4CH1

Chemical Formulae, Equations and Calculations

Principles of Chemistry · 6 question types

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 17% of your exam marks.

stable
Very High
Stable17%

Highest-frequency topic: moles, percentage yield and titration calculations appear on nearly every paper.

Avogadro's law

  • Avogadro's law states that, at the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules
  • Consequence: at the same temperature and pressure, the ratio of any two gases is the same as the volume ratio of those gases

Molar gas volume

  • RTP stands for room temperature and pressure, taken as 20 °C and 1 atm
  • At RTP, one mole of any gas occupies a volume of 24 dm³ (equivalently 24 000 cm³)
  • This value is called the molar gas volume at RTP

The key formula

  • For a gas at RTP:

n=V24(V in dm3)n = \frac{V}{24} \qquad (V \text{ in dm}^3)

  • or equivalently, when V is given in cm³:

n=V24000(V in cm3)n = \frac{V}{24\,000} \qquad (V \text{ in cm}^3)

  • Rearranged:
    • volume (dm³) = moles × 24
    • volume (cm³) = moles × 24 000
Formula triangle linking gas volume in dm³, amount of gas in moles and the molar gas volume 24 dm³/mol at RTP
Source: Calculate Gas Volumes by Save My Exams

Example. What volume does 88 g of CO₂ occupy at RTP?

  • Mr(CO₂) = 44, so moles = 88 / 44 = 2.0 mol
  • Volume = 2.0 × 24 = 48 dm³

Example. Methane burns completely:

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

What volume of oxygen reacts with 100 cm³ of methane, and what total volume of gas products forms? All volumes are measured at the same temperature and pressure.

  • Mole ratio CH₄ : O₂ = 1 : 2, so volume of O₂ needed = 2 × 100 = 200 cm³
  • Total moles of products per mole of CH₄ = 1 (CO₂) + 2 (H₂O) = 3
  • Total volume of products = 3 × 100 = 300 cm³
Worked example

Gas volume at RTP

Calculate the volume of carbon dioxide, measured at RTP, produced when 25 g of calcium carbonate decomposes completely.

CaCO3 → CaO + CO2  (Mr of CaCO3 = 100; molar gas volume = 24 dm³/mol at RTP)

Solution:

  • Moles of CaCO3 = mass ÷ Mr = 25 ÷ 100 = 0.25 mol
  • The mole ratio CaCO3 : CO2 is 1 : 1, so moles of CO2 = 0.25 mol
  • Volume = moles × 24 = 0.25 × 24 = 6 dm³

Mixing units is the main source of dropped marks: use 24 dm³/mol for an answer in dm³, or 24 000 cm³/mol for cm³. If the question wanted cm³, multiply the dm³ answer by 1000 (6 dm³ = 6000 cm³).