The formulae you must know for SPM Chemistry

SPM Chemistry does not give you a formula sheet, so a handful of relationships have to live in your memory, ready to use. The good news is that the list is short and the same formulae reappear across many chapters. Below is the compact set worth knowing cold, grouped by where they turn up, with the meaning of each symbol and the constants you need. Keep the full chemistry calculations hub open beside this as you practise.

The mole, the engine of the whole subject

Almost every calculation runs through the mole. Three routes convert something you can measure into a number of moles:

• From mass: number of moles = mass ÷ molar mass, or n = m / M. Molar mass M is in g mol⁻¹ and equals the relative molecular (or atomic) mass in grams. See moles from mass.
• From particles: number of moles = number of particles ÷ the Avogadro constant. The Avogadro constant is 6.02 x 10²³ mol⁻¹, so number of particles = n × 6.02 x 10²³ mol⁻¹.
• From gas volume: number of moles = volume of gas ÷ molar volume. Molar volume of a gas is 24 dm³ mol⁻¹ at room conditions; at STP it is 22.4 dm³ mol⁻¹. Read the stated conditions and pick the matching value.

A quick way to remember the mass route: cover the quantity you want in the triangle of m, n and M, where m sits on top and n × M sits below.

Concentration and solutions

• Concentration in mol dm⁻³: concentration = number of moles ÷ volume in dm³, or c = n / V. This is molarity. See molarity and concentration.
• Converting between the two concentration units: concentration in g dm⁻³ = concentration in mol dm⁻³ × molar mass. Divide to go the other way.
• Dilution: M₁V₁ = M₂V₂, because the moles of solute do not change when you add water. Volumes may be in cm³ as long as both sides use the same unit.
• Titration ratio: MₐVₐ ÷ MᵦVᵦ = a ÷ b, where a and b are the coefficients of acid and base in the balanced equation. This ties concentration to the mole ratio.

Relative masses and formulae

• Relative molecular mass = the sum of the relative atomic masses of all the atoms in the formula. For example, the relative molecular mass of CO₂ is 12 + (16 × 2) = 44.
• Percentage by mass of an element = (total mass of that element in the formula ÷ molar mass), then multiply by 100 to express it as a percentage.
• Empirical formula is found by dividing each element’s moles by the smallest, giving the simplest whole-number ratio; the molecular formula is a whole-number multiple of it, found from (relative molecular mass ÷ empirical formula mass).

Rate of reaction

• Average rate = change in quantity ÷ time taken. The unit follows the quantity: cm³ s⁻¹ for gas volume, g s⁻¹ for mass, mol dm⁻³ s⁻¹ for concentration.
• Rate at an instant = gradient of the tangent to the graph at that moment. These two are different questions, so read carefully which is asked.

Thermochemistry

• Heat change: Q = mcθ. Here m is the mass of solution in g, c = 4.2 J g⁻¹ °C⁻¹ (take the density of dilute solution as 1 g cm⁻³ so cm³ reads as g), and θ is the temperature change. Q comes out in joules.
• Heat of reaction per mole: ΔH = Q ÷ number of moles, converted to kJ mol⁻¹. Remember the sign: negative for exothermic (temperature rises), positive for endothermic (temperature falls).

The constants and standard values

Keep these exact figures ready, because questions assume them without stating them:

• Avogadro constant: 6.02 x 10²³ mol⁻¹.
• Molar volume of gas: 24 dm³ mol⁻¹ at room conditions (22.4 dm³ mol⁻¹ at STP).
• Specific heat capacity of solution: 4.2 J g⁻¹ °C⁻¹.
• Density of dilute aqueous solution: 1 g cm⁻³.

For units and how the prefixes convert (cm³ to dm³, g to kg), the units and prefixes reference is worth a look; the single most common arithmetic error is a volume left in cm³ when the formula wants dm³, so remember 1000 cm³ = 1 dm³.

How to actually memorise these

Do not copy the list out and hope. Write the mole “map” from a blank page, the three routes into moles and the routes back out, then attempt a mixed calculation and check which link you reached for. A formula you can rebuild under pressure is worth far more than one you have merely read. If your child can recite the formulae but freezes on which to use, that is a pattern-recognition gap, and it is exactly what focused practice fixes. Our teachers teach SPM Chemistry online, one-to-one in English from RM50 an hour, with a paid one-hour trial to start, working real past-paper calculations until choosing the right formula is instant.