C1.284SD0 · Section C1: Principles of chemistry

The mole and chemical calculations

Key learnings

Relative masses, the mole and n = m / Mr
Relative atomic mass ( $A_r$ ) is the average mass of an atom relative to 1/12 of a carbon-12 atom (weighted by isotope abundances); relative formula mass ( $M_r$ ) is the sum of the $A_r$ of every atom in the formula. The mole is the amount of a substance whose mass in grams equals its $M_r$ . The central formula is $n = \dfrac{m}{M_r}$ ( $n$ in mol, $m$ in g). On the 4SD0 Edexcel data sheet every $A_r$ is a whole number except copper ( $63.5$ ) and chlorine ( $35.5$ ).
Reacting-mass route: moles, ratio, mass
To find an unknown mass from a balanced equation: (1) convert the known mass to moles with $n = \dfrac{m}{M_r}$ ; (2) use the balancing numbers (the mole ratio) to find the moles of the required substance; (3) convert back with $m = n \times M_r$ . For $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ the ratio is $1:1$ , so 100 g (1 mol) of $\text{CaCO}_3$ gives 56 g (1 mol) of CaO. The balancing numbers, not the masses, give the ratio — the step most often skipped.
Empirical vs molecular formula; % yield
The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number of atoms in one molecule. From data, divide each element's mass (or %) by its $A_r$ to get moles, then divide all by the smallest for the ratio. Get the molecular formula by scaling the empirical formula by (true $M_r$ ÷ empirical mass). Percentage yield = actual ÷ theoretical × 100, actual on top.

Common mistakes

Drawn from real examiner reports.

Empirical = simplest whole-number ratio

Define the empirical formula as the simplest whole-number ratio of atoms — "the ratio of atoms" is not enough. Do not give the molecular formula when the empirical is asked ( $\text{C}_6\text{H}_{12}\text{O}_6$ instead of $\text{CH}_2\text{O}$ ), and scale a decimal ratio such as 1.33 up to whole numbers rather than rounding to 1 or 2. The final subscripts must be integers.

Nov24 Q11b — a decimal ratio was not scaled to whole numbers; Jun24 1CR Q11d — final subscripts must be integers

Divide mass by Ar before comparing

When finding an empirical formula from masses, divide each element's mass by its relative atomic mass to get moles first — only then is the ratio a ratio of atoms. Taking the masses directly (2.4 g Mg : 1.6 g O) gives the wrong answer; $2.4/24 : 1.6/16 = 1:1$ , so $\text{MgO}$ . Use oxygen's atomic $A_r$ of 16, not 32 (the $M_r$ of $\text{O}_2$ ).

Nov24 Q11b — 32 used as the mass of oxygen instead of the atomic Ar of 16

Bracket subscript multiplies every atom

When working out an $M_r$ , a subscript after a bracket multiplies everything inside it, so $\text{Ca(OH)}_2$ has $M_r = 40 + 2\times(16+1) = 74$ , not $40 + 16 + 1$ . The same applies to the number after a dot in a hydrated salt. Ignoring the bracket multiplier gives an $M_r$ that is too small and feeds an error through the whole calculation.

Use the reacting ratio from the equation

In a reacting-mass calculation, use the balancing numbers as the mole ratio — not the masses, and not a 1:1 assumption. If the equation shows a 2:1 ratio you must multiply (or divide) the moles accordingly; forgetting to scale the moles by the equation's ratio is the top reacting-mass error. Read the coefficients before converting moles between substances.

Jun23 Q10ai — moles compared without the 2:1 ratio; Jun24 1CR Q12ciii — moles not multiplied by two from the equation

Percentage yield can never exceed 100 %

Percentage yield is actual ÷ theoretical × 100, mass over mass with the actual yield on top. Inverting it (theoretical over actual) gives a value above 100 %, which is impossible for a yield and should be an immediate warning to re-check. Watch the divisor and round the final percentage correctly — a mis-rounded or inverted yield is a common lost mark.

Jun24 1CR Q3d — percentage divisor wrong and the percentage mis-rounded

Keep full figures until the end

Do not truncate or round part-way through a calculation — cutting 0.056 to 0.05, or moles to one significant figure, shifts the ratio and loses the final mark. Keep full calculator figures through every intermediate step and round only the final answer. Also avoid applying the reacting ratio twice by mixing the moles method with a simple-ratio method.

Jun23 Q9c — truncating a figure mid-calculation shifted the ratio; Jun23 Q7c — the ratio was applied twice

Exam tips

Show every line: Mr, moles, mass
Set out working line by line — the balanced equation, each $M_r$ , the moles ( $n = m/M_r$ ), the mole ratio, then the mass. Mark schemes give method marks for a correct $M_r$ and moles line even if the final arithmetic slips, so a worked answer still banks marks.
Sanity-check the final answer
Before writing the answer, sense-check it: a percentage yield is at most 100 %; a metal oxide is heavier than the metal it came from; empirical-formula subscripts must be whole numbers; and every quantity needs its unit (mol, g, %). A quick check catches an inverted result.
Match the method to the data given
Read the data you are given and pick the matching route: masses or percentages of each element → empirical formula (divide by $A_r$ ); a mass plus a balanced equation → reacting mass; actual and theoretical masses → percentage yield. The wrong route wastes marks.

Revision notes