Equal moles of any gas occupy equal volumes

Definition

Molar gas volume

The volume occupied by one mole of any gas at a stated temperature and pressure, which is 24 dm³ at room temperature and pressure.

Gas particles are very far apart, so the volume of a gas depends on the number of particles and not on the size of each particle.
At the same temperature and pressure, the volume of a gas is directly proportional to its amount in moles.
1. Doubling the amount of gas doubles its volume, and halving it halves the volume.
At room temperature and pressure, about $20 ∘C20\ ^{\circ}\text{C}$ $20^{\circ} C$ and 1 atmosphere, one mole of any gas occupies $24 dm324\,\text{dm}^3$ $24 dm^{3}$ .
1. This is the same as $cm324\,000\,\text{cm}^3$ , whether the gas is hydrogen, oxygen or carbon dioxide.

Graph showing that the volume of a gas is directly proportional to the number of moles (V = constant × n), represented by a straight line passing through the origin.

Key Idea

One mole of hydrogen weighs $g2.0\,\text{g}$ and one mole of carbon dioxide weighs $g44.0\,\text{g}$ , yet at room temperature and pressure each fills the same $dm324\,\text{dm}^3$ .

Converting gas volume and moles using 24 dm³

To find a gas volume, multiply the amount in moles by $24$ : $(mol)×24\text{volume (dm}^3\text{)} = \text{amount (mol)} \times 24$
To find an amount in moles, divide the gas volume in $dm3\text{dm}^3$ $dm^{3}$ by $24$ .
1. For a volume in $cm3\text{cm}^3$ , divide by $00024\,000$ instead, or convert it into $dm3\text{dm}^3$ first.
So $mol0.25\,\text{mol}$ of carbon dioxide occupies $0.25 \times 24 = 6.0\,\text{dm}^3$ at room temperature and pressure.
The value of $dm324\,\text{dm}^3$ applies only at room temperature and pressure, so a different temperature or pressure gives a different volume.

Example

Calculate the amount in moles of oxygen in $cm3480\,\text{cm}^3$ of the gas at room temperature and pressure.
Convert the volume: $dm3480\,\text{cm}^3 \div 1000 = 0.480\,\text{dm}^3$ .
Amount $\dfrac{0.480}{24} = 0.0200\,\text{mol}$ .

Balancing numbers give the ratio of gas volumes

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

The balancing numbers give the mole ratio, and equal amounts of gases have equal volumes, so for gases they also give the volume ratio.
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O(l)$ , the volumes of methane, oxygen and carbon dioxide are in the ratio $1 : 2 : 1$ .
1. So $cm350\,\text{cm}^3$ of methane reacts with $cm3100\,\text{cm}^3$ of oxygen and makes $cm350\,\text{cm}^3$ of carbon dioxide.
Water is a liquid at room temperature, so it is left out of the gas volume ratio.
In $2H2(g)+O2(g)→2H2O(l)2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$ , $cm340\,\text{cm}^3$ of hydrogen needs half its volume, $cm320\,\text{cm}^3$ , of oxygen.

Common Mistake

Do not use the balancing numbers as a volume ratio for solids, liquids or solutions, because the rule applies only to gases measured at the same temperature and pressure.

Finding gas volume from reactant mass via moles

First convert the mass of reactant into moles by dividing by its $M_r$ .
Next use the mole ratio from the balanced equation to find the amount in moles of gas formed.
Then multiply the amount of gas by $24$ to find its volume in $dm3\text{dm}^3$ at room temperature and pressure.
Give the answer to a sensible number of significant figures, usually matching the data in the question.

Example

Calculate the volume of carbon dioxide at room temperature and pressure when $g5.00\,\text{g}$ of calcium carbonate is heated: $CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
$Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ , so the amount $\dfrac{5.00}{100.1} = 0.04995\,\text{mol}$ .
The ratio $CaCO3:CO2\text{CaCO}_3 : \text{CO}_2$ is $1 : 1$ , so $mol0.04995\,\text{mol}$ of carbon dioxide forms.
Volume $\times 24 = 1.20\,\text{dm}^3$ .

