Quantitative analysis

What you'll learn

How to calculate concentrations in g/dm³ and mol/dm³.
How an accurate acid-alkali titration is carried out and used in calculations.
How to calculate percentage yield and atom economy.
How to use molar gas volume and Avogadro’s law in reacting-volume calculations.

These points are Separate Chemistry content because the specification points end in C. Some of the calculation work below is Higher Tier only; it is flagged gently as we meet it.

The big idea: measuring amounts

Quantitative analysis means using measurements and calculations to find out “how much” of a substance is present or produced.

The central quantity is the amount of substance, usually measured in moles.

Definition

Amount of substance

The amount of substance, symbol $n$ , tells you how many particles are present. It is measured in moles (mol). At GCSE, moles are the link between mass, solution concentration, and gas volume.

The relative formula mass, symbol $M_r$ , is the total of the relative atomic masses in a formula. For example, for NaOH: sodium 23, oxygen 16, hydrogen 1, so $M_r = 40$ .

For a pure substance:

\frac{m}{M_r}

where $m$ is mass in grams.

This map shows the main routes between mass, concentration and gas volume.

Map linking moles to mass, concentration and gas volume at RTP

Key Idea

Moles are the bridge

Most quantitative chemistry questions become easier once you convert the information into moles, then use the balanced equation ratio.

Concentration of solutions

A solution is formed when a solute dissolves in a solvent. The concentration tells you how much solute is dissolved in a certain volume of solution.

You need two common units:

grams per cubic decimetre, g/dm³: mass of solute per dm³ of solution
moles per cubic decimetre, mol/dm³: moles of solute per dm³ of solution

This concentration work in mol/dm³ is Higher Tier only in Separate Chemistry.

For concentration in mol/dm³:

\frac{n}{V}

where $c$ is concentration, $n$ is amount in moles, and $V$ is volume in dm³.

Remember: 1000 cm³ = 1 dm³, so divide cm³ by 1000 to convert to dm³.

To convert between g/dm³ and mol/dm³:

concentration=c×Mr\text{mass concentration} = c \times M_r

and

\frac{\text{mass concentration}}{M_r}

Example

Converting concentration and finding moles

A sodium hydroxide solution has concentration 8.00 g/dm³. Find its concentration in mol/dm³, then find the moles in 250 cm³ of solution. The $M_r$ of NaOH is 40.0.

Convert from g/dm³ to mol/dm³ by dividing by $M_r$ :

$\frac{8.00}{40.0} = 0.200$ mol/dm³.
Convert the volume into dm³:

250 cm³ = 0.250 dm³.
Use $n = c V$ :

$\times 0.250 = 0.0500$ mol.

Common Mistake

Using cm³ directly

In $\frac{n}{V}$ , the volume must be in dm³, not cm³. If you forget to divide by 1000, your answer will be 1000 times too big or too small.

Acid-alkali titration

A titration is a practical method used to find the concentration of a solution by reacting it with a measured volume of another solution.

Definition

Titration

An acid-alkali titration uses a solution of known concentration to neutralise a measured volume of acid or alkali of unknown concentration.

A typical neutralisation reaction is:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

This is one of the Edexcel core practicals, so you need to know the apparatus and the method.

Labelled acid-alkali titration setup with burette, pipette, conical flask, indicator and white tile

Method for an accurate titration

Use a pipette and pipette filler to transfer a fixed volume, often 25.0 cm³, into a conical flask.
Add a few drops of a suitable indicator, such as phenolphthalein or methyl orange.
Fill a burette with the other solution and record the initial burette reading.
Run solution from the burette into the flask while swirling.
Near the endpoint, add solution drop by drop until the first permanent colour change.
Record the final burette reading. The titre is final reading minus initial reading.
Repeat until you get concordant titres.

Tip

Concordant titres

Concordant titres are repeat titres that are very close together, usually within 0.20 cm³. Ignore the rough titre when calculating the mean.

Titration calculations

The balanced equation gives the reacting mole ratio. This Higher Tier Separate Chemistry skill is often the most demanding part of the topic.

Example

Finding an unknown concentration

25.0 cm³ of sodium hydroxide solution is neutralised by 20.60 cm³ of 0.100 mol/dm³ hydrochloric acid.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Find the concentration of the sodium hydroxide.

Calculate the moles of hydrochloric acid used. First convert 20.60 cm³ to 0.02060 dm³:

$\times 0.02060 = 0.002060$ mol.
Use the balanced equation ratio. HCl and NaOH react 1:1, so:

moles of NaOH = 0.002060 mol.
Convert 25.0 cm³ to 0.0250 dm³, then calculate the sodium hydroxide concentration:

$\frac{0.002060}{0.0250} = 0.0824$ mol/dm³.

