Chapter 5. The Mole and Stoichiometry

One mole contains 6.02214076 × 10²³ entities (Avogadro’s constant) [11]. Molar mass (g/mol) equals the formula mass in atomic mass units.

Key conversions: grams ⇄ moles ⇄ particles, using molar mass and Avogadro’s constant.

Balancing equations. Atoms of every element must be equal on both sides. Example: 2H₂ + O₂ → 2H₂O.

Stoichiometric procedure

Balance the equation.

Convert given quantity to moles.

Use the mole ratio from coefficients.

Convert to the desired unit.

Limiting reagent. The reactant that runs out first determines the product amount.

Percent yield = (actual yield / theoretical yield) × 100%.

Worked example. 4.00 g H₂ reacts with excess O₂. Moles H₂ = 4.00 / 2.016 = 1.98 mol. Ratio 2:2 gives 1.98 mol H₂O = 35.7 g.

Empirical formula. Convert percent composition to moles, divide by the smallest, and round to whole-number ratios [3].

Review: How many molecules are in 18.0 g of water?