Calculate the number of moles in 4.0\ g of sodium hydroxide (M_r = 40). [1 mark]

Pearson Edexcel Edexcel 2019-style practice: 25.0\ cm^3 of sodium hydroxide solution was neutralised by 20.0\ cm^3 of 0.100\ mol dm^-3 sulfuric acid. (a) Write the balanced equation. (b) Calculate the concentration of the sodium hydroxide solution.

Use moles of acid, the 1:2 ratio, then concentration. (a) 2NaOH + H_2SO_4 → Na_2SO_4 + 2H_2O (1). (b) n(H_2SO_4) = 0.100 × 20.0/1000 = 2.00 × 10^-3\ mol (1). Ratio H_2SO_4 : NaOH = 1:2, so n(NaOH) = 2 × 2.00 × 10^-3 = 4.00 × 10^-3\ mol (1). Concentration = 4.00 × 10^-325.0/1000 = 0.160\ mol dm^-3 (2).

EnglandChemistrySyllabus dot point

How do we count and measure the particles taking part in chemical reactions?

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

An Edexcel 9CH0 Topic 5 answer covering the mole, empirical and molecular formulae, balanced equations, the ideal gas equation, concentration and titrations, percentage yield and atom economy.

Generated by Claude Opus 4.89 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 answer
Empirical and molecular formulae
Titration calculations
Percentage yield and atom economy
Examples in context
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What this topic is asking

Edexcel Topic 5 wants you to be fluent with the mole, calculate empirical and molecular formulae, balance equations, use the ideal gas equation, perform concentration and titration calculations, and work out percentage yield and atom economy.

The answer

The mole and key relationships

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms. Divide each element's mass (or percentage) by its relative atomic mass, then divide by the smallest result. The molecular formula is a whole-number multiple of the empirical formula, found by comparing the empirical mass to the relative molecular mass.

Titration calculations

A titration finds an unknown concentration by reacting it with a standard solution to an indicator end point. Use $n = c \times V$ for the known solution, apply the mole ratio from the balanced equation, then calculate the unknown concentration.

Percentage yield and atom economy

A high atom economy reduces waste and is favoured in green chemistry; addition reactions have $100\%$ atom economy because there is only one product, whereas substitution and elimination reactions are lower because a by-product also forms.

A full titration and percentage-yield calculation

In a preparation, $5.00\ \text{g}$ of calcium carbonate ( $M_r = 100.1$ ) is reacted with excess hydrochloric acid: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$ . The student isolates $4.85\ \text{g}$ of calcium chloride ( $M_r = 111.1$ ). Find the percentage yield.

step 1: Moles of the limiting reactant

$n(\text{CaCO}_3) = \dfrac{5.00}{100.1} = 0.0500\ \text{mol}$ .

step 2: Theoretical moles of product

The ratio $\text{CaCO}_3 : \text{CaCl}_2$ is $1:1$ , so the theoretical $n(\text{CaCl}_2) = 0.0500\ \text{mol}$ .

step 3: Theoretical mass

Theoretical mass $= 0.0500 \times 111.1 = 5.56\ \text{g}$ .

step 4: Percentage yield

$\text{yield} = \dfrac{\text{actual mass}}{\text{theoretical mass}} \times 100 = \dfrac{4.85}{5.56} \times 100 = 87.2\%$ .

Examples in context

Example 1. Why addition polymerisation is favoured in green chemistry. Making poly(ethene) from ethene is an addition reaction with $100\%$ atom economy, because every atom of the monomer ends up in the polymer with no by-product. Compare this with making an ester by Fischer esterification, where water is lost and the atom economy is lower. Industry increasingly chooses high-atom-economy routes to reduce waste and cost, applying exactly the atom-economy calculation from Topic 5.

Example 2. Standardising solutions in analytical labs. Sodium hydroxide solutions absorb $\text{CO}_2$ from the air and change concentration, so they cannot be trusted as a primary standard. Analysts standardise them by titrating against a stable primary standard, then use $n = c \times V$ and the mole ratio to calculate the true concentration. This routine quality-control step is the titration calculation above carried out in practice.

Try this

Q1. Calculate the number of moles in $4.0\ \text{g}$ of sodium hydroxide ( $M_r = 40$ ). [1 mark]

Cue. $n = \frac{4.0}{40} = 0.10\ \text{mol}$ .

Q2. A compound contains $40\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Determine its empirical formula. [3 marks]

Cue. Divide by $A_r$ : $C = 3.33$ , $H = 6.7$ , $O = 3.33$ ; ratio $1:2:1$ , so $CH_2O$ .

Exam-style practice questions

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

Edexcel 20195 marks

25.0\ \text{cm}^3

of sodium hydroxide solution was neutralised by

20.0\ \text{cm}^3

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

sulfuric acid. (a) Write the balanced equation. (b) Calculate the concentration of the sodium hydroxide solution.

Show worked answer →

Use moles of acid, the $1:2$ ratio, then concentration.

(a) $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ (1).

(b) $n(\text{H}_2\text{SO}_4) = 0.100 \times 20.0/1000 = 2.00 \times 10^{-3}\ \text{mol}$ (1). Ratio $\text{H}_2\text{SO}_4 : \text{NaOH} = 1:2$ , so $n(\text{NaOH}) = 2 \times 2.00 \times 10^{-3} = 4.00 \times 10^{-3}\ \text{mol}$ (1). Concentration $= \dfrac{4.00 \times 10^{-3}}{25.0/1000} = 0.160\ \text{mol dm}^{-3}$ (2).

Edexcel 20214 marks

0.240\ \text{g}

of magnesium was reacted with excess hydrochloric acid and the hydrogen collected. (a) Write the equation. (b) Calculate the volume of hydrogen produced at room temperature and pressure (molar gas volume

= 24.0\ \text{dm}^3\,\text{mol}^{-1}

). (

A_r

= 24.3

Show worked answer →

Find moles of Mg, use the $1:1$ ratio to hydrogen, then multiply by the molar gas volume.

(a) $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$ (1).

(b) $n(\text{Mg}) = 0.240 / 24.3 = 9.88 \times 10^{-3}\ \text{mol}$ (1). Ratio $\text{Mg} : \text{H}_2 = 1:1$ , so $n(\text{H}_2) = 9.88 \times 10^{-3}\ \text{mol}$ (1). Volume $= 9.88 \times 10^{-3} \times 24.0 = 0.237\ \text{dm}^3$ ( $237\ \text{cm}^3$ ) (1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)