Understanding Stoichiometry
Stoichiometry is the arithmetic of chemistry — it tells you exactly how much of each substance reacts and forms. Master the mole and the roadmap, and every calculation becomes systematic.
Every chemical equation must obey the law of conservation of mass: atoms cannot be created or destroyed in a chemical reaction. The same atoms that enter a reaction must appear in the products — just rearranged into different compounds.
An unbalanced equation is chemically incorrect — it implies atoms appear from nowhere or vanish. Every stoichiometric calculation you do depends on the balanced equation. If you skip balancing, every answer downstream will be wrong.
Systematic approach: balance metals first, then nonmetals, then hydrogen, then oxygen. For combustion reactions, always balance oxygen last.
Fe₂O₃ has 3 oxygen atoms. To get an even number, use 2 × Fe₂O₃ = 6 oxygen atoms (= 3 O₂).
b) Balance: Mg + O₂ → MgO
c) Why can you never change subscripts to balance an equation?
a) N₂ + 3H₂ → 2NH₃ (N: 2=2 ✓, H: 6=6 ✓)
b) 2Mg + O₂ → 2MgO (Mg: 2=2 ✓, O: 2=2 ✓)
c) A subscript defines the chemical formula of the substance — changing it creates a different compound entirely. For example, changing H₂O to H₂O₂ would turn water into hydrogen peroxide. You must only change the number of molecules (coefficient), never the identity of the molecules (subscript).
Atoms are extraordinarily small — a single hydrogen atom has a mass of about 1.67 × 10⁻²⁴ grams. You can't weigh individual atoms on a lab balance. The mole solves this problem by grouping 6.022 × 10²³ particles together into one countable unit — just like a "dozen" groups 12 items.
Because atoms combine in fixed whole-number ratios — and the mole translates those ratios directly into laboratory-scale masses. When you measure out 18 g of water, you have exactly 1 mole of H₂O — meaning 6.022 × 10²³ water molecules. The mole makes the invisible world of atoms measurable.
n = m / M (moles = mass ÷ molar mass)
Nₐ = 6.022 × 10²³ particles/mol
M (g/mol) = atomic masses from the periodic table, added up
b) What is the mass of 0.5 mol of Fe? (M = 55.85 g/mol)
c) Why is it useful to work in moles rather than grams when calculating reactions?
a) n = 58 g ÷ 58.44 g/mol ≈ 0.992 mol ≈ 1.0 mol
b) m = 0.5 mol × 55.85 g/mol = 27.9 g
c) Atoms and molecules react in fixed whole-number ratios determined by the balanced equation. Moles let you use those ratios directly. Grams of different substances aren't comparable because different elements have different masses — 1 g of H₂ and 1 g of O₂ are not the same number of molecules. Once everything is in moles, the ratios from the balanced equation apply directly.
The mole ratio is the bridge between two different substances in a reaction. Once you know how many moles of one substance you have, the balanced equation tells you exactly how many moles of any other substance reacted or formed.
Chemistry happens at the level of atoms and molecules — not grams. Converting to moles first lets you use the balanced equation's integer ratios. The roadmap (g → mol → mol → g) just translates from the scale you can measure (grams) to the scale where the chemistry happens (moles) and back.
n(CaCl₂) = 0.499 × (1/1) = 0.499 mol
m(CaCl₂) = 0.499 × 110.98 ≈ 55.4 g
b) 28 g N₂ reacts with excess H₂: N₂ + 3H₂ → 2NH₃. Find mol NH₃ produced. (M(N₂) = 28.02 g/mol)
c) What does the coefficient in a balanced equation tell you about moles?
a) n(Mg) = 10 ÷ 24.31 = 0.411 mol. Ratio Mg:MgO = 2:2 = 1:1. n(MgO) = 0.411 mol. m(MgO) = 0.411 × 40.31 ≈ 16.6 g
b) n(N₂) = 28 ÷ 28.02 ≈ 0.999 mol ≈ 1.00 mol. Ratio N₂:NH₃ = 1:2. n(NH₃) = 1.00 × (2/1) = 2.00 mol
c) The coefficient in front of a substance in the balanced equation tells you how many moles of that substance are involved relative to the other substances. A coefficient of 2 means 2 moles of that substance react or form for every occurrence of the reaction as written. Coefficients are the mole ratios — they are the conversion factor you use in Step 3 of the roadmap.
In a real experiment, you rarely mix reactants in the exact stoichiometric ratio. One reactant will run out first — the limiting reagent. The other reactant is in excess and some of it will remain unreacted.
To make 1 sandwich you need 2 slices of bread and 1 filling. If you have 8 slices of bread and 3 fillings: bread can make 4 sandwiches, filling can make 3 sandwiches. The filling runs out first — filling is the limiting reagent. You can only make 3 sandwiches, and you'll have 2 leftover bread slices.
n(O₂) = 32 ÷ 32.00 = 1.00 mol
From O₂: 1.00 × (2/1) = 2.00 mol H₂O
n(O₂) = 10 ÷ 32.00 = 0.3125 mol
From O₂: 0.3125 × (2/1) = 0.625 mol H₂O
b) What happens to the excess reagent at the end of the reaction?
c) If you have exactly the right mole ratio of reactants, what is the limiting reagent?
a) n(Fe) = 56 ÷ 55.85 = 1.003 mol → mol Fe₂O₃ = 1.003 × (2/4) = 0.501 mol
n(O₂) = 48 ÷ 32.00 = 1.50 mol → mol Fe₂O₃ = 1.50 × (2/3) = 1.00 mol
Fe gives less product → Fe is the limiting reagent
b) The excess reagent is still present as unreacted material at the end of the reaction. Some of it remains in the reaction vessel — it simply wasn't consumed because the limiting reagent ran out before any more could react.
c) When reactants are mixed in exactly the stoichiometric ratio, technically neither is limiting — both run out at exactly the same time. In practice you say there is no limiting reagent, or that the reaction goes to completion perfectly. The concept of a limiting reagent only applies meaningfully when one reactant is not in exact stoichiometric proportion.
Stoichiometry tells you the maximum amount of product that could form — the theoretical yield. In a real lab, you always collect less. Side reactions occur, the reaction may not go to completion, or product is lost during transfer and purification. The amount you actually collect is the actual yield.
It tells you how efficient your reaction was. In industry, a 60% yield means 40% of your starting material was wasted — costing money and generating by-products. Chemists work to optimise conditions (temperature, concentration, reaction time) to push yields higher. In a school lab, yields above 85% are generally considered good.
Theoretical yield: calculated from stoichiometry using limiting reagent.
Actual yield: measured in the lab after the experiment.
Percent yield should always be ≤ 100%.
b) % yield = 95% and actual yield = 19 g. Find the theoretical yield.
c) Can percent yield ever exceed 100%? What would that indicate?
a) % yield = (21 / 25) × 100% = 84%
b) Rearrange: theoretical = actual / (% yield / 100) = 19 / 0.95 = 20 g
c) In theory, percent yield cannot exceed 100% — that would mean more product was collected than the stoichiometry allows, which would violate conservation of mass. In practice, a result above 100% indicates a measurement error: the product may contain impurities (making it appear heavier), the product may not have been fully dried, or there was an error in calculating the theoretical yield. A result above 100% is always a signal to check your calculations and measurements.