Formulae in Organic Chemistry Revision Notes | A Level AQA Chemistry

1. Why Organic Chemists Need Multiple Formula Types

Imagine trying to describe a building using only the number of bricks it contains. You would know the quantity, but nothing about the shape, layout, or structure. Organic molecules face exactly the same problem — a single formula type cannot tell the whole story.

AQA requires you to recognise, interpret, and draw six distinct formula types: empirical, molecular, general, structural, displayed, and skeletal. Each one reveals a different layer of information about an organic compound, and examiners will ask you to switch between them fluently.

The good news is that these formula types form a logical hierarchy — from the most stripped-back ratio right through to a full picture of every bond. Work through them in that order and the distinctions become clear.

2. Empirical and Molecular Formulae

Think of the empirical formula as the reduced fraction of chemistry. Glucose has the molecular formula $C_6H_{12}O_6$, but dividing every subscript by 6 gives the empirical formula $CH_2O$. The ratio 1:2:1 is preserved, but the actual molecule size is lost.

This matters because two completely different compounds can share the same empirical formula. Ethanoic acid ($C_2H_4O_2$) and glucose ($C_6H_{12}O_6$) both reduce to $CH_2O$. The empirical formula alone cannot distinguish them.

The molecular formula gives you the true count of atoms. Ethane is $C_2H_6$, not $CH_3$ — even though $CH_3$ is its empirical formula. To move from empirical to molecular, you need the relative molecular mass ($M_r$), which you can obtain from a mass spectrum or combustion analysis data.

Worked example: Finding the Molecular Formula from Empirical Formula and Mr

A compound has the empirical formula $CH_2$ and a relative molecular mass of 56. Determine its molecular formula.

1. $M_r(CH_2) = 12 + (2 \times 1) = 14$ — Add the relative atomic masses: C = 12, H = 1 each, so 2 H atoms contribute 2.
2. $n = \frac{56}{14} = 4$ — This tells you the molecular formula is 4 times the empirical formula.
3. $(CH_2)_4 \rightarrow C_4H_8$ — Every atom count in the empirical formula is multiplied by 4.
4. $M_r(C_4H_8) = (4 \times 12) + (8 \times 1) = 48 + 8 = 56$ ✓ — The calculated Mr matches the given value, confirming the answer.

Answer: Molecular formula: $C_4H_8$

3. General Formula

Now that you can distinguish empirical from molecular formulae, there is a third type that describes an entire family of compounds rather than one specific molecule.

Alkanes all follow the pattern $C_nH_{2n+2}$. When $n = 1$, you get methane $CH_4$; when $n = 2$, ethane $C_2H_6$; when $n = 3$, propane $C_3H_8$. The general formula captures the structural regularity of the homologous series in a single expression.

Notice that the general formula for alkenes ($C_nH_{2n}$) is the same as the empirical formula pattern for cycloalkanes. This is a reminder that a general formula alone does not identify a unique compound — you still need structural information to distinguish, for example, cyclohexane from hex-1-ene.

4. Structural, Displayed, and Skeletal Formulae

The three formula types above — empirical, molecular, and general — all deal with how many atoms are present. The next three types show how those atoms are connected, which is where organic chemistry really comes alive.

For propan-1-ol, the structural formula is written as $CH_3CH_2CH_2OH$. You can read along the carbon chain from left to right, and the $OH$ group at the end tells you immediately where the functional group sits. This is far more informative than the molecular formula $C_3H_8O$, which gives no positional information at all.

Structural formulae are particularly useful for distinguishing structural isomers. Propan-1-ol ($CH_3CH_2CH_2OH$) and propan-2-ol ($CH_3CH(OH)CH_3$) share the molecular formula $C_3H_8O$, but their structural formulae immediately reveal the different positions of the hydroxyl group.

The displayed formula is the most explicit representation. For ethanol, every C–H bond, the C–C bond, the C–O bond, and the O–H bond are all drawn individually. This makes it invaluable for showing bonding in functional groups and for drawing reaction mechanisms, but it becomes unwieldy for larger molecules.

Skeletal formulae are the language of practising organic chemists. Once you learn to read them, they convey structural information far faster than any other format. The zig-zag convention means each carbon is at a vertex or line-end, and you infer the number of hydrogens on each carbon from the valency rule: carbon forms four bonds, so any 'missing' bonds are to hydrogen.

For butane ($C_4H_{10}$), the skeletal formula is a zig-zag of three lines with four vertices — no H atoms shown on the chain carbons at all. For butan-2-ol, the $OH$ group appears explicitly at the second vertex, and the oxygen's hydrogen is also shown because it is not bonded to carbon.

