Fractional Distillation vs Cracking: From Crude Oil to PVC
Learn why fractional distillation separates existing hydrocarbons while cracking breaks larger molecules into smaller ones. Follow the chemistry from crude oil to ethene, chloroethene, and PVC, with worked examples and HSC-style practice questions.
A refinery wants more ethene to manufacture PVC, but crude oil contains a mixture of hydrocarbons rather than a ready-made supply of pure ethene. If operators simply heat the crude oil and send it through a fractionating column, will they create more ethene? No. Fractional distillation separates molecules that already exist, while cracking breaks carbon-carbon bonds to produce new, smaller molecules. Understanding that difference explains the entire route from crude oil to plastic.
Two processes, two different jobs
| Fractional distillation — separation | Crude oil → heater → fractionating column; hot bottom, cool top. Collect refinery gases, naphtha, kerosene, diesel/gas oil, and residue. Molecules remain intact: sorts existing molecules. |
|---|---|
| Cracking — chemical change | C₂H₄ → C₂H₄Cl₂ → C₂H₃Cl → [–CH₂–CHCl–]ₙ Selected naphtha → high-temperature steam cracker → smaller-hydrocarbon mixture; breaks C–C bonds. Purify ethene → 1,2-dichloroethane → chloroethene → PVC by addition polymerisation. |
01Fractional distillation separates molecules without changing them
Crude oil is a complex mixture of hydrocarbons. These molecules differ in size, structure, and boiling point.
To separate them, a refinery heats the crude oil and feeds it into a fractionating column. The column is hottest near the bottom and progressively cooler towards the top.
Fractionating column: required diagram features
| Apparatus and flow | Heated crude oil enters near the bottom; vapour rises through trays while liquid flows downwards. Temperature decreases upwards: hottest at the bottom, coolest at the top. |
|---|---|
| Fraction outlets | Top to bottom: refinery gases, petrol/naphtha, kerosene, diesel/gas oil, heavy residue. Lower-boiling fractions collect higher; higher-boiling fractions collect lower. |
Here is what happens:
- The heated crude oil produces vapour containing different hydrocarbons.
- The vapour rises through the column and cools.
- Hydrocarbons with higher boiling points tend to condense lower down, where the column is hotter.
- More volatile hydrocarbons, with lower boiling points, can travel further upwards before condensing.
The collected liquids and gases are called fractions. Each fraction contains a range of hydrocarbons with similar boiling points. A fraction is not necessarily a pure substance.
The column also contains trays or packing that allow rising vapour to contact descending liquid repeatedly. This improves the separation. The simple picture of each fraction collecting at one height is useful, but real fractions overlap in composition.
Why does boiling point matter?
Imagine sorting a box of intact bead chains by their properties. Fractional distillation sorts the chains; it doesn’t snap them into shorter pieces. Cracking, which we’ll examine shortly, actually breaks the chains.
The chemical version of this distinction is important. During distillation, molecules change physical state, but their covalent bonds remain intact. Intermolecular attractions are overcome during vaporisation and re-form during condensation.
A lower-boiling hydrocarbon is generally more volatile. Volatility describes how readily a substance vaporises under given conditions.
2. What makes one hydrocarbon more volatile than another?
Consider two straight-chain alkanes: hexane, \(\ce{C6H14}\), and dodecane, \(\ce{C12H26}\).
Hexane has a lower boiling point than dodecane. At the same external pressure, hexane is more volatile, so it tends to travel further up a fractionating column before condensing.
Chain length affects boiling point
| Hexane | C₆H₁₄ Smaller surface area; weaker London dispersion forces; lower boiling point. |
|---|---|
| Dodecane | C₁₂H₂₆ Greater surface area and intermolecular contact; stronger London dispersion forces; higher boiling point. |
Both molecules experience London dispersion forces, a type of intermolecular attraction. These forces arise from temporary fluctuations in electron distribution.
For straight-chain alkanes, increasing chain length generally increases the size and surface area of the molecules. This strengthens their dispersion forces, so more energy is needed to separate neighbouring molecules during boiling.
The general trend is:
- Shorter straight-chain alkane → weaker dispersion forces → lower boiling point → greater volatility.
- Longer straight-chain alkane → stronger dispersion forces → higher boiling point → lower volatility.
