AQA AQA 2019-style practice: A 250\ cm^3 solution contains 5.85\ g of sodium chloride. Calculate its concentration in mol dm^-3, then state the mass of silver chloride precipitated if this solution is reacted with excess silver nitrate. (M_r of NaCl = 58.5, M_r of AgCl = 143.5)

Moles of NaCl = (5.85)/(58.5) = 0.100 mol. Volume in dm^3 = (250)/(1000) = 0.250\ dm^3, so concentration = (0.100)/(0.250) = 0.400\ mol dm^-3. The reaction NaCl + AgNO_3 → AgCl + NaNO_3 is 1:1, so 0.100 mol of AgCl forms. Mass = 0.100 × 143.5 = 14.35\ g. Markers reward the moles, the volume conversion to dm^3, the concentration, the 1:1 ratio, and the final mass.

AQA AQA 2021-style practice: Hydrogen and nitrogen react to form ammonia: N_2 + 3H_2 → 2NH_3. In an industrial run, 28.0\ kg of nitrogen reacts and 27.2\ kg of ammonia is collected. Calculate the percentage yield. (M_r of N_2 = 28.0, M_r of NH_3 = 17.0)

Moles of N_2 = (28000)/(28.0) = 1000 mol. From the 1:2 ratio, theoretical moles of NH_3 = 2000 mol, so theoretical mass = 2000 × 17.0 = 34000\ g = 34.0\ kg. Percentage yield = (27.2)/(34.0) × 100 = 80.0\%. Markers reward the limiting-reagent moles, the use of the 1:2 stoichiometry, and the yield calculated from masses (or moles) consistently.

EnglandChemistrySyllabus dot point

How do chemists count atoms and link mass, moles and volume?

The Avogadro constant and the mole, the ideal gas equation, empirical and molecular formulae, balanced equations and associated calculations, percentage yields and atom economy, and concentrations of solutions.

A focused answer to AQA A-Level Chemistry 3.1.2, covering the mole and Avogadro constant, the ideal gas equation, empirical and molecular formulae, reacting mass and solution calculations, percentage yield and atom economy.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The mole and the Avogadro constant
The ideal gas equation
Empirical and molecular formulae
Reacting masses and the limiting reagent
Yield and atom economy

What this dot point is asking

AQA wants you to use the mole and the Avogadro constant, apply the ideal gas equation, find empirical and molecular formulae, perform reacting mass, gas volume and solution calculations, and work out percentage yield and atom economy.

The mole and the Avogadro constant

A mole is the amount of substance that contains $6.022 \times 10^{23}$ particles, where this number is the Avogadro constant $N_A$ . The "particles" must be specified (atoms, molecules, ions or electrons), because a mole of $\text{O}_2$ contains $N_A$ molecules but $2N_A$ oxygen atoms. The number of particles is $N = n \times N_A$ , which is the bridge between the laboratory scale (grams, moles) and the atomic scale (individual particles).

The relative molecular mass $M_r$ (or relative formula mass) is the sum of the relative atomic masses $A_r$ of every atom in the formula, on the scale where one atom of carbon-12 is exactly 12. Because $A_r$ is a ratio it has no units, so $M_r$ has no units, but a mole of a substance has a mass in grams equal numerically to its $M_r$ (the molar mass, $\text{g mol}^{-1}$ ).

The ideal gas equation

The ideal gas equation $pV = nRT$ links the amount of a gas to its pressure, volume and temperature. AQA marks are routinely lost on units: pressure must be in pascals (so $\text{kPa} \times 1000$ ), volume in cubic metres (so $\text{cm}^3 \times 10^{-6}$ or $\text{dm}^3 \times 10^{-3}$ ), and temperature in kelvin (so degrees Celsius $+ 273$ ). An ideal gas is assumed to have negligible particle volume and no intermolecular forces, which real gases approximate best at high temperature and low pressure.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms. Divide each element's mass (or percentage composition) by its $A_r$ , then divide every result by the smallest of them to get the ratio. If a ratio is not close to a whole number (for example $1:1.5$ ), multiply all of them up until they are. The molecular formula is a whole-number multiple of the empirical formula, found by dividing the true $M_r$ (often given by mass spectrometry) by the empirical formula mass and multiplying the subscripts by that factor.

