International ASChemistryEdexcelRevision Notes1. Structure, Bonding & Introduction to Organic ChemistryAmount of SubstanceReacting Mass Calculations

Reacting Mass Calculations (Edexcel International AS Chemistry): Revision Note

Exam code: XCH11

Richard Boole

Last updated

Reacting Mass Calculations

  • Chemical equations can be used to calculate the moles or masses of reactants and products

  • To do this, information given in the question is used to find the amount in moles of the substances being considered

  • Then, the ratio between the substances is identified using the balanced chemical equation

  • Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses

Worked Example

Example 1

Calculate the maximum mass of magnesium oxide that can be produced by completely burning 7.5 g of magnesium in oxygen.

magnesium + oxygen → magnesium oxide

Answer:

  1. Write the balanced chemical equation:

2Mg (s) + O2 (g) → 2MgO (s)

  1. Determine the relative atomic and formula masses:

    • Magnesium, Mg = 24.3 g mol-1

    • Oxygen, O2 = 32.0 g mol-1

    • Magnesium oxide, MgO = 40.3 g mol-1

  2. Calculate the moles of magnesium used in the reaction:

n(Mg) = fraction numerator 7.5 space straight g over denominator 24.3 space straight g space mol to the power of negative 1 end exponent end fraction = 0.3086 moles

  1. Deduce the number of moles of magnesium oxide, using the balanced chemical equation:

    • 2 moles of magnesium form 2 moles of magnesium oxide

      • The ratio is 1 : 1

    • Therefore, n(MgO) = 0.3086 moles

  2. Calculate the mass of magnesium oxide:

Mass = moles x Mr

Mass = 0.3086 mol x 40.3 g mol-1 = 12.4 g

  1. Therefore, the mass of magnesium oxide produced is 12.4 g

Worked Example

Example 2

Calculate the mass of aluminium, in tonnes, that can be produced from 51 tonnes of aluminium oxide. The equation for the reaction is:

aluminium oxide ⟶ aluminium + oxygen

Answer:

  1. Write the balanced chemical equation:

2Al2O3  ⟶  4Al +  3O2 

  1. Determine the relative atomic and formula masses:

    • Aluminium oxide, Al2O3  = 102.0 g mol-1

    • Aluminium, Al = 27.0 g mol-1

    • Oxygen, O2 = 32.0 g mol-1

  2. Calculate the mass, in g, of aluminium oxide used in the reaction:

51 tonnes x 106 = 51 000 000 g

  1. Calculate the moles of aluminium oxide used in the reaction:

n(Al2O3 ) = fraction numerator 51000000 space straight g over denominator 102.0 space straight g space mol to the power of negative 1 end exponent end fraction = 500 000 mol

  1. Deduce the number of moles of aluminium, using the balanced chemical equation:

    • 2 moles of aluminium oxide form 4 moles of aluminium

      • The ratio is 1:2

    • Therefore, n(Al) = 500 000 x 2 = 1 000 000 mol

  2. Calculate the mass of aluminium:

mass = moles x Mr

mass = 1 000 000 mol x 27.0 g mol-1 = 27 000 000 g

  1. Convert the mass to tonnes:

mass = 27000000 over 10 to the power of 6

mass = 27 tonnes

Examiner Tips and Tricks

As long as you are consistent it doesn't matter whether you work in grams or tonnes or any other mass unit as the reacting masses will always be in proportion to the balanced equation.

Balancing Equations using Reacting Masses

  • If the masses of reactants and products of a reaction are known then we can use them to write a balanced equation for that reaction

  • This is done by converting the masses to moles and simplifying to find the molar ratios

Worked Example

Example 3

A student reacts 1.2 g of carbon with 16.2 g of zinc oxide. The resulting products are 4.4 g of carbon dioxide and 13.0 g of zinc.

Determine the balanced equation for the reaction.

Answer:

  1. Write the unbalanced chemical equation:

C + ZnO ⟶ Zn + CO2

  1. Determine the relative atomic and formula masses:

    • Carbon, C = 12.0 g mol-1

    • Zinc, Zn = 65.4 g mol-1

    • Zinc oxide, ZnO = 81.4 g mol-1

    • Carbon dioxide, CO2 = 44.0 g mol-1

  2. Calculate the moles of each substance

moles = mass over M subscript straight r

n(C) = fraction numerator 1.2 over denominator 12.0 end fraction = 0.1 mol

n(Zn) = fraction numerator 13.0 over denominator 65.4 end fraction = 0.2 mol

n(ZnO) = fraction numerator 16.2 over denominator 81.4 end fraction = 0.2 mol

n(CO2) = fraction numerator 4.4 over denominator 44.0 end fraction = 0.1 mol

  1. Determine the ratio (divide by the smallest value)

C = fraction numerator 0.1 over denominator 0.1 end fraction = 1

Zn = fraction numerator 0.2 over denominator 0.1 end fraction = 2

ZnO = fraction numerator 0.2 over denominator 0.1 end fraction = 2

CO2 = fraction numerator 0.1 over denominator 0.1 end fraction = 1

  1. Apply the values to the unbalanced equation:

1C + 2ZnO ⟶ 2Zn + 1CO2

  1. Simplify:

C + 2ZnO ⟶ 2Zn + CO2

Examiner Tips and Tricks

These questions look hard but they are actually quite easy to do, as long as you follow the steps and organise your work neatly.

Remember: The molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.

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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry Content Creator

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.