Structural formulae

Write C4H10C_4H_{10} on a page and you know butane has four carbons and ten hydrogens — but not which atom touches which. Two different arrangements of those same fourteen atoms give two different substances with different boiling points and different chemistry. A structural formula removes the ambiguity: it shows the actual bonds, atom by atom, so the connectivity — not just the head-count — is on the page. This article builds structural formulae from one non-negotiable rule, carbon's tetravalency, and shows how to move from a molecular formula to a full or condensed structural formula for alkanes, alkenes, alkynes, alcohols and carboxylic acids — exactly the ground this ICSE Class 10 topic covers.

The core insight

A structural formula fixes every atom's bonding partners so that carbon always shows four bonds, hydrogen one, and oxygen two — turning a bare atom count like C4H10C_4H_{10} into an unambiguous map of the molecule.

The core idea

Molecular formula vs structural formula. A molecular formula, such as C4H10C_4H_{10} for butane, is only a tally of atoms — it says nothing about arrangement. A structural formula shows every bond, so it also fixes what kind of molecule you are looking at: two molecules can share a molecular formula yet differ in structure (that is the seed of isomerism, met later in the same unit).

Full vs condensed structural formula. A full (expanded) structural formula draws every single C–C and C–H bond as a line. A condensed structural formula keeps the carbon skeleton and its side groups explicit but leaves bonds to hydrogen implied — this is the form you will use for nearly every answer at this level. Butane, condensed, is written

CH3−CH2−CH2−CH3CH_3-CH_2-CH_2-CH_3

where: each dash marks one C–C bond and every carbon's remaining bonds (to hydrogen) are understood rather than drawn.

The governing valencies. Three numbers control every structure you will ever draw here:

  • carbon is tetravalent — exactly 4 bonds total, to any mixture of C, H or O;
  • hydrogen is monovalent — exactly 1 bond;
  • oxygen is divalent — exactly 2 bonds, which is why –OH and the C=O of –COOH always balance.

A double bond (alkenes) consumes two of a carbon's four valencies at that link; a triple bond (alkynes) consumes three. This is why unsaturated chains carry fewer hydrogens than the saturated chain of the same carbon count.

Reading the class off the molecular formula. The general-formula patterns tell you, before you draw a single bond, what kind of skeleton to expect:

CnH2n+2 (alkane),CnH2n (alkene),CnH2n−2 (alkyne)C_nH_{2n+2}\ (\text{alkane}), \qquad C_nH_{2n}\ (\text{alkene}), \qquad C_nH_{2n-2}\ (\text{alkyne})

where: nn is the number of carbon atoms in the chain. Once you know nn and the class, you know exactly how many double or triple bonds to place and how many hydrogens must fill the rest.

Functional groups. For alcohols the –OH group replaces one hydrogen on a carbon of the chain; for carboxylic acids the –COOH group's carbon is counted as one of the chain's own carbons, doubly bonded to one oxygen and singly bonded to a second oxygen that carries the acidic hydrogen.

Figure
Tetravalent carbon in methane compared with the condensed structural formula of butane
Carbon always shows four bonds — the rule behind every structure methane: 4 bonds on 1 carbon C H H H H 4 bonds to H, none to spare butane, condensed structural formula CH3-CH2-CH2-CH3 bonds to H implied, C-C bonds shown end carbons: 3H each. middle carbons: 2H each
Derivation ABuilding the condensed structural formula of butane from its molecular formula, checking tetravalency at every carbon

The goal is to move from the bare atom count C4H10C_4H_{10} to a fully specified, bond-by-bond condensed structural formula, confirming along the way that every carbon really does show exactly four bonds and that the resulting hydrogen count matches the given molecular formula — the same logic that must be applied to any straight-chain alkane, alkene or alkyne asked for in this topic.

Step 1. Identify the class from the general formula. With n=4n = 4 carbons and 1010 hydrogens, check 2n+2=2(4)+2=102n+2 = 2(4)+2 = 10, which matches — so butane fits CnH2n+2C_nH_{2n+2}, the alkane pattern, meaning every C–C bond in the chain is a single bond.

Step 2. Lay out the carbon skeleton as a straight chain of four carbons: C−C−C−CC-C-C-C. Label them C1C_1 (leftmost) through C4C_4 (rightmost); this fixes only the C–C connectivity, not yet the hydrogens.

Step 3. Fill each carbon's remaining valencies with hydrogen. C1C_1 uses 1 bond to link to C2C_2, leaving 3 bonds free — 3 hydrogens. C2C_2 uses 2 bonds (to C1C_1 and C3C_3), leaving 2 bonds free — 2 hydrogens. By symmetry C3C_3 also takes 2 hydrogens and C4C_4 (bonded only to C3C_3) takes 3 hydrogens.

