Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

What you'll learn

Why equal numbers of moles of gases have equal volumes under the same conditions.
How to use 24 dm³ as the volume of one mole of gas at room temperature and pressure.
How to calculate gas volume from mass and relative formula mass.
How to use balanced equations to calculate volumes of gaseous reactants and products.

The big idea: gases and moles

This Higher Tier Chemistry-only topic links amount of substance to the volume of a gas.

You already know that reactions happen in fixed particle ratios. For example, in a balanced equation, the big numbers in front of formulae tell you the ratio of particles — and therefore the ratio of moles.

For gases, there is a very useful extra shortcut: at the same temperature and pressure, equal amounts in moles of gases occupy the same volume.

Definition

Amount of substance

The amount of substance tells you how many particles are present. Its unit is the mole, written as mol. One mole contains the same number of particles, no matter what the substance is.

Key Idea

Equal moles of gases

Equal amounts in moles of gases occupy the same volume, as long as the gases are at the same temperature and pressure.

Here is the calculation route you will use most often: mass can be converted to moles, moles can be converted to gas volume, and balanced equations give mole ratios that also work as gas volume ratios.

Schematic linking mass, moles, gas volume at RTP, and gaseous reaction volume ratios

Room temperature and pressure, RTP

In GCSE Chemistry, room temperature and pressure is usually shortened to RTP.

At RTP:

temperature = 20°C
pressure = 1 atmosphere
one mole of any gas occupies 24 dm³

Definition

Molar gas volume

The molar gas volume is the volume occupied by one mole of a gas. At room temperature and pressure, the molar gas volume is 24 dm³ per mole.

So at RTP:

\times 24

where:

$V$ = volume of gas in dm³
$n$ = amount of gas in mol

You can also rearrange this:

\frac{V}{24}

Tip

Using cm³

If the question gives gas volumes in cm³, either convert to dm³ first, or use 24 000 cm³ as the volume of one mole of gas at RTP.

Common Mistake

Mixing up dm³ and cm³

24 dm³ is the same as 24 000 cm³. Do not divide a volume in cm³ by 24 unless you have converted it into dm³ first.

Example

Calculating gas volume from moles

What volume is occupied by 0.250 mol of oxygen, O₂(g), at RTP?

Use the RTP gas volume equation because the question asks for the volume of a gas at room temperature and pressure:
$\times 24$
Substitute the amount of oxygen:
$\times 24$
Calculate the volume and include the correct unit:
$\text{ dm}^3$

So 0.250 mol of O₂(g) occupies 6.00 dm³ at RTP.

Linking mass, moles and gas volume

Sometimes the question gives you the mass of a gas, not the amount in moles. You then need two steps:

Convert mass to moles.
Convert moles to gas volume.

Definition

Relative formula mass

The relative formula mass, $M_r$ , is the total relative mass of all the atoms in a formula. For a simple molecule, it is also often called the relative molecular mass.

To find moles from mass:

\frac{m}{M_r}

where:

$n$ = amount in mol
$m$ = mass in g
$M_r$ = relative formula mass

Then, for a gas at RTP:

\times 24

Example

Calculating gas volume from mass

Calculate the volume of 8.8 g of carbon dioxide, CO₂(g), at RTP.
Relative atomic masses: C = 12, O = 16.

Calculate the relative formula mass of carbon dioxide:
$Mr(CO2)=12+(2×16)=44M_r(\text{CO}_2) = 12 + (2 \times 16) = 44$
Convert the mass into moles:
$\frac{8.8}{44} = 0.20 \text{ mol}$
Convert moles into gas volume at RTP:
$\times 24 = 4.8 \text{ dm}^3$

So 8.8 g of CO₂(g) occupies 4.8 dm³ at RTP.

Rearranging the gas volume equation

You may need to change the subject of an equation. This means rearranging it to make a different symbol the one being calculated.

From:

\times 24

you can find amount in moles by dividing both sides by 24:

\frac{V}{24}

Then, if you need mass:

\times M_r

Example

Calculating mass from gas volume

Calculate the mass of carbon dioxide, CO₂(g), in 12.0 dm³ of the gas at RTP.
Relative atomic masses: C = 12, O = 16.

Find the amount in moles from the gas volume:
$\frac{V}{24} = \frac{12.0}{24} = 0.500 \text{ mol}$
Calculate the relative formula mass:
$Mr(CO2)=12+(2×16)=44M_r(\text{CO}_2) = 12 + (2 \times 16) = 44$
Convert moles into mass:
$\times M_r = 0.500 \times 44 = 22.0 \text{ g}$

So 12.0 dm³ of CO₂(g) has a mass of 22.0 g at RTP.

