Counting particles by mass: the mole

The amount of substance containing **6.02 × 10²³** particles (Avogadro's constant, N_{A}).

N_{A} = **6.02 × 10²³ mol⁻¹** — the number of particles in one mole.

The mass of **one mole** of a substance, in **g mol⁻¹**; numerically equal to the relative atomic/formula mass.

$n = \dfrac{m}{M}$ — amount (mol) = mass (g) ÷ molar mass (g mol⁻¹).

$N = n\,N_{A}$ — number of particles = amount (mol) × Avogadro's constant.

Rearrange to $m = nM$ — multiply the amount (mol) by the molar mass.

$M = \dfrac{m}{n}$ — divide the mass by the amount in moles.

2 mol of O atoms = **1.20 × 10²⁴** atoms (each CO_{2} has 2 oxygens).

**g mol⁻¹** (grams per mole).

Forgetting to scale by the **number of that atom or ion in the formula** (e.g. 2 Cl⁻ per MgCl_{2}).

The **simplest whole-number ratio** of the atoms of each element in a compound.

The **actual number** of atoms of each element in one molecule of the compound.

The molecular formula is a **whole-number multiple** of the empirical formula (molecular = empirical × x).

**CH_{2}O** — divide every subscript by 6 to get the simplest ratio.

Treat % as g per 100 g → divide each by A_{r} (n = m/M) → divide by the **smallest** → round / scale to whole numbers.

**n(C) = n(CO_{2})** — every carbon atom ends up in one CO_{2}.

**n(H) = 2 × n(H_{2}O)** — each water molecule contains two H atoms.

By **difference**: subtract the masses of C and H from the sample mass, then divide the leftover by 16.00.

$x = \dfrac{M_{r}}{\text{empirical formula mass}}$, then multiply every subscript by x.

Multiply the **whole ratio** by 2 (for .5) or 3 (for .33) to clear it into whole numbers.

Atoms combine in whole-**number** ratios, which only show up once masses are turned into **moles** (÷ A_{r}).

An **empirical** formula — ionic compounds have no separate molecules, so the formula is the simplest ratio.

How much **solute** is dissolved in a given volume of **solution** — usually in **mol dm⁻³**.

$n = CV$ — amount (mol) = concentration (mol dm⁻³) × volume (**dm³**).

Rearrange to $C = \dfrac{n}{V}$ — divide the amount in moles by the volume in dm³.

**Divide by 1000** (1 dm³ = 1000 cm³). E.g. 250 cm³ = 0.250 dm³.

**Multiply by the molar mass M**: g dm⁻³ = mol dm⁻³ × M.

$C_{1}V_{1} = C_{2}V_{2}$ — the amount of solute is unchanged when you add solvent.

Diluting only adds solvent, so the **moles of solute (n = CV) stay constant**.

**1 mg dm⁻³** (1 part per million) — used for very dilute solutions.

A solution of **precisely known concentration**, made up in a **volumetric flask**.

Forgetting to convert the **volume from cm³ to dm³** (÷ 1000) before using n = CV.

The **total** final volume. Water added = V_{2} − V_{1}.

**Dissolve** the weighed solid → **transfer** to a volumetric flask (rinse beaker in) → **make up** to the mark → **invert** to mix.