This mini-lesson covers Structure 1.4 & 1.5: the mole and Avogadro's constant, molar mass, empirical/molecular formulae, concentration, and the ideal gas (molar volume 22.7 dm³ mol⁻¹ at STP).
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Press Start when you're ready.
Structure 1.4
The mole and Avogadro's constant
The mole (mol) is the SI unit for amount of substance. One mole contains Avogadro's number of particles:
NA = 6.02 × 10²³ mol⁻¹the number of particles in exactly one mole
Moles link the number of particles you can count to the mass you can weigh.
n = m ÷ Mn = amount (mol), m = mass (g), M = molar mass (g mol⁻¹)
Calculate
Moles from mass
1How many moles are in 22.0 g of CO₂ (M = 44.01 g mol⁻¹)?
mol
n = m ÷ M = 22.0 ÷ 44.01.
Structure 1.4
Counting particles
Once you know the moles, multiply by NA to get the number of particles:
N = n × 6.02 × 10²³number of particles = moles × Avogadro's number
Relative molar mass Mr is found by adding the relative atomic masses in a formula (e.g. H₂O = 2(1.01) + 16.00 = 18.02).
Calculate
How many particles?
2How many molecules are in 0.50 mol of a gas? Give your answer as the number that multiplies × 10²³.
× 10²³
N = 0.50 × 6.02 × 10²³, so the coefficient is 0.50 × 6.02.
Structure 1.4
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number. Find the empirical formula by converting masses (or %) to moles and dividing by the smallest.
Example — 40.0% C, 6.7% H, 53.3% O gives moles 3.33 : 6.7 : 3.33 → ratio 1 : 2 : 1 → CH₂O (empirical mass 30.03).
Calculate
Find the molecular formula
3A sugar has empirical formula CH₂O (mass 30.03) and a molar mass of 180.16 g mol⁻¹. How many CH₂O units are in one molecule?
n = molecular mass ÷ empirical mass = 180.16 ÷ 30.03. (Glucose is C₆H₁₂O₆.)
Structure 1.4
Concentration of solutions
Concentration links moles to solution volume:
n = C × VC in mol dm⁻³, V in dm³ (1 dm³ = 1000 cm³)
Always convert cm³ to dm³ by dividing by 1000.
Calculate
Moles in a solution
4How many moles of solute are in 250 cm³ of a 0.100 mol dm⁻³ solution?
mol
V = 250 ÷ 1000 = 0.250 dm³; n = C × V = 0.100 × 0.250.
Structure 1.5
The ideal gas
An ideal gas has point particles with no forces between them. At STP (the 2023 guide uses 100 kPa and 273 K) one mole of any ideal gas occupies the molar volume:
Vm = 22.7 dm³ mol⁻¹ (at STP)so n = V ÷ 22.7 for a gas at STP
More generally the ideal gas equation is PV = nRT (R = 8.31 J K⁻¹ mol⁻¹).
Calculate
Volume of a gas at STP
5What volume does 0.50 mol of an ideal gas occupy at STP?
dm³
V = n × 22.7 = 0.50 × 22.7.
Quick check
What is STP?
?In the 2023 IB guide, standard temperature and pressure (STP) are:
Quick check
Boyle's behaviour
?At constant temperature, if the pressure on a fixed amount of gas is doubled, its volume:
Sort it
Which route to moles?
Tap a clue, then the equation you'd use to get moles.
⚖️ n = m ÷ M
💧 n = C × V
🎈 n = V ÷ 22.7
Match it
Match the quantity to its formula
Tap an item on the left, then its match on the right.
Quantity
Formula
Recap
The big ideas to know
Mole: 6.02 × 10²³ particles; n = m ÷ M; N = n × NA
Formulae: empirical = simplest ratio; molecular = actual number
Solutions: n = C × V (convert cm³ → dm³)
Gases: Vm = 22.7 dm³ mol⁻¹ at STP (100 kPa, 273 K); PV = nRT
You've covered Counting particles — the mole and gases for IB Diploma Chemistry SL. Press Finish to see your score.
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