The big idea: Organic chemists draw the same molecule in several ways, each showing a different amount of detail. The five you must know are:
- Empirical formula — the simplest whole-number ratio of atoms. - Molecular formula — the actual number of each atom in one molecule. - Structural (full) formula — every atom and every bond drawn out. - Condensed formula — atoms grouped in a line, bonds left implied. - Skeletal formula — only the carbon skeleton (and functional groups); the carbon–hydrogen atoms are implied.
Being able to convert between these is the core skill of this micro.
Each corner and each line-end is a carbon; the H atoms on carbon are not drawn. The –OH group is shown because it is not on a carbon-to-carbon bond.
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Skeletal shorthand — two rules: - Every corner and every line-end is a carbon atom. - Each carbon has enough hydrogen atoms (drawn or not) to make four bonds in total — so the H atoms on carbon are never drawn.
Functional groups (like –OH, –COOH) are drawn, because they are not part of the carbon backbone.
The most detailed formula is the structural one (every bond shown); the most compact is the skeletal one. The empirical formula is found by dividing the atom counts of the molecular formula by their highest common factor.
| Representation | What it shows | Ethanoic acid as an example |
|---|---|---|
| Empirical | simplest whole-number ratio of atoms | CH2O |
| Molecular | the actual number of each atom | C2H4O2 |
| Structural (full) | every atom and every bond drawn out | H–C(H)(H)–C(=O)–O–H |
| Condensed | groups written in a line, bonds implied | CH3COOH |
| Skeletal | carbon skeleton as lines; H on C implied; groups shown | a line with –COOH on the end carbon |
Empirical from molecular: For glucose, C6H12O6, divide every subscript by 6:
C6H12O6 → CH_{2}O.
So glucose and ethanoic acid (C2H4O2) share the empirical formula CH2O even though they are very different molecules.
Worked example — read a skeletal formula
The skeletal formula below has four carbons with a C=C between the 2nd and 3rd. Deduce its molecular formula.
Solution
- Count the carbons: four corners/ends → C_{4}.
- Add the hydrogens: the two end carbons need 3 H each (CH3); the two middle carbons share the C=C, so each carries 1 H. Total H = 3 + 1 + 1 + 3 = 8.
- Combine: C4H8 — this is but-2-ene.
Final answer
C4H8 (but-2-ene).
The double line is the C=C double bond between the 2nd and 3rd carbons. No functional group is attached, so only the backbone is drawn.
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Practice with real exam questions
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Same formula, different structure: Structural isomers are molecules with the same molecular formula but a different arrangement of atoms — a different structural formula.
Because the atoms are connected differently, isomers can have different properties (and may even be in different classes of compound).
C4H10 has two structural isomers — a straight chain and a branched chain:
A straight chain of four carbons.
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The same C₄H₁₀ atoms, but the fourth carbon branches off the middle — a different structural isomer.
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Three ways isomers differ: Two molecules can share a molecular formula but differ in:
- chain branching — butane vs 2-methylpropane (C4H10); - position of a group — propan-1-ol vs propan-2-ol (C3H8O); - functional group / class — ethanol (an alcohol) vs methoxymethane (an ether), both C2H6O.
How this is tested: Two question shapes appear again and again:
- Paper 1A (MCQ): 'which is a correct alternative representation of this molecule?' or 'which structure is a structural isomer of …?' - Paper 2: 'draw a structural isomer of molecule X' or 'state the type of formula shown'.
For a 'draw an isomer' mark you must keep the same molecular formula but show a genuinely different connectivity.
Don't lose the easy mark: A drawing that is just the same molecule flipped or rotated earns nothing — it must be a different structure. Count the atoms in your answer to check the molecular formula still matches.
IB-style question — name the representation (a)
(a) The compound propan-1-ol can be written as CH3CH2CH2OH. State the type of formula this is. [1]
How to score the marks
- The atoms are written grouped in a line (CH3, CH2, …) with the bonds not drawn out individually.
- That is the definition of a condensed structural formula (it is not skeletal — the carbons and hydrogens are written, not implied).
Final answer
A condensed (structural) formula.
IB-style question — draw an isomer (b)
(b) Molecule X is butan-1-ol, C4H10O. Draw the structural formula of one structural isomer of X that is also an alcohol. [1]
How to score the marks
- Keep the molecular formula: any answer must still be C4H10O.
- Change the connectivity: move the –OH group, or branch the chain. For example, butan-2-ol (CH3CH(OH)CH2CH3) puts the –OH on the second carbon.
- Check the class: it still has an –OH group, so it is still an alcohol, as required.
Final answer
Butan-2-ol, CH3CH(OH)CH2CH3 (–OH on carbon 2). 2-methylpropan-1-ol or 2-methylpropan-2-ol would also score.