Reacting masses and the limiting reagent

Most quantitative questions follow the same chain: convert the known mass (or volume or concentration) into moles, use the balanced equation's mole ratio to find the moles of the wanted species, then convert back to mass, volume or concentration. When two reactant quantities are given, identify the limiting reagent (the one that runs out first, found by comparing moles divided by coefficients); all yields are based on it, and the other reactant is in excess.

Yield and atom economy

Percentage yield measures how much product is actually obtained against the maximum the equation predicts; it is below 100% because of incomplete reactions, side reactions and losses in handling and purification. Atom economy measures how much of the reactant mass ends up in the wanted product rather than in by-products, so it is a sustainability measure: an addition reaction has 100% atom economy (only one product), while a substitution or elimination reaction is lower because a by-product carries away some mass. Industry prefers high atom economy because it cuts waste and the cost of separating and disposing of by-products.

Worked example: gas volume from a reaction

Calculate the volume of carbon dioxide, in $\text{m}^3$ , produced when $10.0$ g of calcium carbonate decomposes at $298$ K and $100$ kPa. ( $M_r$ of $\text{CaCO}_3 = 100$ )

The decomposition equation is: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

Step 1: Convert mass to moles

We use $n = \frac{m}{M_r}$ to find the amount of calcium carbonate. The mole ratio in the balanced equation is 1:1, so the same number of moles of $\text{CO}_2$ is produced.

n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\ \text{mol} \implies n(\text{CO}_2) = 0.100\ \text{mol}

Step 2: Rearrange and apply the ideal gas equation

The ideal gas equation $pV = nRT$ links the amount of gas to its conditions. We rearrange for $V$ and substitute, taking care with units: pressure must be in Pa (so $100\ \text{kPa} = 100{,}000\ \text{Pa}$ ) and temperature is already in kelvin.

V = \frac{nRT}{p} = \frac{0.100 \times 8.31 \times 298}{100{,}000}

Step 3: Calculate and state the volume

V = \frac{247.6}{100{,}000} = 2.48 \times 10^{-3}\ \text{m}^3

Final answer: $V = 2.48 \times 10^{-3}\ \text{m}^3$ (equivalent to about $2.48\ \text{dm}^3$ ).

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA

250\ \text{cm}^3

solution contains

5.85\ \text{g}

of sodium chloride. Calculate its concentration in

\text{mol dm}^{-3}

, then state the mass of silver chloride precipitated if this solution is reacted with excess silver nitrate. (

M_r

\text{NaCl} = 58.5

M_r

\text{AgCl} = 143.5

)

Show worked answer →

Moles of $\text{NaCl} = \frac{5.85}{58.5} = 0.100$ mol. Volume in $\text{dm}^3 = \frac{250}{1000} = 0.250\ \text{dm}^3$ , so concentration $= \frac{0.100}{0.250} = 0.400\ \text{mol dm}^{-3}$ .

The reaction $\text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3$ is 1:1, so $0.100$ mol of $\text{AgCl}$ forms. Mass $= 0.100 \times 143.5 = 14.35\ \text{g}$ .

Markers reward the moles, the volume conversion to $\text{dm}^3$ , the concentration, the 1:1 ratio, and the final mass.

AQA 20213 marksHydrogen and nitrogen react to form ammonia:

\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3

. In an industrial run,

28.0\ \text{kg}

of nitrogen reacts and

27.2\ \text{kg}

of ammonia is collected. Calculate the percentage yield. (

M_r

\text{N}_2 = 28.0

M_r

\text{NH}_3 = 17.0

)

Show worked answer →

Moles of $\text{N}_2 = \frac{28000}{28.0} = 1000$ mol. From the 1:2 ratio, theoretical moles of $\text{NH}_3 = 2000$ mol, so theoretical mass $= 2000 \times 17.0 = 34000\ \text{g} = 34.0\ \text{kg}$ .

Percentage yield $= \frac{27.2}{34.0} \times 100 = 80.0\%$ .

Markers reward the limiting-reagent moles, the use of the 1:2 stoichiometry, and the yield calculated from masses (or moles) consistently.

Related dot points

Sources & how we know this

AQA A-level Chemistry (7405) specification — AQA (2015)