SQA SQA AH 2019-style practice: In a titration, 25.0 cm^3 of a sodium hydroxide solution is neutralised by 20.0 cm^3 of 0.100 mol l^-1 hydrochloric acid. (a) Calculate the moles of HCl used. (b) Using the 1 : 1 ratio, find the moles of NaOH. (c) Calculate the concentration of the NaOH solution.

Markers reward the moles of acid, the mole ratio, and the concentration. (a) Moles of HCl: $n = c × V = 0.100 × (20.0)/(1000) = 2.00 × 10^-3 mol$ (b) The reaction NaOH + HCl → NaCl + H_2O is 1 : 1, so moles of NaOH = 2.00 × 10^-3 mol. (c) Concentration of NaOH: $c = (n)/(V) = 2.00 × 10^-325.0/1000 = 0.0800 mol l^-1$

SQA SQA AH specimen-style practice: Excess silver nitrate is added to a solution containing chloride ions. The precipitate of silver chloride (GFM = 143.5 g) is filtered, dried and weighed as 1.435 g. (a) Name this type of analysis. (b) Calculate the moles of chloride ions present. (c) State why the silver nitrate is added in excess.

The answer must name gravimetric analysis, find the moles, and justify the excess. (a) This is **gravimetric analysis**: the amount of an ion is found by precipitating it and weighing the dried solid. (b) Moles of silver chloride = 1.435143.5 = 0.0100 mol. The reaction is 1 : 1, so the moles of chloride ions = 0.0100 mol. (c) The silver nitrate is added in **excess** to ensure **all** the chloride ions are precipitated, so the mass of precipitate accounts for the full amount present.

ScotlandChemistrySyllabus dot point

How do we turn measured masses and volumes into amounts of substance?

Stoichiometric calculations from balanced equations, gravimetric analysis from measured masses, volumetric analysis including acid-base, redox, complexometric and back titrations, and the calculation of percentage yield and atom economy.

An SQA Advanced Higher Chemistry answer on stoichiometric calculations, covering the mole and balanced equations, gravimetric analysis from measured masses, volumetric analysis through acid-base, redox, complexometric and back titrations, and the calculation of percentage yield and atom economy from experimental data.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-14

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

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Quick answer

Stoichiometric calculations convert measured quantities into amounts of substance using the mole. The number of moles is mass divided by molar mass for a solid, or concentration times volume for a solution. A balanced equation gives the mole ratio that links one substance to another. In gravimetric analysis the amount of an ion is found by precipitating it, drying and weighing the solid, then converting mass to moles. In volumetric analysis a titration measures the volume of one solution that exactly reacts with a known amount of another; this covers acid-base, redox, complexometric and back titrations. Percentage yield compares the actual product obtained with the theoretical maximum, and atom economy compares the mass of the desired product with the total reactant mass. The key skill is setting up the mole relationship and carrying it accurately through to the answer.

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What this key area is asking
The mole and balanced equations
Gravimetric analysis
Volumetric analysis
Percentage yield and atom economy
Examples in context
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What this key area is asking

The SQA wants you to carry out stoichiometric calculations from balanced equations, gravimetric analysis from measured masses, volumetric analysis from titrations (including redox, complexometric and back titrations), and percentage yield and atom economy calculations. Setting up the mole relationship and carrying it through a titration or gravimetric calculation are reliable exam earners.

The mole and balanced equations

A balanced equation gives the mole ratio between substances. Every calculation follows the same route: convert the measured quantity to moles, use the mole ratio to find the moles of the substance you want, then convert back to a mass, volume or concentration.

Gravimetric analysis

The precipitating reagent is added in excess to ensure complete precipitation, and the solid must be dried to constant mass so that no water is weighed with it.

