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1.5.7Principles of Chemistry

Practical- Determine the Formula of a Metal Oxide — Edexcel International GCSE Science Double Award 4SD0 - Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 30 of 50

📚 Edexcel International GCSE Science Double Award 4SD0 - Chemistry Hub / Topical Revision
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Part 1 of 5Combustion of Magnesium

1.5.7 Practical- Determine the Formula of a Metal Oxide

Burning magnesium in a crucible gives a metal oxide whose mass is measured before and after the reaction.

1
Weigh an empty crucible and lid, add a small piece of magnesium ribbon, and weigh again to find the mass of the metal.
2
Heat strongly, lifting the lid frequently so that air can reach the metal, but without letting smoke of the oxide escape.
3
Continue heating until the mass stops increasing, which shows that all of the magnesium has reacted with oxygen.
4
Allow the crucible to cool, weigh it, and subtract the mass of the empty crucible to find the mass of the oxide.

Part 2 of 5Key Measurements

The formula depends on the masses of metal and oxygen, so each one must be found by subtraction.

Mass of metal
  • Found by subtracting the mass of the crucible from the combined mass of the crucible and the metal at the start.
Mass of oxygen
  • Found by subtracting the mass of the metal from the mass of the metal oxide, since the oxygen has combined with the metal.
Reduction with methane
  • Heating copper(II) oxide in a stream of methane removes the oxygen, leaving copper behind as a metal powder.
End point
  • Heating stops when the black oxide has changed completely to brown copper, which shows that all of the oxygen has been removed.

Part 3 of 5Working Out the Formula

Working Out the Formula
Divide by Ar
  • Divide the mass of each element by its relative atomic mass to find the moles of atoms of each element.
Simplest ratio
  • Divide each mole value by the smaller value and round to whole numbers to give the ratio of metal atoms to oxygen atoms.
Write the formula
  • Write the ratio in the form MO with the subscript, for example MgO when the ratio is 1 : 1, or CuO for copper(II) oxide.
Sources of error
  • Escaping oxide smoke, an incomplete reaction and heating short of constant mass all make the oxygen mass incorrect.

Part 4 of 5Worked Example

Question

  • A 1.20 g sample of magnesium is burned and 2.00 g of magnesium oxide is formed.
  • Find the empirical formula of the oxide.
  • Use Ar values Mg = 24 and O = 16.

  1. Mass of oxygen = 2.00 − 1.20 = 0.80 g.
  2. Moles
    • Mg = 1.20 ÷ 24 = 0.050 mol;
      • O = 0.80 ÷ 16 = 0.050 mol.
  3. Ratio MgO = 0.050 ÷ 0.050 : 0.050 ÷ 0.050 = 1 : 1.

Answer The empirical formula is MgO.

Part 5 of 5Exam Checklist & Tips

1.5.7 – Key Points for the Exam

  • Combustion Burning a substance in oxygen, often releasing heat and light.
  • Constant mass The mass reached when repeated heating and weighing gives the same result.
  • Metal oxide A compound of a metal with oxygen, such as MgO or CuO.
  • Reduction A reaction in which oxygen is removed from a compound.
  • Empirical formula The simplest whole number ratio of the atoms of each element in a compound.
  • Crucible A heat-resistant container used to heat solids strongly.

Common Mistakes

  • Subtracting the crucible mass from the wrong total, which gives an incorrect mass of metal or oxide.
  • Lifting the lid too often or too far, so that oxide smoke escapes and the mass of oxygen is underestimated.
  • Dividing the masses by Mr instead of Ar when working with elements.
  • Stopping the heating before the mass is constant, so the metal may not have fully reacted.

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