1.284CH1 · Section 1: Principles of chemistry

Amount of substance: the mole, masses, concentration and gas volume

Key learnings

The mole and n = m / Mr
The mole (mol) is the unit for amount of substance — one mole contains the Avogadro constant, $6.02 \times 10^{23}$ , of particles. For any pure substance:

$n = \frac{m}{M_r}$

with $n$ in mol, $m$ in g and $M_r$ the relative formula mass (add the $A_r$ of every atom, e.g. $M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100$ ). Rearrange for the unknown: $m = n \times M_r$ and $M_r = \dfrac{m}{n}$ .
Ratios, concentration n/V, gas volume 24n
The balanced equation gives the mole ratio — the bridge in reacting-mass sums. For solutions, concentration is amount per unit volume:

$c = \frac{n}{V} \quad (\text{mol/dm}^3)$

with $V$ in $\text{dm}^3$ (divide $\text{cm}^3$ by 1000); mass concentration $\text{g/dm}^3 = c \times M_r$ . At RTP the molar gas volume is $24\ \text{dm}^3$ per mole, so $V_{\text{gas}} = 24\,n$ . Percentage yield $= \dfrac{\text{actual}}{\text{theoretical}} \times 100$ .

Common mistakes

Drawn from real examiner reports.

Skipping the mole-ratio step

The commonest error is jumping from moles of the known straight to mass of the unknown, without the mole ratio from the balanced equation. Always: (1) write the equation, (2) find moles of the known, (3) multiply by the ratio (unknown : known), (4) convert to mass with $m = n \times M_r$ . Showing the ratio earns the method mark even if the arithmetic later slips.

June 2024 Paper 2C Q6(b)(ii) — candidates lost the first method mark because the mole ratio step (0.05 ÷ 2, the 2:1 ratio) was not shown, even when the final mass was correct.

Atoms balance but the ion is impossible

An equation can balance for atoms and charge yet be chemically wrong if it uses a non-existent ion. The name fixes the charge: lead(II) means $\text{Pb}^{2+}$ , not $\text{Pb}^{4+}$ . Get each ion's charge first (group or Roman numeral), then build the formula and balance. A wrong subscript — $\text{SO}_3$ for sulfate not $\text{SO}_4$ — corrupts the whole calculation.

June 2024 Paper 2C examiner report — an ionic equation that balanced for atoms and charge scored zero because it used Pb(4+); the "(II)" in lead(II) bromide fixes the charge as Pb(2+).

Defining Ar as just "the mass of an atom"

"The mass of an atom" scores zero. The mark-scheme definition of relative atomic mass ( $A_r$ ) has two parts: the average mass of an atom of an element, compared to 1/12 the mass of a carbon-12 atom. It is a weighted average over isotope abundance, which is why values like chlorine's $A_r = 35.5$ are not whole numbers.

Forgetting to convert cm³ to dm³

Concentration uses volume in $\text{dm}^3$ , but burette and pipette volumes are given in $\text{cm}^3$ . Divide a $\text{cm}^3$ volume by 1000 before using $c = n/V$ — $25.0\ \text{cm}^3 = 0.0250\ \text{dm}^3$ . Forgetting this makes the concentration wrong by a factor of 1000. Keep the units consistent all the way through the calculation.

Giving an amount of substance in grams

Amount of substance is measured in moles, never grams. "The amount is 8 g" confuses amount with mass — 8 g is a mass, and you convert it with $n = m/M_r$ to get the amount in mol. Watch the wording: "amount", "moles" and "number of particles" all point to mol; "mass" points to grams.

Confusing mol/dm³ with g/dm³

$\text{mol/dm}^3$ and $\text{g/dm}^3$ are different concentration units. To convert a mol/dm³ concentration to g/dm³, multiply by the relative formula mass: $\text{g/dm}^3 = c \times M_r$ . Read which the question wants — a value in mol/dm³ answered in g/dm³ (or vice versa) loses the mark even when the arithmetic is right.

Exam tips

The five-step mole scaffold
Write every step — marks are for the route, not only the answer: (1) balanced equation; (2) convert the given quantity to moles ( $n = m/M_r$ , $cV$ or $V/24$ ); (3) apply the mole ratio; (4) convert back to what is asked; (5) round to the data's s.f.
Round only at the very end
Keep full calculator figures through every step and round only the final answer — usually to the significant figures of the data (often 3 s.f.). Truncating a mid-step value (0.056 to 0.05, or moles to 1 s.f.) shifts the ratio and loses the accuracy mark.
Rearrange the triangle for the unknown
Decide what is unknown first. From $n = m/M_r$ : for mass use $m = n \times M_r$ ; for $M_r$ use $M_r = m/n$ ; for moles, mass $\div M_r$ . For solutions $n = c \times V$ (V in dm³); for a gas at RTP $n = V/24$ . Pick the rearrangement that isolates what is asked.

Revision notes