Introduction
By now you know that chemists can't count atoms one by one — they're far too small and far too numerous. Instead, we use the mole as a "chemist's dozen" to bridge the gap between the invisible particle world and the measurable lab world (grams, cm³, litres). Today's lesson introduces three essential equations that let you convert between number of particles, mass, and volume of gas, all via the mole. These three formulas are the backbone of almost every stoichiometry calculation you'll meet this year (Talbot, p. 418).
The three key equations
- n = N / Nₐ — relates amount of substance (n, in mol) to the number of particles (N) using the Avogadro constant, Nₐ = 6.02 × 10²³ mol⁻¹
- n = m / M — relates amount of substance to mass (m, in g) using the molar mass (M, in g mol⁻¹)
- n = V / Vₘ — relates amount of substance to the volume of a gas (V) using the molar volume (Vₘ), which is 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa) in the IB data booklet
Each formula can be rearranged to find any one variable if the other two are known — a classic "triangle" relationship.
Worked Example 1 — using n = N/Nₐ
Question: How many moles are there in 3.01 × 10²³ molecules of CO₂?
n = N / Nₐ
n = (3.01 × 10²³) / (6.02 × 10²³)
n = 0.500 mol
Worked Example 2 — using n = m/M
Question: Calculate the amount (in mol) of magnesium oxide, MgO, in 8.06 g of the compound. (M of MgO = 24.31 + 16.00 = 40.31 g mol⁻¹)
n = m / M
n = 8.06 / 40.31
n = 0.200 mol
This mirrors the method Talbot uses when deducing reacting ratios from masses — always start by finding moles with n = m/M before comparing ratios (Talbot, p. 96).
Worked Example 3 — using n = V/Vₘ
Question: What volume of hydrogen gas, H₂, is produced at STP if 0.150 mol of the gas is formed?
n = V / Vₘ
V = n × Vₘ
V = 0.150 × 22.7
V = 3.41 dm³
Worked Example 4 — combining two formulas
Question: How many molecules are there in 11.2 dm³ of oxygen gas, O₂, at STP?
Step 1 — find moles from volume:
n = V / Vₘ = 11.2 / 22.7 = 0.493 mol
Step 2 — convert moles to number of particles:
N = n × Nₐ
N = 0.493 × 6.02 × 10²³
N = 2.97 × 10²³ molecules
Notice how the mole acts as the central hub connecting particles, mass, and gas volume — you can hop between any two quantities as long as you pass through n.
Key takeaways
- The mole (n) is the central unit linking particle number, mass, and gas volume.
- Use n = N/Nₐ when dealing with countable particles (atoms, molecules, ions).
- Use n = m/M when dealing with mass — always calculate M first from the formula.
- Use n = V/Vₘ only for gases, and only at the stated conditions (e.g. STP, Vₘ = 22.7 dm³ mol⁻¹).
- Multi-step problems often require converting through n twice — write out each step clearly to avoid errors.