contents

Chemistry

5.14 - Gas Volume Calculations

10 min

Moles and gas volumes

Reacting gas volumes

Volume ratio shortcut

Why r.t.p. matters

note 1 of 6

The molar gas volume at r.t.p.

Avogadro's law - Equal volumes of any gases at the same temperature and pressure contain the same number of particles. So equal amounts (in moles) of any two gases occupy equal volumes.
r.t.p. (room temperature and pressure) means 20 °C and 1 atmosphere.
At r.t.p., one mole of any gas occupies 24 dm³ (or 24,000 cm³). This is the molar gas volume.

Formula for volume of gas

Volume of gas = amount of gas (mol) × 24 dm³ (or 24,000 cm³)

The 24 dm³ figure applies only at r.t.p. If the temperature or pressure changes, one mole of gas occupies a different volume.

note 2 of 6

Worked example - Volume of a gas from its mass

A reaction gives off 0.15 g of carbon dioxide, CO₂. What volume does this gas occupy at room temperature and pressure? Give your answer in cm³.

Step 1 - Find the relative formula mass of the gas

Relative atomic masses: C = 12, O = 16.

$M_{r} ({CO}_{2}) = 12 + (2 \times 16) = 44$

Step 2 - Convert the mass into an amount in moles

$\begin{matrix} Amount & = \frac{mass}{M_{r}} \\ = \frac{0.15}{44} \\ = 0.003409... mol \end{matrix}$

Step 3 - Convert the amount into a volume

Volume = amount × 24 = 0.003409... × 24 = 0.081818... dm³

Step 4 - Convert dm³ into cm³

0.081818... × 1,000 = 81.818... cm³

Step 5 - Round to significant figures

The mass was given to two significant figures, so round to two: 82 cm³.

note 3 of 6

Gas volumes in a reaction

Where a reaction gives off or uses up a gas, its volume can be calculated from the mass of another substance in the reaction using the balanced equation.

The three-step route

Example - sodium reacting with water:

2Na + 2H₂O → 2NaOH + H₂

Mass to amount - Divide the mass of the known substance by its relative formula mass to find its amount in moles.
Amount to amount - Use the mole ratio from the balancing numbers in the equation to find the amount of gas.
Amount to volume - Multiply the amount of gas by 24 for its volume in dm³ at r.t.p. (or by 24,000 for cm³).

Notes on the calculation:

The other reactant is often provided in excess to make sure all of the limiting reactant reacts.
The calculation also works in reverse: gas volume ÷ 24 gives amount in moles, the mole ratio gives the other substance's amount, and amount × $M_{r}$ gives its mass.

note 4 of 6

Worked example - Gas a reaction produces

0.050 g of aluminium is added to excess dilute hydrochloric acid.

2Al + 6HCl → 2AlCl₃ + 3H₂

What volume of hydrogen is produced at room temperature and pressure? Give your answer in dm³.

Step 1 - Mass to amount

Relative atomic mass: Al = 27.

$\begin{matrix} Amount of Al & = \frac{mass}{A_{r}} \\ = \frac{0.050}{27} \\ = 0.0018518... mol \end{matrix}$

Step 2 - Amount to amount

The equation shows 2 mol of Al produces 3 mol of H₂ (a 2 : 3 ratio):

$\begin{matrix} {Amount of H}_{2} & = 0.0018518... \times \frac{3}{2} \\ = 0.0027777... mol \end{matrix}$

Step 3 - Amount to volume

Volume = amount × 24 = 0.0027777... × 24 = 0.066666... dm³

Step 4 - Round to significant figures

The mass of aluminium was given to two significant figures, so give the answer to two: 0.067 dm³.

note 5 of 6

Volume ratios in an all-gas reaction

When all reactants and products being compared are gases at the same temperature and pressure, their volumes react in the same ratio as the balancing numbers in the equation.
There's no need to calculate amounts in moles or use the molar gas volume - the given volume is scaled directly by the balancing ratio.

The two-step method

Example - ammonia burning in oxygen:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Find the ratio from the equation - Use the balancing numbers in front of the given gas and the wanted gas.
Scale the volume - Multiply by the balancing number of the gas wanted, and divide by the balancing number of the gas given.

