where the candle is burning, in the limewater and in the copper sulphate or cobalt chloride paper.

Here is a fun version of this experiment narrated in a cool continental German(?) accent.  We love it especially the accent!!

Here are the points to get from this experiment:


  • the products of combustion are not just CO2 and H2O though you may have been taught that however you should be able to see that there is also unburnt carbon in the funnel (hence its gone black) and there will have been poisonous carbon monoxide with the CO2 because not enough air was used to burn the candle/tea light.
  • the copper sulphate turning blue is due to the presence of water.  But that could have come just from the air in the room so what should you do to discount this possibility?  (It's unlikely your teacher discussed this point probably because he or she didn't have the time for an additional experiment.)
  • The colour of the candle flame is yellow showing it burns incompletely.
  • The black stuff on the funnel was carbon (or soot) showing again that the candle burnt incompletely. 
  •  If the candle has enough oxygen to burn completely then it will just produce water and carbon dioxide.
  • Here is a word equation for the complete combustion of a hydrocarbon like candle wax:    candle wax      +      oxygen   =     carbon dioxide    +      water
  • So the assumption is that the carbon for the carbon dioxide came from the wax and so did the hydrogen for to form the water: hence wax is formed from hydrogen and carbon but we can't be sure if there is not something else there because this experiment is only set up to detect hydrogen and carbon!!!  Nevertheless, confirming hydrogen and carbon is a pretty good result.  
  • Incomplete combustion of the wax happens because there is just not enough oxygen to burn it.
  • Here is a word equation for the incomplete combustion of a hydrocarbon like wax.          candle wax    +    oxygen   =    carbon   +    carbon dioxide  +   water  +   carbon monoxide
  • You could see if you could balance this symbol equation for the incomplete combustion of a hydrocarbon like butane C4H10 that is used as lighter fuel:                                                  C4H10              +               O2          =       C        +         CO          +       CO2
Activities:

Can you construct complete combustion equations?

Say you are asked to construct the equation for the complete combustion of propane: how do you do that?  (Thankfully they give you the formula of propane C3H8!!!!)
  1. Complete combustion means just two products water and carbon dioxide so we can write the formulas of reactants and products like this:                                                                             C3H8        +          O2              =          CO2      +   H2O    
  2. Next you can see there are three carbon atoms in propane so that gives three CO2 molecules so the equation is now:        C3H8        +          O2              =          3CO2      +   H2O   
  3. Next you can see there are eight hydrogen atoms in propane so as two appear in water we'll divide eight by two (difficult!!!) and have four water molecules on the right like this:        C3H8        +          O2              =          3CO2      +   4H2O  
  4. All that's left is to find how many oxygens there are on the right.  I can count ten atoms so divide by two as two atoms make an oxygen molecule and that gives us five O2 on the left: C3H8        +          5O2              =          3CO2      +   4H2O 
  5. Our finished equation is:                                                                                                                  C3H8        +          5O2              =          3CO2      +   4H2O 
  6. This method always works for these hydrocarbon combustion equations.  The only adjustment you'll have to make is if the hydrocarbon has an even number of carbons ( e.g butane C4H10) in it when you will have to double every final value for each reactant and product  to be left with a whole number of oxygen molecules. Try it and see what I mean.  
So having cracked the complete combustion equations how do you feel about the incomplete combustion equations?  

Say you are asked to construct the equation for the incomplete combustion of pentane (C5H12) into water and carbon monoxide , carbon dioxide and carbon!
  1. Let's lay the symbols down first:                                                                                                       C5H12         +           O2       =         C            +          CO         +          CO2         +    H2O  
  2. Now things get really interesting because there is not one straight answer.  You have five carbons in pentane and three products that all contain carbon.  Where do you put the carbons?  We could have three carbon atoms, one carbon monoxide and one carbon dioxide molecule.  Or we could have two carbon atoms, two carbon monoxide molecules and one carbon dioxide molecule!!  In other words, there are several possible equations so we have just to pick one.  So let's go with this:                                                                                                                             C5H12         +           O2       =         3C            +          CO         +          CO2         +    H2O
  3. Next there are twelve hydrogen atoms in pentane so dividing by two gives us six water molecules.  Like this:                                                                                                                 C5H12         +           O2       =         3C            +          CO         +          CO2         +    6H2O
  4. Lastly, we'll count up the oxygen atoms on the right (nine in all) and divide by two to find the number of oxygen molecules like this:                                                                                          C5H12         +          41/2 O2       =         3C            +          CO         +          CO2      +    6H2O
  5. But that's four and a half oxygen molecules which some of us might not be too happy with as it looks untidy (!!) so let's double up all the quantities of atoms and molecules and we'll all feel safer like this:                                                                                                                                       2C5H12         +        9O2       =         6C            +          2CO         +          2CO2         +    12H2O


Can you come up with some alternative equations for this same reaction?

New Pages on the "Mole" and "Using the Mole" in chemical calculations are here 

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