Monitoring chemical reactions

What you'll learn

How concentration links the amount of solute to the volume of solution.
How titration results can be used to find an unknown concentration.
How gas volumes, reacting masses, percentage yield and atom economy are calculated.
How data can help choose the best reaction pathway.

Some of this section is separate Chemistry content rather than Combined Science; because you are studying J248, you should learn the full set here. I’ll flag the Higher Tier-only parts as we go.

The big idea: reactions are compared using moles

A mole is a unit for the amount of substance. It lets chemists count particles in a practical way, just like a dozen means 12 items — except one mole means a huge number of particles.

The Avogadro constant is the number of particles in one mole: about $6.02 \times 10^{23}$ particles per mole.

Balanced chemical equations compare substances in mole ratios, not mass ratios. For example:

2H₂(g) + O₂(g) → 2H₂O(l)

This means 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

The route through most reacting-amount calculations looks like this.

Flowchart showing how mass, concentration, gas volume and balanced equation ratios connect through moles

Common Mistake

Moles are not grams

A mole is an amount of substance, not a mass. Different substances have different masses per mole because their particles have different relative formula masses.

At Higher Tier only, you may need calculations using the Avogadro constant in standard form.

Example

Using the Avogadro constant to find mass

How much mass is represented by $1.20 \times 10^{22}$ molecules of H₂O?

Convert particles to moles using $\frac{\text{number of particles}}{\text{Avogadro constant}}$ :

\frac{1.20 \times 10^{22}}{6.02 \times 10^{23}} = 1.99 \times 10^{-2}\ \text{mol}

Work out the relative formula mass of water:

M_r(\text{H}_2\text{O}) = 1 + 1 + 16 = 18

Convert moles to mass using $m = nM_r$ :

\times 10^{-2} \times 18 = 0.359\ \text{g}

Concentration of solutions

A solute is the substance dissolved. A solvent is the liquid it dissolves in. A solution is the mixture formed.

Concentration tells you how much solute is dissolved in a certain volume of solution. In this topic, concentration is usually measured in moles per cubic decimetre, written as $dm−3\text{mol dm}^{-3}$ .

The key equation is:

\frac{n}{V}

where:

$c$ is concentration in $dm−3\text{mol dm}^{-3}$
$n$ is amount in mol
$V$ is volume in dm³

So you can rearrange it:

n = c V

and:

\frac{n}{c}

If you are given a mass of solute, first convert mass to moles:

\frac{m}{M_r}

Tip

Volume conversion

Before using $\frac{n}{V}$ , convert cm³ to dm³ by dividing by 1000. For example, 250 cm³ is 0.250 dm³.

A standard solution is a solution with a known concentration, usually made by dissolving an accurate mass of solute and making the solution up to an exact volume.

Example

Calculating mass needed for a solution

What mass of sodium hydroxide, NaOH(s), is needed to make 250 cm³ of 0.200 $dm−3\text{mol dm}^{-3}$ solution? Use $Mr(NaOH)=40.0M_r(\text{NaOH}) = 40.0$ .

Convert the volume into dm³:

250\ \text{cm}^3 = 0.250\ \text{dm}^3

Calculate the amount of NaOH needed:

\times 0.250 = 0.0500\ \text{mol}

Convert moles to mass:

nM_r = 0.0500 \times 40.0 = 2.00\ \text{g}

Titration: finding an unknown concentration

A titration is a practical technique used to find the concentration of a solution by reacting it with a solution of known concentration.

In an acid-alkali titration, a measured volume of one solution is placed in a conical flask with an indicator, a dye that changes colour near the point of neutralisation. The other solution is added from a burette until the endpoint, when the colour just changes. The volume added from the burette is called the titre. This is the required practical skill used in acid/alkali titrations.

Labelled titration apparatus showing burette, conical flask, indicator, endpoint and meniscus reading

Good titration technique includes swirling the flask, adding solution drop by drop near the endpoint, and repeating until you get concordant titres — results close enough to be reliable.

The calculation method is always:

Use $n = c V$ to find moles of the known solution.
Use the balanced equation to get the mole ratio.
Use $\frac{n}{V}$ to find the unknown concentration.

