Use cracking to obtain useful hydrocarbons
Explain why large alkane molecules are cracked and balance a representative thermal cracking equation.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Apply the stated reagents and conditions, balance equations, and keep mechanistic claims within the named examples. Written practice is self-checked
Before you begin
- Use displayed and structural formulas, covalent bond polarity, balanced equations, and the organic reaction terms from Topic 13.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does cracking do to a long-chain hydrocarbon?
What you will learn
- State that cracking breaks larger alkane molecules
- Describe the useful lower-molar-mass products
- Balance a cracking equation
Cracking breaks carbon–carbon bonds
Cracking converts some longer-chain alkanes into smaller molecules when heated over a catalyst such as aluminium oxide. A product mixture can contain shorter alkanes and alkenes.
Crude-oil fractions contain larger hydrocarbons than some markets require. Cracking increases the supply of smaller, useful fuels and reactive alkenes used as chemical feedstocks.
Balance atoms, not product guesses
A cracking equation must conserve each element. C₁₀H₂₂ → C₈H₁₈ + C₂H₄ is balanced and shows one alkane and one alkene product.
The example illustrates a possible product split; it is not a universal single-product rule. Real cracking produces a mixture, so do not claim that every large alkane gives those exact molecules.
Pause and explain
Why are alkenes valuable products of cracking?
Worked example
Balance C₁₀H₂₂ → C₈H₁₈ + C₂H₄ and explain its relevance.
Show the worked solution
- Count carbon atoms: 10 on the left and 8 + 2 on the right.
- Count hydrogen atoms: 22 on the left and 18 + 4 on the right.
- The products are smaller hydrocarbons, including an alkene.
Answer C₁₀H₂₂ → C₈H₁₈ + C₂H₄. Cracking can turn heavier fractions into smaller fuels and alkene feedstocks.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Balance C₁₂H₂₆ → C₉H₂₀ + ? and identify the second product if it is one alkene molecule.
Give me a hint
Subtract the known product atom counts from the reactant counts.
Compare my reasoning
- Carbon remaining is 12 − 9 = 3.
- Hydrogen remaining is 26 − 20 = 6.
- C₃H₆ is propene, an alkene.
C₁₂H₂₆ → C₉H₂₀ + C₃H₆.
Look for these in your work
- Carbon count conserved
- Hydrogen count conserved
- Alkene identified
Explain why a refinery may crack some of its heavier hydrocarbon fraction.
Give me a hint
Link product size to demand and chemical use.
Compare my reasoning
- The feed contains larger hydrocarbons.
- Cracking can make smaller fuel-range molecules and alkenes.
- Alkenes can be used as feedstocks for chemical manufacture.
Cracking helps meet demand for smaller useful fuels and supplies reactive alkenes for further manufacture.
Look for these in your work
- Feed described
- Smaller products named
- Alkene use explained
A proposed cracking equation is C₈H₁₈ → C₅H₁₂ + C₃H₆ + H₂. Diagnose it.
Give me a hint
Count hydrogen atoms on each side before accepting the products.
Compare my reasoning
- The left side contains 18 H atoms.
- The right side contains 12 + 6 + 2 = 20 H atoms.
- Remove H₂ to restore the stated atom balance.
C₈H₁₈ → C₅H₁₂ + C₃H₆ is balanced; the extra H₂ makes the proposed equation unbalanced.
Look for these in your work
- Hydrogen counted
- Imbalance identified
- Correction supplied
Common mistakes
- Calling cracking addition or polymerisation
- Failing to conserve hydrogen atoms
- Claiming a single fixed product for every cracking reaction
What can you explain now?
What does cracking do to a long-chain hydrocarbon?
Compare with the explanation
Breaks it into smaller hydrocarbon molecules
Cracking breaks bonds in a larger hydrocarbon to form smaller products.
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Make it stick
Why does cracking produce both smaller fuel molecules and useful alkenes?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.