Grade 8

Mole, molar mass, solution concentration

The mole counts particles in bulk samples. Use molar mass for mass conversions and molar concentration for the amount of solute per solution volume.

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IntuitionIntuition: particles and change

A pinch of table salt contains an enormous number of ions. Chemists use the mole as a counting unit, much as a dozen counts twelve objects.

Every conversion passes through the mole. Switch to the example to follow 88 g of CO₂ through all the arrows.

SchoolSchool: models and rules

Definition: Key idea

One mole contains exactly $N_A=6.02214076\times10^{23}$ specified entities. Molar mass $M$ is mass per amount, and molar concentration $c$ is amount of solute per volume of solution.

For water, $M(\ce{H2O})\approx18.0\,\mathrm{g\,mol^{-1}}$ . Therefore 36.0 g corresponds to 2.00 mol, independent of whether the sample is liquid or vapour.

Key quantities and formulas
Quantity	Symbol/unit	Conversion
Amount of substance	n, mol	n = m/M
Number of particles	N	N = n N_A
Molar concentration	c, mol/L	c = n/V (final solution volume)

n=\frac{m}{M},\qquad N=nN_A,\qquad c=\frac{n}{V}

N_A=6.02214076\times10^{23}\ \mathrm{mol^{-1}}

Example: Worked example

How many moles are in 9.0 g water? Use M = 18.0 g mol⁻¹.

Solution

n = m/M = 9.0/18.0 = 0.50 mol, containing about 3.01 × 10²³ water molecules.

Example: Second example

How many moles in 44 g CO₂? And how many CO₂ molecules?

Solution

M = 12 + 2×16 = 44 g/mol, so n = 1.0 mol and N = 1.0 × 6.022×10²³ = 6.02×10²³ molecules.

Molar concentration uses the final solution volume, not the volume of solvent poured in. Dissolving 0.50 mol solute and making the solution up to 2.0 L gives $c=0.25\,\mathrm{mol\,L^{-1}}$ .

The mole acts as a hub: mass → moles via molar mass, moles → particles via Avogadro’s constant, moles → solution amount via volume and concentration.

Example: Calculation / application example

How much mass of solid NaOH (M = 40.0 g/mol) is needed to prepare 250 mL of a 0.20 mol/L solution?

Solution

n = c × V = 0.20 mol/L × 0.250 L = 0.050 mol. Mass m = n × M = 0.050 mol × 40.0 g/mol = 2.0 g.