The Mole Concept

The mole links particle counts to measurable amounts. One mole contains Avogadro’s number of particles, \(6.022\times10^{23}\).

• Moles from mass: \(n = \dfrac{m}{M}\) (where \(m\) is mass in g; \(M\) is molar mass in g·mol\(^{-1}\)).
• Moles of a gas at STP: \(n = \dfrac{V}{22.4\ \text{dm}^3}\).
• Solution concentration: \(c = \dfrac{n}{V}\) (with \(V\) in dm\(^3\)).

Balancing Chemical Equations

Coefficients provide the mole ratios used in all stoichiometry. Balance atoms, then check charge (if ionic).

Example: Balance propane combustion:

\(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\)

Empirical & Molecular Formulae

Empirical formula: simplest whole-number ratio. Molecular formula: actual numbers of atoms.

1. Convert % to mass (assume 100 g).
2. Convert mass to moles: \(n=\frac{m}{M}\).
3. Divide by the smallest \(n\) to get ratio.
4. Scale to whole numbers; that’s the empirical formula.
5. Find \(n=\dfrac{M_\text{molecular}}{M_\text{empirical}}\); multiply.

Glucose: empirical CH\(_2\)O, \(M_\text{emp}=30\). If \(M_\text{molecular}=180\), \(n=6\) → C\(_6\)H\(_{12}\)O\(_6\).

Quantitative Aspects of Chemical Change

Start with moles, use the balanced ratio, convert to the asked quantity (mass/volume/concentration).

• Mass ⇄ moles: \(m=nM\)
• Solutions: \(n=cV\)
• Gas volumes at STP: \(V = n \times 22.4\ \text{dm}^3\)

Limiting Reagents & Percentage Yield

Limiting reagent is consumed first; it limits product formed.

% Yield: \(\%\,\text{Yield} = \dfrac{\text{actual}}{\text{theoretical}}\times 100\%\)

Worked Stoichiometry Examples

Example 1: Mass–Mass

Question: What mass of water forms when 4.0 g hydrogen burns in excess oxygen?

Balanced: \(2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}\)

\(n(\text{H}_2)=\frac{4.0}{2.016}\approx 1.98\ \text{mol}\). Ratio H\(_2\):H\(_2\)O = 1:1 ⇒ \(n(\text{H}_2\text{O})\approx 1.98\ \text{mol}\). \(m= nM \approx 1.98 \times 18.015 \approx 35.7\ \text{g}\).

Example 2: Solution Stoichiometry

How many cm³ of 0.200 mol·dm\(^{-3}\) HCl are needed to neutralise 25.0 cm³ of 0.100 mol·dm\(^{-3}\) NaOH?

\(\text{HCl}+\text{NaOH}\to\text{NaCl}+\text{H}_2\text{O}\) (1:1)

\(n(\text{NaOH})=0.100\times0.0250=0.00250\ \text{mol}\). Need same moles of HCl. \(V=\frac{n}{c}=\frac{0.00250}{0.200}=0.0125\ \text{dm}^3=12.5\ \text{cm}^3\).

Example 3: Limiting Reagent

Given 5.00 g Al and 10.0 g Cl\(_2\), find the mass of AlCl\(_3\) formed.

\(2\text{Al}+3\text{Cl}_2\to2\text{AlCl}_3\)

\(n(\text{Al})=5.00/26.98=0.185\ \text{mol}\); \(n(\text{Cl}_2)=10.0/70.90=0.141\ \text{mol}\).

Needed Cl\(_2\) for 0.185 mol Al: \(\tfrac{3}{2}\times 0.185=0.278\ \text{mol}\) but only 0.141 available ⇒ Cl\(_2\) limiting.

From 0.141 mol Cl\(_2\): \(n(\text{AlCl}_3)=\tfrac{2}{3}\times 0.141=0.0940\ \text{mol}\). Mass \(=0.0940\times 133.34\approx 12.5\ \text{g}\).

Acids & Bases

Brønsted–Lowry: acids donate H\(^+\); bases accept H\(^+\).

• Strong acid/base: nearly complete ionisation (HCl, HNO\(_3\); NaOH).
• Weak acid/base: partial ionisation (CH\(_3\)COOH; NH\(_3\)).

pH: \( \mathrm{pH}=-\log[H^+] \). At 25 °C, neutral water has pH ≈ 7.

Applications that connect to high-value careers: water treatment services, pharmaceuticals, fertiliser production.

Acid–Base Example

Find the pH of \(1.0\times10^{-3}\) mol·dm\(^{-3}\) HCl.

\([H^+]=1.0\times10^{-3}\Rightarrow \mathrm{pH}=3.00\).

Redox Reactions

Oxidation: loss of electrons. Reduction: gain of electrons.

Assign oxidation numbers to see what changes.

Example: \( \text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu} \).

• Zn: 0 → +2 (oxidised)
• Cu\(^{2+}\): +2 → 0 (reduced)

Energy Changes (Enthalpy, Endo/Exothermic)

Enthalpy change: \(\Delta H = H_\text{products}-H_\text{reactants}\).

• Exothermic: \(\Delta H<0\) (releases heat). Combustion, neutralisation.
• Endothermic: \(\Delta H>0\) (absorbs heat). Thermal decomposition.

Calorimetry: \( q = mc\Delta T \). For reaction enthalpy per mole: \( \Delta H = -\frac{q}{n} \) (sign depends on convention).

Worked Energy Examples

Example 1: Neutralisation Enthalpy (Calorimetry)

50.0 cm³ of 1.0 mol·dm\(^{-3}\) HCl reacts with 50.0 cm³ of 1.0 mol·dm\(^{-3}\) NaOH in a coffee-cup calorimeter. Temperature rises from 20.0 °C to 26.7 °C. Find \(\Delta H\) (kJ·mol\(^{-1}\)) for the reaction, assuming solution density \(=1.00\) g·cm\(^{-3}\) and \(c=4.18\) J·g\(^{-1}\)·°C\(^{-1}\).

1. Total mass \(m\approx 100.0\) g; \(\Delta T=6.7\) °C.
2. \(q=mc\Delta T=100.0\times 4.18\times 6.7\approx 2.80\times 10^3\) J \(=2.80\) kJ (released).
3. Moles limiting = HCl = \(1.0\times 0.0500=0.0500\) mol.
4. \(\Delta H \approx \frac{-2.80}{0.0500}=-56\ \text{kJ·mol}^{-1}\).

Example 2: Endothermic/Exothermic Identification

\(\text{NH}_4\text{NO}_3(s)\to \text{NH}_4^+(aq)+\text{NO}_3^-(aq)\) causes a temperature drop in water. Since the solution cools, the process is endothermic (\(\Delta H>0\)).