Conditions

Every substance compared must be a gas - If any substance is a solid, liquid, or aqueous solution, you must calculate using moles.
All volumes must be at the same temperature and pressure.

note 6 of 6

Worked example - Gas volume from a volume ratio

8.00 cm³ of butane, C₄H₁₀, is burned in oxygen. All of the butane reacts.

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)

What volume of oxygen does the butane react with, measured at the same temperature and pressure?

Step 1 - Read the ratio from the equation

The equation shows 2 volumes of C₄H₁₀ react with 13 volumes of O₂.

Step 2 - Scale the volume by the ratio

$\begin{matrix} {Volume of O}_{2} & = 8.00 \times \frac{13}{2} \\ = 52.0 {cm}^{3} \end{matrix}$

key terms

TermDefinition

Avogadro's lawThe law that equal volumes of any gases at the same temperature and pressure contain the same number of particles, so equal amounts in moles of gases occupy equal volumes.

Room temperature and pressure (r.t.p.)A temperature of 20 °C and a pressure of 1 atmosphere.

Molar gas volumeThe volume occupied by one mole of any gas at room temperature and pressure, which is 24 dm³.

end of notes

Key points to remember

contents

Chemistry

5.14 - Gas Volume Calculations

10 min

Moles and gas volumes

Reacting gas volumes

Volume ratio shortcut

Why r.t.p. matters

note 1 of 6

The molar gas volume at r.t.p.

Avogadro's law - Equal volumes of any gases at the same temperature and pressure contain the same number of particles. So equal amounts (in moles) of any two gases occupy equal volumes.
r.t.p. (room temperature and pressure) means 20 °C and 1 atmosphere.
At r.t.p., one mole of any gas occupies 24 dm³ (or 24,000 cm³). This is the molar gas volume.

Formula for volume of gas

Volume of gas = amount of gas (mol) × 24 dm³ (or 24,000 cm³)

The 24 dm³ figure applies only at r.t.p. If the temperature or pressure changes, one mole of gas occupies a different volume.

note 2 of 6

Worked example - Volume of a gas from its mass

A reaction gives off 0.15 g of carbon dioxide, CO₂. What volume does this gas occupy at room temperature and pressure? Give your answer in cm³.

Step 1 - Find the relative formula mass of the gas

Relative atomic masses: C = 12, O = 16.

$M_{r} ({CO}_{2}) = 12 + (2 \times 16) = 44$

Step 2 - Convert the mass into an amount in moles

$\begin{matrix} Amount & = \frac{mass}{M_{r}} \\ = \frac{0.15}{44} \\ = 0.003409... mol \end{matrix}$

Step 3 - Convert the amount into a volume

Volume = amount × 24 = 0.003409... × 24 = 0.081818... dm³

Step 4 - Convert dm³ into cm³

0.081818... × 1,000 = 81.818... cm³

Step 5 - Round to significant figures

The mass was given to two significant figures, so round to two: 82 cm³.

note 3 of 6

Gas volumes in a reaction

Where a reaction gives off or uses up a gas, its volume can be calculated from the mass of another substance in the reaction using the balanced equation.

The three-step route

Example - sodium reacting with water:

2Na + 2H₂O → 2NaOH + H₂

Mass to amount - Divide the mass of the known substance by its relative formula mass to find its amount in moles.
Amount to amount - Use the mole ratio from the balancing numbers in the equation to find the amount of gas.
Amount to volume - Multiply the amount of gas by 24 for its volume in dm³ at r.t.p. (or by 24,000 for cm³).

Notes on the calculation:

The other reactant is often provided in excess to make sure all of the limiting reactant reacts.
The calculation also works in reverse: gas volume ÷ 24 gives amount in moles, the mole ratio gives the other substance's amount, and amount × $M_{r}$ gives its mass.

note 4 of 6

Worked example - Gas a reaction produces

0.050 g of aluminium is added to excess dilute hydrochloric acid.

2Al + 6HCl → 2AlCl₃ + 3H₂

What volume of hydrogen is produced at room temperature and pressure? Give your answer in dm³.

Step 1 - Mass to amount

Relative atomic mass: Al = 27.

