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3.1.1Organic Chemistry

Formulae of organic compounds — Edexcel AS Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 14 of 42

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Part 1 of 5Organic Compounds and Hydrocarbons

3.1.1 Formulae of organic compounds
Organic chemistry
  • This is the study of carbon compounds, which are numerous because carbon forms strong covalent bonds with itself.
Chains
  • Carbon atoms can join in long chains, branches and rings, giving an almost unlimited number of different molecules.
Hydrocarbon
  • A hydrocarbon contains carbon and hydrogen atoms only, so ethanol is not a hydrocarbon because it contains oxygen.
Saturated
  • A saturated hydrocarbon has only single carbon-carbon bonds, as in propane.
Unsaturated
  • An unsaturated hydrocarbon has at least one carbon-carbon double or triple bond, as in propene.

Part 2 of 5Types of Formula

Types of Formula
Empirical
  • The simplest whole-number ratio of atoms of each element in the compound, such as CH₂ for ethene.
Molecular
  • The actual number of atoms of each element in one molecule, such as C₄H₈.
General
  • A formula that fits a whole homologous series, such as CₙH₂ₙ₊₂ for the alkanes.
Structural
  • Shows the arrangement of atoms group by group without drawing every bond, such as CH₃CH₂CH₃.
Displayed
  • Shows every atom and every bond in the molecule, drawn flat on the page.
TypeExample for butane
EmpiricalC₂H₅
MolecularC₄H₁₀
StructuralCH₃CH₂CH₂CH₃
SkeletalZig-zag line with 3 bonds

Part 3 of 5Skeletal Formulae and Drawing Rules

1
A skeletal formula is a simplified displayed formula in which the carbon atoms and the hydrogen atoms bonded to them are not written.
2
Each corner or end of a line represents a carbon atom, so a zig-zag with three lines shows a four-carbon chain.
3
Hydrogen atoms bonded to carbon are left out, but atoms of other elements, such as O or Cl, are always drawn.
4
Functional groups are drawn in full, such as OH for an alcohol, to show where the group is attached.
5
Always check that each carbon has four bonds in total, filling in the missing ones as implied hydrogen atoms.

Part 4 of 5Worked Example

Question

  • A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has a relative molecular mass of 56.
  • Find its empirical and molecular formulae.
  • (Ar: C 12.0, H 1.0.)

  1. Divide by ArC = 85.7 ÷ 12.0 = 7.14 and H = 14.3 ÷ 1.0 = 14.3.
  2. Divide by the smaller number (7.14): C : H = 1 : 2, so the empirical formula is CH₂.
  3. Mass of CH₂ = 14.0, so 56 ÷ 14.0 = 4.
  4. Multiply the empirical formula by 4.

Answer Empirical formula CH₂, molecular formula C₄H₈

Part 5 of 5Exam Checklist & Tips

3.1.1 – Key Points for the Exam

  • Hydrocarbon A compound containing only carbon and hydrogen.
  • Empirical formula The simplest whole-number ratio of atoms in a compound.
  • Molecular formula The actual numbers of each type of atom in one molecule.
  • Displayed formula A formula showing all atoms and all bonds.
  • Skeletal formula A simplified formula showing the carbon skeleton and functional groups only.

Common Mistakes

  • Writing the empirical formula of butane as C₄H₁₀ rather than C₂H₅.
  • Forgetting that every corner and end of a skeletal formula is a carbon atom bearing enough hydrogens to make four bonds.
  • Calling any compound containing hydrogen and carbon a hydrocarbon, even if it has oxygen or halogens as well.
  • Leaving out hydrogen atoms attached to oxygen in a skeletal formula, where the O–H must be shown.

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