Moles and Gas Volumes [H]

This lesson covers the relationship between moles and gas volumes as required by the AQA GCSE Chemistry specification (4.3.3). This is Higher Tier only content. You need to understand the concept of molar volume, be able to calculate the volume of a gas from the number of moles (and vice versa), and use this in reacting mass and gas volume calculations.

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Molar Volume of a Gas

At room temperature and pressure (RTP) — defined as 20 degrees Celsius and 1 atmosphere (1 atm) — one mole of any gas occupies a volume of 24 dm3 (or 24,000 cm3). This is called the molar volume.

Condition	Temperature	Pressure	Molar Volume
RTP (room temperature and pressure)	20 degrees C	1 atm	24 dm3/mol (24,000 cm3/mol)

This value applies to all gases, regardless of the identity of the gas. One mole of hydrogen (H2), one mole of oxygen (O2), one mole of carbon dioxide (CO2), and one mole of chlorine (Cl2) all occupy 24 dm3 at RTP.

Exam Tip: The value of the molar volume (24 dm3 at RTP) will be given to you in the exam. You do not need to memorise it, but you must know how to use it. The key formula is: volume = moles x 24.

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Key Equations

The relationship between moles and gas volume at RTP is:

Volume of gas (dm3) = moles x 24

Moles = volume of gas (dm3) / 24

If the volume is given in cm3:

Volume of gas (cm3) = moles x 24,000

Moles = volume of gas (cm3) / 24,000

Formula	When to Use
volume (dm3) = moles x 24	To find the volume of a gas in dm3 from moles
moles = volume (dm3) / 24	To find moles from volume in dm3
volume (cm3) = moles x 24,000	To find the volume of a gas in cm3 from moles
moles = volume (cm3) / 24,000	To find moles from volume in cm3

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Worked Examples: Moles to Volume

Example 1

Calculate the volume, in dm3, of 0.5 moles of carbon dioxide at RTP.

• Volume = moles x 24 = 0.5 x 24 = 12 dm3

Example 2

Calculate the volume, in cm3, of 0.1 moles of hydrogen gas at RTP.

• Volume = moles x 24,000 = 0.1 x 24,000 = 2,400 cm3

Example 3

What volume (in dm3) does 2 moles of oxygen gas occupy at RTP?

• Volume = 2 x 24 = 48 dm3

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Worked Examples: Volume to Moles

Example 4

How many moles of gas are in 6 dm3 of nitrogen at RTP?

• Moles = volume / 24 = 6 / 24 = 0.25 mol

Example 5

How many moles are in 480 cm3 of chlorine gas at RTP?

• Moles = volume / 24,000 = 480 / 24,000 = 0.02 mol

Exam Tip: Always check whether the volume is given in dm3 or cm3. If in dm3, divide by 24. If in cm3, divide by 24,000. Getting this wrong is the most common mistake in gas volume calculations.

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Gas Volume Calculation Flow

graph TD
    A["Start: Identify what you know"] --> B{"Volume or Moles?"}
    B -->|"Know moles, find volume"| C["volume = moles x 24 dm3"]
    B -->|"Know volume, find moles"| D["moles = volume / 24 dm3"]
    C --> E{"Units needed?"}
    D --> F["Use moles in further calculations"]
    E -->|"dm3"| G["Answer in dm3"]
    E -->|"cm3"| H["Multiply by 1000 to get cm3"]

    style A fill:#3498db,color:#fff
    style B fill:#e67e22,color:#fff
    style C fill:#27ae60,color:#fff
    style D fill:#27ae60,color:#fff

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Combining Gas Volumes with Reacting Masses

You can combine gas volume calculations with reacting mass calculations. The method is:

1. Write the balanced equation.
2. Calculate moles of the known substance (from mass or volume).
3. Use the mole ratio to find moles of the gas.
4. Convert moles of gas to volume using volume = moles x 24.

Worked Example 6

What volume of hydrogen gas (in dm3) is produced when 4.8 g of magnesium reacts with excess hydrochloric acid at RTP?

Balanced equation: Mg + 2HCl --> MgCl2 + H2

Step 1: moles of Mg = 4.8 / 24 = 0.2 mol

Step 2: From the equation, Mg : H2 = 1 : 1, so moles of H2 = 0.2 mol

Step 3: Volume of H2 = 0.2 x 24 = 4.8 dm3

Worked Example 7

What volume of carbon dioxide (in cm3) is produced when 5 g of calcium carbonate reacts with excess hydrochloric acid at RTP?

Balanced equation: CaCO3 + 2HCl --> CaCl2 + H2O + CO2

Step 1: Mr of CaCO3 = 100. moles = 5 / 100 = 0.05 mol

Step 2: From the equation, CaCO3 : CO2 = 1 : 1, so moles of CO2 = 0.05 mol

Step 3: Volume of CO2 = 0.05 x 24,000 = 1,200 cm3

Exam Tip: When a question gives you a mass and asks for a volume of gas, you need to go through moles as an intermediate step. The sequence is always: mass --> moles --> volume. Never try to skip the moles step.

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Calculating Mass from Gas Volume

You can also work backwards — from the volume of a gas to the mass of a solid reactant or product.

Worked Example 8

240 cm3 of hydrogen gas is collected at RTP from the reaction of zinc with hydrochloric acid. What mass of zinc reacted?

Balanced equation: Zn + 2HCl --> ZnCl2 + H2

Step 1: moles of H2 = 240 / 24,000 = 0.01 mol

Step 2: From the equation, Zn : H2 = 1 : 1, so moles of Zn = 0.01 mol

Step 3: mass of Zn = 0.01 x 65 = 0.65 g

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Gas Volume Ratios

When comparing volumes of gases at the same temperature and pressure, the volume ratio is the same as the mole ratio in the balanced equation.

For example: N2 + 3H2 --> 2NH3