Stoichiometry & The Mole: The Language of Chemical Quantities

Stoichiometry is the quantitative backbone of chemistry. It provides the mathematical tools to calculate exactly how much of each substance reacts and how much product forms — essential for pharmaceutical manufacturing, industrial chemistry, and exam success.

The Mole: Counting Atoms

  • One mole = 6.022 × 10²³ particles (Avogadro's number, Nₐ).
  • 1 mol of carbon (Ar = 12) = 12 g = 6.022 × 10²³ atoms.
  • 1 mol of water (Mr = 18) = 18 g = 6.022 × 10²³ molecules.

Molar Mass: The Bridge Between Mass and Moles

  • Molar mass (M) = the mass of one mole of a substance in g/mol.
  • Fundamental equation: n = m/M (moles = mass ÷ molar mass).
  • Examples: M(H₂O) = 18 g/mol; M(CaCO₃) = 100 g/mol; M(H₂SO₄) = 98 g/mol.

Molar Concentration: Solutions

  • c = n/V (mol/dm³). Always convert cm³ to dm³ by dividing by 1000.
  • c₁V₁ = c₂V₂ for dilutions (moles of solute are conserved).
  • Titration: known c and V of one solution → calculate unknown concentration using stoichiometry.

Balancing Chemical Equations

  • Balance one element at a time — leave H and O for last.
  • Use whole-number coefficients only. Check: count every atom on each side.
  • Example: 4Al + 3O₂ → 2Al₂O₃.

Reacting Masses & Limiting Reagents

  • The molar ratio from the balanced equation gives the reacting mass ratio.
  • The limiting reagent is the reactant that runs out first and determines maximum product.
  • Identify by converting each reactant to moles and dividing by its stoichiometric coefficient.

Percentage Yield & Atom Economy

  • % yield = (actual yield ÷ theoretical yield) × 100. Always ≤ 100%.
  • Atom economy = (molar mass of desired product ÷ total molar mass of all products) × 100%.
  • High atom economy = less waste = better sustainability.

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