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Chemistry · 3. Stoichiometry and mole calculations

Amount of substance, moles and solution concentrations

Master conversions between mass, gas volume (modern 24.79 dm³ standard), solution concentration and reaction stoichiometry.

40 minutesContent version: 2.1
Start the interactive pathwayOpen the virtual experiment

Driving question

How do we calculate exact stoichiometric quantities so reactants combine completely without waste?

Curriculum coverage

Vietnam Science 8, Chemistry 10; IGCSE Stoichiometry; A-Level Amount of substance.

Model scope and limitations

Modern curricula use 25°C, 1 bar (24.79 dm³/mol replacing legacy 22.4 L/mol at 0°C, 1 atm); calculations assume ideal gas behavior.

Learning objectives

  • Calculate moles from mass (m/M), molar gas volume at r.t.p. (V/24.79) and concentration (c · V).
  • Identify the limiting reactant and calculate unreacted excess.
  • Calculate percentage yield (%) and purity of chemical samples.

Prerequisite knowledge

  • Relative atomic/molecular mass and chemical formulae.
  • Balancing basic chemical equations.

Core knowledge

The Mole and Avogadro Constant

One mole contains 6.022 × 10²³ elementary particles. Molar mass (M) is numerically equal to relative formula mass with unit g/mol.

Limiting and Excess Reactants

The reactant with the smallest ratio of moles to stoichiometric coefficient is exhausted first, determining theoretical yield.

Solution Concentrations

Molarity c = n / V (mol/dm³). Mass percentage C% = (m_solute / m_solution) × 100%. Connected by density: c = (10 · C% · D) / M.

Worked example

Dissolve 5.6 g Fe (0.1 mol) in 200 cm³ 1.5 M HCl (0.3 mol). Fe + 2HCl → FeCl₂ + H₂. Mole ratio: Fe (0.1/1) < HCl (0.3/2 = 0.15) ⇒ Fe is limiting; HCl is in excess by 0.1 mol. Volume of H₂ = 0.1 × 24.79 = 2.479 dm³.

Misconceptions to avoid

  • Total solution volume is not always the exact sum of component volumes due to packing contractions.
  • Stoichiometric coefficients represent molar ratios, NOT direct mass ratios.

Virtual experiment procedure: Stoichiometry and reacting masses

  1. Accurately weigh solid reagents using an electronic balance.
  2. Dissolve and make up to the calibration mark in a volumetric flask.
  3. Calculate moles based on the balanced stoichiometric or ionic equation.

Safety and cautions

  • Wear eye protection during exothermic dissolution; always add acid slowly to water, never water to acid.

Evidence to collect

  • Data table recording initial masses, calculated moles, identified limiting reactant and percentage yield.
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Diagnostic check

1. What volume does 1 mole of any gas occupy at modern standard ambient conditions (25°C, 1 bar)?

  1. 24.79 dm³ (litres)
  2. 22.4 L
  3. 1.0 L
Show answer and explanation

Correct answer: 24.79 dm³ (litres)

Modern curricula define standard ambient molar gas volume as 24.79 dm³.

2. Dissolving 4 g NaOH (M = 40 g/mol) in water to make 500 cm³ gives what molarity?

  1. 0.2 M
  2. 0.1 M
  3. 0.4 M
Show answer and explanation

Correct answer: 0.2 M

n = 4/40 = 0.1 mol; c = 0.1 mol / 0.5 dm³ = 0.2 mol/dm³.

3. The coefficients in N₂ + 3H₂ → 2NH₃ indicate what ratio?

  1. Ratio of moles and gas volumes
  2. Direct mass ratio in grams
  3. Density ratio
Show answer and explanation

Correct answer: Ratio of moles and gas volumes

Coefficients represent mole ratios and gas volumes at identical T and P.

Application practice

1. 0.2 mol Al reacts with 0.3 mol O₂: 4Al + 3O₂ → 2Al₂O₃. Which is limiting?

  1. Al is limiting
  2. O₂ is limiting
  3. Both react completely
Show answer and explanation

Correct answer: Al is limiting

Al ratio = 0.2/4 = 0.05 < O₂ ratio = 0.3/3 = 0.10. Al is the limiting reactant.

2. Theoretical yield is 10 g, but only 8 g is recovered. What is the percentage yield?

  1. 80%
  2. 125%
  3. 20%
Show answer and explanation

Correct answer: 80%

Yield = (8 / 10) × 100% = 80%.

3. To prepare 100 cm³ of 0.5 M CuSO₄ using CuSO₄·5H₂O (M = 250 g/mol), weigh what mass?

  1. 12.5 g
  2. 8.0 g
  3. 25.0 g
Show answer and explanation

Correct answer: 12.5 g

n = 0.1 × 0.5 = 0.05 mol; m = 0.05 × 250 = 12.5 g.

Academic references

  1. Cambridge IGCSE Chemistry 0620 — official syllabus and resources
  2. Cambridge AS & A Level Chemistry 9701 — official syllabus
  3. OpenStax Chemistry 2e — peer-reviewed general chemistry
  4. IUPAC Color Books — authoritative chemical terminology and nomenclature
  5. BIPM — SI Brochure, 9th edition

This lesson is maintained against the cited sources below. Simulations are learning models and do not replace supervised physical-laboratory safety procedures.

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