Step 4. Write the condensed structural formula: CH3−CH2−CH2−CH3CH_3-CH_2-CH_2-CH_3. Verify tetravalency at C2C_2: 1 bond to C1C_1 + 1 bond to C3C_3 + 2 bonds to H = 4 bonds total, satisfying carbon's valency. Verify the hydrogen total: 3+2+2+3=103+2+2+3 = 10, matching C4H10C_4H_{10} exactly, so the structure is consistent with the given molecular formula. Verified: hydrogen count 3+2+2+3=103+2+2+3=10 matches C4H10C_4H_{10}; every carbon counted at exactly 4 bonds.

Worked example

Question: Draw the condensed structural formula of propane (C3H8C_3H_8) and verify that every carbon atom is tetravalent.

Step 1. Match the general formula: n=3n=3 gives 2n+2=82n+2 = 8, matching C3H8C_3H_8, so propane is a saturated, straight-chain alkane with only single bonds.

Step 2. Lay out three carbons in a chain: C−C−CC-C-C. The two end carbons each use 1 bond to their single neighbour, leaving 3 bonds free for hydrogen; the middle carbon uses 2 bonds (one to each neighbour), leaving 2 bonds free.

Step 3. Fill in hydrogens accordingly: end carbons get 3 H each, the middle carbon gets 2 H, giving the condensed structural formula CH3−CH2−CH3CH_3-CH_2-CH_3.

Step 4. Check both conditions. Hydrogen total: 3+2+3=83+2+3 = 8, matching C3H8C_3H_8. Tetravalency at the middle carbon: 1 bond to the left C + 1 bond to the right C + 2 bonds to H = 4 bonds. Answer: CH3−CH2−CH3CH_3-CH_2-CH_3, with every carbon confirmed tetravalent.

Competitive stretch

Extend the same check to ethanoic acid, CH3COOHCH_3COOH. The carboxyl carbon is doubly bonded to one oxygen and singly bonded to a second oxygen that itself bonds to hydrogen, plus singly bonded to the methyl carbon: 11 (to CH3CH_3) +2+ 2 (double bond to O) +1+ 1 (to –OH oxygen) =4= 4 bonds, tetravalent as required. Each oxygen shows exactly 2 bonds — the double-bonded O has both bonds to the same carbon, and the –OH oxygen has one bond to C and one to H — so oxygen's divalency is satisfied throughout. Full structural formula: CH3−C(=O)−OHCH_3-C(=O)-OH, conventionally written CH3COOHCH_3COOH.

Common traps

  • Drawing a carbon with 3 or 5 total bonds instead of 4 — always recount every bond (C, H, or O, with a double bond counted as 2 and a triple bond as 3) before finalising a structure.
  • Forgetting that a double or triple bond uses more than one of a carbon's four valencies, so that carbon carries fewer hydrogens than a saturated one at the same position.
  • Answering "draw the structure" with just the molecular formula (C4H8C_4H_8) instead of showing the actual bond arrangement — the two are not interchangeable in this topic.
  • Leaving an oxygen atom with the wrong bond count in –OH or –COOH — oxygen must always show exactly 2 bonds, never 1 or 3.

Ask the tutor: Want to practise converting molecular formulae into condensed structural formulae for alkanes, alkenes, alcohols and carboxylic acids, with your tetravalency count checked step by step?

Why can two molecules share the same molecular formula but be different substances

Because a molecular formula only counts atoms, not their arrangement. If the same set of atoms can be joined in more than one valid way while still respecting each atom's valency, you get different substances with the same molecular formula but different structural formulae — different bond arrangements mean different properties, such as boiling point. This idea, called isomerism, builds directly on structural formulae and is usually the next topic after this one.

Do I have to draw every single hydrogen atom on an exam

Not usually. The condensed structural formula, where bonds to hydrogen are implied rather than drawn, is accepted for straight-chain alkanes, alkenes, alkynes, alcohols and carboxylic acids at this level — it is faster and less error-prone. Only draw the full expanded formula, every bond shown, if the question explicitly asks for it. Either way, the underlying valency count at each carbon must still work out to 4.

How do I know how many hydrogens to put on each carbon without counting bonds one by one every time

Use the general formula to fix the total hydrogen count first, then distribute it using the pattern: end carbons of a saturated straight chain each get 3 H, every interior carbon gets 2 H, because interior carbons use two of their four bonds linking to two neighbours. For a double or triple bond, subtract 1 or 2 hydrogens respectively from each carbon at that link compared to the saturated case, since those valencies are used up by the extra bond instead.

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