Using balanced equations for gas volumes

Balanced equations show the ratio of moles reacting and being produced.

For gases, if temperature and pressure are the same, the mole ratio is also the gas volume ratio.

For example:

N2(g)+3H2(g)→2NH3(g)\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}

This means:

1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia.
Therefore, 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia.

So 10 cm³ of N₂(g) would react with 30 cm³ of H₂(g) to form 20 cm³ of NH₃(g), assuming all gases are measured at the same temperature and pressure.

Key Idea

Gas volume ratios

For gases at the same temperature and pressure, the balancing numbers in the equation give the reacting volume ratio directly.

Example

Calculating product volume from reactant volume

Carbon monoxide reacts with oxygen to form carbon dioxide:

2CO(g)+O2(g)→2CO2(g)2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}

What volume of CO₂(g) is made when 150 cm³ of O₂(g) reacts completely with excess CO(g)? All gases are measured at the same temperature and pressure.

Use the balanced equation to find the gas volume ratio:
$O2:CO2=1:2\text{O}_2 : \text{CO}_2 = 1 : 2$
Compare the given oxygen volume with the ratio. The equation says 1 volume of O₂(g) makes 2 volumes of CO₂(g), so multiply by 2:
$\text{ cm}^3 \times 2 = 300 \text{ cm}^3$
State the volume of product:

The volume of CO₂(g) produced is 300 cm³.

When to use volume ratios directly

You can use the balanced equation as a gas volume ratio when:

the substances you are comparing are gases
the gases are measured under the same temperature and pressure
the equation is balanced

You do not need to convert to moles first if you are only comparing gas volumes in the same reaction under the same conditions.

Common Mistake

Only for gases

Balanced equation coefficients can be used as volume ratios for gases only. Do not use them as volume ratios for solids, liquids or aqueous solutions.

Example

Finding a reactant gas volume

Ammonia is made in the Haber process:

N2(g)+3H2(g)→2NH3(g)\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}

Calculate the volume of H₂(g) needed to make 80 dm³ of NH₃(g), assuming all gases are measured at the same temperature and pressure.

Use the balanced equation to identify the relevant ratio:
$H2:NH3=3:2\text{H}_2 : \text{NH}_3 = 3 : 2$
Work out the scale factor from 2 volumes of NH₃(g) to 80 dm³:
$80 \div 2 = 40$
Apply the same scale factor to hydrogen:
$\times 40 = 120 \text{ dm}^3$

So 120 dm³ of H₂(g) is needed.

Combining mass and gas volume in reactions

Some questions mix the two techniques. A common route is:

Use the balanced equation to get a mole ratio.
Convert mass to moles, or volume to moles.
Use the ratio.
Convert back to volume if needed.

For example, if the question gives a mass of a reactant and asks for the volume of a gaseous product, you cannot jump straight from grams to dm³ using the equation ratio. First, convert grams to moles.

Common Mistake

Using mass ratios as volume ratios

The balancing numbers give mole ratios. For gases at the same temperature and pressure, they also give volume ratios. They do not directly give mass ratios.

Example

Calculating gas volume from a reacting mass

Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide:

CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g)\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

Calculate the volume of CO₂(g) produced at RTP when 5.00 g of calcium carbonate reacts completely.
Relative formula mass of CaCO₃ = 100.

Convert the mass of calcium carbonate into moles:
$moln(\text{CaCO}_3) = \frac{5.00}{100} = 0.0500 \text{ mol}$
Use the balanced equation ratio. The ratio of CaCO₃ to CO₂ is 1 : 1, so:
$moln(\text{CO}_2) = 0.0500 \text{ mol}$
Convert moles of carbon dioxide into volume at RTP:
$\times 24 = 1.20 \text{ dm}^3$

So 1.20 dm³ of CO₂(g) is produced at RTP.

Exam technique

In the exam

Check the units first: if the volume is in cm³, convert to dm³ or use 24 000 cm³ per mole.
If all the relevant substances are gases at the same temperature and pressure, use the balanced equation as a volume ratio.
If a mass is given, convert mass to moles before using the balanced equation ratio.

Self review

Check yourself

What volume does one mole of any gas occupy at RTP?
Why can the balancing numbers in an equation be used as gas volume ratios?
A gas volume is given in cm³. What must you be careful about before using 24 dm³ per mole?

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

What you'll learn

Why equal numbers of moles of gases have equal volumes under the same conditions.
How to use 24 dm³ as the volume of one mole of gas at room temperature and pressure.
How to calculate gas volume from mass and relative formula mass.
How to use balanced equations to calculate volumes of gaseous reactants and products.

The big idea: gases and moles

This Higher Tier Chemistry-only topic links amount of substance to the volume of a gas.