Example

Finding an unknown concentration

25.0 cm³ of NaOH(aq) is titrated with 0.100 $dm−3\text{mol dm}^{-3}$ HCl(aq). The mean titre is 22.50 cm³. Find the concentration of NaOH(aq).

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

Calculate the moles of HCl used:

22.50\ \text{cm}^3 = 0.02250\ \text{dm}^3

\times 0.02250 = 0.002250\ \text{mol}

Use the equation ratio. HCl and NaOH react 1:1, so:

n(\text{NaOH}) = 0.002250\ \text{mol}

Calculate the concentration of NaOH:

\frac{n}{V} = \frac{0.002250}{0.0250} = 0.0900\ \text{mol dm}^{-3}

Gas volumes and moles

At room temperature and pressure, called RTP, one mole of any gas occupies 24 dm³. This is the molar gas volume used at GCSE.

So:

\times 24

and:

\frac{V}{24}

where gas volume is in dm³.

Common Mistake

Gas volume shortcut

The 24 dm³ per mole shortcut only applies to gases at room temperature and pressure. Do not use it for liquids, solids or solution volumes.

Example

Calculating a gas volume from mass

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

What volume of CO₂(g) forms from 2.00 g of CaCO₃(s)? Use $Mr(CaCO3)=100M_r(\text{CaCO}_3) = 100$ .

Convert the mass of calcium carbonate into moles:

\frac{m}{M_r} = \frac{2.00}{100} = 0.0200\ \text{mol}

Use the balanced equation. The ratio CaCO₃ : CO₂ is 1:1, so:

n(\text{CO}_2) = 0.0200\ \text{mol}

Convert moles of gas into volume at RTP:

\times 24 = 0.0200 \times 24 = 0.480\ \text{dm}^3

This is 480 cm³.

Theoretical yield and percentage yield

The theoretical yield is the maximum amount of product expected from the calculation, assuming the reaction goes perfectly.

The actual yield is the amount of product actually made in the experiment.

The percentage yield compares these:

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

Yields are often below 100% because reactions may be incomplete, product may be lost during transfer or purification, or side reactions may happen.

Common Mistake

Mole ratios before masses

Use the balanced equation ratio with moles, not grams. The safe route is always mass → moles → mole ratio → moles → mass.

Example

Calculating theoretical mass and percentage yield

5.00 g of CaCO₃(s) is heated. The actual mass of CaO(s) collected is 2.50 g.

CaCO₃(s) → CaO(s) + CO₂(g)

Use $Mr(CaCO3)=100M_r(\text{CaCO}_3) = 100$ and $Mr(CaO)=56M_r(\text{CaO}) = 56$ .

Convert the reactant mass into moles:

n(\text{CaCO}_3) = \frac{5.00}{100} = 0.0500\ \text{mol}

Use the equation ratio. CaCO₃ and CaO react 1:1, so:

n(\text{CaO}) = 0.0500\ \text{mol}

Calculate the theoretical mass of CaO:

nM_r = 0.0500 \times 56 = 2.80\ \text{g}

Calculate the percentage yield:

\text{percentage yield} = \frac{2.50}{2.80} \times 100 = 89.3\%

Atom economy

Atom economy measures how much of the mass of the reactants becomes the desired product. It is about reducing waste.

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

If there are numbers in front of formulae in the balanced equation, multiply each $M_r$ by its coefficient.

Key Idea

Yield vs atom economy

Percentage yield tells you how much product you actually got compared with the maximum possible. Atom economy tells you how much reactant mass ends up in the desired product rather than unwanted products.

Example

Calculating atom economy

Hydrochloric acid reacts with sodium hydroxide to make sodium chloride:

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

Calculate the atom economy for making NaCl. Use $Mr(HCl)=36.5M_r(\text{HCl}) = 36.5$ , $Mr(NaOH)=40.0M_r(\text{NaOH}) = 40.0$ and $Mr(NaCl)=58.5M_r(\text{NaCl}) = 58.5$ .

