Core definitions and the mole concept

Foundational Particle and Formula Definitions

Chemistry rests on precise definitions of particles and formulae, because every later calculation depends on knowing exactly what each term means.

An atom is the smallest particle of an element that retains the chemical properties of that element.

An element is a substance made of only one type of atom, where every atom shares the same number of protons.

An ion is a charged particle formed when an atom or group of atoms loses or gains one or more electrons. Cations are positive ions formed by electron loss; anions are negative ions formed by electron gain.

A molecule is a group of two or more atoms held together by covalent bonds and behaving as a discrete unit.

A compound is a substance containing two or more different elements chemically bonded together in fixed proportions.

The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound.

The molecular formula gives the actual number of atoms of each element present in one molecule of a compound.

For example, glucose has the molecular formula C₆H₁₂O₆ but the empirical formula CH₂O.

Ionic compounds such as NaCl are described by a formula unit, since they form giant lattices rather than discrete molecules.

The Mole and Avogadro Constant

The mole (mol) is the SI unit for amount of substance, defined as the number of particles equal to the number of atoms in exactly 12 g of carbon-12.

This number, called the Avogadro constant (L), equals $6.02\times {10}^{23}$ particles per mole.

The mole therefore acts as a counting unit for atoms, molecules, ions or formula units, in the same way that “dozen” counts twelve items.

Key Equations

Number of particles from moles:

$$N=nL$$

Variables: (N) = number of particles (dimensionless); (n) = amount of substance (mol); (L) = Avogadro constant = $6.02\times {10}^{23} mo{l}^{-1}$.

Moles from mass:

$$n=\frac{m}{{M}_{r}}$$

Variables: (n) = amount of substance (mol); (m) = mass (g); $({M}_{r})$ = molar mass (g mol⁻¹).

To find the number of particles in a given mass, first calculate moles using $n=\frac{m}{{M}_{r}}$, then multiply by (L). This two-step approach connects laboratory-measurable mass to particle-level counts.

Worked Example: Particles in a Sample of Water

Scenario

Calculate the number of water molecules in 9.00 g of pure water (H₂O).

Equation used

$$n=\frac{m}{{M}_{r}}$$

Given

$$m=9.00 g$$

$${M}_{r}({H}_{2}O)=18.0 g mo{l}^{-1}$$

Working — moles of water

$$n=\frac{9.00}{18.0}$$

$$n=0.500 mol$$

Equation used

$$N=nL$$

Substitution:

$$N=0.500\times 6.02\times {10}^{23}$$

Answer

$$N=3.01\times {10}^{23}$$

Interpretation

Just 9.00 g of water contains around $3\times {10}^{23}$ discrete H₂O molecules, illustrating how a small mass corresponds to an enormous particle count.

Examiner InsightQuote the Avogadro constant to the same precision as the data, typically $6.02\times {10}^{23}$ mol⁻¹. Round only the final answer.

Exam TipKeep three significant figures throughout unless told otherwise.

Balanced Full and Ionic Equations

A chemical equation must show conservation of mass and charge, with state symbols indicating the physical state of every species involved.

A balanced equation has equal numbers of each type of atom on both sides; the total charge on both sides must also match.

The state symbols are (s) for solid, (l) for liquid, (g) for gas and (aq) for aqueous (dissolved in water).

For example, the reaction of sodium with water is written as:

An ionic equation shows only the species that take part in the reaction, omitting spectator ions which appear unchanged on both sides.

For example, when silver nitrate reacts with sodium chloride solution:

The Na⁺ and NO₃⁻ ions are spectator ions because they remain in solution unchanged throughout.

To write an ionic equation: split all aqueous strong electrolytes into their ions, leave solids, liquids and gases unsplit, then cancel any species appearing identically on both sides.

MisconceptionMany students forget that charges must balance in an ionic equation, not just atoms. An equation balancing atoms but not charges scores zero.

Exam TipAlways check both sides of an ionic equation for total atoms AND total charge.

Relative Masses, Molar Mass and Parts per Million

All atomic masses are measured on the carbon-12 scale, where one atom of ¹²C is assigned a mass of exactly 12 atomic mass units.

The relative atomic mass $({A}_{r})$ of an element is the weighted mean mass of its atoms compared with 1/12 the mass of a ¹²C atom. It is “weighted” because elements exist as mixtures of isotopes in fixed proportions, and $({A}_{r})$ accounts for both isotope masses and natural abundances.

