VCE · Chemistry · Stoichiometry & the Molelive from the app
Empirical & molecular formulae
The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of each atom in one molecule. Percentage composition only ever gives you the empirical formula, so methanal (CH₂O), ethanoic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) all share 40.0% carbon. A measured molar mass tells you the whole-number multiplier n that turns CH₂O into the real molecule.
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chemistry · stoichiometry & the mole · empirical & molecular formulaedrag it · it is yours
the one idea
why this one carries the topic.
The mole is a counting word for things too small to count — it turns a mass you can weigh into particles you can't see. Every problem is the same move: mass to moles, follow the equation's ratio, back to what was asked. One bridge, crossed in different directions.
The top half turns 40.0% C, 6.7% H and 53.3% O into grams, then moles, then a 1 : 2 : 1 ratio: CH₂O. Raising the molar mass from 30 to 180 g/mol grows one molecule from methanal to glucose, while the percentages above never change.
what examiners catch — A common slip is rounding a ratio such as 1 : 1.5 to 1 : 2 instead of multiplying through to 2 : 3, or dividing percentages by each other instead of by molar masses. Marks go to showing grams → moles → divide by the smallest, then n = molar mass ÷ empirical formula mass.
what you leave with
five things, not forty.
Percentage composition by mass: %element = (mass of that element in one formula ÷ molar mass) × 100, so CH₂O is 12.01 ÷ 30.03 × 100 = 40.0% carbon.
Empirical formula from percentages: treat them as grams in 100 g, divide each by its molar mass (moles = m ÷ MM), then divide every mole amount by the smallest.
If a mole ratio lands near 1.5 or 1.33, multiply every number by 2 or 3 to reach whole numbers rather than rounding.
Molecular formula: n = molar mass ÷ empirical formula mass, then multiply every subscript by n; 180.2 ÷ 30.03 = 6 turns CH₂O into C₆H₁₂O₆.
Avogadro's constant (6.022 × 10²³ mol⁻¹) turns moles into particles: 100 g of glucose is 0.555 mol, or 3.34 × 10²³ molecules.
The same grams-to-moles step drives limiting reagent reactions: compare moles against the balanced equation, never grams.
what's underneath
nothing here is a standalone fact.
knowhere maps every concept to what it rests on and what rests on it — 1 underneath this one, 1 built on top. Each one says why, in a sentence, not as an arrow on a diagram.
this conceptempirical & molecular formulaeThe top half turns 40.0% C, 6.7% H and 53.3% O into grams, then moles, then a 1 : 2 : 1 ratio: CH₂O.
sits under itmolar massbecause formula determination IS mole-ratio arithmetic
built on itwater of hydration: heating a salt to constant massbecause Finding x is the same move as an empirical formula: masses to moles to the simplest whole-number ratio.
the rest of stoichiometry & the mole
5 more, same treatment.
Each one is its own knowscape in the app — built for how a particular student takes things in, not one explanation handed to everybody.
limiting reagentsin the appmeasuring gasesin the appmolar massin the appthe molelive →stoichiometry of combustionin the app
this is one of 865
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