Mole Concept and Stoichiometry — MDCAT Chemistry Lesson

MDCAT Chemistry · curriculum topic: Introduction to Fundamental Concepts of Chemistry

Learning objective

By the end of this lesson you should be able to convert confidently between mass, moles, number of particles and gas volume at STP, and use a balanced equation to find reacting masses.

One definition, three conversions

A mole is the amount of substance containing as many particles as the Avogadro constant, NA ≈ 6.022 × 10²³ mol⁻¹ (defined exactly as 6.02214076 × 10²³ mol⁻¹). Mole calculations are built from three conversions applied in some order, so learn the three rather than memorising question types.

  • Mass to moles: n = m / M, where m is mass in grams and M is molar mass in g mol⁻¹.
  • Moles to particles: N = n × NA ≈ n × 6.022 × 10²³ (atoms, molecules, ions or formula units — read the question carefully).
  • Moles of gas to volume: one mole of an ideal gas occupies about 22.4 dm³ (22.4 litres) at 0 °C and 1 atm — the older STP convention used for every 22.4 L value on this page. On the 1 bar standard-state convention the molar volume is about 22.7 dm³, so check which conditions a question states.

Getting molar mass right

Molar mass is the sum of the atomic masses in the formula: H₂O is (2 × 1) + 16 = 18 g mol⁻¹, CO₂ is 12 + (2 × 16) = 44 g mol⁻¹, NaOH is 23 + 16 + 1 = 40 g mol⁻¹, and CaCO₃ is 40 + 12 + 48 = 100 g mol⁻¹.

The molar mass step is worth double-checking: a miscounted subscript or an ignored water of crystallisation makes every later step wrong even when the method is perfect.

From a balanced equation to reacting masses

A balanced equation gives a mole ratio, not a mass ratio. The reliable route is always the same: convert the quantity you are given into moles, apply the mole ratio from the equation, then convert back into whatever the question asks for.

  • Balance the equation before using any coefficient.
  • Convert the given mass or volume into moles.
  • Multiply by the mole ratio (product coefficient ÷ reactant coefficient).
  • Convert the result to the required quantity — mass, particles or volume at STP.
  • If quantities for two reactants are given, identify the limiting reactant first and base the whole calculation on it.

Worked example — reacting masses

Hydrogen burns in oxygen: 2H₂ + O₂ → 2H₂O. What mass of water is produced when 4.0 g of hydrogen reacts completely with excess oxygen? (H = 1, O = 16)

  1. Molar mass of H₂ = 2 g mol⁻¹, so moles of H₂ = 4.0 / 2 = 2.0 mol.
  2. Mole ratio from the balanced equation: 2 mol H₂ gives 2 mol H₂O, a 1 : 1 ratio, so 2.0 mol of water is formed.
  3. Molar mass of H₂O = (2 × 1) + 16 = 18 g mol⁻¹.
  4. Mass of water = n × M = 2.0 × 18 = 36 g. (Oxygen is in excess, so no limiting-reactant step is needed.)

Answer: 36 g of water is produced.

Practice questions

Question 1. How many moles are present in 11 g of CO₂? (C = 12, O = 16)

  • A. 0.25 mol
  • B. 0.5 mol
  • C. 1 mol
  • D. 2 mol

Answer: A. 0.25 mol — M(CO₂) = 12 + 32 = 44 g mol⁻¹, so n = m/M = 11/44 = 0.25 mol.

Question 2. The number of molecules in 0.5 mol of oxygen gas (O₂) is approximately:

  • A. 3.01 × 10²³
  • B. 6.02 × 10²³
  • C. 1.20 × 10²⁴
  • D. 22.4 × 10²³

Answer: A. 3.01 × 10²³ — N = n × NA = 0.5 × 6.022 × 10²³ ≈ 3.01 × 10²³ molecules. Note that monatomic gases such as helium or argon exist as separate atoms, not molecules, so this molecule count applies to a molecular gas like O₂.

Question 3. What is the mass of 2 mol of NaOH? (Na = 23, O = 16, H = 1)

  • A. 40 g
  • B. 60 g
  • C. 80 g
  • D. 120 g

Answer: C. 80 g — M(NaOH) = 23 + 16 + 1 = 40 g mol⁻¹, so m = n × M = 2 × 40 = 80 g.

Question 4. The volume occupied by one mole of an ideal gas at 0 °C and 1 atm is:

  • A. 1 dm³
  • B. 11.2 dm³
  • C. 22.4 dm³
  • D. 44.8 dm³

Answer: C. 22.4 dm³ — One mole of an ideal gas occupies about 22.4 dm³ at 0 °C and 1 atm — the molar volume on that STP convention. On the 1 bar convention it is about 22.7 dm³.

Question 5. For 2H₂ + O₂ → 2H₂O, how many moles of water form when 3 mol of O₂ reacts with excess hydrogen?

  • A. 1.5 mol
  • B. 3 mol
  • C. 6 mol
  • D. 9 mol

Answer: C. 6 mol — The equation gives 1 mol O₂ → 2 mol H₂O, so 3 mol O₂ produces 6 mol of water when hydrogen is in excess.

Common mistakes

  • Using coefficients as a mass ratio. Coefficients give a mole ratio; convert to moles first.
  • Confusing moles of molecules with moles of atoms — 1 mol of O₂ contains 2 mol of oxygen atoms.
  • Applying 22.4 dm³ per mole to liquids and solids, or to gases at conditions other than 0 °C and 1 atm.
  • Skipping the limiting-reactant check when quantities of both reactants are given.

Take the numericals further

The mole concept underpins many Chemistry calculations. The MDCAT Complete Preparation Bundle covers 21 Chemistry topics with explained MCQs, alongside Biology, Physics, English and logical reasoning.

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