Volumetric analysis

Redox titration calculation

A solution of iron(II) ions is titrated against $0.0200 \text{ mol l}^{-1}$ acidified potassium permanganate, of which $24.0 \text{ cm}^3$ is needed. The reaction is $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$ . Calculate the moles of iron(II) present.

step 1: Find the moles of permanganate

n(\text{MnO}_4^-) = c \times V = 0.0200 \times \frac{24.0}{1000} = 4.80 \times 10^{-4} \text{ mol}

step 2: Apply the mole ratio

The equation shows $1$ mole of permanganate reacts with $5$ moles of iron(II), so:

n(\text{Fe}^{2+}) = 5 \times 4.80 \times 10^{-4} = 2.40 \times 10^{-3} \text{ mol}

step 3: State the result

There are $2.40 \times 10^{-3} \text{ mol}$ of iron(II) ions in the titrated sample. The $1 : 5$ ratio is the key step, so reading it correctly from the balanced equation is essential.

Percentage yield and atom economy

The efficiency of a preparation is judged by two calculations:

\text{percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical mass of product}} \times 100

\text{atom economy} = \frac{\text{mass of desired product}}{\text{total mass of reactants}} \times 100

Percentage yield reflects losses during the reaction and purification, while atom economy reflects how much of the reactant mass is built into the desired product.

Examples in context

Stoichiometric calculations are the quantitative heart of analysis. Water authorities use gravimetric analysis to measure sulfate by precipitating barium sulfate, and use complexometric titration to measure the hardness (calcium and magnesium content) of water. Redox titrations measure the iron content of an ore or a tablet, and the vitamin C content of fruit juice. Back titrations measure the calcium carbonate in an indigestible solid such as an antacid tablet or eggshell, by reacting it with a known excess of acid and titrating the leftover acid. In synthesis, percentage yield and atom economy decide whether a route is economic and green, which is exactly the analysis carried out in the project.

Try this

Q1. State the two expressions for the number of moles, for a solid and for a solution. [2 marks]

Cue. $n = m/M$ for a solid; $n = c \times V$ for a solution.

Q2. State why a precipitating reagent is added in excess in gravimetric analysis. [1 mark]

Cue. To ensure all of the ion is precipitated, so the full amount is weighed.

Q3. $20.0 \text{ cm}^3$ of $0.150 \text{ mol l}^{-1}$ acid is needed in a $1 : 1$ titration. Calculate the moles of acid. [1 mark]

Cue. $n = 0.150 \times 20.0/1000 = 3.00 \times 10^{-3} \text{ mol}$ .

Exam-style practice questions

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

SQA AH 20194 marksIn a titration,

25.0 \text{ cm}^3

of a sodium hydroxide solution is neutralised by

20.0 \text{ cm}^3

0.100 \text{ mol l}^{-1}

hydrochloric acid. (a) Calculate the moles of HCl used. (b) Using the

1 : 1

ratio, find the moles of NaOH. (c) Calculate the concentration of the NaOH solution.

Show worked answer →

Markers reward the moles of acid, the mole ratio, and the concentration.

(a) Moles of HCl:

n = c \times V = 0.100 \times \frac{20.0}{1000} = 2.00 \times 10^{-3} \text{ mol}

(b) The reaction $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ is $1 : 1$ , so moles of NaOH $= 2.00 \times 10^{-3} \text{ mol}$ .

c = \frac{n}{V} = \frac{2.00 \times 10^{-3}}{25.0/1000} = 0.0800 \text{ mol l}^{-1}

SQA AH specimen3 marksExcess silver nitrate is added to a solution containing chloride ions. The precipitate of silver chloride (

GFM = 143.5 \text{ g}

) is filtered, dried and weighed as

1.435 \text{ g}

. (a) Name this type of analysis. (b) Calculate the moles of chloride ions present. (c) State why the silver nitrate is added in excess.

Show worked answer →

The answer must name gravimetric analysis, find the moles, and justify the excess.

(a) This is gravimetric analysis: the amount of an ion is found by precipitating it and weighing the dried solid.

(b) Moles of silver chloride $= \dfrac{1.435}{143.5} = 0.0100 \text{ mol}$ . The reaction is $1 : 1$ , so the moles of chloride ions $= 0.0100 \text{ mol}$ .

(c) The silver nitrate is added in excess to ensure all the chloride ions are precipitated, so the mass of precipitate accounts for the full amount present.

Related dot points

Sources & how we know this

SQA Advanced Higher Chemistry Course Specification — SQA (2019)