Add the relative formula masses of the reactants:

36.5 + 40.0 = 76.5

Identify the desired product mass:

M_r(\text{NaCl}) = 58.5

Calculate atom economy:

\text{atom economy} = \frac{58.5}{76.5} \times 100 = 76.5\%

Choosing a reaction pathway

This final decision-making skill is Higher Tier only.

In industry, the “best” reaction is not always the one with the highest atom economy. Chemists may compare:

atom economy
percentage yield
rate of reaction
equilibrium position, meaning whether a reversible reaction favours products or reactants
usefulness of by-products
safety, cost and waste disposal

Example

Choosing a reaction pathway

A company can make the same product by two routes.

Route A has atom economy 100%, yield 45%, and a slow rate. Route B has atom economy 70%, yield 90%, a fast rate, and a useful by-product.

Compare useful product from the data. Route A keeps all reactant atoms in theory, but its low yield gives roughly $100%×45%=45%100\% \times 45\% = 45\%$ useful output. Route B gives roughly $70%×90%=63%70\% \times 90\% = 63\%$ useful output.
Compare practical factors. Route B is faster, so it can make product more quickly, and its by-product is useful rather than waste.
Choose Route B for large-scale production, unless the question gives another priority such as minimising all by-products.

Exam technique

In the exam

Start every calculation by identifying what you are converting into moles: mass, solution concentration or gas volume.
Use the balanced equation only after you have moles; the big numbers give mole ratios.
Keep units in your working, convert cm³ to dm³, and round your final answer to a sensible number of significant figures.

Self review

Check yourself

Why must solution volumes usually be converted from cm³ to dm³ before using $n = c V$ ?
What is the difference between percentage yield and atom economy?
In the equation 2H₂(g) + O₂(g) → 2H₂O(l), how would the mole ratio affect a gas volume calculation?

Monitoring chemical reactions

What you'll learn

How concentration links the amount of solute to the volume of solution.
How titration results can be used to find an unknown concentration.
How gas volumes, reacting masses, percentage yield and atom economy are calculated.
How data can help choose the best reaction pathway.

Some of this section is separate Chemistry content rather than Combined Science; because you are studying J248, you should learn the full set here. I’ll flag the Higher Tier-only parts as we go.

The big idea: reactions are compared using moles

A mole is a unit for the amount of substance. It lets chemists count particles in a practical way, just like a dozen means 12 items — except one mole means a huge number of particles.

The Avogadro constant is the number of particles in one mole: about $6.02 \times 10^{23}$ particles per mole.

Balanced chemical equations compare substances in mole ratios, not mass ratios. For example:

2H₂(g) + O₂(g) → 2H₂O(l)

This means 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

The route through most reacting-amount calculations looks like this.

Flowchart showing how mass, concentration, gas volume and balanced equation ratios connect through moles

Common Mistake

Moles are not grams

A mole is an amount of substance, not a mass. Different substances have different masses per mole because their particles have different relative formula masses.

At Higher Tier only, you may need calculations using the Avogadro constant in standard form.

Example

Using the Avogadro constant to find mass

How much mass is represented by $1.20 \times 10^{22}$ molecules of H₂O?

Convert particles to moles using $\frac{\text{number of particles}}{\text{Avogadro constant}}$ :

\frac{1.20 \times 10^{22}}{6.02 \times 10^{23}} = 1.99 \times 10^{-2}\ \text{mol}

Work out the relative formula mass of water:

M_r(\text{H}_2\text{O}) = 1 + 1 + 16 = 18

Convert moles to mass using $m = nM_r$ :

\times 10^{-2} \times 18 = 0.359\ \text{g}

Concentration of solutions

A solute is the substance dissolved. A solvent is the liquid it dissolves in. A solution is the mixture formed.

The key equation is:

\frac{n}{V}

where:

$c$ is concentration in $dm−3\text{mol dm}^{-3}$
$n$ is amount in mol
$V$ is volume in dm³

So you can rearrange it:

n = c V

and:

\frac{n}{c}

If you are given a mass of solute, first convert mass to moles:

\frac{m}{M_r}

Tip

Volume conversion

Before using $\frac{n}{V}$ , convert cm³ to dm³ by dividing by 1000. For example, 250 cm³ is 0.250 dm³.