The relative molecular mass $({M}_{r})$ is the sum of the $({A}_{r})$ values of all atoms shown in the molecular formula. For example, ${M}_{r}({H}_{2}O)=(2\times 1.0)+16.0=18.0$.

The relative formula mass is used for ionic compounds and other giant structures, since they do not exist as discrete molecules. It is calculated in the same way: by adding the $({A}_{r})$ values of every atom in the formula unit. For example, ${M}_{r}(NaCl)=23.0+35.5=58.5$.

The molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. Numerically the molar mass equals the relative molecular or formula mass, but it carries the unit g mol⁻¹, whereas $({M}_{r})$ is a dimensionless ratio.

Parts per million (ppm) is a unit for very dilute concentrations, defined as the number of mass units of a substance per million mass units of the total mixture. For atmospheric gases this is calculated as:

$$ppm=\frac{\text{mass of component}}{\text{total mass}}\times {10}^{6}$$

For example, atmospheric CO₂ at around 420 ppm corresponds to 420 g of CO₂ per million grams of air.

Worked Example: Calculating Mr and ppm

Scenario

Calculate (a) the relative formula mass of ammonium sulfate, (NH₄)₂SO₄, and (b) the ppm of a pollutant when 0.080 g is present in 4000 g of air.

Part (a) — relative formula mass:

Sum of $({A}_{r})$ values:

$${M}_{r}=(2\times 14.0)+(8\times 1.0)+32.1+(4\times 16.0)$$

$${M}_{r}=28.0+8.0+32.1+64.0$$

$${M}_{r}=132.1$$

Part (b) — ppm calculation:

Equation used

$$ppm=\frac{\text{mass of component}}{\text{total mass}}\times {10}^{6}$$

Substitution:

$$ppm=\frac{0.080}{4000}\times {10}^{6}$$

Answer

$$ppm=20$$

Final answers: (a) ${M}_{r}=132.1$; (b) 20 ppm.

Interpretation

Ppm is convenient for trace amounts where mol dm⁻³ would give awkwardly small numbers, especially for atmospheric gases.

Concentration in mol dm⁻³ and g dm⁻³

Concentration measures how much solute is dissolved per unit volume of solution, allowing chemists to compare solutions on a quantitative basis.

The standard SI volume for concentration is the cubic decimetre (dm³), which is identical to one litre (1 dm³ = 1000 cm³).

Key Equations

Concentration in moles per dm³:

$$c=\frac{n}{V}$$

Variables: (c) = concentration (mol dm⁻³); (n) = amount of solute (mol); (V) = volume of solution (dm³).

Conversion between mol dm⁻³ and g dm⁻³:

$${c}_{g d{m}^{-3}}={c}_{mol d{m}^{-3}}\times {M}_{r}$$

Volumes in cm³ must be converted to dm³ before substitution by dividing by 1000.

A concentration in g dm⁻³ tells you the mass of solute per litre of solution, while mol dm⁻³ gives the moles per litre. The two are interconverted through the molar mass, since one mole always corresponds to a fixed mass.

Worked Example: Concentration of a Sodium Hydroxide Solution

Scenario

4.00 G of NaOH is dissolved in distilled water and made up to 250 cm³ of solution. Calculate the concentration in mol dm⁻³ and g dm⁻³.

Equation used

$$n=\frac{m}{{M}_{r}}$$

Given

$$m=4.00 g$$

$${M}_{r}(NaOH)=23.0+16.0+1.0=40.0 g mo{l}^{-1}$$

Working — moles of NaOH

$$n=\frac{4.00}{40.0}$$

$$n=0.100 mol$$

Volume conversion (cm³ to dm³):

$$V=\frac{250}{1000}$$

$$V=0.250 d{m}^{3}$$

Equation used

$$c=\frac{n}{V}$$

Substitution:

$$c=\frac{0.100}{0.250}$$

$$c=0.400 mol d{m}^{-3}$$

Equation used

$${c}_{g d{m}^{-3}}={c}_{mol d{m}^{-3}}\times {M}_{r}$$

Substitution:

$${c}_{g d{m}^{-3}}=0.400\times 40.0$$

Answer

$$c=0.400 mol d{m}^{-3}=16.0 g d{m}^{-3}$$

Interpretation

The same solution can be expressed in two equivalent units; the link is the molar mass, which converts the count of particles into a mass per litre.

Examiner InsightA common error is leaving volume in cm³ when applying $c=\frac{n}{V}$. Always convert to dm³ first by dividing by 1000.

Exam TipWrite the unit conversion explicitly to avoid losing the method mark.