If the question asks for an unknown volume instead, work out the moles needed first, then rearrange to $\frac{n}{c}$ and convert dm³ back to cm³ if required.

Percentage yield

The theoretical yield is the maximum mass of product predicted by the balanced equation. The actual yield is the mass you really obtain in the experiment.

Percentage yield compares the actual yield with the theoretical yield:

yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Actual yield is usually less than theoretical yield because:

the reaction may be incomplete
product may be lost during filtering, transferring or drying
unwanted side reactions may make different products

Example

Calculating percentage yield

A reaction has a theoretical yield of 7.50 g, but only 6.30 g is collected. Calculate the percentage yield.

Identify the two masses:

actual yield = 6.30 g, theoretical yield = 7.50 g.
Substitute into the percentage yield equation:

$yield=6.307.50×100=84.0\text{percentage yield} = \frac{6.30}{7.50} \times 100 = 84.0$ %.
Interpret the result: the experiment produced 84.0% of the maximum possible mass.

Common Mistake

A yield over 100%

A percentage yield above 100% usually means the product is wet, impure, or the mass was measured incorrectly. It does not mean the reaction made more atoms than were available.

Atom economy

Atom economy measures how efficiently atoms in the reactants are turned into the desired product.

Definition

Atom economy

The atom economy of a reaction is the percentage of the total mass of reactants that ends up in the desired product.

reactants×100\text{atom economy} = \frac{\text{relative formula mass of desired product(s)}}{\text{total relative formula mass of reactants}} \times 100

Use the balanced equation, including any big numbers in front of formulae.

Example

Calculating atom economy

Calcium carbonate decomposes to make calcium oxide.

CaCO₃(s) → CaO(s) + CO₂(g)

The desired product is CaO. Calculate the atom economy. Use $M_r$ values: CaCO₃ = 100, CaO = 56.

Identify the desired product and its $M_r$ :

desired product = CaO, so $M_r = 56$ .
Identify the total $M_r$ of the reactants:

reactant = CaCO₃, so total reactant $M_r = 100$ .
Calculate atom economy:

$economy=56100×100=56\text{atom economy} = \frac{56}{100} \times 100 = 56$ %.

Common Mistake

Mixing up yield and atom economy

Percentage yield uses the actual mass made. Atom economy uses only the balanced equation and relative formula masses.

Choosing a reaction pathway

This is Higher Tier only. In industry, chemists choose reaction pathways by comparing several factors:

atom economy: how much waste is made
yield: how much desired product is actually obtained
rate: how quickly product is made
equilibrium position: in a reversible reaction, whether the mixture contains more products or reactants at equilibrium
usefulness of by-products: whether other products can be sold or reused
cost, energy use, safety and environmental impact

Example

Choosing a reaction pathway

Two pathways make the same product.

Pathway A has atom economy 90%, yield 55%, is fast, but makes a harmful by-product.
Pathway B has atom economy 70%, yield 85%, is slower, but makes a useful by-product.

Decide which pathway may be better for industry.

Compare the rough useful product made per 100 g of reactants using atom economy and yield:

Pathway A: $90 \times 0.55 = 49.5$ g useful product.

Pathway B: $70 \times 0.85 = 59.5$ g useful product.
Compare practical factors: A is faster, but its harmful by-product may increase disposal costs and environmental damage.
Make a justified choice: B may be better if the slower rate is acceptable, because it gives more useful product overall and its by-product is useful.

Molar volume of gases

This Higher Tier Separate Chemistry content uses the fact that gases occupy predictable volumes.

The molar volume of a gas is the volume occupied by one mole of molecules of any gas at a stated temperature and pressure. At room temperature and pressure, often called RTP, one mole of any gas occupies:

24 dm³
or 24000 cm³

The value will be provided in the exam if needed.

For gas volume in dm³ at RTP:

Vgas=n×24V_{\text{gas}} = n \times 24

and

\frac{V_{\text{gas}}}{24}

Example

Calculating gas volume from a mass

Magnesium reacts with excess hydrochloric acid.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the volume of hydrogen made at RTP from 0.120 g of magnesium. Use $A_r$ of Mg = 24.0.

Convert magnesium mass to moles:

$\frac{0.120}{24.0} = 0.00500$ mol.
Use the balanced equation ratio. Mg:H₂ is 1:1, so:

moles of H₂ = 0.00500 mol.
Convert moles of hydrogen to volume at RTP:

$Vgas=0.00500×24=0.120V_{\text{gas}} = 0.00500 \times 24 = 0.120$ dm³ = 120 cm³.