A standard solution is a solution with a known concentration, usually made by dissolving an accurate mass of solute and making the solution up to an exact volume.

Example

Calculating mass needed for a solution

What mass of sodium hydroxide, NaOH(s), is needed to make 250 cm³ of 0.200 $dm−3\text{mol dm}^{-3}$ solution? Use $Mr(NaOH)=40.0M_r(\text{NaOH}) = 40.0$ .

Convert the volume into dm³:

250\ \text{cm}^3 = 0.250\ \text{dm}^3

Calculate the amount of NaOH needed:

\times 0.250 = 0.0500\ \text{mol}

Convert moles to mass:

nM_r = 0.0500 \times 40.0 = 2.00\ \text{g}

Titration: finding an unknown concentration

A titration is a practical technique used to find the concentration of a solution by reacting it with a solution of known concentration.

Labelled titration apparatus showing burette, conical flask, indicator, endpoint and meniscus reading

Good titration technique includes swirling the flask, adding solution drop by drop near the endpoint, and repeating until you get concordant titres — results close enough to be reliable.

The calculation method is always:

Use $n = c V$ to find moles of the known solution.
Use the balanced equation to get the mole ratio.
Use $\frac{n}{V}$ to find the unknown concentration.

Example

Finding an unknown concentration

25.0 cm³ of NaOH(aq) is titrated with 0.100 $dm−3\text{mol dm}^{-3}$ HCl(aq). The mean titre is 22.50 cm³. Find the concentration of NaOH(aq).

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

Calculate the moles of HCl used:

22.50\ \text{cm}^3 = 0.02250\ \text{dm}^3

\times 0.02250 = 0.002250\ \text{mol}

Use the equation ratio. HCl and NaOH react 1:1, so:

n(\text{NaOH}) = 0.002250\ \text{mol}

Calculate the concentration of NaOH:

\frac{n}{V} = \frac{0.002250}{0.0250} = 0.0900\ \text{mol dm}^{-3}

Gas volumes and moles

At room temperature and pressure, called RTP, one mole of any gas occupies 24 dm³. This is the molar gas volume used at GCSE.

So:

\times 24

and:

\frac{V}{24}

where gas volume is in dm³.

Common Mistake

Gas volume shortcut

The 24 dm³ per mole shortcut only applies to gases at room temperature and pressure. Do not use it for liquids, solids or solution volumes.

Example

Calculating a gas volume from mass

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

What volume of CO₂(g) forms from 2.00 g of CaCO₃(s)? Use $Mr(CaCO3)=100M_r(\text{CaCO}_3) = 100$ .

Convert the mass of calcium carbonate into moles:

\frac{m}{M_r} = \frac{2.00}{100} = 0.0200\ \text{mol}

Use the balanced equation. The ratio CaCO₃ : CO₂ is 1:1, so:

n(\text{CO}_2) = 0.0200\ \text{mol}

Convert moles of gas into volume at RTP:

\times 24 = 0.0200 \times 24 = 0.480\ \text{dm}^3

This is 480 cm³.

Theoretical yield and percentage yield

The theoretical yield is the maximum amount of product expected from the calculation, assuming the reaction goes perfectly.

The actual yield is the amount of product actually made in the experiment.

The percentage yield compares these:

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

Yields are often below 100% because reactions may be incomplete, product may be lost during transfer or purification, or side reactions may happen.

Common Mistake

Mole ratios before masses

Use the balanced equation ratio with moles, not grams. The safe route is always mass → moles → mole ratio → moles → mass.

Example

Calculating theoretical mass and percentage yield

5.00 g of CaCO₃(s) is heated. The actual mass of CaO(s) collected is 2.50 g.

CaCO₃(s) → CaO(s) + CO₂(g)

Use $Mr(CaCO3)=100M_r(\text{CaCO}_3) = 100$ and $Mr(CaO)=56M_r(\text{CaO}) = 56$ .