There is one qualification. Chain length is a useful guide, not a universal rule for ranking every hydrocarbon. Molecular branching also affects surface contact and boiling point, while pressure affects boiling temperatures.
Worked example: Which hydrocarbon condenses higher?
At atmospheric pressure, hexane has a boiling point of approximately \(69\,^\circ\mathrm{C}\), while dodecane boils at approximately \(216\,^\circ\mathrm{C}\). Which would tend to condense higher in a fractionating column, and do either of the molecules change identity?
Step 1
Hexane has the lower boiling point, so it is more volatile under the same pressure.
Step 2
The upper column is cooler. Hexane can remain in the vapour phase at temperatures where dodecane is more likely to condense.
Step 3
Hexane tends to condense higher in the column, while dodecane tends to condense lower down.
Neither molecule has been chemically changed. The process separates existing hydrocarbons according to differences in volatility.
03Cracking breaks larger hydrocarbons into smaller molecules
Now consider a different problem. A refinery has a supply of heavy hydrocarbons, but it wants more small molecules for making useful chemicals.
Fractional distillation cannot solve that problem by itself. It can separate the hydrocarbons already present, but it cannot turn a large hydrocarbon into ethene.
Cracking can.
Cracking is a chemical process in which larger hydrocarbon molecules break apart to form smaller molecules. It requires high temperatures, and some cracking processes also use a catalyst to help the reaction occur.
The crucial difference is that cracking breaks covalent bonds, particularly carbon-carbon bonds. The atoms are rearranged into new molecules.
One illustrative cracking pathway
- One decane molecule
C₁₀H₂₂; CH₃–(CH₂)₆–CH₂ | CH₂–CH₃
The vertical bar marks the C–C bond selected for cleavage; hydrogen atoms rearrange.
- One octane + one ethene
C₈H₁₈ + C₂H₄
Octane: CH₃–(CH₂)₆–CH₃. Ethene: CH₂=CH₂, showing its C=C double bond.
For example, one possible cracking pathway for decane is:
\[
\ce{C10H22 -> C8H18 + C2H4}
\]
The equation is balanced:
- Carbon: \(10 = 8 + 2\).
- Hydrogen: \(22 = 18 + 4\).
One decane molecule produces one octane molecule and one ethene molecule in this illustrative pathway.
Notice that ethene, \(\ce{C2H4}\), contains a carbon-carbon double bond. It is an alkene. Octane, \(\ce{C8H18}\), is an alkane, with only single bonds between its carbon atoms.
Real cracking produces a mixture of products, not just octane and ethene. The equation represents one possible reaction pathway, rather than a complete description of an industrial cracker.
Why do refineries need both processes?
Crude oil does not naturally provide every hydrocarbon in the quantities industry needs. A refinery may have a surplus of heavier fractions while demand for smaller hydrocarbons is higher.
Fractional distillation separates the available mixture. Cracking changes the molecules to produce a different mixture that includes smaller hydrocarbons, often including alkenes.
For industrial ethene production, a common route is steam cracking, which uses very high temperatures and steam to process a hydrocarbon feed such as naphtha or ethane. The ethene must then be separated from the other products.
Worked example: Balance a cracking pathway
Decane, \(\ce{C10H22}\), follows one possible cracking pathway to produce octane, \(\ce{C8H18}\), and ethene. Write the balanced equation and explain what it tells you.
Step 1
\[
\ce{C10H22 -> C8H18 + C2H4}
\]
Step 2
The products contain \(8 + 2 = 10\) carbon atoms, matching the reactant.
Step 3
The products contain \(18 + 4 = 22\) hydrogen atoms, again matching the reactant.
Step 4
The equation is already balanced. It shows that cracking can produce both an alkane and an alkene from a larger alkane. It does not mean that all decane molecules follow this particular pathway, or that an industrial cracker produces only these two substances.
04From ethene to chloroethene and PVC
Producing ethene is only one stage of the manufacturing route. Ethene is not the monomer used to make PVC directly.
The route considered here has three main chemical stages.
Stage 1: Convert ethene into 1,2-dichloroethane.
Ethene reacts with chlorine. Chlorine adds across the carbon-carbon double bond to form 1,2-dichloroethane.
\[
\ce{CH2=CH2 + Cl2 -> ClCH2CH2Cl}
\]
Stage 2: Produce chloroethene.