For gases, there is a very useful extra shortcut: at the same temperature and pressure, equal amounts in moles of gases occupy the same volume.

Definition

Amount of substance

The amount of substance tells you how many particles are present. Its unit is the mole, written as mol. One mole contains the same number of particles, no matter what the substance is.

Key Idea

Equal moles of gases

Equal amounts in moles of gases occupy the same volume, as long as the gases are at the same temperature and pressure.

Schematic linking mass, moles, gas volume at RTP, and gaseous reaction volume ratios

Room temperature and pressure, RTP

In GCSE Chemistry, room temperature and pressure is usually shortened to RTP.

At RTP:

temperature = 20°C
pressure = 1 atmosphere
one mole of any gas occupies 24 dm³

Definition

Molar gas volume

The molar gas volume is the volume occupied by one mole of a gas. At room temperature and pressure, the molar gas volume is 24 dm³ per mole.

So at RTP:

\times 24

where:

$V$ = volume of gas in dm³
$n$ = amount of gas in mol

You can also rearrange this:

\frac{V}{24}

Tip

Using cm³

If the question gives gas volumes in cm³, either convert to dm³ first, or use 24 000 cm³ as the volume of one mole of gas at RTP.

Common Mistake

Mixing up dm³ and cm³

24 dm³ is the same as 24 000 cm³. Do not divide a volume in cm³ by 24 unless you have converted it into dm³ first.

Example

Calculating gas volume from moles

What volume is occupied by 0.250 mol of oxygen, O₂(g), at RTP?

Use the RTP gas volume equation because the question asks for the volume of a gas at room temperature and pressure:
$\times 24$
Substitute the amount of oxygen:
$\times 24$
Calculate the volume and include the correct unit:
$\text{ dm}^3$

So 0.250 mol of O₂(g) occupies 6.00 dm³ at RTP.

Linking mass, moles and gas volume

Sometimes the question gives you the mass of a gas, not the amount in moles. You then need two steps:

Convert mass to moles.
Convert moles to gas volume.

Definition

Relative formula mass

The relative formula mass, $M_r$ , is the total relative mass of all the atoms in a formula. For a simple molecule, it is also often called the relative molecular mass.

To find moles from mass:

\frac{m}{M_r}

where:

$n$ = amount in mol
$m$ = mass in g
$M_r$ = relative formula mass

Then, for a gas at RTP:

\times 24

Example

Calculating gas volume from mass

Calculate the volume of 8.8 g of carbon dioxide, CO₂(g), at RTP.
Relative atomic masses: C = 12, O = 16.

Calculate the relative formula mass of carbon dioxide:
$Mr(CO2)=12+(2×16)=44M_r(\text{CO}_2) = 12 + (2 \times 16) = 44$
Convert the mass into moles:
$\frac{8.8}{44} = 0.20 \text{ mol}$
Convert moles into gas volume at RTP:
$\times 24 = 4.8 \text{ dm}^3$

So 8.8 g of CO₂(g) occupies 4.8 dm³ at RTP.

Rearranging the gas volume equation

You may need to change the subject of an equation. This means rearranging it to make a different symbol the one being calculated.

From:

\times 24

you can find amount in moles by dividing both sides by 24:

\frac{V}{24}

Then, if you need mass:

\times M_r

Example

Calculating mass from gas volume

Calculate the mass of carbon dioxide, CO₂(g), in 12.0 dm³ of the gas at RTP.
Relative atomic masses: C = 12, O = 16.

Find the amount in moles from the gas volume:
$\frac{V}{24} = \frac{12.0}{24} = 0.500 \text{ mol}$
Calculate the relative formula mass:
$Mr(CO2)=12+(2×16)=44M_r(\text{CO}_2) = 12 + (2 \times 16) = 44$
Convert moles into mass:
$\times M_r = 0.500 \times 44 = 22.0 \text{ g}$

So 12.0 dm³ of CO₂(g) has a mass of 22.0 g at RTP.

Using balanced equations for gas volumes

Balanced equations show the ratio of moles reacting and being produced.

For gases, if temperature and pressure are the same, the mole ratio is also the gas volume ratio.

For example:

N2(g)+3H2(g)→2NH3(g)\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}

This means:

1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia.
Therefore, 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia.

So 10 cm³ of N₂(g) would react with 30 cm³ of H₂(g) to form 20 cm³ of NH₃(g), assuming all gases are measured at the same temperature and pressure.

Key Idea

Gas volume ratios

For gases at the same temperature and pressure, the balancing numbers in the equation give the reacting volume ratio directly.

