Fractional Distillation and Properties of Hydrocarbons

Crude oil straight from the ground is almost useless. It is a mixture of a huge number of hydrocarbons of many different sizes, so it does not burn cleanly or serve a single purpose. To make it useful, it is separated into fractions.

A fraction is a group of hydrocarbons of similar molecular size, and therefore similar boiling point, separated from crude oil together.

Molecules of a similar size behave in a similar way, so each fraction can be turned into a specific product such as a fuel. The separation relies on one key idea: different-sized hydrocarbons boil at different temperatures. Larger molecules have stronger forces between them, so they need more energy and a higher temperature to boil.

Fractional distillation separates the mixture using evaporation and condensation in a tall fractionating column that is hot at the bottom and cooler at the top.

The process, step by step:

1. Crude oil is heated until most of it evaporates into a mixture of vapours.
2. The vapours pass into the bottom of the column and rise up through it.
3. As the vapours rise, they cool down, because the column gets colder towards the top.
4. Each vapour condenses back to a liquid when it reaches the level where the temperature matches its boiling point.
5. The liquids are collected as separate fractions at different heights.

Large molecules have high boiling points, so they condense low down where it is hot. Small molecules have low boiling points, so they rise higher before condensing. Gases that do not condense at all leave from the very top.

The two processes to name in the exam are evaporation (heating the oil into vapour) and condensation (cooling the vapour back into liquid fractions at different levels of the column).

Once separated, the fractions are processed into two broad kinds of product: fuels and feedstock for the petrochemical industry.

Use	Examples
Fuels	petrol, diesel, kerosene, heavy fuel oil, liquefied petroleum gas (LPG)
Petrochemical products	solvents, lubricants, polymers, detergents

The petrochemical industry uses hydrocarbon fractions as feedstock, the raw material fed into reactions that build new substances. Because the carbon atoms in these molecules can bond in so many ways, they form families of similar compounds, giving a vast range of useful products from a single starting mixture.

You are expected to know these named fuels and product types, but not the names of any other fractions.

Three properties of the fractions change in a predictable way as the hydrocarbon molecules get larger. Learning the direction of each trend is a common exam requirement.

Property	As molecules get larger…	Why
Boiling point	increases	stronger intermolecular forces need more energy to overcome
Viscosity	increases (thicker, flows less easily)	larger molecules are harder to move past one another
Flammability	decreases (harder to ignite)	large molecules vaporise less easily, so ignite less readily

So small molecules like those in LPG and petrol have low boiling points, flow easily and catch fire readily, which makes them good fuels. Large molecules like those in bitumen have high boiling points, are thick and tar-like, and are hard to ignite.

The three trends to recall as molecule size increases: boiling point up, viscosity up, flammability down.

Burning a hydrocarbon in plenty of oxygen is complete combustion. Both the carbon and the hydrogen in the fuel are oxidised, and energy is released. The only products are carbon dioxide and water.

Complete combustion: hydrocarbon + oxygen → carbon dioxide + water (+ energy).

Worked example — write the balanced equation for the complete combustion of propane, C₃H₈.

Start by placing the carbon and hydrogen products. Three carbons give 3 CO₂; eight hydrogens give 4 H₂O:

${\text{C}}_3{\text{H}}_8+{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_2\text{O}$

Now count oxygen on the right: $3\times 2=6$ in the CO₂, plus $4\times 1=4$ in the water, giving 10 oxygen atoms. That needs 5 O₂ molecules on the left:

${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_2\text{O}$

Final check: 3 C, 8 H and 10 O on each side. Balanced.

A dependable routine for combustion: balance carbon first, then hydrogen, then finish with oxygen, because oxygen is on its own as O₂ and so is easiest to adjust last.

Worked example — balance the complete combustion of octane, C₈H₁₈, a component of petrol.

Carbon: 8 carbons give 8 CO₂. Hydrogen: 18 hydrogens give 9 H₂O.

${\text{C}}_8{\text{H}}_{18}+{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}$

Count oxygen on the right: $(8\times 2)+(9\times 1)=16+9=25$ oxygen atoms. That is an odd number, so 12.5 O₂ would balance it. To keep whole numbers, double every formula in the equation:

$2{\text{C}}_8{\text{H}}_{18}+25{\text{O}}_2\to 16{\text{CO}}_2+18{\text{H}}_2\text{O}$

Check: left has 16 C, 36 H, 50 O; right has 16 C, 36 H, and $(16\times 2)+18=50$ O. Balanced.

When balancing oxygen leaves you needing a half, double everything in the equation to clear the fraction.

1. Getting the boiling point and column position the wrong way round

Large molecules have high boiling points and condense low down in the hot part of the column. Small molecules have low boiling points and rise higher before condensing. Do not swap these.

2. Saying fractional distillation is a chemical reaction

It is a physical separation. No bonds within the hydrocarbon molecules are broken, so no new substances are made. It only relies on the different boiling points of the molecules already present.

3. Reversing the flammability trend

As molecules get larger, flammability decreases. Big molecules are harder to ignite, not easier. Only boiling point and viscosity increase with size.

4. Writing carbon monoxide or soot for complete combustion

Complete combustion produces only carbon dioxide and water. Carbon monoxide and soot (carbon) are products of incomplete combustion, when oxygen is limited, and should not appear in a complete-combustion equation.

5. Leaving a combustion equation unbalanced for oxygen

Always finish by counting the oxygen atoms on the right and matching them with O₂ on the left. If that gives a half, double every formula in the equation to keep whole-number multipliers.