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Combustion of Hydrocarbons Chemistry Tutorial

Key Concepts

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Complete Combustion of Hydrocarbons

Any hydrocarbon will combust in excess oxygen to produce carbon dioxide gas and water vapour.

For the complete combustion of a hydrocarbon:

⚛ oxygen gas is the reactant in excess

⚛ hydrocarbon is the limiting reagent

We can write a general word equation for the complete combustion of any hydrocarbon as shown below:

hydrocarbon + excess oxygen gas → carbon dioxide gas + water vapour

Hydrocarbons include alkanes, alkenes and alkynes, so we can say that:

⚛ any alkane combusts in excess oxygen to produce carbon dioxide gas and water vapour

alkane + excess oxygen gas → carbon dioxide gas + water vapour

⚛ any alkene combusts in excess oxygen to produce carbon dioxide gas and water vapour

alkene + excess oxygen gas → carbon dioxide gas + water vapour

⚛ any alkyne combusts in excess oxygen to produce carbon dioxide gas and water vapour

alkyne + excess oxygen gas → carbon dioxide gas + water vapour

Example: Complete Combustion of Methane

Methane, CH4(g), is a hydrocarbon. It is a compound made up only of the elements carbon (C) and hydrogen (H).

Methane is a gas at room temperature and pressure. It is a common component of natural gas which is used as a fuel.

Methane combusts in excess oxygen to produce carbon dioxide gas (CO2(g)) and water vapour (H2O(g)).

Combustion in excess oxygen is referred to as complete combustion.

We can write a balanced chemical equation to represent the complete combustion of methane gas as shown below:

  1. Write the word equation for the complete combustion of methane:

    general equation: reactants → products
    word equation: methane + oxygen gas → carbon dioxide gas + water vapour

  2. Write the molecular formula for each reactant and product in the word equation:

    Reactants   Products
    methane :

    oxygen gas :

    		CH4(g)

    O2(g)

    		  carbon dioxide gas :

    water vapour :

    		CO2(g)

    H2O(g)

  3. Write the unbalanced chemical equation by substituting the molecular formula for the name of each reactant and product in the word equation:

    general equation: reactants → products
    word equation: methane + oxygen gas → carbon dioxide gas + water vapour
    unbalanced chemical equation : CH4(g) + O2(g) → CO2(g) + H2O(g)

  4. Balance the chemical equation:

    unbalanced chemical equation : CH4(g) + O2(g) → CO2(g) + H2O(g)  
    No. C atoms : 1     = 1     C atoms balanced
    No. H atoms :   4     ≠       2 H atoms NOT balanced
    Need to multiply the number of water molecules by 2 to balance the hydrogen atoms. Then, check the balance of this new chemical equation as shown below.
      CH4(g) + O2(g) → CO2(g) + 2H2O(g)  
    No. C atoms : 1     = 1     C atoms balanced
    No. H atoms :   4     =       4 H atoms balanced
    No. O atoms :     2 ≠   2 +     2 O atoms NOT balanced
    Need to multiply the number of oxygen molecules by 2 to balance the oxygen atoms. Then check the balance of this new chemical equation as shown below:
      CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)  
    No. C atoms : 1     = 1     C atoms balanced
    No. H atoms :   4     =       4 H atoms balanced
    No. O atoms :     4 =   2 +     2 O atoms balanced

  5. The balanced chemical equation for the complete combustion of methane gas is:

    CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

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Incomplete Combustion of Hydrocarbons

If there is insufficient oxygen gas present for the combustion of a hydrocarbon to produce the most oxidised form of carbon which is carbon dioxide gas, we refer to the reaction as the incomplete combustion of the hydrocarbon.

For the incomplete combustion of a hydrocarbon:

⚛ oxygen gas is the limiting reagent

⚛ hydrocarbon is the reactant in excess

The incomplete combustion of a hydrocarbon usually produces a "sooty" flame due to the presence of carbon (C), or soot, as a product of the incomplete combustion reaction.

The hydrogen in the hydrocarbon will be oxidised to water, H2O, but the carbon in the hydrocarbon may, or may not, be oxidised to carbon monoxide gas (CO(g)).

Example: Incomplete Combustion of Methane

In a particular experiment, excess methane gas (CH(g)) was combusted in limited oxygen gas to produce soot (solid carbon) and water vapour.

We can write a balanced chemical equation to represent this incomplete combustion of methane in this experiment as shown below:

  1. Write the word equation for the incomplete combustion of methane:

    general equation: reactants → products
    word equation: methane + oxygen gas → solid carbon + water vapour

  2. Write the molecular formula for each reactant and product in the word equation:

    Reactants   Products
    methane :

    oxygen gas :

    		CH4(g)

    O2(g)

    		  solid carbon :

    water vapour :

    		C(s)

    H2O(g)

  3. Write the unbalanced chemical equation by substituting the formula for the name of each reactant and product in the word equation:

    general equation: reactants → products
    word equation: methane + oxygen gas → solid carbon + water vapour
    unbalanced chemical equation : CH4(g) + O2(g) → C(s) + H2O(g)

  4. Balance the chemical equation:

    unbalanced chemical equation : CH4(g) + O2(g) → C(s) + H2O(g)  
    No. C atoms : 1     = 1     C atoms balanced
    No. H atoms :   4     ≠       2 H atoms NOT balanced
    Need to multiply the number of water molecules by 2 to balance the hydrogen atoms. Then check the balance of the new equation:
      CH4(g) + O2(g) → C(s) + 2H2O(g)  
    No. C atoms : 1     = 1     C atoms balanced
    No. H atoms :   4     =       4 H atoms balanced
    No. O atoms :     2 =         2 O atoms balanced

  5. The balanced chemical equation for the incomplete combustion of methane gas in this experiment is :

    CH4(g) + O2(g) → C(s) + 2H2O(g)

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Footnotes:

(1) Combustion of organic mixtures like you find in wood is not always a simple case of oxidation of the carbon and hydrogen present in the substance, rather, a large number of complex chemical reactions takes place, one of which is pyrolysis in which large molecules are fragmented into smaller molecules in the absence of oxygen. The pyrolysis of wood produces small gaseous molecules which then react with oxygen above the surface of the wood, giving rise to the flames we see.