Heating 1,2-dichloroethane causes it to eliminate hydrogen chloride, \(\ce{HCl}\), forming chloroethene.
\[
\ce{ClCH2CH2Cl -> CH2=CHCl + HCl}
\]
Chloroethene is also known as vinyl chloride. It is the monomer used to manufacture poly(chloroethene), commonly called polyvinyl chloride or PVC.
Stage 3: Polymerise the chloroethene.
Chloroethene undergoes addition polymerisation. Its carbon-carbon double bond opens, allowing many monomer molecules to join into a long carbon chain.
\[
\ce{n CH2=CHCl -> [-CH2-CHCl-]_{n}}
\]
Here, \(n\) represents a large number of repeating monomer units. The chlorine atoms remain attached to the polymer chain as substituents.
From ethene to PVC
- Chlorine addition
CH₂=CH₂ + Cl₂ → ClCH₂–CH₂Cl
Ethene forms 1,2-dichloroethane as the C=C bond becomes a single bond.
- Heating
ClCH₂–CH₂Cl → CH₂=CHCl + HCl
Heating eliminates HCl to produce chloroethene (vinyl chloride).
- Addition polymerisation
n CH₂=CHCl → [–CH₂–CH(Cl)–]ₙ
Many monomers join as C=C bonds open, forming a carbon backbone with Cl substituents retained.
There are two distinctions worth keeping clear.
First, cracking is used upstream to help produce ethene from larger hydrocarbons. The later conversion of ethene into chloroethene involves separate chemical reactions.
Second, polymerisation joins chloroethene molecules; it does not create ethene or add chlorine to the monomer. The chlorine is introduced earlier in the manufacturing route.
The processes side by side
| Feature | Fractional distillation | Cracking |
|---|---|---|
| Main purpose | Separate a mixture | Produce smaller molecules |
| What changes? | Physical state and composition of each collected fraction | Chemical identity of the hydrocarbons |
| Are covalent bonds broken? | No, not during the intended separation | Yes, bonds within hydrocarbons are broken |
| Products | Fractions containing mixtures of existing hydrocarbons | A mixture containing newly formed, smaller hydrocarbons |
| Can it produce more ethene from a large hydrocarbon? | No | Yes, depending on the feed and reaction conditions |
A refinery may use distillation again after cracking to separate the resulting mixture. That does not undo the cracking reaction or create extra ethene. It separates and concentrates the molecules that the cracker has already produced.
04Questions and solutions
Question 1
Crude oil is heated and fed into a fractionating column. A gas fraction collects near the top, while a heavier fraction collects lower down.
State the property that allows these fractions to be separated, and explain why the process does not break the hydrocarbons apart.
Solution 1
The fractions are separated mainly by their boiling points, which reflect differences in volatility, and the hydrocarbons remain chemically unchanged.
The column has a temperature gradient, with the hottest region at the bottom. Lower-boiling hydrocarbons can remain in the vapour phase at lower temperatures and travel further up the column before condensing. Higher-boiling hydrocarbons tend to condense lower down.
Vaporisation overcomes intermolecular attractions, not the covalent bonds holding each molecule together. Each collected fraction therefore contains existing hydrocarbons with similar boiling ranges rather than newly formed molecules.
Question 2
A refinery increases the temperature of its crude-oil feed. More vapour enters the fractionating column, but analysis shows that the same hydrocarbon species are present as before.
An operator claims that the higher temperature must have increased ethene production.
Explain why the evidence does not support this claim. Also explain what would happen to the separation if the temperature gradient in the column became weaker, and identify the process needed to produce ethene from larger hydrocarbons.
Solution 2
The evidence indicates increased vaporisation, not the formation of new ethene molecules.
Heating can cause more of the existing hydrocarbons to enter the vapour phase. However, if the molecular species remain unchanged, there is no evidence that cracking has occurred. A larger vapour flow does not necessarily mean a larger ethene yield.
A weaker temperature gradient would generally make the separation less effective. The differences in where hydrocarbons condense would become less useful, so the collected fractions could have more overlapping compositions.
To produce ethene from larger hydrocarbons, the refinery needs cracking. This process breaks carbon-carbon bonds and forms smaller molecules. A subsequent separation stage can then isolate ethene from the other products.
