Bonding properties, balanced equations and mole calculations

Properties linked to bonding and structure

Understanding why materials behave as they do requires knowing their bonding type and structure:

Metallic bonding: a lattice of positive metal ions (cations) surrounded by a "sea" of delocalised electrons. The electrostatic attraction between the cations and the electron sea is the metallic bond. Metals conduct electricity because these electrons are free to move; metals are malleable because layers of ions can slide without breaking bonds.

Writing and balancing chemical equations

A word equation names reactants and products. A symbol equation uses chemical formulae. A balanced equation has the same number of atoms of each element on both sides.

Steps to balance:

1. Write the unbalanced equation.
2. Count atoms of each element on each side.
3. Add coefficients (numbers in front of formulae) — never change the subscripts inside a formula.
4. Recount until balanced.
5. Add state symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous.

Example: combustion of propane C₃H₈ + O₂ → CO₂ + H₂O (unbalanced) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (balanced)

Check: C: 3=3 ✓ H: 8=8 ✓ O: 10=10 ✓

Ionic equations: cancel spectator ions (ions that appear unchanged on both sides). E.g. for any acid–base neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l)

The mole concept

The mole is the SI unit of amount of substance. One mole contains 6.02 × 10²³ particles (Avogadro's number, Nₐ).

Molar mass M: mass of one mole of a substance in g mol⁻¹. Numerically equal to the relative formula mass (Mr).

Core equations:

• n = m ÷ M (moles = mass ÷ molar mass)
• m = n × M (mass = moles × molar mass)

Mr calculation: add together the Ar values for all atoms in the formula. Example: H₂SO₄ → 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g mol⁻¹

Reacting masses

Use the balanced equation to find the ratio of moles, then convert to masses.

Example: How many grams of water form when 4 g of hydrogen burns? 2H₂ + O₂ → 2H₂O Ratio: 2 mol H₂ : 2 mol H₂O → 1:1 Moles H₂ = 4 ÷ 2 = 2 mol Moles H₂O = 2 mol Mass H₂O = 2 × 18 = 36 g

Concentration and solutions

Concentration (c) = moles ÷ volume: c = n ÷ V, where V is in dm³ (litres). 1 dm³ = 1 litre = 1000 cm³. To convert cm³ → dm³: divide by 1000.

Rearrangements:

• n = c × V
• c = n ÷ V
• V = n ÷ c

Example: What mass of NaOH (M = 40 g mol⁻¹) is needed to make 250 cm³ of 0.1 mol/dm³ solution? n = 0.1 × (250/1000) = 0.025 mol m = 0.025 × 40 = 1.0 g

Percentage yield and atom economy

Percentage yield = (actual yield ÷ theoretical yield) × 100 %

Atom economy = (mass of desired product ÷ total mass of all products from equation) × 100 %

High atom economy is desirable in industrial chemistry for sustainability and cost-effectiveness. WJEC examiners frequently ask you to justify why a process has a high or low atom economy.

WJEC required practicals relevant to U1.2

• Preparing a standard solution from a primary standard (e.g. anhydrous sodium carbonate): dissolve, transfer quantitatively to volumetric flask, make up to the mark.
• Titration to determine concentration: precisely measure volume of acid to neutralise alkali; calculate using n = c × V.

Common examiner traps

1. Changing subscripts to balance: never alter the formula (e.g. writing H₃ instead of H₂). Only change coefficients.
2. Forgetting state symbols: WJEC often awards a mark specifically for state symbols on the balanced equation.
3. Volume in the wrong units: concentration formula uses dm³. A common error is plugging in cm³ directly. Always divide cm³ by 1000.
4. Confusing Mr with Ar: Mr is for a compound (sum of all Ar values); Ar is for an element.