Working back from gas volume to reactant mass

Divide the gas volume in $dm3\text{dm}^3$ by $24$ to find the amount of gas in moles.
Use the mole ratio to find the amount in moles of the reactant.
Multiply by the reactant's $A_{r}$ or $M_{r}$ to find its mass in grams.
1. In $Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ , $cm3180\,\text{cm}^3$ of hydrogen is $mol\dfrac{0.180}{24} = 0.00750\,\text{mol}$ .
2. The ratio $Mg:H2\text{Mg} : \text{H}_2$ is $1 : 1$ , so the mass of magnesium that reacted is $0.00750 \times 24.3 = 0.182\,\text{g}$ .

Exam technique

When a question gives a gas volume in $cm3\text{cm}^3$ , convert it into $dm3\text{dm}^3$ before dividing by $24$ , and show that step, because it is the step most often missed.

Self review

What volume does one mole of any gas occupy at room temperature and pressure?
What is the volume of $mol0.150\,\text{mol}$ of nitrogen at room temperature and pressure?
How many moles of gas are in $cm31200\,\text{cm}^3$ at room temperature and pressure?
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ , what volume of oxygen reacts with $cm330\,\text{cm}^3$ of methane?
Why is water left out of the volume ratio in this equation?

Equal moles of any gas occupy equal volumes

Definition

Molar gas volume

The volume occupied by one mole of any gas at a stated temperature and pressure, which is 24 dm³ at room temperature and pressure.

Gas particles are very far apart, so the volume of a gas depends on the number of particles and not on the size of each particle.
At the same temperature and pressure, the volume of a gas is directly proportional to its amount in moles.
1. Doubling the amount of gas doubles its volume, and halving it halves the volume.
At room temperature and pressure, about $20 ∘C20\ ^{\circ}\text{C}$ $20^{\circ} C$ and 1 atmosphere, one mole of any gas occupies $24 dm324\,\text{dm}^3$ $24 dm^{3}$ .
1. This is the same as $cm324\,000\,\text{cm}^3$ , whether the gas is hydrogen, oxygen or carbon dioxide.

Graph showing that the volume of a gas is directly proportional to the number of moles (V = constant × n), represented by a straight line passing through the origin.

Key Idea

One mole of hydrogen weighs $g2.0\,\text{g}$ and one mole of carbon dioxide weighs $g44.0\,\text{g}$ , yet at room temperature and pressure each fills the same $dm324\,\text{dm}^3$ .

Converting gas volume and moles using 24 dm³

To find a gas volume, multiply the amount in moles by $24$ : $(mol)×24\text{volume (dm}^3\text{)} = \text{amount (mol)} \times 24$
To find an amount in moles, divide the gas volume in $dm3\text{dm}^3$ $dm^{3}$ by $24$ .
1. For a volume in $cm3\text{cm}^3$ , divide by $00024\,000$ instead, or convert it into $dm3\text{dm}^3$ first.
So $mol0.25\,\text{mol}$ of carbon dioxide occupies $0.25 \times 24 = 6.0\,\text{dm}^3$ at room temperature and pressure.
The value of $dm324\,\text{dm}^3$ applies only at room temperature and pressure, so a different temperature or pressure gives a different volume.

Example

Calculate the amount in moles of oxygen in $cm3480\,\text{cm}^3$ of the gas at room temperature and pressure.
Convert the volume: $dm3480\,\text{cm}^3 \div 1000 = 0.480\,\text{dm}^3$ .
Amount $\dfrac{0.480}{24} = 0.0200\,\text{mol}$ .