Question 3
One possible cracking pathway is:
\[
\ce{C10H22 -> C8H18 + C2H4}
\]
A \(14.2\ \mathrm{g}\) sample of decane has exactly \(50.0\%\) of its molecules follow this pathway. Ignore any ethene produced by other pathways.
Calculate the mass of ethene produced. Use \(A_\mathrm{r}(\mathrm{C})=12.0\) and \(A_\mathrm{r}(\mathrm{H})=1.0\).
Solution 3
The stated pathway produces \(1.40\ \mathrm{g}\) of ethene.
First, calculate the molar mass of decane. Here, \(M\) represents molar mass, \(n\) represents amount in moles, and \(m\) represents mass.
\[
\begin{aligned}
M(\ce{C10H22})
&=10(12.0)+22(1.0)\\
&=142\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]
Calculate the amount of decane in the original sample:
\[
\begin{aligned}
n(\ce{C10H22})
&=\frac{14.2\ \mathrm{g}}
{142\ \mathrm{g\,mol^{-1}}}\\
&=0.100\ \mathrm{mol}
\end{aligned}
\]
Only \(50.0\%\) follows the specified pathway:
\[
\begin{aligned}
n(\ce{C10H22}\text{ reacting})
&=0.500(0.100)\\
&=0.0500\ \mathrm{mol}
\end{aligned}
\]
The balanced equation gives a \(1:1\) mole ratio between decane reacting by this pathway and ethene produced. Therefore,
\[
n(\ce{C2H4})=0.0500\ \mathrm{mol}
\]
The molar mass of ethene is:
\[
M(\ce{C2H4})=2(12.0)+4(1.0)
=28.0\ \mathrm{g\,mol^{-1}}
\]
Finally,
\[
\begin{aligned}
m(\ce{C2H4})
&=nM\\
&=0.0500\ \mathrm{mol}
\times 28.0\ \mathrm{g\,mol^{-1}}\\
&=1.40\ \mathrm{g}
\end{aligned}
\]
The sample produces \(1.40\ \mathrm{g}\) of ethene through the specified pathway. An industrial cracker’s overall ethene yield would depend on the full set of reactions, the feed composition, and the operating conditions.
Question 4
A cracker produces a gas mixture containing ethene, ethane, and propene. A student claims that the ethene can be polymerised directly to form PVC.
Assess this claim. Outline the route from ethene to PVC, including balanced equations for the chemical reactions and the polymerisation step. Explain why the gas mixture must be separated before the downstream reactions.
Solution 4
The claim is incorrect: ethene is not the monomer used to make PVC directly. Ethene must first be separated from the cracker’s product mixture and converted into chloroethene.
The sequence is as follows.
Step 1: Separate the ethene.
The cracker produces several hydrocarbons. A separation stage isolates and purifies the ethene needed for the next reaction. This stage changes the composition of the stream, not the chemical identity of the ethene.
Step 2: React ethene with chlorine.
\[
\ce{CH2=CH2 + Cl2 -> ClCH2CH2Cl}
\]
The product is 1,2-dichloroethane. Chlorine adds across ethene’s carbon-carbon double bond.
Step 3: Convert 1,2-dichloroethane into chloroethene.
\[
\ce{ClCH2CH2Cl -> CH2=CHCl + HCl}
\]
Heating causes the molecule to eliminate hydrogen chloride, forming chloroethene.
Step 4: Polymerise the chloroethene.
\[
\ce{n CH2=CHCl -> [-CH2-CHCl-]_{n}}
\]
The double bond opens and the monomers join to form PVC.
The key point is that each stage has a different job. Separation isolates ethene, chemical reactions convert it into chloroethene, and addition polymerisation joins the chloroethene molecules. Ethene alone cannot form PVC through the stated polymerisation step because it does not contain the chlorine substituent present in the PVC repeat unit.
Question 5
A refinery has fractionated crude oil and cracked a suitable hydrocarbon fraction. After separating the cracking products, a downstream plant receives \(1.00\ \mathrm{mol}\) of pure ethene.
