Moles and Gas Volumes
This topic covers the relationship between the amount of an ideal gas in moles and the volume it occupies at a specific temperature and pressure. It is a fundamental concept for solving stoichiometry problems that involve gaseous reactants or products.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- One mole of any ideal gas occupies the same volume as one mole of any other ideal gas under identical conditions of temperature and pressure.
- At Room Temperature and Pressure (rtp), this volume is 24 dm³ (or 24,000 cm³). This value will be provided if needed.
- The relationship allows for direct conversion between the volume of a gas and the number of moles, which can then be used in stoichiometric calculations.
- For reactions involving only gases at constant temperature and pressure, the ratio of their volumes is equal to the ratio of their moles in the balanced chemical equation.
Formulae
moles of gas = volume / molarvolume Use this to calculate the amount of gas (in moles) from a given volume, or vice versa, provided the conditions (and thus the molar volume) are known.
Definitions
- Molar Gas Volume
- The volume occupied by one mole of an ideal gas at a specified temperature and pressure. At rtp, this is 24 dm³mol⁻¹.
- Room Temperature and Pressure (rtp)
- A standard set of conditions used for comparing gas properties, typically defined as around 20-25°C and 1 atmosphere of pressure.
Worked example
Ammonia is produced by the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). What mass of nitrogen gas is required to produce 4.8 dm³ of ammonia, assuming the reaction goes to completion at room temperature and pressure? (Molar gas volume at rtp = 24 dm³mol⁻¹; Ar values: N = 14)
- 1
Step 1:
Calculate the moles of ammonia (NH₃) produced.
Moles = volume / molar volume = 4.8 dm³ / 24 dm³mol⁻¹ - 2
Step 2:
Simplify the fraction:
4.8 / 24 = 48 / 240 = 1/5 = 0.2 mol of NH₃ - 3
Step 3:
Use the stoichiometry from the balanced equation to find the moles of nitrogen (N₂).
The ratio N₂ :
NH₃ is 1:2.
Therefore, moles of N₂ = 0.2 mol / 2 = 0.1 mol.
- 4
Step 4:
Calculate the molar mass of N₂.
Mr(N₂) = 2 × 14 = 28 g/mol - 5
Step 5:
Calculate the mass of N₂ required.
Mass = moles × Mr = 0.1 mol × 28 g/mol
Answer: 2.8 g
Common mistakes
- ×Applying the molar volume of 24 dm³mol⁻¹ when the question specifies conditions other than rtp. This value is not universal.
- ×Making unit conversion errors between cm³ and dm³. Remember 1 dm³ = 1000 cm³. Always ensure your volume and molar volume units match before calculating.
- ×Incorrectly applying the volume-to-mole relationship for substances that are not in the gaseous state in the reaction.
No-calculator tips
- ✓Before dividing, check if the volume given is a simple multiple or fraction of the molar volume (e.g., 24 dm³). For example, 6 dm³ is 1/4 of 24 dm³, so it's 0.25 moles.
- ✓Simplify fractions before converting to decimals. For instance, 3.6 / 24 becomes 36 / 240. Both are divisible by 12, giving 3 / 20, which is easier to work with (0.15).
- ✓For volumes given in cm³, dividing by 1000 to get dm³ (e.g., 1200 cm³ → 1.2 dm³) often creates numbers that are easier to handle in subsequent fraction simplification.