Example

Calculating product volume from reactant volume

Carbon monoxide reacts with oxygen to form carbon dioxide:

2CO(g)+O2(g)→2CO2(g)2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}

What volume of CO₂(g) is made when 150 cm³ of O₂(g) reacts completely with excess CO(g)? All gases are measured at the same temperature and pressure.

Use the balanced equation to find the gas volume ratio:
$O2:CO2=1:2\text{O}_2 : \text{CO}_2 = 1 : 2$
Compare the given oxygen volume with the ratio. The equation says 1 volume of O₂(g) makes 2 volumes of CO₂(g), so multiply by 2:
$\text{ cm}^3 \times 2 = 300 \text{ cm}^3$
State the volume of product:

The volume of CO₂(g) produced is 300 cm³.

When to use volume ratios directly

You can use the balanced equation as a gas volume ratio when:

the substances you are comparing are gases
the gases are measured under the same temperature and pressure
the equation is balanced

You do not need to convert to moles first if you are only comparing gas volumes in the same reaction under the same conditions.

Common Mistake

Only for gases

Balanced equation coefficients can be used as volume ratios for gases only. Do not use them as volume ratios for solids, liquids or aqueous solutions.

Example

Finding a reactant gas volume

Ammonia is made in the Haber process:

N2(g)+3H2(g)→2NH3(g)\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}

Calculate the volume of H₂(g) needed to make 80 dm³ of NH₃(g), assuming all gases are measured at the same temperature and pressure.

Use the balanced equation to identify the relevant ratio:
$H2:NH3=3:2\text{H}_2 : \text{NH}_3 = 3 : 2$
Work out the scale factor from 2 volumes of NH₃(g) to 80 dm³:
$80 \div 2 = 40$
Apply the same scale factor to hydrogen:
$\times 40 = 120 \text{ dm}^3$

So 120 dm³ of H₂(g) is needed.

Combining mass and gas volume in reactions

Some questions mix the two techniques. A common route is:

Use the balanced equation to get a mole ratio.
Convert mass to moles, or volume to moles.
Use the ratio.
Convert back to volume if needed.

Common Mistake

Using mass ratios as volume ratios

The balancing numbers give mole ratios. For gases at the same temperature and pressure, they also give volume ratios. They do not directly give mass ratios.

Example

Calculating gas volume from a reacting mass

Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide:

CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g)\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

Calculate the volume of CO₂(g) produced at RTP when 5.00 g of calcium carbonate reacts completely.
Relative formula mass of CaCO₃ = 100.

Convert the mass of calcium carbonate into moles:
$moln(\text{CaCO}_3) = \frac{5.00}{100} = 0.0500 \text{ mol}$
Use the balanced equation ratio. The ratio of CaCO₃ to CO₂ is 1 : 1, so:
$moln(\text{CO}_2) = 0.0500 \text{ mol}$
Convert moles of carbon dioxide into volume at RTP:
$\times 24 = 1.20 \text{ dm}^3$

So 1.20 dm³ of CO₂(g) is produced at RTP.

Exam technique

In the exam

Check the units first: if the volume is in cm³, convert to dm³ or use 24 000 cm³ per mole.
If all the relevant substances are gases at the same temperature and pressure, use the balanced equation as a volume ratio.
If a mass is given, convert mass to moles before using the balanced equation ratio.

Self review

Check yourself

What volume does one mole of any gas occupy at RTP?
Why can the balancing numbers in an equation be used as gas volume ratios?
A gas volume is given in cm³. What must you be careful about before using 24 dm³ per mole?

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

What you'll learn

The big idea: gases and moles

Room temperature and pressure, RTP

Linking mass, moles and gas volume

Rearranging the gas volume equation

Using balanced equations for gas volumes

When to use volume ratios directly

Combining mass and gas volume in reactions

Recap questions

You've reached the end

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

What you'll learn

The big idea: gases and moles

Room temperature and pressure, RTP

Linking mass, moles and gas volume

Rearranging the gas volume equation

Using balanced equations for gas volumes

When to use volume ratios directly

Combining mass and gas volume in reactions

Recap questions

You've reached the end

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

What you'll learn

The big idea: gases and moles

Room temperature and pressure, RTP

Linking mass, moles and gas volume

Rearranging the gas volume equation

Using balanced equations for gas volumes

When to use volume ratios directly

Combining mass and gas volume in reactions

Recap questions

You've reached the end

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only) Revision Guide

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

What you'll learn

The big idea: gases and moles

Room temperature and pressure, RTP

Linking mass, moles and gas volume

Rearranging the gas volume equation

Using balanced equations for gas volumes

When to use volume ratios directly

Combining mass and gas volume in reactions

Recap questions

You've reached the end

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only) Revision Guide