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First exams 2025

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Reacting Mass Calculations (SL IB Chemistry)

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Philippa

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Philippa

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Reacting Mass Calculations

  • The number of moles of a substance can be found by using the following equation:

               number space of space moles space equals space fraction numerator mass space of space substance space in space grams over denominator molar space mass space open parentheses straight g space mol to the power of negative 1 end exponent close parentheses end fraction

  • It is important to be clear about the type of particle you are referring to when dealing with moles
    • E.g. 1 mole of CaF2 contains one mole of CaF2 formula units, but one mole of Ca2+ and two moles of F- ions

Reacting masses

  • The masses of reactants are useful to determine how much of the reactants exactly react with each other to prevent waste
  • To calculate the reacting masses, the chemical equation is required
  • This equation shows the ratio of moles of all the reactants and products, also called the stoichiometry, of the reaction
  • To find the mass of products formed in a reaction the following pieces of information are needed:
    • The mass of the reactants
    • The molar mass of the reactants
    • The balanced equation

Worked example

Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen.

magnesium (s) + oxygen (g) → magnesium oxide (s)

Answer:

Step 1: The symbol equation is:

2Mg(s)    +     O2(g)            2MgO(s)

Step 2: The relative atomic masses are:

magnesium : 24.31         oxygen : 16.00         

Step 3: Calculate the moles of magnesium used in reaction

    number space of space moles space equals space fraction numerator 6.0 space straight g over denominator 24.31 space straight g space mol to the power of negative 1 end exponent end fraction equals space 0.25 space mol

Step 4: Find the ratio of magnesium to magnesium oxide using the balanced chemical equation

  Magnesium Magnesium oxide
Mol 2 2
Ratio 1 1
Change in mol -0.25 +0.25

Therefore, 0.25 mol of MgO is formed

Step 5: Find the mass of magnesium oxide

mass = mol x M

mass = 0.25 mol x 40.31 g mol-1

mass = 10.08 g

Therefore, mass of magnesium oxide produced is 10 g (2 sig figs)

Worked example

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

2Al2O3  ⟶  4Al +  3O2 

Answer:

Step 1: Calculate the moles if aluminium oxide used

mass of Al2O3 in g = 51 x 10= 51,000,000 g

moles = fraction numerator 51 comma 000 comma 000 space straight g over denominator 101.96 space straight g space mol to the power of negative 1 end exponent end fraction= 500,196.16 mol

Step 2: Find the ratio of Al2O3 to Al using the molar ratio from the balanced equation

2Al2O3 : 4Al

Ratio is thus 1 : 2

So 500,196.16 mol moles of Al2O3 produces 100,0392.31 moles of Al

Step 3: Calculate mass of Al

mass = Moles x Mr

mass = 1,000,392.31 mol x 26.98 g mol-1 = 26,990,584.54 g

Step 4: Convert mass from grams to tonnes

fraction numerator 26 comma 990 comma 584.54 space straight g over denominator 10 to the power of 6 end fraction= 26.99 tonnes

Examiner Tip

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.

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Philippa

Author: Philippa

Expertise: Chemistry

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.