1.34C-1.35C - Concentration and gas volume calculations | Edexcel IGCSE Chemistry

Chemists often measure a volume in the lab, but balanced equations compare amounts of substance in moles. In this lesson, the same idea appears twice: a solution volume can be linked to concentration, and a gas volume can be linked to the molar volume of a gas at room temperature and pressure. The main skill is keeping the units under control before using the numbers.

Moles as the bridge

A balanced chemical equation gives a ratio of amounts of substance. It does not directly compare cm³ of solution with cm³ of gas, or a measured solution volume with a formula mass. That is why many calculation questions have a hidden middle step: convert the measured quantity into moles, use the mole ratio, then convert to the quantity asked for.

Amount of substance

Amount of substance is measured in moles, mol. In chemical calculations, moles let you compare the numbers of particles or formula units reacting.

For this lesson, there are two volume-to-mole links:

Measured quantity	Link to amount of substance
Volume of a solution	Use concentration in mol/dm³.
Volume of a gas at rtp	Use the molar gas volume, 24 dm³ or 24 000 cm³ per mole.

The word volume is therefore not enough on its own. Always ask: is it a volume of solution, or a volume of gas? A solution volume is used with concentration. A gas volume at room temperature and pressure is used with molar volume.

Concentration in solution

The concentration of a solution tells you how much solute is dissolved in a certain volume of the final solution. For Edexcel concentration calculations in this cluster, concentration is measured in mol/dm³.

Concentration

Concentration is the amount of dissolved solute per unit volume of solution.

[DIAGRAM: asset_name: Concentration and gas volume calculations - diagram 01; asset_slug: c12_concentration_and_gas_volume_calculations__diagram_01; recommended_method: image_gen; description: Monochrome NovaLearn-style schematic showing three solution vessels ordered by increasing concentration: 1 mol in 2 dm3, 1 mol in 1 dm3, and 2 mol in 1 dm3. Includes the relationship concentration = amount / volume and unit mol/dm3.]
Diagram

Solution concentration

c = \frac{n}{V}

n = cV

V = \frac{n}{c}

where $c$ is concentration in mol/dm³, $n$ is amount in mol, and $V$ is volume of solution in dm³.

The volume in the formula must be in dm³, not cm³. If the question gives cm³, divide by 1000 first:

Volume in cm³	Volume in dm³
25.0 cm³	0.0250 dm³
100 cm³	0.100 dm³
250 cm³	0.250 dm³
1000 cm³	1.000 dm³

Worked example: calculate the concentration of a solution that contains 0.0200 mol of sodium chloride in 250 cm³ of solution.

First convert the volume:

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

Then substitute:

c = \frac{0.0200}{0.250} = 0.0800\ \mathrm{mol}\,\mathrm{dm}^{-3}

Notice that the volume is the volume of the solution, not just the volume of water added. The solute and solvent together make the final solution volume used in the calculation.

Solution stoichiometry

Once you have an amount in mol, you can use the balanced equation. This is where many marks are lost: students calculate the moles in one solution correctly, then forget to use the equation ratio before finding the second concentration or amount.

Use this route for a solution-reaction calculation:

Write or use the balanced equation.
Convert any solution volumes from cm³ to dm³.
Use $n = cV$ to find the known amount in mol.
Use the balanced-equation ratio to find the required amount in mol.
Use $c = n/V$ or $V = n/c$ if the question asks for another solution concentration or volume.

Worked example: 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide neutralises 20.0 cm³ of hydrochloric acid. Calculate the concentration of the hydrochloric acid.

The equation is:

\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}

The ratio of NaOH to HCl is 1:1.

25.0\ \mathrm{cm}^{3} = 0.0250\ \mathrm{dm}^{3}

n(\mathrm{NaOH}) = cV = 0.100 \times 0.0250 = 0.00250\ \mathrm{mol}

So:

n(\mathrm{HCl}) = 0.00250\ \mathrm{mol}

Now convert the acid volume:

20.0\ \mathrm{cm}^{3} = 0.0200\ \mathrm{dm}^{3}

c(\mathrm{HCl}) = \frac{0.00250}{0.0200} = 0.125\ \mathrm{mol}\,\mathrm{dm}^{-3}

This is not a lesson on titration technique. The calculation works because the equation and volumes are given; the assessed skill here is amount, volume and concentration.

