G3 Science Chemistry Formula List
G3 Science Chemistry calculation relationships with units, reacting ratios, worked examples and the correct course scope.
On this page
A calculation and formula reference for G3 Science Chemistry (K326 / K328). Read the relationship, units and condition together. The balanced equation controls a reacting ratio; a formula subscript controls the composition of one particle.
Mass, amount and chemical formulae
| Relationship | Symbols and units | When to use it |
|---|---|---|
| Mᵣ = ∑ Aᵣ | Mᵣ and Aᵣ have no unit | Include every subscript and bracket multiplier in the formula. |
| n = m/M; m = nM | n mol, m g, M g/mol | Match the mass unit to the molar-mass unit; this does not require a reaction ratio. |
| mean Aᵣ = ∑(A_(r,i) fᵢ) | Fractional isotope abundance fᵢ has no unit; ∑ fᵢ = 1 | For percentage abundances, divide each percentage by 100. |
| total positive charge + total negative charge = 0 | Ionic charges in units of elementary charge | Choose the smallest whole-number ion ratio; never change an ion’s own formula. |
| CₙH₂ₙ₊₂; CₙH₂ₙ | n is the number of carbon atoms | Acyclic alkanes / acyclic alkenes with one double bond; not all hydrocarbons. |
Particles, gases and solutions
| Relationship | Symbols and units | When to use it |
|---|---|---|
| x = Mᵣ(compound)/Mᵣ(empirical formula) | Whole-number multiplier x; relative masses have no unit | Multiply every empirical-formula subscript by x to obtain the molecular formula; first find the simplest mole ratio. |
| N = nN_A | Particle count N; N_A in mol⁻¹ | Specify the counted entity; N_A ≈ 6.02 × 10²³ mol⁻¹ for calculations. |
| n(A)/a = n(B)/b | Moles; equation coefficients a and b | For aA → bB, use a balanced equation and the limiting reactant. |
| n = V/Vₘ | Gas volume V and molar gas volume Vₘ in matching units | At r.t.p. the school reference is Vₘ = 24 dm³/mol; use the supplied value. |
| c = n/V; n = cV | c mol/dm³; solution volume V dm³ | Convert cm³ to dm³ before substitution: divide by 1000. |
| cₘₐₛₛ = m/V = cM | Mass concentration g/dm³; M g/mol | Distinguish mass concentration from molar concentration. |
| c₁V₁ = c₂V₂ | Matching concentration and volume units | Dilution of the same solute with no reaction; not the general titration equation. |
| c_AV_A/a = c_BV_B/b | Concentrations mol/dm³ and volumes dm³ | Titration at the reacting ratio a:b; equal volumes or equal concentrations need not result. |
Following a reaction
| Relationship | Symbols and units | When to use it |
|---|---|---|
| average rate = Δquantity/Δ t | Examples: cm³/s or g/s | State the measured quantity; gradient of a tangent gives instantaneous rate. |
Reaction patterns to recognise
These patterns help identify products; they are not balanced equations. Write the species and balance atoms and charge for the actual reaction.
- Dilute non-oxidising acid + suitably reactive metal → salt + hydrogen.
- Acid + metal oxide or hydroxide → salt + water.
- Acid + carbonate → salt + water + carbon dioxide.
- Complete hydrocarbon combustion → carbon dioxide + water.
A quick application: the reacting ratio
For H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l), 25.0 cm³ of 0.100 mol/dm³ acid contains 0.100 × 0.0250 = 2.50 × 10⁻³ mol. It needs twice that amount of sodium hydroxide: 5.00 × 10⁻³ mol. At 0.200 mol/dm³, the alkali volume is n/c = 0.0250 dm³ = 25.0 cm³.
The equal volumes here come from both the 1:2 reacting ratio and the chosen concentrations. “c₁V₁ = c₂V₂” would give the wrong answer for this titration.
G3 Science stopping point
Percentage yield, percentage purity and R_f calculations are not required. Gas volumes are calculated at the stated common conditions using molar gas volume, without gas-law changes of temperature or pressure. pH is interpreted without logarithmic calculations.
Find the explanation behind a relationship on the course hub, then use the topic check with this sheet closed.