LEARNING OBJECTIVES
By the end of this chapter, you should be able to:
- write formulae and balanced equations with state symbols
- calculate relative masses and reacting masses
- use moles, particles and molar gas volume
- calculate solution concentrations and titration quantities
- find empirical and molecular formulae, yield, purity and limiting reactants
THE BIG IDEA
Use formulae, balanced equations and the mole to calculate chemical quantities accurately.
A balanced equation is a counting statement. Its coefficients give mole ratios between reactants and products.
Most calculations follow the same route: convert the known quantity to moles, use the equation ratio, then convert to the required quantity.
SECTION 01
Formulae, equations and state symbols
A molecular formula gives the number and type of atoms in one molecule. An empirical formula gives the simplest whole-number ratio.
Balance equations by changing coefficients, never by changing correct formulae. Use (s), (l), (g) and (aq) to show physical states.
An ionic equation removes spectator ions and shows the particles that change, such as H⁺(aq) + OH⁻(aq) → H₂O(l).
KEY IDEAS
- Total atoms and total charge must balance.
- Use ion charges to deduce ionic formulae.
- Spectator ions appear unchanged on both sides of a full ionic equation.
SECTION 02
Relative masses and reacting masses
Relative molecular mass, Mᵣ, is the sum of Aᵣ values in a molecule. The same calculation is called relative formula mass for an ionic compound.
Balanced coefficients provide reacting-mass proportions even before the mole is introduced.
Mass of magnesium oxide formed
- Equation: 2Mg + O₂ → 2MgO.
- Mᵣ(Mg) = 24 and Mᵣ(MgO) = 40.
- The 2:2 ratio means 24 g Mg forms 40 g MgO.
- For 6.0 g Mg, multiply by 40/24.
Answer: 10.0 g of MgO.
SECTION 03
Moles, particles and gases
One mole contains 6.02 × 10²³ particles, the Avogadro constant. Molar mass has the same numerical value as Aᵣ or Mᵣ but units g/mol.
At room temperature and pressure, one mole of any gas occupies 24 dm³.
Particles in 9.0 g of water
- Mᵣ(H₂O) = 18.
- Moles = 9.0 ÷ 18 = 0.50 mol.
- Molecules = 0.50 × 6.02 × 10²³.
Answer: 3.01 × 10²³ water molecules.
SECTION 04
Solutions and titrations
Concentration measures solute per volume of solution. Convert cm³ to dm³ by dividing by 1000.
In a titration, use the pipetted volume and mean concordant titre to find moles. Apply the equation ratio before finding the unknown concentration or volume.
Moles in a measured solution
- 25.0 cm³ = 0.0250 dm³.
- Concentration = 0.160 mol/dm³.
- Moles = concentration × volume.
Answer: 0.00400 mol of solute.
SECTION 05
Formulae, limiting reactants and percentages
For an empirical formula, convert each element mass or percentage to moles, divide by the smallest and scale to whole numbers. Use molar mass to convert empirical to molecular formula.
The limiting reactant produces the smaller possible amount of product and is completely used. Other reactants are in excess.
Percentage yield compares actual with theoretical product; purity compares desired substance with the total sample; percentage composition finds an element's mass share.
Empirical formula from composition
- A compound contains 40.0% C, 6.7% H and 53.3% O.
- Mole ratios: 40.0/12 = 3.33, 6.7/1 = 6.7, 53.3/16 = 3.33.
- Divide by 3.33 to get approximately 1:2:1.
Answer: The empirical formula is CH₂O.
QUICK CHAPTER SUMMARY
The ideas to carry forward
- Balance equations before calculating.
- Convert known quantities to moles.
- Use equation coefficients as mole ratios.
- One mole of gas occupies 24 dm³ at r.t.p.
- Percentage calculations must use the correct reference quantity.