Gas volume at rtp

At room temperature and pressure, Edexcel uses this molar gas volume:

Molar gas volume at rtp

1\ \mathrm{mol\ gas} = 24\ \mathrm{dm}^{3} = 24000\ \mathrm{cm}^{3}

This value applies to gases at the same temperature and pressure. The identity of the gas does not change the volume occupied by one mole under these conditions: one mole of hydrogen, oxygen or carbon dioxide all occupy 24 dm³ at rtp.

[DIAGRAM: asset_name: Concentration and gas volume calculations - diagram 02; asset_slug: c12_concentration_and_gas_volume_calculations__diagram_02; recommended_method: image_gen; description: Monochrome NovaLearn-style gas syringe and calculation pathway showing 1 mol gas = 24 dm3 = 24 000 cm3 at rtp, with separate routes for gas volume in dm3 and gas volume in cm3 plus the reminder 1 dm3 = 1000 cm3.]
Diagram

Use the version that matches the units in the question:

Gas volume unit	Amount in mol	Gas volume from amount
dm³	$n = V/24$	$V = n \times 24$
cm³	$n = V/24000$	$V = n \times 24000$

Worked example 1: calculate the volume of 0.250 mol of carbon dioxide at rtp.

V = n \times 24 = 0.250 \times 24 = 6.00\ \mathrm{dm}^{3}

This is the same as:

6.00\ \mathrm{dm}^{3} = 6000\ \mathrm{cm}^{3}

Worked example 2: calculate the amount of oxygen in 120 cm³ of oxygen gas at rtp.

n = \frac{120}{24000} = 0.00500\ \mathrm{mol}

Do not use 22.4 dm³ in this course for these questions. Edexcel specifies 24 dm³ and 24 000 cm³ at rtp.

Combined calculation route

Some questions combine both parts of this lesson. A solution concentration may give the amount of a reactant, the equation ratio gives the amount of gas produced, and the molar gas volume gives the gas volume at rtp.

Worked example: 50.0 cm³ of 0.100 mol/dm³ hydrochloric acid reacts with excess magnesium. Calculate the volume of hydrogen produced at rtp, in cm³.

The equation is:

\mathrm{Mg} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2}

Find the amount of HCl:

50.0\ \mathrm{cm}^{3} = 0.0500\ \mathrm{dm}^{3}

n(\mathrm{HCl}) = 0.100 \times 0.0500 = 0.00500\ \mathrm{mol}

Use the equation ratio:

2\ \mathrm{mol\ HCl} : 1\ \mathrm{mol\ H_2}

n(\mathrm{H_2}) = 0.00500 / 2 = 0.00250\ \mathrm{mol}

Convert moles of gas to volume in cm³:

V(\mathrm{H_2}) = 0.00250 \times 24000 = 60.0\ \mathrm{cm}^{3}

Common calculation traps are predictable:

Trap	How to avoid it
Using cm³ directly in $n = cV$	Convert solution volume to dm³ first.
Dividing a cm³ gas volume by 24	Use 24 000 for cm³, or convert to dm³ first.
Ignoring the balanced equation	Use the mole ratio after finding the first amount.
Rounding too early	Keep full calculator values until the final answer.
Leaving off the unit	State mol, mol/dm³, dm³ or cm³ as appropriate.

Convert measured volumes into moles with the correct unit relationship, use the balanced-equation ratio, then convert back only if the question asks for a volume or concentration.

Use that route in the final check below.

Last Accessed

Principles of chemistry

Inorganic chemistry

Physical chemistry

Organic chemistry

1.34C-1.35C - Concentration and gas volume calculations

Moles as the bridge

Concentration in solution

Solution stoichiometry

Gas volume at rtp

Combined calculation route

Learn faster with Nova