This chapter covers how chemists count particles by mass and volume, write formulas and balanced equations, and work out how much of a substance takes part in a reaction. It is one of the most calculation-heavy parts of SPM Chemistry.
How do the skills depend on each other?
Mole work is a chain. A weak link early in the chain spoils everything after it.
- Converting between mass, moles and particles is the first step.
- Finding empirical and molecular formulas uses it to turn masses into a simple mole ratio.
- Balancing chemical equations gives the mole ratio.
- Solving reacting-mass and gas-volume calculations combines steps 1 and 3.
- Calculating concentration from supplied data adds volume of solution.
- Identifying the limiting quantity compares two amounts.
A further group of pages, conservation reasoning across a reaction, asks why these calculations work at all.
One example from mass to concentration
A student dissolves 4.0 g of sodium hydroxide in water to make 250 cm³ of solution. The relative formula mass of NaOH is 40, so the moles are 4.0 ÷ 40 = 0.10 mol.
The volume is 250 cm³, which is 0.250 dm³. The concentration is 0.10 ÷ 0.250 = 0.40 mol dm⁻³. That is three steps, each from a different lesson, joined by one unit conversion.
Who should start where?
A student new to moles should begin with the first lesson and do the practice set after each step. A student who already converts masses easily can move to balancing and concentration.
Use the mole and stoichiometry steps tool to check your layout. The atoms and ions in matter and atomic structure supply the relative atomic masses. For a teacher to find the exact step that breaks, see online one-to-one Chemistry tuition.