$\begin{matrix} Amount of Al & = \frac{mass}{A_{r}} \\ = \frac{0.050}{27} \\ = 0.0018518... mol \end{matrix}$

Step 2 - Amount to amount

The equation shows 2 mol of Al produces 3 mol of H₂ (a 2 : 3 ratio):

$\begin{matrix} {Amount of H}_{2} & = 0.0018518... \times \frac{3}{2} \\ = 0.0027777... mol \end{matrix}$

Step 3 - Amount to volume

Volume = amount × 24 = 0.0027777... × 24 = 0.066666... dm³

Step 4 - Round to significant figures

The mass of aluminium was given to two significant figures, so give the answer to two: 0.067 dm³.

note 5 of 6

Volume ratios in an all-gas reaction

When all reactants and products being compared are gases at the same temperature and pressure, their volumes react in the same ratio as the balancing numbers in the equation.
There's no need to calculate amounts in moles or use the molar gas volume - the given volume is scaled directly by the balancing ratio.

The two-step method

Example - ammonia burning in oxygen:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Find the ratio from the equation - Use the balancing numbers in front of the given gas and the wanted gas.
Scale the volume - Multiply by the balancing number of the gas wanted, and divide by the balancing number of the gas given.

Conditions

Every substance compared must be a gas - If any substance is a solid, liquid, or aqueous solution, you must calculate using moles.
All volumes must be at the same temperature and pressure.

note 6 of 6

Worked example - Gas volume from a volume ratio

8.00 cm³ of butane, C₄H₁₀, is burned in oxygen. All of the butane reacts.

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)

What volume of oxygen does the butane react with, measured at the same temperature and pressure?

Step 1 - Read the ratio from the equation

The equation shows 2 volumes of C₄H₁₀ react with 13 volumes of O₂.

Step 2 - Scale the volume by the ratio

$\begin{matrix} {Volume of O}_{2} & = 8.00 \times \frac{13}{2} \\ = 52.0 {cm}^{3} \end{matrix}$

key terms

TermDefinition

Avogadro's lawThe law that equal volumes of any gases at the same temperature and pressure contain the same number of particles, so equal amounts in moles of gases occupy equal volumes.

Room temperature and pressure (r.t.p.)A temperature of 20 °C and a pressure of 1 atmosphere.

Molar gas volumeThe volume occupied by one mole of any gas at room temperature and pressure, which is 24 dm³.

end of notes

5.14 - Gas Volume Calculations

The molar gas volume at r.t.p.

Formula for volume of gas

Worked example - Volume of a gas from its mass

Step 1 - Find the relative formula mass of the gas

Step 2 - Convert the mass into an amount in moles

Step 3 - Convert the amount into a volume

Step 4 - Convert dm3 into cm3

Step 5 - Round to significant figures

Gas volumes in a reaction

The three-step route

Notes on the calculation:

Worked example - Gas a reaction produces

Step 1 - Mass to amount

Step 2 - Amount to amount

Step 3 - Amount to volume

Step 4 - Round to significant figures

Volume ratios in an all-gas reaction

The two-step method

Conditions

Worked example - Gas volume from a volume ratio

Step 1 - Read the ratio from the equation

Step 2 - Scale the volume by the ratio

Key points to remember

5.14 - Gas Volume Calculations

The molar gas volume at r.t.p.

Formula for volume of gas

Worked example - Volume of a gas from its mass

Step 1 - Find the relative formula mass of the gas

Step 2 - Convert the mass into an amount in moles

Step 3 - Convert the amount into a volume

Step 4 - Convert dm3 into cm3

Step 5 - Round to significant figures

Gas volumes in a reaction

The three-step route

Notes on the calculation:

Worked example - Gas a reaction produces

Step 1 - Mass to amount

Step 2 - Amount to amount

Step 3 - Amount to volume

Step 4 - Round to significant figures

Volume ratios in an all-gas reaction

The two-step method

Conditions

Worked example - Gas volume from a volume ratio

Step 1 - Read the ratio from the equation

Step 2 - Scale the volume by the ratio

Key points to remember

Step 4 - Convert dm³ into cm³

Step 4 - Convert dm³ into cm³