Avogadro’s law and gas volume ratios

Avogadro’s law says that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules.

This means that for gases at the same conditions, the balanced equation coefficients give the reacting volume ratio.

For example:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

This tells you that 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia, as long as all gases are measured at the same temperature and pressure.

Example

Using gas volume ratios

In the Haber reaction, what volume of ammonia can form from 90 cm³ of hydrogen, with nitrogen in excess?

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Use the gas ratio from the balanced equation:

H₂:NH₃ = 3:2.
Scale from 90 cm³ of hydrogen:

volume of NH₃ = $90 \times \frac{2}{3} = 60$ cm³.
Keep the same volume unit because this is a gas volume ratio question:

ammonia volume = 60 cm³.

Common Mistake

Gas ratios need matching conditions

Only compare gas volumes directly when the gases are measured at the same temperature and pressure, and only use equation coefficients for substances that are gases.

Exam technique

In the exam

Convert volumes carefully: cm³ to dm³ by dividing by 1000, unless you are using 24000 cm³ for gas molar volume.
Start reacting-mass and titration questions by calculating moles, then use the balanced equation ratio.
For titrations, use the mean of concordant titres only, not the rough titre.
Do not confuse percentage yield with atom economy: yield uses experimental mass, atom economy uses the equation.
Give answers to a sensible number of significant figures and always include units.

Self review

Check yourself

Why must titration volumes in cm³ usually be converted to dm³ before using $\frac{n}{V}$ ?
What are three reasons why actual yield is usually less than theoretical yield?
How does Avogadro’s law let you use a balanced equation to compare gas volumes?

Quantitative analysis

What you'll learn

How to calculate concentrations in g/dm³ and mol/dm³.
How an accurate acid-alkali titration is carried out and used in calculations.
How to calculate percentage yield and atom economy.
How to use molar gas volume and Avogadro’s law in reacting-volume calculations.

These points are Separate Chemistry content because the specification points end in C. Some of the calculation work below is Higher Tier only; it is flagged gently as we meet it.

The big idea: measuring amounts

Quantitative analysis means using measurements and calculations to find out “how much” of a substance is present or produced.

The central quantity is the amount of substance, usually measured in moles.

Definition

Amount of substance

The relative formula mass, symbol $M_r$ , is the total of the relative atomic masses in a formula. For example, for NaOH: sodium 23, oxygen 16, hydrogen 1, so $M_r = 40$ .

For a pure substance:

\frac{m}{M_r}

where $m$ is mass in grams.

This map shows the main routes between mass, concentration and gas volume.

Map linking moles to mass, concentration and gas volume at RTP

Key Idea

Moles are the bridge

Most quantitative chemistry questions become easier once you convert the information into moles, then use the balanced equation ratio.

Concentration of solutions

A solution is formed when a solute dissolves in a solvent. The concentration tells you how much solute is dissolved in a certain volume of solution.

You need two common units:

grams per cubic decimetre, g/dm³: mass of solute per dm³ of solution
moles per cubic decimetre, mol/dm³: moles of solute per dm³ of solution

This concentration work in mol/dm³ is Higher Tier only in Separate Chemistry.

For concentration in mol/dm³:

\frac{n}{V}

where $c$ is concentration, $n$ is amount in moles, and $V$ is volume in dm³.

Remember: 1000 cm³ = 1 dm³, so divide cm³ by 1000 to convert to dm³.

To convert between g/dm³ and mol/dm³:

concentration=c×Mr\text{mass concentration} = c \times M_r

and

\frac{\text{mass concentration}}{M_r}

Example

Converting concentration and finding moles

A sodium hydroxide solution has concentration 8.00 g/dm³. Find its concentration in mol/dm³, then find the moles in 250 cm³ of solution. The $M_r$ of NaOH is 40.0.

Convert from g/dm³ to mol/dm³ by dividing by $M_r$ :

$\frac{8.00}{40.0} = 0.200$ mol/dm³.
Convert the volume into dm³:

250 cm³ = 0.250 dm³.
Use $n = c V$ :

$\times 0.250 = 0.0500$ mol.

Common Mistake

Using cm³ directly

In $\frac{n}{V}$ , the volume must be in dm³, not cm³. If you forget to divide by 1000, your answer will be 1000 times too big or too small.

Acid-alkali titration

A titration is a practical method used to find the concentration of a solution by reacting it with a measured volume of another solution.

