3.1.2Physical Chemistry
Types of Formulae — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 65 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5Formulae That Count Atoms
Some formulae tell only the numbers of atoms, not their arrangement.
Empirical formula
- Empirical formula The simplest whole-number ratio of atoms of each element in a compound, such as CH₂ for ethene.
Molecular formula
- Molecular formula The actual number of atoms of each element in one molecule, such as C₄H₈ for butene.
General formula
- General formula An algebraic formula for a homologous series, such as CₙH₂ₙ₊₂ for alkanes.
Relationship
- Relationship The molecular formula is a whole-number multiple of the empirical formula, found using relative molecular mass.
Part 2 of 5Formulae That Show Structure
Formulae That Show Structure
Structural formula
- Structural formula Shows the minimum detail needed to unambiguously show the arrangement of atoms, such as CH₃CH(OH)CH₃ for propan-2-ol.
Displayed formula
- Displayed formula Shows all atoms and all covalent bonds, including the O–H bond in an alcohol.
Skeletal formula
- Skeletal formula Each line end or vertex represents a carbon atom, and functional groups such as –OH are written explicitly.
| Type | What it shows | Example (butane) |
|---|---|---|
| Structural | Atoms grouped to show the arrangement of carbons | CH₃CH₂CH₂CH₃ |
| Displayed | Every atom and every bond drawn | Full diagram with all C–H and C–C bonds |
| Skeletal | Carbon chain as zigzag lines, hydrogens omitted | Zigzag with 3 bonds |
Part 3 of 5Choosing and Converting
1
Use the empirical formula when only the composition by mass is known from analysis data.
2
Divide each percentage by the relative atomic mass, then divide by the smallest result, to find the ratio of atoms.
3
Compare the empirical formula mass with the relative molecular mass to find the multiple, giving the molecular formula.
4
Use displayed formulae when asked to show every bond, and skeletal formulae to draw larger molecules efficiently.
5
Use structural formulae to distinguish isomers, because they show how the atoms are connected.
Part 4 of 5Worked Example
Question
- A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass.
- Its relative molecular mass is 56.
- Find the empirical and molecular formulae.
- Moles C = 85.7 ÷ 12.0 = 7.14; moles H = 14.3 ÷ 1.0 = 14.3.
- Divide by the smaller value (7.14): C = 1.00 and H = 2.00, so the ratio is 1 : 2.
- Empirical formula = CH₂, with empirical mass 12 + 2 = 14.
- 56 ÷ 14 = 4, so multiply the empirical formula by 4.
Answer
- Empirical formula CH₂;
- molecular formula C₄H₈
Part 5 of 5Exam Checklist & Tips
3.1.2 – Key Points for the Exam
- Empirical formula The simplest whole-number ratio of atoms of each element in a compound.
- Molecular formula The actual number of atoms of each element in a molecule.
- Displayed formula A diagram showing every atom and every covalent bond in a molecule.
- Skeletal formula A simplified diagram in which carbon atoms are vertices and hydrogens on carbons are omitted.
Common Mistakes
- Giving the empirical formula when the molecular formula is requested.
- Omitting hydrogen atoms attached to non-carbon atoms, such as the H in an –OH group, in skeletal formulae.
- Drawing bonds to the wrong atom, such as O–H written as C–H–O in a displayed formula.
- Writing CH₃CH₂OH as an empirical formula, when it is a structural formula.
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