1.4 The mole

Learning objectives

1.4.1Mole (mol)• SI unit of amount of substance• One mole = Avogadro constant number of entities• Convert between amount of substance and number of specified particles 1.4.2Relative atomic mass (Ar) and relative formula mass (Mr)• Compared to ¹²C standard• Determine relative formula mass from relative atomic masses 1.4.3Molar mass (M)• Units: g mol⁻¹• Relationships: particles ↔ moles ↔ mass• Solve mass, mole, and particle problems using n = m/M 1.4.4Empirical and molecular formulas• Empirical: simplest ratio of atoms• Molecular: actual number of atoms in molecule• Interconvert percentage composition and empirical formula• Determine molecular formula from empirical formula and molar mass 1.4.5Molar concentration• Determined by amount of solute and volume of solution• Solve concentration, amount, and volume problems; square brackets mean molar concentration 1.4.6Avogadro's law• Equal volumes of gases (same T, P) contain equal numbers of molecules• Mole ratio calculations with gas volumes• Use gas volume ratios from balanced equations

The mole is the SI unit for amount of substance. One mole contains the Avogadro constant number of specified entities, such as atoms, molecules, ions, or electrons.

$N = nNₐ where Nₐ = 6.022 × 10²³ mol⁻¹$

Identify which entity the question asks for, then multiply the amount in moles by Avogadro's constant and by the number of those entities in each formula unit or molecule when needed.

Specify the entity before calculating. One mole of H₂O contains one mole of molecules, two moles of H atoms and one mole of O atoms; multiplying by Nₐ without the formula-unit multiplier answers a different particle question.

1 mark

How many ions are present in 0.20 mol of $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ ?

Relative atomic mass is a weighted mean relative to one twelfth of the mass of a carbon-12 atom. Relative formula mass is the sum of the relative atomic masses represented in the formula.

$M_r = Σ(A_r × subscript)$

Relative atomic mass and relative formula mass are ratios on the carbon-12 scale, so they have no units. Apply every subscript, including waters of crystallization or repeated ions.

For Ca(OH)₂, include both bracketed groups: Mᵣ = Aᵣ(Ca) + 2[Aᵣ(O) + Aᵣ(H)]. Keep Mᵣ dimensionless; attach g mol⁻¹ only when the same numerical total is used as a molar mass.

1 mark

Define the term relative atomic mass $\left(A_{\mathrm{r}}\right)$ .

Molar mass is the mass of one mole of a substance, measured in g mol⁻¹. It connects mass, amount, and particle number in a conversion chain.

$n = m/M and m = nM$

To reach particles from mass, divide mass by molar mass to obtain moles, then use the Avogadro conversion and count the requested entities if the formula contains more than one.

Use units to choose the direction: 9.0 g of H₂O divided by 18.0 g mol⁻¹ gives 0.50 mol. To find molecules, continue from moles to nNₐ; do not multiply mass directly by the Avogadro constant.

2 marks

Calculate the number of hydrogen atoms in 1.00 g of propan-2-ol.

An empirical formula gives the simplest whole-number ratio of atoms. A molecular formula gives the actual number of each atom in a molecule.

Step	Operation
1	Convert each composition value to moles
2	Divide all mole values by the smallest
3	Multiply to reach the simplest whole-number ratio
4	Write the empirical formula
5	Divide molecular molar mass by empirical-formula mass and multiply every subscript by that integer

Keep the empirical ratio simplest, and make sure the molecular-formula multiplier is a whole number consistent with the given molar mass.

A composition of 40.0% C, 6.7% H and 53.3% O gives the simplest ratio CH₂O after division by atomic masses. If the molar mass is 180 g mol⁻¹, compare it with the empirical-formula mass 30 to obtain the multiplier 6 and molecular formula C₆H₁₂O₆.

Do not round a ratio such as 1 : 1.50 : 1 to 1 : 2 : 1. Preserve the calculated values and multiply every ratio by the same small integer: ×2 converts halves, while ×3 can resolve values close to thirds such as 1.33 or 1.67. Round only after the common multiplier produces values consistent with the data precision.

3 marks

4.32 g of the compound was combusted completely in oxygen and produced 9.49 g of $\mathrm{CO}_{2}$ and 5.18 g of $\mathrm{H}_{2} \mathrm{O}$ .
Determine the empirical formula of the compound, using sections 1 and 7 of the data booklet.

Molar concentration is the amount of solute in moles per cubic decimetre of solution. Square brackets can denote molar concentration.

$n = VC therefore C = n/V$

Use V in dm³ when calculating C in mol dm⁻³. Rearrange the same relationship to find amount or volume, and keep the solution volume distinct from the solute mass.

Convert volume before substitution: 500 cm³ = 0.500 dm³, so 0.250 mol in that solution gives C = 0.500 mol dm⁻³. The denominator is the final solution volume, not the volume or mass of solute alone.

$For dilution with no solute loss or reaction: n_before = n_after, so c₁V₁ = c₂V₂$

Adding solvent increases the final solution volume while the amount of solute stays constant, so concentration decreases. Use the final solution volume—not the solvent volume added—and do not apply c₁V₁ = c₂V₂ when solute reacts or is removed.

2 marks

Calculate the molar concentration of the resulting solution of lithium hydroxide.

At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules. Under the same conditions, gas-volume ratios follow mole ratios.

$V_gas = nV_m$

Balance the equation, identify the limiting gas amount when needed, apply the coefficient ratio to gas volumes at the same temperature and pressure, then convert to the requested volume units.

For N₂ + 3H₂ → 2NH₃ at one temperature and pressure, one gas volume of N₂ requires three equal gas volumes of H₂ and forms two of NH₃. This direct volume ratio works because the gases share conditions; otherwise convert through moles.

2 marks

$140.0 \mathrm{~cm}^{3}$ of ethyne are reacted with $160.0 \mathrm{~cm}^{3}$ of hydrogen in a container at 420 K and $1.00 \times 10^{5} \mathrm{~Pa}$ . Ethane is the only product.

Determine the maximum possible volume of ethane formed under these conditions of temperature and pressure.

Retrieve the quantitative chain: count entities with mN_A, sum relative masses from formulae, convert mass with n=m/M, derive formula ratios, use n=VC for solutions, and apply gas-volume ratios at the same temperature and pressure.

Before finalising, check the requested entity, formula subscripts, units, dm³ conversion, balanced-equation coefficients, and any limiting reactant.

Next Topic

—1.5 Ideal gasesRead

1.4 The mole

Learning objectives

The Mole and Particle Number

Relative Atomic and Formula Mass

Molar Mass Conversions

Empirical and Molecular Formulae

Molar Concentration

Avogadro's Law and Gas Volumes

The Mole Summary

Next Topic

Converting Moles to Entities

Applying Relative Masses

Solving Molar-Mass Problems

Deriving Chemical Formulae

Calculating Concentration

Using Gas-Volume Ratios