Definition

Titration

An acid-alkali titration uses a solution of known concentration to neutralise a measured volume of acid or alkali of unknown concentration.

A typical neutralisation reaction is:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

This is one of the Edexcel core practicals, so you need to know the apparatus and the method.

Labelled acid-alkali titration setup with burette, pipette, conical flask, indicator and white tile

Method for an accurate titration

Use a pipette and pipette filler to transfer a fixed volume, often 25.0 cm³, into a conical flask.
Add a few drops of a suitable indicator, such as phenolphthalein or methyl orange.
Fill a burette with the other solution and record the initial burette reading.
Run solution from the burette into the flask while swirling.
Near the endpoint, add solution drop by drop until the first permanent colour change.
Record the final burette reading. The titre is final reading minus initial reading.
Repeat until you get concordant titres.

Tip

Concordant titres

Concordant titres are repeat titres that are very close together, usually within 0.20 cm³. Ignore the rough titre when calculating the mean.

Titration calculations

The balanced equation gives the reacting mole ratio. This Higher Tier Separate Chemistry skill is often the most demanding part of the topic.

Example

Finding an unknown concentration

25.0 cm³ of sodium hydroxide solution is neutralised by 20.60 cm³ of 0.100 mol/dm³ hydrochloric acid.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Find the concentration of the sodium hydroxide.

Calculate the moles of hydrochloric acid used. First convert 20.60 cm³ to 0.02060 dm³:

$\times 0.02060 = 0.002060$ mol.
Use the balanced equation ratio. HCl and NaOH react 1:1, so:

moles of NaOH = 0.002060 mol.
Convert 25.0 cm³ to 0.0250 dm³, then calculate the sodium hydroxide concentration:

$\frac{0.002060}{0.0250} = 0.0824$ mol/dm³.

If the question asks for an unknown volume instead, work out the moles needed first, then rearrange to $\frac{n}{c}$ and convert dm³ back to cm³ if required.

Percentage yield

The theoretical yield is the maximum mass of product predicted by the balanced equation. The actual yield is the mass you really obtain in the experiment.

Percentage yield compares the actual yield with the theoretical yield:

yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Actual yield is usually less than theoretical yield because:

the reaction may be incomplete
product may be lost during filtering, transferring or drying
unwanted side reactions may make different products

Example

Calculating percentage yield

A reaction has a theoretical yield of 7.50 g, but only 6.30 g is collected. Calculate the percentage yield.

Identify the two masses:

actual yield = 6.30 g, theoretical yield = 7.50 g.
Substitute into the percentage yield equation:

$yield=6.307.50×100=84.0\text{percentage yield} = \frac{6.30}{7.50} \times 100 = 84.0$ %.
Interpret the result: the experiment produced 84.0% of the maximum possible mass.

Common Mistake

A yield over 100%

A percentage yield above 100% usually means the product is wet, impure, or the mass was measured incorrectly. It does not mean the reaction made more atoms than were available.

Atom economy

Atom economy measures how efficiently atoms in the reactants are turned into the desired product.

Definition

Atom economy

The atom economy of a reaction is the percentage of the total mass of reactants that ends up in the desired product.

reactants×100\text{atom economy} = \frac{\text{relative formula mass of desired product(s)}}{\text{total relative formula mass of reactants}} \times 100

Use the balanced equation, including any big numbers in front of formulae.

Example

Calculating atom economy

Calcium carbonate decomposes to make calcium oxide.

CaCO₃(s) → CaO(s) + CO₂(g)

The desired product is CaO. Calculate the atom economy. Use $M_r$ values: CaCO₃ = 100, CaO = 56.

Identify the desired product and its $M_r$ :

desired product = CaO, so $M_r = 56$ .
Identify the total $M_r$ of the reactants:

reactant = CaCO₃, so total reactant $M_r = 100$ .
Calculate atom economy:

$economy=56100×100=56\text{atom economy} = \frac{56}{100} \times 100 = 56$ %.

Common Mistake

Mixing up yield and atom economy

Percentage yield uses the actual mass made. Atom economy uses only the balanced equation and relative formula masses.

Choosing a reaction pathway

This is Higher Tier only. In industry, chemists choose reaction pathways by comparing several factors:

atom economy: how much waste is made
yield: how much desired product is actually obtained
rate: how quickly product is made
equilibrium position: in a reversible reaction, whether the mixture contains more products or reactants at equilibrium
usefulness of by-products: whether other products can be sold or reused
cost, energy use, safety and environmental impact

Example

Choosing a reaction pathway

Two pathways make the same product.

