One mole of any gas occupies the same volume under the same conditions. At room temperature and pressure, questions normally give this as 24 dm³, which is 24 000 cm³, per mole.
This lesson sits inside concentration, gas and yield calculations. It reuses the moles-first chain from solution volumes, but with a gas volume in place of a solution volume.
Why can volume stand in for moles?
At a fixed temperature and pressure, the space a gas takes up depends on how many particles it has, not on what kind they are. So 1 mol of hydrogen and 1 mol of carbon dioxide occupy the same volume under the same conditions.
That gives two conversions. Volume in dm³ = moles × 24, and moles = volume in dm³ ÷ 24. If the volume is in cm³, divide by 1000 first, or use 24 000 cm³ per mole.
What are the steps?
- Check the conditions. Note the molar volume the question supplies and use that number.
- Find the moles of the substance you know, using mass ÷ molar mass or concentration × volume.
- Use the equation ratio to get moles of the gas.
- Convert moles of gas to volume by multiplying by 24 dm³/mol.
- Give the unit the question asks for. Multiply dm³ by 1000 for cm³.
Worked example
The numbers are invented for practice. 0.120 g of magnesium reacts completely with excess dilute hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen forms at room temperature and pressure? Take the molar volume as 24 dm³/mol.
Step 1, moles of Mg: Aᵣ of Mg = 24, so moles = 0.120 ÷ 24 = 0.00500 mol.
Step 2, ratio: Mg : H₂ is 1 : 1, so moles of H₂ = 0.00500 mol.
Step 3, volume: 0.00500 × 24 = 0.120 dm³.
Step 4, in cm³: 0.120 × 1000 = 120 cm³.
Check: 0.00500 mol × 24 000 cm³/mol = 120 cm³. Both routes agree.
What is the mistake to watch for?
A frequent slip is to ignore the equation ratio because the question is about gas.
Mistaken working: 0.0200 mol of hydrogen peroxide decomposes: 2H₂O₂ → 2H₂O + O₂. The student wrote moles of O₂ = 0.0200 mol.
The ratio H₂O₂ : O₂ is 2 : 1, not 1 : 1.
The correct working is 0.0200 ÷ 2 = 0.0100 mol of O₂, which is 0.0100 × 24 = 0.240 dm³, or 240 cm³. Write the ratio as two numbers on their own line every time, even when it looks like 1 : 1.
Check yourself
Use 24 dm³/mol for each question, then open the answers.
1. What volume, in cm³, does 0.0300 mol of carbon dioxide occupy at room temperature and pressure?
Show answer
0.0300 × 24 = 0.720 dm³, which is 720 cm³.
2. A sample of gas has a volume of 48 cm³ at room temperature and pressure. How many moles is that?
Show answer
48 ÷ 1000 = 0.048 dm³. Moles = 0.048 ÷ 24 = 0.0020 mol.
3. In 2H₂ + O₂ → 2H₂O, 60 cm³ of hydrogen reacts completely with oxygen, both gases measured under the same conditions. What volume of oxygen is needed?
Show answer
H₂ : O₂ is 2 : 1, and equal conditions mean volumes follow the same ratio. So oxygen = 60 ÷ 2 = 30 cm³.
Where does this lead next?
Next, calculate a percentage yield, where you compare a measured result with the amount the equation predicts. The mole and equation-ratio tutor can help you check a ratio step, and the equation balance reasoning trainer confirms the equation you start from.
Gas questions tend to fail at the ratio step rather than the volume step, which is easy to miss when you mark your own work. A teacher in online one-to-one Chemistry tuition can read your working line by line and point to exactly where it drifted.