Skip to notes
wise otter.Chemistry
Contents
9476 / 2027
The Mole Concept and Stoichiometry

Lesson 4 of 5

Gas volumes and hydrocarbon combustion

Use volume ratios only when temperature and pressure match.

A-Level 9476 (2026-2027)

  • I can construct balanced chemical equations.Syllabus

    Syllabus 9476, 6(f). Construct balanced chemical equations.

  • I can calculate reacting amounts in solids, gases and solutions.Syllabus

    Syllabus 9476, 6(g). Calculate reacting amounts in solids, gases and solutions.

  • I can deduce stoichiometric relationships from measurements.Syllabus

    Syllabus 9476, 6(h). Deduce stoichiometric relationships from measurements.

At the same temperature and pressure, gas volumes follow mole ratios

Water may condense; do not count a liquid as part of the final gas volume.

For gases behaving ideally, V is proportional to n at fixed temperature and pressure. The balanced coefficients therefore also give gas-volume ratios under those conditions. If conditions differ, use pV = nRT first. Do not insert a memorised molar gas volume when the stated temperature and pressure require a different value.

Complete combustion of a hydrocarbon obeys CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O. The carbon dioxide amount gives x; oxygen consumption then gives y. For an oxygen-containing fuel CxHyOz, the oxygen coefficient becomes x + y/4 - z/2.

Worked exampleInfer a hydrocarbon from gas-volume changes

At the same temperature and pressure, 10.0 cm3 of a gaseous hydrocarbon burns completely using 35.0 cm3 O2 and producing 20.0 cm3 CO2. Find the formula.

Try it first. Then show the worked stepsHide the worked steps
  1. The CO2/fuel volume ratio is 2, so x = 2.
  2. The O2/fuel ratio is 3.5, so x + y/4 = 3.5.
  3. Substitute x = 2: y/4 = 1.5 and y = 6.
Answer

C2H6; 2C2H6 + 7O2 → 4CO2 + 6H2O.

Try it step by step

Work it out in steps: Carbon atoms, x, then Hydrogen atoms, y, then Formula.

  1. Problem 1. At the same temperature and pressure, 10.0 cm3 of a gaseous hydrocarbon CxHy burns completely in 50.0 cm3 of O2 and gives 30.0 cm3 of CO2. Use CxHy + (x + y/4)O2 -> xCO2 + (y/2)H2O. Find x and y as whole numbers.
  2. Problem 2. At the same temperature and pressure, 20.0 cm3 of a gaseous hydrocarbon burns completely in 60.0 cm3 of O2 and gives 40.0 cm3 of CO2. Find x and y as whole numbers, then the formula.
  3. Problem 3. At the same temperature and pressure, 15.0 cm3 of a gaseous hydrocarbon burns completely in 97.5 cm3 of O2 and gives 60.0 cm3 of CO2. Find the number of carbon and hydrogen atoms as whole numbers, then the formula.
Show the answer

Problem 1.

Carbon atoms, x: CO2 / fuel = 30.0/10.0 = 3, so x = 3.

Hydrogen atoms, y: O2 / fuel = 50.0/10.0 = 5 = x + y/4, so y/4 = 2 and y = 8.

Formula: The hydrocarbon is C3H8, propane.

Problem 2.

Carbon atoms, x: CO2 / fuel = 40.0/20.0 = 2, so x = 2.

Hydrogen atoms, y: O2 / fuel = 60.0/20.0 = 3 = 2 + y/4, so y = 4.

Formula: The hydrocarbon is C2H4, ethene.

Problem 3.

Carbon atoms, x: CO2 / fuel = 60.0/15.0 = 4, so x = 4.

Hydrogen atoms, y: O2 / fuel = 97.5/15.0 = 6.5 = 4 + y/4, so y/4 = 2.5 and y = 10.

Formula: The hydrocarbon is C4H10, butane.

Check your understanding

In the example, 10.0 cm3 of fuel burns with 50.0 cm3 of O2. After cooling so the water condenses, what dry gas volume remains?

  1. 65.0 cm3: the CO2, the excess O2 and the steam made
  2. 20.0 cm3: just the CO2, since the oxygen is used up
  3. 35.0 cm3: 15.0 cm3 of unused O2 plus 20.0 cm3 of CO2
Show the answer

35.0 cm3: 15.0 cm3 of unused O2 plus 20.0 cm3 of CO2

Measured at the original temperature and pressure, the dry gas is the excess O2 and the CO2 made. Condensed water is liquid, so it is not counted. Absorbing the CO2 would leave 15.0 cm3.