Worked example: Drawing the Three Structural Formula Types for Butanoic Acid

Draw the structural formula, displayed formula, and skeletal formula for butanoic acid ($C_3H_7COOH$, i.e. $C_4H_8O_2$).

1. 4 carbons in total; terminal $COOH$ group on C-1 — Butanoic acid belongs to the carboxylic acid series. The $COOH$ group is always at the end of the chain and counts as one of the four carbons.
2. $CH_3CH_2CH_2COOH$ — Group each carbon with its attached hydrogens, reading left to right along the chain. The carboxyl group is written as $COOH$ at the end.
3. Show all bonds: three C–C bonds, seven C–H bonds, one C=O bond, one C–O bond, one O–H bond — Every single bond is a line, the double bond in $C=O$ is a double line. No atom or bond is omitted. Total bonds drawn: 13 individual bond lines.
4. Zig-zag of three C–C lines; $COOH$ written explicitly at the right-hand end — The three CH$_2$/CH$_3$ carbons form the zig-zag skeleton. The carboxyl carbon is part of the $COOH$ label. Hydrogen atoms on the chain carbons are not shown.
5. 3 zig-zag carbons + 1 carboxyl carbon = 4 C; inferred H: $CH_3$ (3H) + $CH_2$ (2H) + $CH_2$ (2H) + $COOH$ (1H) = 8 H; 2 O from $COOH$ — Atom count from skeletal formula matches molecular formula $C_4H_8O_2$. This cross-check confirms the skeletal formula is correct.

Answer: Structural: $CH_3CH_2CH_2COOH$; Displayed: all 13 bonds drawn explicitly; Skeletal: three-line zig-zag with $COOH$ at the terminus.

5. Comparing All Six Formula Types

You now have all six formula types in your toolkit. Before moving to exam technique, it is worth placing them side by side so the distinctions are sharp and retrievable under pressure.

Notice that propan-1-ol and propan-2-ol share the same empirical formula, molecular formula, and general formula. Only the structural, displayed, and skeletal formulae can distinguish them — which is precisely why these three representations are so important in organic chemistry.

6. Drawing Practice: Structural, Displayed, and Skeletal Formulae

Knowing the definitions is necessary but not sufficient — AQA will ask you to produce drawings. The most reliable approach is to follow a consistent three-step method for each formula type.

Bridge: the fastest way to avoid formula-type errors is to ask what the representation is trying to show: atom ratio, full atom count, family pattern, or exact connectivity.

For skeletal formulae in particular, students frequently make two errors: forgetting to show the $OH$ hydrogen in alcohols, and drawing the wrong number of line segments for the carbon chain. The worked example below addresses both.

Worked example: Drawing Structural, Displayed, and Skeletal Formulae for Pent-2-ene

Draw the structural formula, displayed formula, and skeletal formula for pent-2-ene ($C_5H_{10}$, double bond between C-2 and C-3).

1. 5 carbons in a chain: C1–C2=C3–C4–C5 — The name 'pent-2-ene' tells you: 5 carbons (pent-), one double bond (-ene), starting at C-2 (2-).
2. C1: $CH_3$ (3H); C2: $CH=$ (1H); C3: $=CH$ (1H); C4: $CH_2$ (2H); C5: $CH_3$ (3H). Total H: 3+1+1+2+3 = 10 ✓ — Each carbon forms 4 bonds. Subtract bonds to other carbons (and the double bond counts as 2) to find the number of H atoms. The total of 10 H matches $C_5H_{10}$.
3. $CH_3CH=CHCH_2CH_3$ — Read along the chain. The double bond is shown as $=$ between C-2 and C-3. Each carbon group is written with its H atoms.
4. Draw all 5 C atoms, all 10 H atoms, 3 C–C single bonds (C1–C2, C3–C4, C4–C5), 1 C=C double bond (C2=C3, drawn as 2 lines), and 10 C–H bonds — Every bond is a line. The double bond between C-2 and C-3 is drawn as two parallel lines, contributing 2 bond lines. Count: 3 C–C single bond lines + 2 lines for C=C + 10 C–H bond lines = 15 bond lines total.
5. Zig-zag of 4 line segments; double line between the 2nd and 3rd segments (C2=C3); no H atoms shown on the chain — 5 carbons require 4 line segments. The double bond appears as a double line at the second segment. No H atoms are drawn — they are inferred from valency.

Answer: Structural: $CH_3CH=CHCH_2CH_3$; Displayed: 15 bond lines with C2=C3 as a double bond (2 lines); Skeletal: 4-segment zig-zag with a double line at the second segment.

7. Exam Technique and Common Mistakes

With all six formula types covered and drawing practice complete, the final step is to translate this knowledge into mark-winning exam responses. AQA questions on formulae tend to fall into three categories: identification, conversion, and drawing — and each has its own pitfalls.