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*Revision Materials and Past Papers* 2.1.1 Atomic structure and isotopes 2.1.2 Compounds, formulae and equations 2.1.3 Amount of substance 2.1.4 Acids 2.1.5 Redox 2.2.1 Electron structure 2.2.2 Bonding and structure 3.1.1 Periodicity 3.1.2 Group 2 3.1.3 The halogens 3.1.4 Qualitative analysis 3.2.1 Enthalpy 3.2.2 Reaction Rates 3.2.3 Chemical equilibrium 4.1 Basic concepts and hydrocarbons 4.1.2 Alkanes 4.1.3 Alkenes 4.2.1 Alcohols 4.2.2 Haloalkanes 4.2.3 Organic synthesis 4.2.4 Analytical techniques 5.1.1 How fast? 5.1.2 How far? 5.1.3 Acids, bases and buffers 5.2.1 Lattice enthalpy 5.2.2 Enthalpy and entropy 5.2.3 Redox and electrode potentials 5.3.1 Transition elements 5.3.2 Qualitative analysis 6.1.1 Aromatic compounds 6.1.2 Carbonyl compounds 6.1.3 Carboxylic acids and esters 6.2.1 Amines 6.2.2 Amino acids, amides and chirality 6.2.3 Polyesters and polyamides 6.2.4 Carbon–carbon bond formation 6.2.5 Organic synthesis 6.3.1 Chromatography and qualitative analysis 6.3.2 Spectroscopy Required Practicals

2.1.3 Amount of substance

Determination of formulaePercentage yields and atom economyThe Mole and Calculation of reacting masses, gas volumes and mole concentrations

Amount of Substance, Determination of Formulae

Specification Reference 2.1.3 (b)–(d)

Quick Notes

  • Empirical formula = simplest whole number ratio of atoms in a compound
  • Molecular formula = actual number of atoms in a molecule
  • Hydrated Compounds
    • Anhydrous = no water present
    • Hydrated = contains water of crystallisation

Full Notes

Understanding Empirical and Molecular Formulae

The empirical formula shows the simplest whole number ratio of atoms in a compound, while the molecular formula represents the actual number of atoms in a molecule.

Example Glucose (C6H12O6)

Empirical formula = CH2O (simplest ratio 1:2:1)
Molecular formula = C6H12O6 (actual composition)

Determining the Empirical Formula

To find the empirical formula of a compound from mass or percentage composition:

  1. Convert (percentage) mass of each element to moles using: Moles = Mass (g) ÷ Atomic mass (Ar)
  2. Divide all mole values by the smallest number of moles.
  3. Round to the nearest whole number to get the simplest ratio.
  4. Write the empirical formula.
Worked Example

Find the empirical formula for the compound with a composition by mass of C 52.2%, H 13.0% and O 34.8%.

OCR (A) A-Level Chemistry worked example showing calculation of empirical formula from mass percentages of C, H and O.

Determining the Molecular Formula

To find the molecular formula, use:

n = Molecular mass ÷ Empirical mass

Multiply the empirical formula by n to get the molecular formula.

Worked Example

The empirical formula of a compound is CH2O, and its molar mass is 180 g/mol. Find the molecular formula.

  1. Calculate the empirical formula mass.
  2. Find n.
  3. Multiply empirical formula by n.
OCR (A) A-Level Chemistry worked example showing CH2O empirical formula and calculation to obtain molecular formula C6H12O6.

Water of Crystallisation

Hydrated compounds contain water molecules as part of their crystal structure. The water molecules are trapped between particles in the solid.

Water of crystallisation refers to this water, shown in the formula as “•xH2O”, where x is the moles of water in the solid compared to moles of compound.

For Example Copper(II) sulfate

A common hydrated form of copper(II) sulfate is CuSO4•5H2O (s).
OCR (A) A-Level Chemistry diagram showing hydrated copper(II) sulfate CuSO4·5H2O.This formula tells us that for every one mole of CuSO4 in the solid crystal, there are also 5 moles of H2O molecules.

Anhydrous compounds contain no water.

Example Anhydrous copper(II) sulfate

CuSO4(s) is anhydrous as the formula contains no water of crystallisation.

OCR (A) A-Level Chemistry diagram showing anhydrous copper(II) sulfate CuSO4.

To find the formula of a hydrated salt:

Summary