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Class XI · Chemistry · Unit 1 · Lecture

Stoichiometry

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.

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 bundleHow many
a pair2
a dozen (eggs)12
a gross144
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.
  • 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 carbon1 mole = 6.02 × 10²³ C atoms
18 g of water1 mole = 6.02 × 10²³ H₂O molecules
44 g of CO₂1 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.
  • 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.
  • 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.

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 %.
  • 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
  • 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.
  • 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. Next: 📝 Practice · 🧪 Virtual lab
⚛ Live panelStoichiometry
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