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The Mole

Count particles using the mole and Avogadro's constant.

  • Define and explain The Mole in your own words
  • Use key terms such as mole accurately
  • Apply what you have learned to new examples and questions
  • Avoid the common mistakes learners make with this topic

Chemists cannot count atoms one by one, so they weigh them instead — the mole links mass to numbers of particles. With it you can predict exactly how much product a reaction should make, and judge how efficient real reactions are. In this chapter you will do the maths that industry relies on.

This lesson focuses on The Mole: count particles using the mole and Avogadro's constant.

Definition: The Mole

Count particles using the mole and Avogadro's constant.

Key ideas

The mole connects mass and particles

One mole of any substance contains 6.02 × 10²³ particles and has a mass in grams equal to its relative formula mass — so 18 g of water is one mole. The relationship moles = mass ÷ Mr lets you convert between the two in seconds. This is the key that unlocks every reacting-mass calculation.

Use the equation's molar ratio

A balanced equation gives the ratio of moles that react: in 2Mg + O₂ → 2MgO, 2 moles of Mg make 2 moles of MgO, a 1:1 ratio. The method is always the same: convert the known mass to moles, use the ratio to find moles of the unknown, then convert back to mass. If a reactant runs out first, it is the limiting reactant and it alone sets the maximum product.

Key term — mole: The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant).

Worked example: The Mole

Calculate the number of moles in 10.0 g of sodium hydroxide, NaOH. (Ar: Na = 23.0, H = 1.0, O = 16.0).

Mr = 23.0 + 16.0 + 1.0 = 40.0. Moles = 10.0 ÷ 40.0 = 0.250 mol.

Answer: Mr = 23.0 + 16.0 + 1.0 = 40.0. Moles = 10.0 ÷ 40.0 = 0.250 mol.

Common mistakes
  • Using the mass ratio instead of the mole ratio Grams do not react in the equation's ratio — moles do. Always convert masses to moles first, apply the ratio from the balanced equation, then convert back to grams.
  • Forgetting to double the oxygen in O₂ Mr of O₂ is 32, not 16 — the molecule has two atoms. The same trap applies to H₂, N₂ and the halogens: read the formula, not just the element.

Practice

In the worked example, what mass of oxygen reacts with the 12.0 g of magnesium?
Use the 2:1 ratio of Mg to O₂.

Moles of Mg = 0.494; moles of O₂ = 0.494 ÷ 2 = 0.247 mol; mass = 0.247 × 32.0 = 7.90 g. (Check: 12.0 + 7.90 = 19.9 g of product.).

5.84 g of NaCl is dissolved to make 1.00 dm³ of solution. What is its concentration? (Mr of NaCl = 58.4)
Convert grams to moles first.

Moles = 5.84 ÷ 58.4 = 0.100 mol; concentration = 0.100 ÷ 1.00 = 0.100 mol/dm³.

A reaction should theoretically make 8.40 g of product but only 6.72 g is obtained. Calculate the percentage yield.
(actual ÷ theoretical) × 100.

(6.72 ÷ 8.40) × 100 = 80.0%.

Calculate the concentration of a solution containing 0.20 mol of solute in 0.50 dm³ of solution.
Concentration = moles ÷ volume.

0.20 ÷ 0.50 = 0.40 mol/dm³.

Quick check

The Mole — quick check

Which of these best defines "mole"?

The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant).

Explain why a chemist might prefer a reaction with 100% atom economy even if its percentage yield is lower.

High atom economy means most reactant atoms end up in the desired product rather than waste, saving raw materials and reducing waste disposal — even if some product is lost in practice.
Key takeaways
  • The Mole: count particles using the mole and Avogadro's constant.
  • The mole connects mass and particles: One mole of any substance contains 6.02 × 10²³ particles and has a mass in grams equal to its relative formula mass — so 18 g of water is one mole.
  • concentration: Moles of solute per cubic decimetre of solution, in mol/dm³.
  • Watch out for: using the mass ratio instead of the mole ratio
Masses in Reactions →Back to Grade 11 chaptersRelated resources →