The complete lecture — every counting idea behind stoichiometry shown through an everyday object you already know. Scroll down; the live panel on the right keeps pace, and the narration moves with you.
1 — The mole: a chemist's bundle
Atoms and molecules are far too small and far too many to count one by one. So chemists do exactly what a shopkeeper does with eggs — they count in bundles. A grocer's bundle is the dozen (12); the chemist's bundle is the mole.
- Mole — the gram-atomic, gram-molecular or gram-formula mass of any substance (atoms, molecules or ions) that contains 6.02 × 10²³ particles.
Just as "one dozen" always means twelve — whether of eggs, doughnuts or pencils — "one mole" always means the same enormous count, no matter the substance.
| Everyday bundle | How many |
| a pair | 2 |
| a dozen (eggs) | 12 |
| a gross | 144 |
| a mole (particles) | 6.02 × 10²³ |
Why bundle at all? Single atoms are invisible. By packaging them into moles, masses we can actually weigh on a balance map directly onto numbers of particles.
2 — Avogadro's number
- Avogadro's number (Nₐ) — the number of particles (atoms, ions or molecules) in one mole of any substance = 6.02 × 10²³, regardless of its chemical nature.
You would never count a sack of rice grain by grain — you weigh it, because you know roughly what one grain weighs. Chemists use the same trick. Because one mole of a substance has a known mass, weighing replaces counting: weigh out the molar mass in grams and you have automatically counted out 6.02 × 10²³ particles.
Key formulasmoles = given mass (g) / molar mass
particles = moles × 6.02 × 10²³
| Weigh out… | and you have |
| 12 g of carbon | 1 mole = 6.02 × 10²³ C atoms |
| 18 g of water | 1 mole = 6.02 × 10²³ H₂O molecules |
| 44 g of CO₂ | 1 mole = 6.02 × 10²³ CO₂ molecules |
The leap: a dozen is 12; the chemist's "dozen" is 6.02 × 10²³ — the same idea, scaled to the size of atoms.
3 — Molar mass & converting grams ⇄ moles ⇄ particles
- Atomic / molecular / formula mass — the mass of an atom, molecule or formula unit in a.m.u. H = 1, O = 16, so H₂O = 18 a.m.u.
- Molar mass — the same number, now read in grams per mole. 1 mole of H₂O weighs 18 g.
Think of a kitchen balance with three matched dials. Put grams on the pan; divide by the molar mass and the second dial reads moles; multiply by Avogadro's number and the third dial reads particles. Every numerical in this chapter is one trip across this scale.
The three conversionsmoles = mass / molar mass
moles = particles / 6.02 × 10²³
moles = volume / 22.4 dm³
grams → moles → particles
0.3 g C → 0.3 / 12 = 0.025 mol
→ 0.025 × 6.02×10²³ = 1.505 × 10²² atoms
a.m.u = dalton (Da) = 1/12 the mass of a carbon-12 atom = 1.66 × 10⁻²⁴ g.
4 — Molar volume
- Molar volume — the volume of 1 mole of a gas at standard temperature (273 K) and pressure (1 atm). For all ideal gases it is 22.4 dm³.
Imagine a standard moving box of fixed size — 22.4 litres. Pour in one mole of hydrogen and it just fills the box. Empty it, pour in one mole of carbon dioxide — heavier, bigger molecules — and it fills exactly the same box. Volume cares about the number of molecules, not their size or mass.
Avogadro's law: equal volumes of gases at the same temperature and pressure contain an equal number of molecules — not equal mass.
MCQ trap: 1 dm³ = 1 L = 1000 cm³, so 22.4 dm³ is the same as 22 400 cm³.
5 — Percentage composition
Cut any compound open and ask: what share of its mass does each element own? The answer is a pie chart — the same way you might split a pizza bill by who ate what.
Percentage of an element% = (mass of element in 1 mole / molar mass) × 100
water (H₂O, molar mass 18)
%H = (2/18)×100 = 11.1 %
%O = (16/18)×100 = 88.9 %
carbon dioxide (CO₂, molar mass 44)
%C = (12/44)×100 = 27.3 % · %O = 72.7 %
Check: the slices of every compound must add up to 100 %.
6 — Empirical & molecular formula
- Empirical formula — the simplest whole-number ratio of atoms. CH₂O.
- Molecular formula — the actual number of atoms in one molecule. Glucose, C₆H₁₂O₆.
Think of a patterned floor. The empirical formula is the single repeating tile — the smallest pattern that captures the ratio. The molecular formula is the whole floor: the tile laid down n times.
Linked by nMolecular = n × Empirical · n = molecular mass / empirical mass
% → empirical (the tile)
40% C, 6.7% H, 53.3% O → ÷ atomic mass → 3.33 : 6.7 : 3.33
÷ smallest → 1 : 2 : 1 → CH₂O
tile → whole floor
CH₂O has mass 30; molecular mass 180 → n = 180/30 = 6
6 × CH₂O = C₆H₁₂O₆
7 — Limiting reactant
- Limiting reactant — the ingredient that runs out first; it decides how much product forms.
- Excess reactant — whatever is left over on the counter.
A burger needs 2 buns + 1 patty. You can have a freezer full of patties, but the moment you run out of buns the line stops — the buns limit the order. In 2H₂ + O₂ → 2H₂O it is exactly the same: the reactant that runs out first caps the water made; the rest is excess.
Find it by amount ÷ coefficient — the smaller value limits. Change the sliders in the live panel and press Build to watch one ingredient run out.
Recipe link: a burger is to buns + patties what a molecule of water is to H₂ + O₂ — the smallest available batch sets the yield.
8 — Percentage yield & recap
- Theoretical yield — the maximum product the balanced equation predicts.
- Practical (actual) yield — what you really collect after the run.
- Percentage yield — actual as a share of theoretical; the efficiency score.
A factory plans for a full crate of product, but spillage, side-reactions and filtering losses mean it never quite gets there. Percentage yield is its report card — marks scored out of marks possible.
Percentage yield% yield = (practical / theoretical) × 100
worked
theoretical 15.648 g O₂, collected 14.9 g
% yield = (14.9 / 15.648) × 100 = 95.2 %
Recap: mole = 6.02×10²³ • weigh to count • grams ⇄ moles ⇄ particles • 22.4 dm³ gas box • % composition • empirical → molecular • limiting reactant • % yield.
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