Pathway A has atom economy 90%, yield 55%, is fast, but makes a harmful by-product.
Pathway B has atom economy 70%, yield 85%, is slower, but makes a useful by-product.

Decide which pathway may be better for industry.

Compare the rough useful product made per 100 g of reactants using atom economy and yield:

Pathway A: $90 \times 0.55 = 49.5$ g useful product.

Pathway B: $70 \times 0.85 = 59.5$ g useful product.
Compare practical factors: A is faster, but its harmful by-product may increase disposal costs and environmental damage.
Make a justified choice: B may be better if the slower rate is acceptable, because it gives more useful product overall and its by-product is useful.

Molar volume of gases

This Higher Tier Separate Chemistry content uses the fact that gases occupy predictable volumes.

24 dm³
or 24000 cm³

The value will be provided in the exam if needed.

For gas volume in dm³ at RTP:

Vgas=n×24V_{\text{gas}} = n \times 24

and

\frac{V_{\text{gas}}}{24}

Example

Calculating gas volume from a mass

Magnesium reacts with excess hydrochloric acid.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the volume of hydrogen made at RTP from 0.120 g of magnesium. Use $A_r$ of Mg = 24.0.

Convert magnesium mass to moles:

$\frac{0.120}{24.0} = 0.00500$ mol.
Use the balanced equation ratio. Mg:H₂ is 1:1, so:

moles of H₂ = 0.00500 mol.
Convert moles of hydrogen to volume at RTP:

$Vgas=0.00500×24=0.120V_{\text{gas}} = 0.00500 \times 24 = 0.120$ dm³ = 120 cm³.

Avogadro’s law and gas volume ratios

Avogadro’s law says that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules.

This means that for gases at the same conditions, the balanced equation coefficients give the reacting volume ratio.

For example:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

This tells you that 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia, as long as all gases are measured at the same temperature and pressure.

Example

Using gas volume ratios

In the Haber reaction, what volume of ammonia can form from 90 cm³ of hydrogen, with nitrogen in excess?

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Use the gas ratio from the balanced equation:

H₂:NH₃ = 3:2.
Scale from 90 cm³ of hydrogen:

volume of NH₃ = $90 \times \frac{2}{3} = 60$ cm³.
Keep the same volume unit because this is a gas volume ratio question:

ammonia volume = 60 cm³.

Common Mistake

Gas ratios need matching conditions

Only compare gas volumes directly when the gases are measured at the same temperature and pressure, and only use equation coefficients for substances that are gases.

Exam technique

In the exam

Convert volumes carefully: cm³ to dm³ by dividing by 1000, unless you are using 24000 cm³ for gas molar volume.
Start reacting-mass and titration questions by calculating moles, then use the balanced equation ratio.
For titrations, use the mean of concordant titres only, not the rough titre.
Do not confuse percentage yield with atom economy: yield uses experimental mass, atom economy uses the equation.
Give answers to a sensible number of significant figures and always include units.

Self review

Check yourself

Why must titration volumes in cm³ usually be converted to dm³ before using $\frac{n}{V}$ ?
What are three reasons why actual yield is usually less than theoretical yield?
How does Avogadro’s law let you use a balanced equation to compare gas volumes?

Quantitative analysis

What you'll learn

The big idea: measuring amounts

Concentration of solutions

Acid-alkali titration

Method for an accurate titration

Titration calculations

Percentage yield

Atom economy

Choosing a reaction pathway

Molar volume of gases

Avogadro’s law and gas volume ratios

Recap questions

You've reached the end

Quantitative analysis

What you'll learn

The big idea: measuring amounts

Concentration of solutions

Acid-alkali titration

Method for an accurate titration

Titration calculations

Percentage yield

Atom economy

Choosing a reaction pathway

Molar volume of gases

Avogadro’s law and gas volume ratios

Recap questions

You've reached the end

Quantitative analysis

What you'll learn

The big idea: measuring amounts

Concentration of solutions

Acid-alkali titration

Method for an accurate titration

Titration calculations

Percentage yield

Atom economy

Choosing a reaction pathway

Molar volume of gases

Avogadro’s law and gas volume ratios

Recap questions

You've reached the end

Quantitative analysis Revision Guide

Quantitative analysis

What you'll learn

The big idea: measuring amounts

Concentration of solutions

Acid-alkali titration

Method for an accurate titration

Titration calculations

Percentage yield

Atom economy

Choosing a reaction pathway

Molar volume of gases

Avogadro’s law and gas volume ratios

Recap questions

You've reached the end

Quantitative analysis Revision Guide