Some Basic Concepts in Chemistry and Mole Concept

1The mole is a count

A mole is a count, not a mass and not a volume: 6.022 × 10²³ of whatever entity you name. Naming the entity is the whole discipline. One mole of Na₂SO₄ is one mole of formula units, but three moles of ions and seven moles of atoms, and a question asking for the number of particles has not finished asking.

The same trap wears a chemical hat. One mole of oxygen means 1 mol of O₂ if you mean the gas and 1 mol of O if you mean the atom, and those differ by a factor of two in mass and by a factor of two in count. Nothing about a molar mass tells you how many particles are present — only the amount in moles does.

Figure. One amount of substance, counted three ways: 0.100 mol of formula units, 0.300 mol of ions, 0.700 mol of atoms. Read the subscripts first, then decide which count the question wants.

2The two bridges: molar mass and molar volume

Particles cannot be counted directly, so every problem reaches the mole through one of two bridges. Mass is the unconditional one: n = m/M holds for a solid, a liquid or a gas, at any temperature and any pressure. Volume is the conditional one: n = V/22.4 with V in litres, and only for a gas behaving ideally at 273 K and 1 atm.

Molar masses are rarely integers because they are isotopic averages. Chlorine is 35.45 g/mol, the 35.5 everyone quotes, because natural chlorine is about 76% ³⁵Cl and 24% ³⁷Cl — not because any single chlorine atom weighs 35.5 u. The conditional bridge carries a gift in exchange for its conditions: since 22.4 L holds one mole whatever the gas is, equal volumes of two gases at the same temperature and pressure hold equal numbers of molecules. Change the pressure standard and the number moves with it — at 273 K and 1 bar, the current IUPAC standard, the molar volume is 22.7 L.

Figure. Every conversion in this chapter passes through the middle node. Mass reaches it unconditionally; volume reaches it only for a gas, and only at the stated temperature and pressure. There is no edge straight from mass to volume — going that way without stopping at moles is the commonest wasted step in the chapter.

3Empirical and molecular formula

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is a whole-number multiple of it. Percentage compositions and combustion data hand you masses, and masses are not ratios of atoms until each is divided by an atomic mass. Equal masses of two elements are never equal numbers of atoms unless the elements happen to weigh the same.

Figure. The two bars are drawn to scale, so the molecular mass really is six empirical units long. Finding n is measuring one bar against the other; the empirical formula fixes the unit, the molar mass fixes how many of them there are.

The procedure

1. Percentages to molesTake 100 g of the compound, so each percentage reads directly as a mass in grams; divide each by its atomic mass.
2. Divide by the smallestThe smallest quotient becomes 1 and the others read off against it.
3. Clear the fractionA ratio of 1 : 1.5 is 2 : 3, not 1 : 2. Multiply through by 2, 3 or 4 — never round a genuine fraction away.
4. Scale to the molar massn = molar mass ÷ empirical formula mass, and the molecular formula is the empirical formula taken n times.

4Stoichiometry and the limiting reagent

A balanced equation is a statement about moles and about nothing else. When two reactants are both quantified, one runs out first and fixes every product amount; the other is simply left over at the end. Find it by dividing each reactant's moles by its own stoichiometric coefficient — the smallest quotient is how far the reaction can actually run.

Comparing raw masses is meaningless and comparing raw moles is right only when the coefficients happen to be equal. The quotient has a physical meaning worth holding on to: it is the extent of reaction each reactant could support if it were the only limit.

Figure. Both bars measure the same thing — how far the reaction can run — so they can be laid against each other directly. Raw moles cannot be: 1.5 mol of H₂ and 0.906 mol of O₂ are not the same currency until each has been divided by its coefficient.

How it works

1. Balance firstThe coefficients are the whole method, so an unbalanced equation gives a confidently wrong limiting reagent.
2. Moles, then divideConvert every given mass to moles, then divide each amount by its coefficient.
3. Smallest quotient winsThat reactant is limiting, and its quotient is the extent of reaction for everything in the equation.
4. Account for the restMultiply the extent by each other coefficient to find what was consumed, and subtract to find the excess.

