Chemistry 20 · Stoichiometry
The mole is a counting unit
A balanced equation counts particles, and a balance measures grams. The mole is the only thing that connects them, which is why every quantitative question in chemistry passes through it.
- 1. The words, first
- 2. The mole, in three directions
- 3. Mole ratios and mass stoichiometry
- 4. Limiting reagent, gases and solutions
- 5. What costs marks
The words, first
The idea: Amount, mass and concentration are three different quantities. Being sloppy about which one a number is causes more wrong answers than any arithmetic error.
| Word | What it means |
|---|---|
| Mole | 6.02 × 10²³ particles. A count, like a dozen, just much larger. |
| Avogadro's number | 6.02 × 10²³ per mole. |
| Molar mass (M) | The mass of one mole, in g/mol. Read off the periodic table and added up. |
| Amount (n) | The number of moles. Not the mass, and not the number of particles. |
| Mole ratio | The ratio of coefficients in a balanced equation. A ratio of particles, never of grams. |
| Limiting reagent | The reactant that runs out first and therefore fixes how much product can form. |
| Excess reagent | The one left over. If a question says excess, that reactant is not limiting and you need not check it. |
| Theoretical yield | The mass the stoichiometry predicts. |
| Actual yield | What was really obtained. Percent yield is actual ÷ theoretical × 100. |
The mole, in three directions
The idea: Mass, amount and number of particles convert into each other through two relationships. Getting into moles is always the first step.
n = m/M · N = n × 6.02 × 10²³ · n = cV · n = V/Vm for a gas
Worked example. 36.0 g of water: M = 18.02 g/mol, so n = 36.0/18.02 ≈ 2.00 mol. And 0.250 mol of NaCl: m = (0.250)(58.44) ≈ 14.6 g. If the answer for a quarter of a mole is not about a quarter of the molar mass, the division went the wrong way.
Mole ratios and mass stoichiometry
The idea: Coefficients count particles. Convert to moles, apply the ratio, convert back — never apply a ratio to grams.
Worked example. How much O₂ reacts with 2.0 mol of Al, given 4Al + 3O₂ → 2Al₂O₃?
n(O₂) = 2.0 mol Al × (3 mol O₂ / 4 mol Al) = 1.5 mol
Writing the ratio as a fraction with the wanted substance on top keeps it the right way up automatically.
Mass to mass. Burning 1.00 mol of CH₄: CH₄ + 2O₂ → CO₂ + 2H₂O is 1:1 for CO₂, so 1.00 mol forms, and m = (1.00)(44.01) ≈ 44.0 g.
Why balancing first matters. 2H₂ + O₂ → 2H₂O is 2:1 in moles but 1:8 in grams. Reading the coefficients as masses produces answers that are wrong by a large factor and look reasonable.
Limiting reagent, gases and solutions
The idea: When both reactant amounts are given, one of them runs out first. Work out how much product each could make on its own; the smaller answer wins.
Worked example. 2.0 mol H₂ with 1.5 mol O₂, by 2H₂ + O₂ → 2H₂O.
- 2.0 mol H₂ would need 1.0 mol O₂ — and 1.5 mol is available, so hydrogen runs out first.
- H₂ : H₂O is 2:2, so 2.0 mol of water forms, and 0.5 mol of O₂ is left over.
When a question says excess, it is telling you which reactant is limiting, so no comparison is needed. 0.0500 mol AgNO₃ with excess NaCl gives 0.0500 mol AgCl, or (0.0500)(143.32) ≈ 7.17 g.
Gas stoichiometry. The mole ratio still does the work; molar volume converts in or out. Decomposing 0.500 mol CaCO₃ gives 0.500 mol CO₂, which at SATP is (0.500)(24.8) = 12.4 L.
Solution stoichiometry. Same idea with n = cV at each end — which is exactly what a titration calculation is.
What costs marks
The idea: Everything here is a variation on one habit: get into moles first.
- Applying a mole ratio to grams. Convert to moles first, always.
- Not balancing the equation. The coefficients are the ratio; if they are wrong, everything after is wrong.
- Forgetting to check for a limiting reagent when two amounts are given.
- Rounding molar masses too early. Carry the figures through and round at the end.