Calculate the number of moles in 11.0\ g of carbon dioxide. [1 mark]

Calculate the atom economy for producing hydrogen by Zn + H_2SO_4 → ZnSO_4 + H_2. [2 marks]

OCR OCR 2019-style practice: A 25.0\ cm^3 sample of sodium hydroxide solution was neutralised by 22.40\ cm^3 of 0.100\ mol dm^-3 hydrochloric acid. (a) Write the equation. (b) Calculate the concentration of the sodium hydroxide solution in mol dm^-3.

(a) NaOH + HCl → NaCl + H_2O (1). (b) Moles of HCl = c × V = 0.100 × 22.401000 = 2.24 × 10^-3\ mol (1). The 1:1 ratio gives moles of NaOH = 2.24 × 10^-3\ mol (1). Concentration = nV = 2.24 × 10^-325.0/1000 = 0.0896\ mol dm^-3 (2). Markers reward the equation, moles of acid, the mole ratio, and the final concentration to 3 s.f.

OCR OCR 2021-style practice: Ethanol (C_2H_5OH) is oxidised to ethanoic acid (CH_3COOH). 9.20\ g of ethanol gave 9.00\ g of ethanoic acid. (a) Calculate the theoretical yield of ethanoic acid. (b) Calculate the percentage yield.

(a) Moles of ethanol = 9.2046.0 = 0.200\ mol (1). The 1:1 ratio gives 0.200\ mol ethanoic acid, so theoretical mass = 0.200 × 60.0 = 12.0\ g (1). (b) Percentage yield = 9.0012.0 × 100 = 75.0\% (2). Markers reward the moles of ethanol, the theoretical mass via the mole ratio, and the percentage yield.

EnglandChemistrySyllabus dot point

How does the mole let us count particles and predict the masses, volumes and concentrations in a reaction?

The Avogadro constant and the mole, molar mass, the ideal gas equation, empirical and molecular formulae, concentration and titration calculations, and percentage yield and atom economy.

An OCR H432 module 2 answer covering the Avogadro constant, molar mass, the ideal gas equation, empirical and molecular formulae, concentration, titrations, percentage yield and atom economy.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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What this topic is asking
The mole and molar mass
Gas volumes and the ideal gas equation
Empirical and molecular formulae
Concentration and titrations
Percentage yield and atom economy
Examples in context
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What this topic is asking

OCR specification point 2.1.3 wants you to use the mole and the Avogadro constant, work with molar mass and the ideal gas equation, find empirical and molecular formulae, carry out concentration and titration calculations, and evaluate reactions using percentage yield and atom economy. This is the calculation engine reused throughout H432.

The mole and molar mass

The core relationships are:

Gas volumes and the ideal gas equation

At a stated temperature and pressure, equal volumes of gases contain equal numbers of moles. OCR uses the ideal gas equation:

Unit conversions matter: $1\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^3$ , $1\ \text{dm}^3 = 1 \times 10^{-3}\ \text{m}^3$ , $1\ \text{kPa} = 1000\ \text{Pa}$ , and $T(\text{K}) = \theta(^{\circ}\text{C}) + 273$ .

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number of atoms. Divide each element's percentage (or mass) by its $A_r$ , then divide by the smallest result.

Concentration and titrations

Concentration links moles and volume by $n = c \times V$ (with $V$ in $\text{dm}^3$ ). A titration uses a known concentration to find an unknown one, working through the balanced mole ratio.

