The big idea: Atoms are far too small and too numerous to count one by one, so chemists count them in moles.
One mole is simply a fixed number of particles — 6.02 × 10²³ of them. That number is Avogadro's constant, NA.
Think of it like 'a dozen', just much bigger: a dozen = 12 things; a mole = 6.02 × 10²³ things.
Three quantities, one idea: Every mole calculation links three things:
- the number of particles, N - the amount in moles, n - the mass in grams, m
The molar mass M (in g mol⁻¹) is the mass of one mole — read it off the periodic table as the relative atomic/formula mass.
To go from an amount in moles to an actual number of particles, multiply by Avogadro's constant.
- number of particles (atoms, molecules or ions)
- amount of substance (mol)
- Avogadro's constant = 6.02 × 10²³ mol⁻¹
Worked example — counting atoms
A sample contains 0.250 mol of carbon dioxide, CO2. How many oxygen atoms does it contain?
Solution
- Formula first — find the number of CO2 molecules:
- Substitute the values:
- Work it out — these are CO2 molecules:
- Each CO2 molecule has 2 oxygen atoms, so multiply by 2:
Final answer
3.01 × 10²³ oxygen atoms.
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The molar mass M links mass and amount. Rearrange the given equation depending on what you are asked for.
- amount of substance (mol)
- mass of the sample (g)
- molar mass (g mol⁻¹)
Worked example — mass from moles
Calculate the mass of 0.50 mol of water, H2O. (M(H2O) = 18.02 g mol⁻¹.)
Solution
- Start from the given equation and rearrange for mass — formula first:
- Substitute the values:
- Work it out — keep the unit:
Final answer
m = 9.0 g.
Worked example — molar mass from mass and moles
0.20 mol of an unknown compound has a mass of 11.7 g. Calculate its molar mass.
Solution
- Rearrange the given equation for molar mass — formula first:
- Substitute:
- Work it out:
Final answer
M = 58.5 g mol⁻¹ (this is sodium chloride, NaCl).
How this is tested: The mole is the foundation of every quantitative question in the course.
- Paper 1A (MCQ): a one-step 'how many particles / what mass' calculation. - Paper 2: a part-marks calculation, often 'find the amount, then the mass' or 'how many ions'.
The classic trap: forgetting to multiply by the number of that atom/ion in the formula (e.g. 2 nitrate ions per Mg(NO3)2).
Three easy marks: (1) Write the formula before the numbers. (2) Carry the unit on every line. (3) Check whether they ask for molecules, atoms or ions — then scale by the formula.
IB-style question — magnesium nitrate (a)
A sample contains 0.150 mol of magnesium nitrate, Mg(NO3)2. (a) Calculate the mass of the sample. (M = 148.33 g mol⁻¹.)
Solution
- Formula first — mass from amount:
- Substitute:
- Work it out:
Final answer
m = 22.2 g.
IB-style question — magnesium nitrate (b)
(b) Calculate the number of nitrate ions, NO3⁻, present in the same 0.150 mol sample.
Solution
- There are 2 nitrate ions in each Mg(NO3)2, so first find the amount of nitrate ions:
- Formula first — number of ions:
- Substitute and solve:
Final answer
1.81 × 10²³ nitrate ions.