Interactive Chemistry Worksheets for Students
Hydrocarbons are compounds that contain the elements hydrogen and carbon. They are obtained by the fractional distillation of crude oil.
Hydrocarbons can be divided into three main groups, alkanes, alkenes and alkynes.
Examples of hydrocarbons are propane, C3H8 and octane, C8H18
|Carbon atoms||Alkane, CnH2n+2||Alkene, CnH2n||Alkyne, CnH2n-2|
|2||ethane, C2H6||ethene, C2H4||ethyne, C2H2|
|3||propane, C3H8||propene, C3H6||propyne, C3H4|
|4||butane, C4H10||butene, C4H8||butyne, C4H6|
|5||pentane, C5H12||pentene, C5H10||pentyne, C5H8|
|6||hexane, C6H14||hexene, C6H12||hexyne, C6H10|
|7||heptane, C7H16||heptene, C7H14||heptyne, C7H12|
|8||octane, C8H18||octene, C8H16||octyne, C8H14|
|9||nonane, C9H20||nonene, C9H18||nonyne, C9H16|
|10||decane, C10H22||decene, C10H20||decyne, C10H18|
Combustion is a self-sustaining chemical reaction in which heat is produced by the burning of a fuel with oxygen.
Note: In Senior Chemistry fuels can be burnt in other substances like chlorine. These substances act in the same way as oxygen and are called oxidizing agents.
A helpful aid in balancing combustion equations is to balance each element in alphabetical order, C H O
Balance the carbon atoms first, hydrogen atoms second and oxygen atoms last, C H O
1. Pentane + oxygen carbon dioxide + water
C5H12 + O2 CO2 + H2O
|Balance C first||C5H12 + O2 5CO2 + H2O||5 C atoms are needed|
|Balance H second||C5H12 + O2 5CO2 + 6H2O||12 H atoms are needed|
|Balance O third||C5H12 + 3O2 5CO2 + 6H2O||6 O atoms are needed|
The balanced equation for the complete combustion of pentane is
C5H12 + 3O2 5CO2 + 6H2O
2. Butane + oxygen carbon dioxide + water
C4H10 + O2 CO2 + H2O
|Balance C first||C4H10 + O2 4CO2 + H2O||4 C atoms are needed|
|Balance H second||C4H10 + O2 4CO2 + 5H2O||10 H atoms are needed|
|Balance O third||C4H10 + 5.5O2 4CO2 + 6H2O||11 O atoms are needed|
The balanced equation can be left as a fraction as chemists deal with quantities called moles.
C4H10 + 5.5O2 4CO2 + 6H2O
The 5.5 O2 is not 5½ molecules of O2. It represents 5.5 moles of oxygen.
If you don't like working in halves then double each number in front the formula
2C4H10 + 11O2 8CO2 + 12H2O
3. Ethanol + oxygen carbon dioxide + water
C2H5OH + O2 CO2 + H2O
|Balance C first||C2H5OH + O2 2CO2 + H2O||5 C atoms are needed|
|Balance H second||C2H5OH + O2 2CO2 + 3H2O||6 H atoms are needed|
|Balance O third||C2H5OH + 3O2 2CO2 + 3H2O||7 O atoms are needed|
The balanced equation for the combustion of ethanol is
C2H5OH + 3O2 2CO2 + 3H2O
Complete combustion is the burning of hydrocarbons in sufficient oxygen to produce carbon dioxide and water as products.
A common example of complete combustions is the blue flame of a Bunsen burner. The air hole is open.
Hydrocarbon + Oxygen Carbon dioxide + water
Propane + oxygen carbon dioxide + water
C3H8 + 5O2 3CO2 + 4H2O
Incomplete combustion is the burning of a hydrocarbon in insufficient oxygen producing carbon (soot), carbon monoxide and water as products.
A common example of the incomplete combustion of hydrocarbons is the yellow flame of a Bunsen burner. The air hole is closed.
Hydrocarbon + Oxygen Carbon + Carbon monoxide + water
When writing equations with incomplete combustion it is advisable to include only one carbon product otherwise there will be multiple solutions to the equation.
Multiple carbon products produce balanced equations with multiple solutions
The incomplete combustion of propane, C3H8 produces two possible balanced equations.
C3H8 + 3 O2 2CO + C + 4H2O
C3H8 + 2.5 O2 CO + 2C + 4H2O
Hydrocarbon + Oxygen Carbon monoxide + water
Hydrocarbon + Oxygen Carbon + water
When there is only one carbon product there is only one solution to each equation.
If the product is only carbon monoxide CO, there is only one solution to the balanced equation.
C3H8 + 4 O2 3 CO + 5H2O
If the product is only soot C, there is only one solution to the balanced equation.
C3H8 + 2 O2 3C + 4H2O