A back-titration calculation

$2.50\ \text{g}$ of impure calcium carbonate was reacted with $50.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ HCl (an excess). The unreacted acid needed $18.0\ \text{cm}^3$ of $0.500\ \text{mol dm}^{-3}$ NaOH. Find the percentage purity of the calcium carbonate.

step 1: Total moles of acid added

$n(\text{HCl}) = c \times V = 1.00 \times \dfrac{50.0}{1000} = 0.0500\ \text{mol}$ .

step 2: Moles of excess acid

$n(\text{NaOH}) = 0.500 \times \dfrac{18.0}{1000} = 9.00 \times 10^{-3}\ \text{mol}$ . The $1:1$ neutralisation means excess HCl $= 9.00 \times 10^{-3}\ \text{mol}$ .

step 3: Moles of acid that reacted with the carbonate

$0.0500 - 9.00 \times 10^{-3} = 0.0410\ \text{mol}$ . Since $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$ , moles of $\text{CaCO}_3 = \dfrac{0.0410}{2} = 0.0205\ \text{mol}$ .

step 4: Mass and percentage purity

Mass $= n \times M = 0.0205 \times 100.1 = 2.05\ \text{g}$ . Purity $= \dfrac{2.05}{2.50} \times 100 = 82.0\%$ .

Percentage yield and atom economy

Examples in context

Example 1. Airbag chemistry. Sodium azide decomposes to nitrogen gas: $2\text{NaN}_3 \rightarrow 2\text{Na} + 3\text{N}_2$ . Engineers use $pV = nRT$ to work out the mass of azide needed to inflate a bag to a set volume at a given temperature and pressure within milliseconds, a direct application of the gas equation and the mole ratio.

Example 2. Atom economy in industry. Manufacturing ethene oxide directly from ethene and oxygen has a far higher atom economy than older multi-step routes that produced large amounts of by-product. Comparing atom economies guides chemists toward cleaner, cheaper processes, exactly the "green chemistry" judgement OCR rewards.

Try this

Q1. Calculate the number of moles in $11.0\ \text{g}$ of carbon dioxide. [1 mark]

Cue. $M_r = 44.0$ , so $n = \frac{11.0}{44.0} = 0.250\ \text{mol}$ .

Q2. Calculate the atom economy for producing hydrogen by $\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2$ . [2 marks]

Cue. $\frac{M_r(\text{H}_2)}{M_r(\text{ZnSO}_4) + M_r(\text{H}_2)} \times 100 = \frac{2.0}{161.5 + 2.0} \times 100 = 1.22\%$ .

Exam-style practice questions

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

OCR 20195 marksA

25.0\ \text{cm}^3

sample of sodium hydroxide solution was neutralised by

22.40\ \text{cm}^3

0.100\ \text{mol dm}^{-3}

hydrochloric acid. (a) Write the equation. (b) Calculate the concentration of the sodium hydroxide solution in

\text{mol dm}^{-3}

Show worked answer →

(a) $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ (1).

(b) Moles of HCl $= c \times V = 0.100 \times \dfrac{22.40}{1000} = 2.24 \times 10^{-3}\ \text{mol}$ (1). The $1:1$ ratio gives moles of NaOH $= 2.24 \times 10^{-3}\ \text{mol}$ (1).

Concentration $= \dfrac{n}{V} = \dfrac{2.24 \times 10^{-3}}{25.0/1000} = 0.0896\ \text{mol dm}^{-3}$ (2).

Markers reward the equation, moles of acid, the mole ratio, and the final concentration to 3 s.f.

OCR 20214 marksEthanol (

\text{C}_2\text{H}_5\text{OH}

) is oxidised to ethanoic acid (

\text{CH}_3\text{COOH}

9.20\ \text{g}

of ethanol gave

9.00\ \text{g}

of ethanoic acid. (a) Calculate the theoretical yield of ethanoic acid. (b) Calculate the percentage yield.

Show worked answer →

(a) Moles of ethanol $= \dfrac{9.20}{46.0} = 0.200\ \text{mol}$ (1). The $1:1$ ratio gives $0.200\ \text{mol}$ ethanoic acid, so theoretical mass $= 0.200 \times 60.0 = 12.0\ \text{g}$ (1).

(b) Percentage yield $= \dfrac{9.00}{12.0} \times 100 = 75.0\%$ (2).

Markers reward the moles of ethanol, the theoretical mass via the mole ratio, and the percentage yield.

Related dot points

Sources & how we know this

OCR A-Level Chemistry A (H432) specification — OCR (2015)