Convert the reactant mass into moles:

n(\text{CaCO}_3) = \frac{5.00}{100} = 0.0500\ \text{mol}

Use the equation ratio. CaCO₃ and CaO react 1:1, so:

n(\text{CaO}) = 0.0500\ \text{mol}

Calculate the theoretical mass of CaO:

nM_r = 0.0500 \times 56 = 2.80\ \text{g}

Calculate the percentage yield:

\text{percentage yield} = \frac{2.50}{2.80} \times 100 = 89.3\%

Atom economy

Atom economy measures how much of the mass of the reactants becomes the desired product. It is about reducing waste.

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

If there are numbers in front of formulae in the balanced equation, multiply each $M_r$ by its coefficient.

Key Idea

Yield vs atom economy

Example

Calculating atom economy

Hydrochloric acid reacts with sodium hydroxide to make sodium chloride:

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

Calculate the atom economy for making NaCl. Use $Mr(HCl)=36.5M_r(\text{HCl}) = 36.5$ , $Mr(NaOH)=40.0M_r(\text{NaOH}) = 40.0$ and $Mr(NaCl)=58.5M_r(\text{NaCl}) = 58.5$ .

Add the relative formula masses of the reactants:

36.5 + 40.0 = 76.5

Identify the desired product mass:

M_r(\text{NaCl}) = 58.5

Calculate atom economy:

\text{atom economy} = \frac{58.5}{76.5} \times 100 = 76.5\%

Choosing a reaction pathway

This final decision-making skill is Higher Tier only.

In industry, the “best” reaction is not always the one with the highest atom economy. Chemists may compare:

atom economy
percentage yield
rate of reaction
equilibrium position, meaning whether a reversible reaction favours products or reactants
usefulness of by-products
safety, cost and waste disposal

Example

Choosing a reaction pathway

A company can make the same product by two routes.

Route A has atom economy 100%, yield 45%, and a slow rate. Route B has atom economy 70%, yield 90%, a fast rate, and a useful by-product.

Compare useful product from the data. Route A keeps all reactant atoms in theory, but its low yield gives roughly $100%×45%=45%100\% \times 45\% = 45\%$ useful output. Route B gives roughly $70%×90%=63%70\% \times 90\% = 63\%$ useful output.
Compare practical factors. Route B is faster, so it can make product more quickly, and its by-product is useful rather than waste.
Choose Route B for large-scale production, unless the question gives another priority such as minimising all by-products.

Exam technique

In the exam

Start every calculation by identifying what you are converting into moles: mass, solution concentration or gas volume.
Use the balanced equation only after you have moles; the big numbers give mole ratios.
Keep units in your working, convert cm³ to dm³, and round your final answer to a sensible number of significant figures.

Self review

Check yourself

Why must solution volumes usually be converted from cm³ to dm³ before using $n = c V$ ?
What is the difference between percentage yield and atom economy?
In the equation 2H₂(g) + O₂(g) → 2H₂O(l), how would the mole ratio affect a gas volume calculation?

Monitoring chemical reactions

What you'll learn

The big idea: reactions are compared using moles

Concentration of solutions

Titration: finding an unknown concentration

Gas volumes and moles

Theoretical yield and percentage yield

Atom economy

Choosing a reaction pathway

Recap questions

You've reached the end

Monitoring chemical reactions

What you'll learn

The big idea: reactions are compared using moles

Concentration of solutions

Titration: finding an unknown concentration

Gas volumes and moles

Theoretical yield and percentage yield

Atom economy

Choosing a reaction pathway

Recap questions

You've reached the end

Monitoring chemical reactions

What you'll learn

The big idea: reactions are compared using moles

Concentration of solutions

Titration: finding an unknown concentration

Gas volumes and moles

Theoretical yield and percentage yield

Atom economy

Choosing a reaction pathway

Recap questions

You've reached the end

Monitoring chemical reactions Revision Guide

Monitoring chemical reactions

What you'll learn

The big idea: reactions are compared using moles

Concentration of solutions

Titration: finding an unknown concentration

Gas volumes and moles

Theoretical yield and percentage yield

Atom economy

Choosing a reaction pathway

Recap questions

You've reached the end

Monitoring chemical reactions Revision Guide