Deduce molecular and empirical formulas from structures
Count atoms in a structural, displayed or skeletal formula and reduce a molecular formula to its empirical ratio.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the published functional-group scope and nomenclature limits; draw formula and mechanism evidence with correct connectivity, electron movement and stereochemistry. Written practice is self-checked
Before you begin
- Use covalent structures, valency, molecular formulas, lone pairs, and single, double and triple bonds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does a molecular formula report?
What you will learn
- Count carbon and heteroatoms from a structure
- Complete implicit carbon valencies
- Deduce molecular and empirical formulas
Count atoms without changing the connectivity
In a displayed formula, count every symbol. In a condensed formula, each repeated CH₃, CH₂ or CH group gives the stated atoms. In a skeletal formula, each endpoint and vertex is carbon; add omitted C-bound H atoms to make carbon valency four.
Heteroatoms and their attached hydrogens are shown explicitly. Check that the final atom counts satisfy the bonds and valencies in the supplied structure.
Molecular and empirical formulas answer different questions
The molecular formula gives the actual number of each type of atom in one molecule. The empirical formula gives the simplest whole-number ratio; divide all subscripts by their greatest common divisor.
A structure may have the same molecular formula as another compound but different connectivity. Formula deduction does not by itself identify a unique isomer.
Pause and explain
What is the empirical formula of C₄H₈O₂?
Worked example
Deduce molecular and empirical formulas for ethanol, CH₃CH₂OH.
Show the worked solution
- Count two carbon atoms and six hydrogens, including the O–H hydrogen.
- There is one oxygen atom, so the molecular formula is C₂H₆O.
- The subscripts have no common factor above 1, so this is also the empirical formula.
Answer Molecular formula C₂H₆O; empirical formula C₂H₆O.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Count atoms and give the molecular formula of CH₃CH₂CH₂OH.
Give me a hint
Count the carbons and hydrogens in each group, including O–H.
Compare my reasoning
- There are three carbon atoms.
- Hydrogen count is 3 + 2 + 2 + 1 = 8.
- One oxygen gives C₃H₈O.
C₃H₈O.
Look for these in your work
- Carbon count
- All hydrogens included
- Oxygen included
Find the empirical formula of C₆H₁₂.
Give me a hint
Divide subscripts by their greatest common divisor.
Compare my reasoning
- The greatest common divisor of 6 and 12 is 6.
- Divide both subscripts by 6.
- The simplest ratio is CH₂.
CH₂.
Look for these in your work
- Common factor found
- All subscripts reduced
- Simplest ratio stated
A skeletal formula has five carbon vertices/ends and only single bonds in an open chain. Give its alkane molecular formula and explain the count.
Give me a hint
Apply CₙH₂ₙ₊₂ for an open-chain alkane.
Compare my reasoning
- There are five carbon atoms, so n = 5.
- Substitute into CₙH₂ₙ₊₂.
- The formula is C₅H₁₂.
C₅H₁₂.
Look for these in your work
- Carbon vertices counted
- General formula selected
- Hydrogen count calculated
Common mistakes
- Forgetting the hydrogen attached to O or N
- Treating an empirical formula as the only possible molecular formula
- Counting a skeletal carbon twice at a branch
What can you explain now?
What does a molecular formula report?
Compare with the explanation
The actual number of each atom in a molecule
A molecular formula gives the actual atom counts.
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Make it stick
Why can two different organic structures share the same molecular formula?
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