3.2.1Additional lessons
Representing Formulas of Organic Compounds — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 44 of 98
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 6Types of Formula
Different formulas show different amounts of information about the same molecule.
Empirical
- Empirical The simplest whole-number ratio of atoms of each element, such as CH₂ for ethene and cyclohexane.
Molecular
- Molecular The actual number of atoms of each element in one molecule, such as C₄H₁₀ for butane.
Full structural
- Full structural A diagram showing every atom and every bond, so the arrangement of atoms is completely clear.
Condensed structural
- Condensed structural A shortened line formula that shows groups of atoms, such as CH₃CH₂CH₂CH₃ for butane.
Part 2 of 6Skeletal and 3D Formulas
Skeletal and 3D Formulas
Skeletal
- Skeletal Lines show the carbon backbone, each corner and end is a carbon atom, and hydrogens on carbon are not drawn.
Functional groups
- Functional groups Atoms other than carbon and hydrogen, such as O in –OH, are written out in a skeletal formula.
Wedge bonds
- Wedge bonds Solid wedges show bonds coming forward and hashed wedges show bonds going back in 3D representations.
Choosing
- Choosing Use skeletal formulas for large molecules and full structural formulas to show all bonds clearly.
| Compound | Molecular | Condensed structural | Empirical |
|---|---|---|---|
| Butane | C₄H₁₀ | CH₃CH₂CH₂CH₃ | C₂H₅ |
| Ethanol | C₂H₆O | CH₃CH₂OH | C₂H₆O |
| Ethanoic acid | C₂H₄O₂ | CH₃COOH | CH₂O |
| Ethene | C₂H₄ | CH₂=CH₂ | CH₂ |
Part 3 of 6From Data to Formula
1
Divide the mass or percentage of each element by its relative atomic mass to get the amount in moles.
2
Divide each amount by the smallest to get a ratio.
3
Multiply to turn the ratio into whole numbers; this gives the empirical formula.
4
Divide the molar mass by the empirical formula mass to find the multiple.
5
Multiply the empirical formula by this multiple to get the molecular formula.
Part 4 of 6Drawing Conventions
Four bonds
- Four bonds Each carbon atom in an organic compound forms four covalent bonds, counting a double bond as two.
Hydrogens
- Hydrogens In skeletal formulas calculate the number of hydrogens on each carbon by making up four bonds.
Isomers
- Isomers Different structures can share one molecular formula, so the molecular formula is not enough to define a compound.
Repeating groups
- Repeating groups Use brackets in condensed formulas for repeated groups, as in CH₃(CH₂)₃CH₃.
Part 5 of 6Worked Example
Question
- A compound contains 40.0% C, 6.7% H and 53.3% O by mass.
- Its molar mass is 60 g mol⁻¹.
- Find its empirical and molecular formulas.
- Moles in 100 g: C 40.0 ÷ 12.01 = 3.33; H 6.7 ÷ 1.01 = 6.63; O 53.3 ÷ 16.00 = 3.33.
- Divide by 3.33: C 1, H 2, O 1, so the empirical formula is CH₂O (mass 30).
- 60 ÷ 30 = 2, so the molecular formula is C₂H₄O₂.
Answer
- Empirical formula CH₂O;
- molecular formula C₂H₄O₂.
Part 6 of 6Exam Checklist & Tips
3.2.1 – Key Points for the Exam
- Empirical formula The simplest whole-number ratio of atoms in a compound.
- Molecular formula The actual numbers of atoms of each element in a molecule.
- Structural formula A formula showing how atoms are arranged and bonded.
- Skeletal formula A simplified diagram showing the carbon skeleton and functional groups only.
Common Mistakes
- Confusing empirical and molecular formulas, such as writing CH₂O for a compound of molar mass 60.
- Counting a corner of a skeletal formula as a hydrogen atom.
- Giving a carbon atom five bonds.
- Drawing a condensed formula that does not show how groups are connected.
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