Stoichiometry & the Mole
High SchoolCounting atoms by weighing
Atoms are too small to count individually, so chemists count in moles. One mole is Avogadro's number of particles: NA = 6.022 × 10²³ mol⁻¹. The molar mass M (g mol⁻¹) equals the relative atomic/molecular mass in grams, providing the bridge between mass (weighable) and number of particles (countable).
A balanced chemical equation obeys the law of conservation of mass — atoms are neither created nor destroyed, so each element must appear in equal numbers on both sides. The coefficients give the mole ratio in which substances react, the heart of all stoichiometric calculation. The limiting reagent is the reactant that runs out first and caps the product; the others are in excess. Percentage yield compares actual to theoretical product.
The mole as a translator
Think of the mole as a chemist's "dozen" — a fixed count so large that a workable mass contains it. A recipe says "2 cups flour + 1 egg", but you cannot buy atoms by the cup; the mole lets a balanced equation like 2H₂ + O₂ → 2H₂O be read as "2 moles of hydrogen react with 1 mole of oxygen to give 2 moles of water." The coefficients are ratios of particles, and moles convert those particle ratios into gram quantities you can actually weigh out.
The limiting-reagent idea is everyday logic: if you have 10 slices of bread and 3 slices of cheese, you can make only 3 sandwiches — cheese limits you, and 4 bread slices are left over. In chemistry you must compare reactants in moles, adjusted by their coefficients, never by raw mass.
Worked example — mass of product from a reaction
What mass of water forms when 4.0 g of hydrogen burns in excess oxygen? (M: H₂ = 2.0, H₂O = 18.0 g mol⁻¹.)
- Write and balance the equation: 2H₂ + O₂ → 2H₂O.
- Convert the known mass to moles: n(H₂) = 4.0 g ÷ 2.0 g mol⁻¹ = 2.0 mol.
- Apply the mole ratio from the equation: H₂ : H₂O = 2 : 2 = 1 : 1, so n(H₂O) = 2.0 mol.
- Convert moles of product to mass: m = n × M = 2.0 mol × 18.0 g mol⁻¹ = 36 g.
- Answer: 36 g of water. (Oxygen was in excess, so hydrogen was limiting.)
Every stoichiometry problem is the same three-step path: mass → moles → (mole ratio) → moles → mass. Balance first; convert to moles always; let the equation's coefficients do the bridging.
How many molecules are in 9.0 g of water?
n = 9.0 ÷ 18.0 = 0.50 mol. N = 0.50 × 6.022 × 10²³ = 3.0 × 10²³ molecules.
Balance: C₃H₈ + O₂ → CO₂ + H₂O.
Balance C then H then O. 3 carbons → 3 CO₂; 8 hydrogens → 4 H₂O. Oxygen on the right: 3(2) + 4(1) = 10 O atoms = 5 O₂. Result: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
2.0 mol of N₂ reacts with 2.0 mol of H₂ (N₂ + 3H₂ → 2NH₃). Which is limiting?
The ratio required is N₂ : H₂ = 1 : 3. To consume 2.0 mol N₂ you would need 6.0 mol H₂, but only 2.0 mol is present — so H₂ is limiting. The 2.0 mol H₂ needs just 0.67 mol N₂, leaving N₂ in excess; product NH₃ = 2.0 × (2/3) = 1.33 mol.
A reaction has a theoretical yield of 25.0 g but produces 21.0 g. Find the percentage yield.
% yield = (21.0 ÷ 25.0) × 100 = 84.0%. Losses to side reactions, incomplete reaction, or transfer typically keep real yields below 100%.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The process, step by step
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1What mass of MgO forms when 4.86 g of Mg burns fully? (2Mg + O2 -> 2MgO)
- Molar mass Mg = 24.3, so moles Mg = 4.86 / 24.3 = 0.200 mol.
- Mole ratio Mg:MgO is 2:2 = 1:1, so moles MgO = 0.200 mol.
- Molar mass MgO = 40.3, so mass = 0.200 x 40.3 = 8.06 g.
Example 2How many molecules are in 0.50 mol of water?
- One mole contains 6.02e23 particles (Avogadro number).
- Multiply: 0.50 mol x 6.02e23 molecules/mol.
- That gives 3.01e23 molecules.
Now you try
Work each one out first, then tap to reveal the worked answer.