Moles and stoichiometry: study guide

Key ideas

The mole links the mass you weigh to the number of particles. Stoichiometry then uses the coefficients of a balanced equation to convert between amounts of reactants and products.

• Moles are mass divided by molar mass: n = m ÷ M. One mole contains 6.022 × 10²³ particles.
• Balance the equation first. The coefficients give the ratio of moles, not of grams.
• The limiting reagent is the one that runs out first. Convert each reactant to moles, divide by its coefficient, and the smallest value limits the reaction.
• Percent yield compares what you actually got with the theoretical maximum from the limiting reagent.

Key equations

• n = m ÷ M
• N = n × 6.022 × 10²³
• percent yield = actual ÷ theoretical × 100
• mass % = mass of part ÷ total mass × 100

Worked example

For 2 H₂ + O₂ → 2 H₂O, mix 4.0 g H₂ and 40.0 g O₂. n(H₂) = 4.0 ÷ 2.016 = 1.98 mol and n(O₂) = 40.0 ÷ 32.00 = 1.25 mol. The ratio needs 0.99 mol O₂ for the H₂, and 1.25 mol is available, so H₂ limits. Water formed: 1.98 mol × 18.02 g/mol = 35.7 g. If 30.0 g is collected, the percent yield is 30.0 ÷ 35.7 = 84 %.

Common mistakes

• Using grams directly in the mole ratio from the equation.
• Not balancing the equation before comparing coefficients.
• Choosing the reactant with the smaller mass as the limiting one, instead of the one with the smaller number of moles per coefficient.

Practice

Test yourself with practice questions on this topic: Moles and stoichiometry