Molecular size sets a hydrocarbon's boiling point, viscosity and flammability

1. A hydrocarbon molecule becomes larger as it gains more carbon and hydrogen atoms.
2. As molecular size increases, the boiling point increases, so a larger hydrocarbon must be heated more before it boils.
3. As molecular size increases, the viscosity increases, so larger hydrocarbons are thicker and flow less easily.
4. As molecular size increases, the flammability decreases, so larger hydrocarbons are harder to ignite.

Why smaller hydrocarbons make more convenient fuels

1. A useful fuel must reach where it burns, mix with oxygen and ignite easily.
2. Smaller hydrocarbons have lower boiling points, lower viscosity and higher flammability.
3. Their lower boiling points let them vaporise more readily so they mix with air.
4. Their lower viscosity lets them flow easily through pipes and fuel systems.
5. Their higher flammability lets them ignite more easily.
6. Larger hydrocarbons need more heating to vaporise, flow less easily and are harder to light.
7. These trends decide which hydrocarbons are chosen for which fuels.

Complete combustion oxidises both the carbon and the hydrogen

1. Combustion is a reaction in which a substance burns in oxygen.
2. Burning a hydrocarbon fuel releases energy to the surroundings.
3. In complete combustion there is plenty of oxygen for the fuel to burn fully.
4. The carbon is oxidised to carbon dioxide.
5. The hydrogen is oxidised to water.
6. The general reaction is: hydrocarbon + oxygen $\to$ carbon dioxide + water.

Balancing a combustion equation: adjust coefficients, never formulae

1. Write the correct formula of the hydrocarbon, then $O_2$, then the products ${CO}_2$ and $H_{2}O$.
2. Balance the carbon atoms using a number in front of ${CO}_2$.
3. Balance the hydrogen atoms using a number in front of $H_{2}O$.
4. Balance the oxygen atoms last using a number in front of $O_2$.
5. A coefficient is a number placed in front of a formula to show how many of that particle react.
6. If a coefficient comes out as a fraction, multiply the whole equation to clear it.
7. Never change the small subscript numbers inside a formula.

Worked example and final checks for a combustion equation

1. For ethane, start with $C_2{H}_6+O_2\to {CO}_2+H_{2}O$.
2. Two carbons need $2{CO}_2$, and six hydrogens need $3H_{2}O$.
3. The products now hold seven oxygen atoms, giving $\frac{7}{2}{O}_2$.
4. Doubling every coefficient gives $2C_2{H}_6(g)+7O_2(g)\to 4{CO}_2(g)+6H_{2}O(l)$.
5. Check the carbon, hydrogen and oxygen atoms match on both sides.
6. Check that no formula was altered and the coefficients are the smallest whole-number ratio.