1.4 The mole

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.

How it is assessed

Questions ask for the number of specified atoms or ions in a stated amount of a molecular or ionic substance.

Command terms

determine

What earns marks

Multiply moles by the Avogadro constant and count the requested entity per molecule or formula unit before selecting or reporting the answer.

Watch for

Counting formula units instead of the requested atoms or ions, or omitting the entity multiplicity in the formula.

Representative question

[Maximum number: 1]

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

$0.20 \times 1 \times 6 \times 10^{23}$

$0.20 \times 2 \times 6 \times 10^{23}$

$0.20 \times 3 \times 6 \times 10^{23}$

$0.20 \times 7 \times 6 \times 10^{23}$

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.

How it is assessed

Questions ask for the precise carbon-12-relative definition of relative atomic mass or for a formula mass obtained by summing relative atomic masses.

Command terms

define / determine

What earns marks

Refer to the weighted mean mass of an atom relative to the carbon-12 reference, and include every formula subscript in the mass sum without attaching units to the relative value.

Watch for

Using the mass of an element instead of an atom in the definition, or omitting subscripts when summing a hydrated formula.

Representative question

[Maximum number: 1]

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.

How it is assessed

Questions determine molar mass from mass and amount or determine the number of specified atoms from a sample mass.

Command terms

determine

What earns marks

Use n=m/M or m=nM with g mol⁻¹, then multiply by the Avogadro constant and the number of requested atoms per molecule when required.

Watch for

Using mass divided by moles in the wrong direction, or stopping at moles when the question asks for atoms.

Representative question

[Maximum number: 2]

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

《 $\frac{1.00 \mathrm{~g}}{(12.01 \times 3+1.01 \times 8+16.00) \mathrm{g} \mathrm{mol}^{-1}}=$ 》 0.0166 «mol $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}$ » « $0.0166 \mathrm{~mol} \times 6.02 \times 10^{23}$ molecules $\mathrm{mol}^{-1} \times 8$ atoms molecule ${ }^{-1}=$ » $8.01 \times 10^{22}$ «atoms of hydrogen»

Marking guidance:

Accept answers in the range $7.99 \times 10^{22}$ to $8.19 \times 10^{22}$ .

Award [2] for correct final answer.

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.

How it is assessed

Questions derive an empirical formula from composition or scale an empirical formula to the molecular formula using molar mass.

Command terms

determine

What earns marks

Show conversion to moles, the simplest whole-number ratio, and the final formula; for molecular formula, use the integer molar-mass multiplier on every subscript.

Watch for

Rounding mole ratios before reaching whole numbers, or multiplying only one subscript when converting to the molecular formula.

Representative question

[Maximum number: 3]

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.

（a）

$n(\mathrm{C}) \ll=n\left(\mathrm{CO}_{2}\right)=\frac{9.49 \mathrm{~g}}{44.01 \mathrm{~g} \mathrm{~mol}^{-1}}>=0.216<\mathrm{mol}>$

AND

$n(\mathrm{H}) ~ « 2 n\left(\mathrm{H}_{2} \mathrm{O}\right)=2 \times \frac{5.18 \mathrm{~g}}{18.02 \mathrm{~g} \mathrm{~mol}^{-1}} »=0.575<\mathrm{mol} \gg$

OR
$H=0.581 \text { 《g》 }$

AND
$C=2.59 \text { «g»v }$ $\begin{aligned} & \text { 《 } m(\mathrm{O})=4.32-(m(\mathrm{C})+m(\mathrm{H})) & =4.32-(2.59+0.581)=1.15 \mathrm{~g} \| \end{aligned}$ $n(\mathrm{O})=\frac{1.15 \mathrm{~g}}{16.00 \mathrm{~g} \mathrm{~mol}^{-1}}$

／ 0.0718 «mol»
$<n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{O})=0.216: 0.575: 0.0718=3: 8: 1>$ $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$

M2 for finding mass of oxygen

M3 for finding empirical formula Award［3］for correct final answer．

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.

How it is assessed

Questions calculate concentration from solute amount and solution volume, including a titration context.

Command terms

calculate

What earns marks

Obtain moles when needed, divide by the stated equivalence or solution volume in dm³, and report mol dm⁻³ with the correct significant figures.

Watch for

Using cm³ without conversion to dm³, or dividing by solute mass instead of solution volume.

Representative question

[Maximum number: 2]

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

$n(\mathrm{Li})=\frac{0.200\ \mathrm{g}}{6.94\ \mathrm{g\ mol^{-1}}} =0.0288\ \mathrm{mol}$ $n(\mathrm{LiOH})=n(\mathrm{Li})=0.0288\ \mathrm{mol}$ $[\mathrm{LiOH}] =\frac{0.0288\ \mathrm{mol}}{0.5000\ \mathrm{dm^3}} =0.0576\ \mathrm{mol\ dm^{-3}}$

Award [2] for the correct final answer.

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.

How it is assessed

Questions apply balanced-equation mole ratios to gas volumes at fixed conditions or convert a reaction amount to gas volume at STP.

Command terms

determine / calculate

What earns marks

Identify the limiting reactant where relevant, preserve the balanced-equation ratio, and use the stated molar volume or gas equation with consistent units.

Watch for

Using a mass ratio instead of the balanced gas-volume ratio, or ignoring the limiting gas before calculating product volume.

Representative question

[Maximum number: 2]

$140.0 \mathrm{~cm}^{3}$ 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.

hydrogen is the limiting reactant $\mathrm{OR} 2 \mathrm{H}_{2}: 1 \mathrm{C}_{2} \mathrm{H}_{6} \vee$ $80.0 \mathrm{~cm}^{3} \text { V }$

Marking guidance:

Award [2] for correct final answer.

Accept answers expressed in $d m^{3}$ .

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.

Objective notes

6 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 particlesView 1.4.2Relative atomic mass (Ar) and relative formula mass (Mr)• Compared to ¹²C standard• Determine relative formula mass from relative atomic massesView 1.4.3Molar mass (M)• Units: g mol⁻¹• Relationships: particles ↔ moles ↔ mass• Solve mass, mole, and particle problems using n = m/MView 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 massView 1.4.5Molar concentration• Determined by amount of solute and volume of solution• Solve concentration, amount, and volume problems; square brackets mean molar concentrationView 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 equationsView

1.4 The mole

The Mole and Particle Number

Converting Moles to Entities

Assessment in practice

Relative Atomic and Formula Mass

Applying Relative Masses

Assessment in practice

Molar Mass Conversions

Solving Molar-Mass Problems

Assessment in practice

Empirical and Molecular Formulae

Deriving Chemical Formulae

Assessment in practice

Molar Concentration

Calculating Concentration

Assessment in practice

Avogadro's Law and Gas Volumes

Using Gas-Volume Ratios

Assessment in practice

The Mole Summary

Objective notes

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