Balanced equations, yield and atom economyAQA A-Level Chemistry: Revision notes

A balanced equation has the same number of each type of atom, and the same total charge, on both sides. To balance an unfamiliar reaction, write the correct formulae first, then change only the coefficients (the big numbers in front), never the subscripts.

Example: aluminium burning in oxygen: 4Al + 3O₂ → 2Al₂O₃.

An ionic equation shows only the species that change. Write the full equation with state symbols, split aqueous ionic compounds into ions, and cancel the spectator ions that appear unchanged on both sides.

Neutralisation: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) becomes H⁺(aq) + OH⁻(aq) → H₂O(l), because Na⁺ and Cl⁻ are spectators.

Every calculation follows the same path:

1. Convert the data to amount in mol (n = m/M for a mass; n = cV for a solution, with V in dm³)
2. Use the mole ratio from the balanced equation
3. Convert the answer: mass = n × M; concentration = n/V; volume of solution = n/c

Titration example: 25.0 cm³ of NaOH needs 22.40 cm³ of 0.100 mol dm⁻³ HCl. n(HCl) = 0.100 × 0.02240 = 2.24 × 10⁻³ mol. The ratio is 1 : 1, so c(NaOH) = 2.24 × 10⁻³ / 0.0250 = 0.0896 mol dm⁻³.

For gases, find the amount with the ideal gas equation, pV = nRT (SI units: Pa, m³, K), then use the equation's ratio.

Example: 0.243 g of magnesium (Mr = 24.3) reacts completely with acid: Mg + 2HCl → MgCl₂ + H₂.

• n(Mg) = 0.243 / 24.3 = 0.0100 mol = n(H₂)
• At 298 K and 100 kPa: $V = \frac{nRT}{p} = \frac{0.0100 \times 8.31 \times 298}{100000}$ = 2.48 × 10⁻⁴ m³ = 248 cm³

Percentage yield compares the product actually obtained with the maximum possible:

$\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100$

The theoretical mass comes from the limiting reagent and the mole ratio.

Percentage atom economy measures how much of the mass of the reactants becomes the desired product:

$\text{atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all reactants}} \times 100$

Use the coefficients: for C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ the desired product mass is 2 × 46.0 = 92.0, so the atom economy is 92.0 / 180.0 × 100 = 51.1%.

A reaction can have a high yield and a low atom economy, because yield does not account for by-products.

Processes with a high atom economy are better for society and industry:

• Economic: less waste to treat or dispose of, fewer raw materials and less energy used per tonne of product, and by-products that can be sold
• Ethical: conserves finite resources for future generations
• Environmental: less pollution and waste, with fewer harmful by-products

An addition reaction, such as C₂H₄ + H₂O → C₂H₅OH, has a 100% atom economy.

Making a standard solution (e.g. 250 cm³ of 0.0500 mol dm⁻³ Na₂CO₃, which needs 1.33 g):

1. Weigh the solid accurately by difference
2. Dissolve in a beaker in less than 250 cm³ of distilled water
3. Transfer through a funnel to a volumetric flask, rinsing the beaker and funnel into it
4. Make up to the mark with distilled water, using a dropper for the last drops (bottom of the meniscus on the line)
5. Stopper and invert several times

Titration: rinse the burette with the acid, fill it (below the tap), pipette 25.0 cm³ of the other solution into a conical flask, add indicator, and titrate until the end point. Repeat until you have concordant titres (within 0.10 cm³) and take their mean.