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3.2.1Additional lessons

Representing Formulas of Organic Compounds — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 44 of 98

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Part 1 of 6Types of Formula

3.2.1 Representing Formulas of Organic Compounds

Different formulas show different amounts of information about the same molecule.

Empirical
  • The simplest whole-number ratio of atoms of each element, such as CH₂ for ethene and cyclohexane.
Molecular
  • The actual number of atoms of each element in one molecule, such as C₄H₁₀ for butane.
Full structural
  • A diagram showing every atom and every bond, so the arrangement of atoms is completely clear.
Condensed structural
  • A shortened line formula that shows groups of atoms, such as CH₃CH₂CH₂CH₃ for butane.

Part 2 of 6Skeletal and 3D Formulas

Skeletal and 3D Formulas
Skeletal
  • Lines show the carbon backbone, each corner and end is a carbon atom, and hydrogens on carbon are not drawn.
Functional groups
  • Atoms other than carbon and hydrogen, such as O in –OH, are written out in a skeletal formula.
Wedge bonds
  • Solid wedges show bonds coming forward and hashed wedges show bonds going back in 3D representations.
Choosing
  • Use skeletal formulas for large molecules and full structural formulas to show all bonds clearly.
CompoundMolecularCondensed structuralEmpirical
ButaneC₄H₁₀CH₃CH₂CH₂CH₃C₂H₅
EthanolC₂H₆OCH₃CH₂OHC₂H₆O
Ethanoic acidC₂H₄O₂CH₃COOHCH₂O
EtheneC₂H₄CH₂=CH₂CH₂

Part 3 of 6From Data to Formula

1
Divide the mass or percentage of each element by its relative atomic mass to get the amount in moles.
2
Divide each amount by the smallest to get a ratio.
3
Multiply to turn the ratio into whole numbers; this gives the empirical formula.
4
Divide the molar mass by the empirical formula mass to find the multiple.
5
Multiply the empirical formula by this multiple to get the molecular formula.

Part 4 of 6Drawing Conventions

Four bonds
  • Each carbon atom in an organic compound forms four covalent bonds, counting a double bond as two.
Hydrogens
  • In skeletal formulas calculate the number of hydrogens on each carbon by making up four bonds.
Isomers
  • Different structures can share one molecular formula, so the molecular formula is not enough to define a compound.
Repeating groups
  • Use brackets in condensed formulas for repeated groups, as in CH₃(CH₂)₃CH₃.

Part 5 of 6Worked Example

Question

  • A compound contains 40.0% C, 6.7% H and 53.3% O by mass.
  • Its molar mass is 60 g mol⁻¹.
  • Find its empirical and molecular formulas.

  1. Moles in 100 g: C 40.0 ÷ 12.01 = 3.33; H 6.7 ÷ 1.01 = 6.63; O 53.3 ÷ 16.00 = 3.33.
  2. Divide by 3.33: C 1, H 2, O 1, so the empirical formula is CH₂O (mass 30).
  3. 60 ÷ 30 = 2, so the molecular formula is C₂H₄O₂.

Answer

  • Empirical formula CH₂O;
    • molecular formula C₂H₄O₂.

Part 6 of 6Exam Checklist & Tips

3.2.1 – Key Points for the Exam

  • Empirical formula The simplest whole-number ratio of atoms in a compound.
  • Molecular formula The actual numbers of atoms of each element in a molecule.
  • Structural formula A formula showing how atoms are arranged and bonded.
  • Skeletal formula A simplified diagram showing the carbon skeleton and functional groups only.

Common Mistakes

  • Confusing empirical and molecular formulas, such as writing CH₂O for a compound of molar mass 60.
  • Counting a corner of a skeletal formula as a hydrogen atom.
  • Giving a carbon atom five bonds.
  • Drawing a condensed formula that does not show how groups are connected.

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