alkanes chemical properties

Higher and foundation tier

Chemically the alkanes are pretty dull and uninteresting. The only real use for them is as fuels. They are used to fuel the world! They provide fuels for cars, trains, ships, planes and pretty much any other form of transport that you can think of! Consider the following equations:

1. Carbon burns in air to produce carbon dioxide gas.

carbon(s) + oxygen(g) → carbon dioxide(g)
C(s) + O2(g) → CO2(g)

2. hydrogen burns in air to produce hydrogen oxide (water).
hydrogen(g) + oxygen(g) → hydrogen oxide(g)
2H2(g) + O2(g) → 2H2O(g)
Since alkanes are hydrocarbons and hydrocarbons contain only the elements carbon and hydrogen, then when a hydrocarbon is burned the products should be carbon dioxide and water. Essentially when you burn or combust a substance you take each element in turn that is present in the compound and join it with oxygen.

3. Methane (CH4) for example is a hydrocarbon, it is a compound which contains only the elements carbon and hydrogen, so when it burns in air it will produce carbon dioxide and hydrogen oxide (water vapour).
methane(g) + oxygen(g) → carbon dioxide(g) + hydrogen oxide(g)
CH4 + 2O2(g) → CO2(g) + 2H2O(g)

Complete and incomplete combustion

The gas we use at home for cooking and to heat our homes is methane. Methane is also used in the science lab as the fuel for Bunsen burners. Similarly if we burn the hydrocarbon pentane, C5H12 , identical products will be obtained since this molecule like methane contains only carbon and hydrogen.

pentane(g) + oxygen(g) → carbon dioxide(g) + hydrogen oxide(g)
C5H12 + 8O2(g) → 5CO2(g) + 6H2O(g)
The only difference between burning pentane and methane is that because pentane is a larger molecule which contains more carbon and hydrogen it requires more oxygen. Sometimes this can be a problem if you try to burn a fuel inside sealed containers such as engines or boiler where there may be a lack of oxygen. Burning hydrocarbons in air where there is plenty of oxygen available results in the production of carbon dioxide gas and water vapour. This is called complete combustion. Burning hydrocarbons where there is a limited supply of oxygen is called incomplete combustion and different products are obtained e.g.
C5H12 + 8O2(g) → 5CO2(g) + 6H2O(g)   complete combustion of pentane
C5H12 + 51/2O2(g) → 5CO(g) + 6H2O(g)   incomplete combustion of pentane
For the complete combustion of pentane 8 moles of oxygen gas are required for every mole of pentane, when only 5.5 moles of oxygen are available the poisonous gas carbon monoxide is produced instead of carbon dioxide.

Example 2 - the complete and incomplete combustion of methane gas.
CH4 + 2O2(g) → CO2(g) +   2H2O(g)   complete combustion of methane
CH4 + 11/2O2(g) → CO(g) +   2H2O(g)   incomplete combustion of methane
The incomplete combustion of methane produces carbon monoxide whereas when methane is burned in a plentiful supply of oxygen carbon dioxide is produced. Similarily the equations below show the products of the complete and incomplete combustion of propane.
C3H8 + 5O2(g) → 3CO2(g) + 4H2O(g)   complete combustion of propane
C3H8 + 31/2O2(g) → 3CO(g) + 4H2O(g)   incomplete combustion of propane
It is a similar story when propane is burned, the complete combustion of a hydrocarbon fuel like propane produces carbon dioxide and water vapour as the products whereas the incomplete combustion produces the toxic gas carbon monoxide and water vapour.

Note, in some of the equations above, for example in the incomplete combustion of methane and propane, I have used 11/2 and 31/2 moles of oxygen to balance these equations. In some textbooks you may see that instead of having decimals or fractions in equations they simply double the number of moles in the equations. This will remove the decimals or fractions in the equations, but the choice is yours as to whether to leave them as I have written the equations or double everything to remove the fractions!

incomplete combustion of diesel fuel in a bus If the hydrocarbons are burned in a very limited supply of air it is possible that no carbon monoxide will be produced, instead solid carbon or soot is produced e.g. The equations below show the equations for the complete and incomplete combustion of propane when various amounts of oxygen are avaialble.

propane + oxygen → carbon (soot) + hydrogen oxide   incomplete combustion of propane
C3H8 + 3O2(g) → 3CO(g) + 4H2O(g)   incomplete combustion of propane
C3H8 + 2O2(g) → 3C(s) + 4H2O(g)   incomplete combustion of propane
(Hint- To balance these equations, however many carbon atoms are in the hydrocarbon that is the number of moles of soot produced, however many moles of hydrogen are in the hydrocarbon half this number of moles of water will be produced).

It is very likely that when the hydrocarbon fuels undergo combustion that a combination of carbon dioxide, carbon monoxide and soot are produced, it all depends on the amount of oxygen gas available. The equations below show the complete and incomplete combustion of hexane:
hexane + oxygen → carbon (soot) + hydrogen oxide   incomplete combustion of hexane
C6H14 + 61/2O2(g) → 6CO(g) + 7H2O(g)   incomplete combustion of hexane
C6H14 + 31/2O2(g) → 6C(s) + 7H2O(g)  incomplete combustion of hexane

Testing for the products of combustion

As discussed above the products for the complete combustion of a hydrocarbon are carbon dioxide and water. The apparatus below can be set-up to collect the products of combustion and to test them.

testing for the products of combustion

The products of combustion, carbon dioxide and water vapour leave the candle flame and are drawn into the funnel by a pump. Here the vapours enter the first boiling tube. This boiling tube is surrounded by ice chilled water, this will cause the water vapour from the combustion to condense and collect as liquid water inside the boiling tube. This can then be tested with blue cobalt chloride paper (test for water- blue cobalt chloride paper turns pink). The carbon dioxide from the combustion will pass through this first test-tube and enter the second boiling tube. This tube is filled with limewater. The limewater will turn chalky or milky due to the presence of carbon dioxide.

As mentioned above the main use for alkanes is as fuels. Many fuels not only contain alkanes but also the element sulfur. Sulfur is a yellow solid in its elemental form. It forms an acidic non-metal oxide when it is burned:

sulfur(s) + oxygen(g) → sulfur dioxide(g)
S(s) + O2(g) → SO2(g)
Sulfur dioxide gas will dissolve in water to form an acid, sulfurous acid. We can modify the apparatus set-up above to test for an acidic gas by simply adding another boiling tube filled with universal indicator solution. If any acidic gases are present; such as sulfur dioxide then the universal indicator solution will turn red. This new set-up is shown below:

testing for acidic gases

Key Points

Practice questions

Check your understanding - Questions on combustion of alkanes.