3.3 The mole and the Avogadro constant
- Syllabus
- 0620–2026–2027
- Topic
- 3.3
- Level
- —
Learning objectives
3.3.1Concentration can be measured in g/dm3• State: concentration can be measured in g/dm³ or mol/dm³3.3.2Mole, mol, is the unit of amount of• State: the mole, mol, is the unit of amount of substance and that one mole contains 6.02 × 1023 particles, e.g. atoms, ions, molecules; this number is the Avogadro constant3.3.3Amount of substance relation: amount• Use amount of substance relation: amount (mol) = mass (g) / molar mass (g/mol), to calculate amount, mass, molar mass, relative mass and particle number using Avogadro constant3.3.4Molar gas volume, taken as 24 dm3 at• Use the molar gas volume, taken as 24 dm³ at room temperature and pressure, r.t.p., in calculations involving gases3.3.5Stoichiometric reacting masses• Calculate stoichiometric reacting masses, limiting reactants, volumes of gases at r.t.p., volumes of solutions and concentrations of solutions expressed in g/dm³ and mol/dm³, including conversion between cm 3 and dm³3.3.6Experimental data from a titration to• Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution3.3.7Empirical formulae and molecular• Calculate empirical formulae and molecular formulae, given appropriate data3.3.8Percentage yield, percentage• Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data
State concentration units
Concentration tells how much solute is present per unit volume of solution.
| Unit | Meaning |
|---|---|
| g/dm³ | grams of solute in each dm³ of solution |
| mol/dm³ | moles of solute in each dm³ of solution |
Use the final solution volume, not merely the volume of solvent added. Since 1 dm³ = 1000 cm³, divide cm³ by 1000 before using mol/dm³.
A concentration value is incomplete without its unit; g/dm³ and mol/dm³ are not interchangeable without the solute's molar mass.
Define the mole and Avogadro constant
The mole, mol, is the unit of amount of substance. One mole contains 6.02 × 10²³ specified particles; this number is the Avogadro constant, Nₐ.
| Substance description | Particle counted |
|---|---|
| element such as He | atoms |
| molecular substance such as CO₂ | molecules |
| ionic compound such as NaCl | formula units; ions can then be counted from the formula |
One mole of MgCl₂ contains 1 mol of Mg²⁺ ions and 2 mol of Cl⁻ ions, so it contains 2 × 6.02 × 10²³ chloride ions.
Always name the particle. One mole of molecules does not necessarily contain one mole of atoms.
Link mass, amount and particle number
Molar mass connects a measured mass to amount of substance; the Avogadro constant then connects amount to particle count.
| Find | Relation |
|---|---|
| amount, n | n = mass ÷ molar mass |
| mass | mass = n × molar mass |
| molar mass | molar mass = mass ÷ n |
| particles, N | N = n × 6.02 × 10²³ |
| amount from particles | n = N ÷ (6.02 × 10²³) |
For 22.0 g CO₂, n = 22.0/44.0 = 0.500 mol, so molecules = 0.500 × 6.02 × 10²³ = 3.01 × 10²³.
Use molar mass in g/mol with mass in g. Then apply any atoms or ions per formula unit after finding the number of formula units.
Use molar gas volume at r.t.p.
At room temperature and pressure, r.t.p., one mole of any gas occupies 24 dm³, which is 24 000 cm³.
| Given | Amount of gas |
|---|---|
| volume in dm³ | n = V ÷ 24 |
| volume in cm³ | n = V ÷ 24 000 |
| amount n | V = n × 24 dm³, or n × 24 000 cm³ |
0.0200 mol CO₂ occupies 0.0200 × 24 = 0.480 dm³, or 480 cm³, at r.t.p.
The 24 dm³ value applies to gases at r.t.p.; do not use it for liquids or solids, and keep cm³ and dm³ consistent.
Solve multi-step stoichiometry
Convert every available quantity to moles, use the balanced-equation ratio, then convert the required moles into the requested mass, gas volume or solution quantity.
| Quantity | Convert to or from moles |
|---|---|
| mass | n = m/M |
| gas at r.t.p. | n = V/24 dm³ |
| solution | n = cV, with V in dm³ |
| g/dm³ concentration | mass concentration = molar concentration × molar mass |
For two reactants, calculate n ÷ equation coefficient for each. The smaller value identifies the limiting reactant; use it to calculate the maximum product. The other reactant is in excess.
Convert cm³ to dm³ by dividing by 1000. After using the mole ratio, convert back to the unit requested.
Never compare reactant masses or moles directly when their equation coefficients differ; compare coefficient-adjusted amounts.
Calculate from titration data
At the titration end-point, the measured portions have reacted in the exact mole ratio shown by the balanced equation.
| Step | Action |
|---|---|
| 1 | choose the concordant titre and convert cm³ to dm³ |
| 2 | calculate known moles with n = cV |
| 3 | use the balanced-equation coefficient ratio |
| 4 | calculate unknown concentration with c = n/V, or unknown volume with V = n/c |
25.0 cm³ of 0.0500 mol/dm³ KOH contains 0.00125 mol. For H₂SO₄ + 2KOH, acid moles are 0.000625 mol; if its titre is 20.0 cm³, c = 0.000625/0.0200 = 0.03125 mol/dm³.
Use the equation ratio between solutes, not the ratio of burette and pipette volumes. Volumes may differ even at exact neutralisation.
Calculate empirical and molecular formulae
An empirical formula is the simplest whole-number mole ratio of elements. A molecular formula is a whole-number multiple of the empirical formula.
| Step | Action |
|---|---|
| 1 | use masses directly, or assume 100 g for percentage data |
| 2 | divide each element's mass by its Aᵣ |
| 3 | divide every mole value by the smallest |
| 4 | multiply all ratios if needed to obtain whole numbers |
Find the empirical-formula mass, then multiplier = Mᵣ ÷ empirical-formula mass. Multiply every empirical subscript by this same integer.
For a hydrate, calculate moles of anhydrous salt and water separately; divide both by the salt moles to obtain salt : water = 1 : x.
Do not round a ratio such as 1:1.5 directly; multiply all ratios by 2. A molecular multiplier must be a whole number.
Calculate yield, composition and purity
Each percentage compares a selected part with its correct whole, then multiplies by 100.
| Quantity | Calculation |
|---|---|
| percentage yield | actual product ÷ theoretical product × 100 |
| percentage composition by mass | mass contribution of element in formula ÷ Mᵣ × 100 |
| percentage purity | mass of pure substance ÷ total mass of impure sample × 100 |
Find theoretical yield from the limiting reactant and balanced equation before applying the yield formula. Actual and theoretical quantities must use the same unit.
In NH₄NO₃, nitrogen contributes 2 × 14 = 28 to Mᵣ 80, so nitrogen composition = 28/80 × 100 = 35%.
Yield measures process success, composition measures a formula's mass fraction, and purity measures a sample. Their numerators and denominators are not interchangeable.