Skip to content

Ideal gas equation notes and lesson for A Level chemistry (free PDF download)

  • by
Ideal gas equation notes and lesson for A Level chemistry (free PDF download)

This chemistry grades 11-12 lesson on ideal gas equation notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on ideal gas equation and check yourself with the quiz.

Quick answer: The amount of substance, in mol, links mass, particle number, solution volume and gas volume. Use n = m/M, n = cV and pV = nRT with SI units, find empirical formulae from mole ratios, and judge a reaction’s efficiency with atom economy, which compares the useful product’s mass with all products.

Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.1.2

Ideal gas equation notes: the key ideas

How do you calculate the amount of substance?

The mole links the masses we weigh to the particles that react. All amount calculations start from one quantity, n, measured in mol.

Definition

One mole is the amount of substance that contains 6.02 × 10²³ particles (the Avogadro constant). The molar mass M is the mass of one mole in g mol⁻¹, equal numerically to the relative formula mass.

Formula

n = m / M   (m in g, M in g mol⁻¹)
n = c × V   (c in mol dm⁻³, V in dm³)
n = N / 6.02 × 10²³   (N = number of particles)

Map with amount of substance in mol at the centre linked to mass, number of particles, solution concentration and gas pressure volume temperatureamount nin molmass m / gn = m / Mparticles Nn = N / 6.02 × 10²³solutionn = c × V (dm³)gasn = pV / RT
Every route to the mole: mass, particles, solutions and gases.
Worked example 1

A farmer in Kenya dissolves 8.40 g of sodium hydrogencarbonate, NaHCO₃, in 250 cm³ of water. Find its concentration.
M = 23.0 + 1.0 + 12.0 + 48.0 = 84.0 g mol⁻¹. n = 8.40 ÷ 84.0 = 0.100 mol.
c = 0.100 mol ÷ 0.250 dm³ = 0.400 mol dm⁻³.

How do you find an empirical and a molecular formula?

Definition

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms in one molecule, a whole-number multiple of the empirical formula.

Flow chart from masses or percentages through moles and ratio to the empirical formula and then the molecular formulaMasses orpercentagesDivide byAr: molesDivide by thesmallestWhole-numberratio: empiricalScale byM ÷ M(emp)Example: 40.0 % C, 6.7 % H, 53.3 % O → CH₂O → C₆H₁₂O₆ (M = 180)
Five steps from composition data to the molecular formula.
Worked example 2

A sugar contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen by mass, with M = 180 g mol⁻¹.
In 100 g: C = 40.0 ÷ 12.0 = 3.33 mol; H = 6.7 ÷ 1.0 = 6.7 mol; O = 53.3 ÷ 16.0 = 3.33 mol.
Divide by 3.33: C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O (mass 30.0).
180 ÷ 30.0 = 6, so the molecular formula is C₆H₁₂O₆.

Common mistake

If a ratio comes out as 1 : 1.5, do not round to 1 : 2. Multiply every number by 2 to get 2 : 3. Only round when the value is within about 0.05 of a whole number.

How does the ideal gas equation work?

An ideal gas follows the equation pV = nRT. At a fixed pressure, doubling the kelvin temperature doubles the volume, as the graph shows.

Graph of gas volume against kelvin temperature, a straight line through the origin reaching 1.66 dm cubed at 400 kelvin010020030040001234temperature T / Kvolume V / dm³1.66 dm³ at 400 K0.0500 mol at 100 kPa
At constant pressure and amount, V is proportional to T (kelvin).
Formula

pV = nRT
p = pressure in Pa, V = volume in m³, n = amount in mol, T = temperature in K, R = 8.31 J K⁻¹ mol⁻¹.

ConvertRule
°C to Kadd 273
cm³ to m³multiply by 10⁻⁶
dm³ to m³multiply by 10⁻³
kPa to Pamultiply by 10³
Worked example 3

Find the volume of 0.250 mol of gas at 27 °C and 101 kPa.
T = 300 K, p = 101 000 Pa.
V = nRT ÷ p = (0.250 × 8.31 × 300) ÷ 101 000 = 6.17 × 10⁻³ m³ = 6.17 dm³.

What is atom economy?

Definition

Atom economy = (relative formula mass of the desired product ÷ sum of the relative formula masses of all products) × 100 %. It measures how much of the reactant mass ends up in the useful product.

Fermentation, C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, gives (2 × 46.0) ÷ 180.0 × 100 = 51.1 %, because carbon dioxide is also produced. Atom economy differs from percentage yield, which compares the actual mass obtained with the maximum possible mass.

Lina’s tip

Write the unit of every number as you substitute. If the units of an ideal gas problem do not give m³ at the end, you forgot to convert cm³, dm³ or kPa.

Key points to remember
  • n = m / M; n = c × V; n = N / 6.02 × 10²³
  • pV = nRT with p in Pa, V in m³, T in K, R = 8.31 J K⁻¹ mol⁻¹
  • Empirical formula: divide masses by Ar, then by the smallest value
  • Molecular formula = empirical formula × (M ÷ empirical mass)
  • Atom economy = Mr desired product ÷ Mr all products × 100 %

Moles, formulae, gas volumes and atom economy: frequently asked questions

What is the difference between empirical and molecular formula?

The empirical formula is the simplest whole-number ratio of atoms, such as CH₂. The molecular formula is the real number of atoms in a molecule, such as C₄H₈. You need the molar mass to go from one to the other.

Why must temperature be in kelvin in pV = nRT?

The volume of a gas is proportional to the absolute temperature. A 0 °C value would give zero volume, which is wrong. Always add 273 to the Celsius value.

Why must volume be in m³ and pressure in Pa?

The gas constant R = 8.31 J K⁻¹ mol⁻¹ uses SI units. One pascal times one cubic metre equals one joule, so the units cancel correctly only in Pa and m³.

Does a high atom economy mean a high yield?

No. Atom economy compares the masses in the balanced equation, whatever happens in the real experiment. A reaction can have a perfect atom economy yet a low percentage yield if it is incomplete or if product is lost.

How do you get the number of atoms in a sample?

Find the amount in mol, then multiply by 6.02 × 10²³. For atoms inside molecules, multiply by the number of atoms of that element in each formula unit.

Physics and chemistry textbooks to download in PDF, from grade 6 to AP Physics and AP Chemistry

Follow us