5Molarity, molality and mole fraction

Molarity and molality differ by one word each, and it is the word that matters. Molarity is moles of solute per litre of solution — the volume of the finished mixture, read off a flask. Molality is moles of solute per kilogram of solvent — the mass of the liquid you started with, read off a balance. A volume expands on warming and a mass does not, so molarity falls as a solution is heated while molality, mole fraction and mass per cent do not move at all.

In dilute aqueous solution the two nearly coincide, because a litre of solution is close to a kilogram of water, and that coincidence is exactly why the distinction gets missed. In a concentrated solution it collapses: 98% sulphuric acid is 18.4 M and 5 × 10² m, a factor of twenty-seven apart, because almost none of the sample is solvent.

Figure. One 100 g sample of 98% acid, drawn to scale. Molarity divides by the volume of the whole bar; molality divides by the mass of the two-gram sliver at its right-hand end. Same sample, different denominators — which is the entire reason 18.4 and 5 × 10² can describe the same bottle.

6Reading a bottle label: M = 10xd/Mm

Reagent bottles are labelled by mass per cent and density; calculations want molarity. Rather than rebuild the 100 g basis on every question, build it once symbolically and keep the result. One hundred grams of solution contains x grams of solute, which is x/Mm moles, and occupies 100/d millilitres, which is 0.1/d litres. Dividing the one by the other gives M = 10xd/Mm.

The 10 is not a fudge factor. It is 1000 mL per litre divided by the 100 g of the basis, and knowing that is what lets you rearrange the relation instead of memorising it.

Figure. Bottle labels give mass per cent and density; molarity needs moles per litre. Build the 100 g basis once: solute moles x/Mm over volume 100/d mL becomes M = 10xd/Mm after the litre conversion.

Where the formula comes from

1. Fix a basisTake exactly 100 g of solution, so the mass per cent x reads directly as x grams of solute.
2. Moles on topn = x/Mm, with Mm the molar mass of the solute.
3. Litres underneathThose 100 g occupy 100/d mL, that is 0.1/d L. Here d is the density of the solution, never the density of the solvent.
4. DivideM = (x/Mm) ÷ (0.1/d), which tidies to 10xd/Mm.

7Dilution and mixing

Adding solvent changes the volume and leaves the amount of solute alone, and that is the whole of M₁V₁ = M₂V₂ — a statement that n is conserved, written in units of molarity times volume. Mixing two solutions of the same solute is the same statement with more terms: add the moles, add the volumes, divide.

Nothing here is a special property of dilution. It is bookkeeping on a quantity that cannot change, which is why the relation works in any consistent pair of units — millilitres on both sides are as good as litres.

Figure. Width is volume and height is concentration, so the area of each rectangle is the number of moles. Four times the width against a quarter of the height leaves the area untouched — that is all M₁V₁ = M₂V₂ says, and it is why the relation never needs the solute's identity.

8Equivalents, n-factor and normality

An equivalent is one mole of reacting capacity: one mole of H⁺ donated or accepted, or one mole of electrons transferred. Multiplying an amount by its n-factor converts moles into capacity, so N = M × n-factor, and at any end point the two reactants have exchanged equal capacity — N₁V₁ = N₂V₂ — whatever the coefficients in the balanced equation happen to be.

That is the convenience normality buys, and the price is printed on the same receipt: n-factor belongs to the reaction, not to the formula. The same bottle of KMnO₄ is 5 equivalents per mole in acid, 3 in neutral solution and 1 in strong alkali, so a normality quoted without a reaction is not yet a number.

Figure. n-factor belongs to the reaction, not the formula: the same H₂SO₄ is 2 when both protons titrate and 1 when the stop is bisulphate; the same KMnO₄ is 5, 3 or 1 by medium. Normality is M × n for that reaction.