Assume the following reactions go to completion, with no material lost:
\[
\ce{C2H4 + Cl2 -> C2H4Cl2}
\]
\[
\ce{C2H4Cl2 -> C2H3Cl + HCl}
\]
All the chloroethene then undergoes addition polymerisation.
Use these molar masses:
- \(M(\ce{C2H4})=28.0\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{Cl2})=71.0\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{C2H3Cl})=62.5\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{HCl})=36.5\ \mathrm{g\,mol^{-1}}\)
An engineer compares the initial \(28.0\ \mathrm{g}\) of ethene with the \(62.5\ \mathrm{g}\) of PVC repeat-unit material produced and claims that polymerisation has created \(34.5\ \mathrm{g}\) of matter.
Explain why this conclusion is wrong. Track the mass through the entire downstream route, identify where the additional element enters the product, and explain what a separator placed after cracking can and cannot change about the ethene supply.
Solution 5
No mass is created during polymerisation. The PVC contains chlorine introduced earlier in the process, while hydrogen leaves the route in hydrogen chloride.
Step 1: Calculate the mass entering the first reaction.
One mole of ethene reacts with one mole of chlorine:
\[
\ce{C2H4 + Cl2 -> C2H4Cl2}
\]
The total mass of the reactants is:
\[
\begin{aligned}
m_{\text{reactants}}
&=28.0\ \mathrm{g}+71.0\ \mathrm{g}\\
&=99.0\ \mathrm{g}
\end{aligned}
\]
The reaction produces \(1.00\ \mathrm{mol}\) of 1,2-dichloroethane. Its molar mass is \(99.0\ \mathrm{g\,mol^{-1}}\), so the product has a mass of \(99.0\ \mathrm{g}\).
Step 2: Track the mass during chloroethene production.
The 1,2-dichloroethane decomposes according to:
\[
\ce{C2H4Cl2 -> C2H3Cl + HCl}
\]
One mole produces one mole of chloroethene and one mole of hydrogen chloride.
The combined product mass is:
\[
\begin{aligned}
m_{\text{products}}
&=62.5\ \mathrm{g}+36.5\ \mathrm{g}\\
&=99.0\ \mathrm{g}
\end{aligned}
\]
The total mass remains unchanged. Some of the chlorine introduced in the first reaction leaves in \(\ce{HCl}\), along with hydrogen.
Step 3: Track the mass during polymerisation.
One mole of chloroethene contains enough material to form \(62.5\ \mathrm{g}\) of PVC repeat-unit material, assuming complete polymerisation and no losses.
Addition polymerisation joins the monomers without producing a small-molecule by-product. The \(62.5\ \mathrm{g}\) of chloroethene therefore becomes \(62.5\ \mathrm{g}\) of polymer repeat-unit material.
Step 4: Check the overall mass balance.
The total mass entering the downstream process is:
\[
28.0\ \mathrm{g}+71.0\ \mathrm{g}
=99.0\ \mathrm{g}
\]
The total mass leaving it is:
\[
62.5\ \mathrm{g}+36.5\ \mathrm{g}
=99.0\ \mathrm{g}
\]
There is no violation of conservation of mass.
The apparent increase from \(28.0\ \mathrm{g}\) of ethene to \(62.5\ \mathrm{g}\) of PVC is \(34.5\ \mathrm{g}\). This is the net result of replacing one hydrogen atom in each ethene-derived unit with a chlorine atom. Approximately \(35.5\ \mathrm{g}\) of chlorine is retained in the PVC, while \(1.0\ \mathrm{g}\) of hydrogen leaves in the hydrogen chloride. The net mass increase is \(35.5-1.0=34.5\ \mathrm{g}\).
Step 5: Decide what separation can achieve.
A separator placed after cracking can increase the purity of the ethene stream and, if recovery improves, increase the amount of pure ethene delivered to the PVC plant. It cannot increase the total amount of ethene chemically produced by the cracker.
If the cracker produces too little ethene, the solution must involve the cracking process, such as its feed composition or operating conditions, rather than relying on separation to create molecules that are not there.
The useful next step is to follow the carbon-carbon double bond through the chemistry. In ethene, it allows chlorine to add across the bond; in chloroethene, it allows monomers to join during polymerisation. Tracking which bonds remain intact, which bonds break, and which new bonds form makes it much easier to distinguish separation, cracking, addition reactions, and polymerisation.