Chemistry 20 you still need
Chemistry 30 assumes the calculations from Chemistry 20 without ever teaching them again. Moles, molar mass, concentration and mole ratios run underneath every unit — thermochemistry, titration, electrochemistry — so a gap here shows up everywhere.
Why these skills keep reappearing
Chemistry 30 is four units of new chemistry built on one page of old arithmetic. Every titration, every calorimetry question and every electrochemistry calculation ends in the same place: convert to moles, use the ratio from the balanced equation, convert back. Nothing on the exam teaches that again, and nothing works without it.
What the mole actually does is translate between what you can measure and what reacts. A balance reads grams, a burette reads millilitres, an ammeter reads amps — but atoms react in whole numbers of particles. The mole is the exchange rate, which is why every formula on this page has moles on one side of it.
It is also where marks quietly disappear. Alberta's own commentary lists molar mass, mole ratios that are not 1:1, and rounding among the things students get wrong — not because the chemistry is hard, but because the arithmetic underneath it was never made automatic.
Where it turns up
- Dosing medicine — a pharmacist converting a prescribed dose in milligrams per kilogram into millilitres of a solution is doing c = n ÷ V with the stakes turned up.
- Water treatment — chlorine and fluoride are dosed in parts per million, which is milligrams per litre — the same conversion as a dilution problem.
- Industry — a plant that overshoots a reagent by 5% wastes that 5% every batch, so limiting reagent calculations are simply money.
- Everything after this — any chemistry, biology, nursing or engineering program assumes you can do this without thinking about it.
- Why it matters
- 1. The mole, and what it converts
- 2. Concentration and dilution
- 3. Mole ratios and limiting reagents
- 4. Significant figures and units
The mole, and what it converts
The idea: you can weigh grams and measure litres, but reactions happen in whole particles. The mole is the exchange rate between the two, and every calculation in this course passes through it.
| To go from | To moles | Back again |
|---|---|---|
| Mass (g) | n = m ÷ M | m = nM |
| Concentration and volume | n = cV | c = n ÷ V |
| Gas at STP | n = V ÷ 22.4 L/mol | V = 22.4n |
| Number of particles | n = N ÷ 6.02 × 10²³ | N = 6.02 × 10²³ n |
| Charge (electrochemistry) | mol e⁻ = q ÷ 96 500 | q = 96 500 × mol e⁻ |
Molar mass, done carefully
Add the molar mass of every atom in the formula, counting subscripts and anything in brackets.
Ca(OH)₂: 40.08 + 2(16.00 + 1.01) = 40.08 + 34.02 = 74.10 g/mol. The bracket multiplies both the oxygen and the hydrogen — missing that is the single most common molar mass error.
The map: whatever you are given, convert it to moles; use the mole ratio from the balanced equation; convert moles into whatever the question wants. Almost every stoichiometry question in Chemistry 30 is that one sentence.
Concentration and dilution
The idea: concentration is moles per litre. Diluting changes the volume and the concentration but never the number of moles you already had.
- c = n ÷ V, with V in litres. Millilitres divided by 1000 first — this is where most lost marks in titration questions come from.
- Dilution: c₁V₁ = c₂V₂, because the moles of solute are unchanged.
- Parts per million is milligrams per litre for dilute water solutions.
Worked: preparing a dilute solution
What volume of 2.00 mol/L HCl is needed to make 250.0 mL of 0.150 mol/L HCl?
c₁V₁ = c₂V₂ → (2.00)(V₁) = (0.150)(0.2500 L) → V₁ = 0.01875 L = 18.8 mL.
Sense check: the stock is about thirteen times more concentrated, so you need about a thirteenth of the volume.
Mole ratios and limiting reagents
The idea: the coefficients in a balanced equation are a ratio of particles, not of grams. That ratio is the only bridge between one substance and another.
Worked: a titration ratio that is not 1:1
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
If 18.0 mL of 0.100 mol/L NaOH neutralizes 25.0 mL of sulfuric acid:
- n(NaOH) = (0.0180 L)(0.100 mol/L) = 1.80 × 10⁻³ mol
- Ratio: 1 mol of acid per 2 mol of base, so n(H₂SO₄) = 9.00 × 10⁻⁴ mol
- c = 9.00 × 10⁻⁴ ÷ 0.0250 L = 0.0360 mol/L
Skipping step 2 doubles the answer, and it is the step that gets skipped.
Limiting reagent, quickly
Convert both amounts to moles, divide each by its coefficient, and the smaller number is the limiting reagent. It decides how much product forms; the other reactant is left over.
Significant figures and units
The idea: on a machine-scored exam, a right method with a wrong rounding earns nothing at all. This is where careful students still lose marks.
- Multiplying or dividing: the answer carries as many significant figures as the least precise value used.
- Adding or subtracting: match decimal places instead.
- Round once, at the end. Rounding an intermediate value and carrying it forward is how answers drift into the wrong box.
- In a pH, only the decimals count. A pH of 2.72 carries two significant figures.
- Check the unit asked for: kJ against kJ/mol, mol/L against g/L, °C against K. A numerical-response box does not care that you understood the chemistry.