Balancing numbers give the ratio of gas volumes

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

The balancing numbers give the mole ratio, and equal amounts of gases have equal volumes, so for gases they also give the volume ratio.
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O(l)$ , the volumes of methane, oxygen and carbon dioxide are in the ratio $1 : 2 : 1$ .
1. So $cm350\,\text{cm}^3$ of methane reacts with $cm3100\,\text{cm}^3$ of oxygen and makes $cm350\,\text{cm}^3$ of carbon dioxide.
Water is a liquid at room temperature, so it is left out of the gas volume ratio.
In $2H2(g)+O2(g)→2H2O(l)2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$ , $cm340\,\text{cm}^3$ of hydrogen needs half its volume, $cm320\,\text{cm}^3$ , of oxygen.

Common Mistake

Do not use the balancing numbers as a volume ratio for solids, liquids or solutions, because the rule applies only to gases measured at the same temperature and pressure.

Finding gas volume from reactant mass via moles

First convert the mass of reactant into moles by dividing by its $M_r$ .
Next use the mole ratio from the balanced equation to find the amount in moles of gas formed.
Then multiply the amount of gas by $24$ to find its volume in $dm3\text{dm}^3$ at room temperature and pressure.
Give the answer to a sensible number of significant figures, usually matching the data in the question.

Example

Calculate the volume of carbon dioxide at room temperature and pressure when $g5.00\,\text{g}$ of calcium carbonate is heated: $CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
$Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ , so the amount $\dfrac{5.00}{100.1} = 0.04995\,\text{mol}$ .
The ratio $CaCO3:CO2\text{CaCO}_3 : \text{CO}_2$ is $1 : 1$ , so $mol0.04995\,\text{mol}$ of carbon dioxide forms.
Volume $\times 24 = 1.20\,\text{dm}^3$ .

Working back from gas volume to reactant mass

Divide the gas volume in $dm3\text{dm}^3$ by $24$ to find the amount of gas in moles.
Use the mole ratio to find the amount in moles of the reactant.
Multiply by the reactant's $A_{r}$ or $M_{r}$ to find its mass in grams.
1. In $Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ , $cm3180\,\text{cm}^3$ of hydrogen is $mol\dfrac{0.180}{24} = 0.00750\,\text{mol}$ .
2. The ratio $Mg:H2\text{Mg} : \text{H}_2$ is $1 : 1$ , so the mass of magnesium that reacted is $0.00750 \times 24.3 = 0.182\,\text{g}$ .

Exam technique

When a question gives a gas volume in $cm3\text{cm}^3$ , convert it into $dm3\text{dm}^3$ before dividing by $24$ , and show that step, because it is the step most often missed.

Self review

What volume does one mole of any gas occupy at room temperature and pressure?
What is the volume of $mol0.150\,\text{mol}$ of nitrogen at room temperature and pressure?
How many moles of gas are in $cm31200\,\text{cm}^3$ at room temperature and pressure?
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ , what volume of oxygen reacts with $cm330\,\text{cm}^3$ of methane?
Why is water left out of the volume ratio in this equation?

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Converting gas volume and moles using 24 dm³

Balancing numbers give the ratio of gas volumes

Finding gas volume from reactant mass via moles

Working back from gas volume to reactant mass

Review 5.1.5 Molar gas volume and volumes of gases in reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Converting gas volume and moles using 24 dm³

Balancing numbers give the ratio of gas volumes

Finding gas volume from reactant mass via moles

Working back from gas volume to reactant mass

Review 5.1.5 Molar gas volume and volumes of gases in reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Converting gas volume and moles using 24 dm³

Balancing numbers give the ratio of gas volumes

Finding gas volume from reactant mass via moles

Working back from gas volume to reactant mass

Review 5.1.5 Molar gas volume and volumes of gases in reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.5 Molar gas volume and volumes of gases in reactions Revision Guide

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Converting gas volume and moles using 24 dm³

Balancing numbers give the ratio of gas volumes

Finding gas volume from reactant mass via moles

Working back from gas volume to reactant mass

Review 5.1.5 Molar gas volume and volumes of gases in reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.5 Molar gas volume and volumes of gases in reactions Revision Guide