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ASChemistryAQARevision Notes1. Physical Chemistry1.3 The Mole, Avogadro & The Ideal Gas Equation1.3.2 Reacting Masses

Reacting Masses (AQA AS Chemistry)

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Reacting Masses

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

number of mol = fraction numerator bold mass bold space bold of bold space bold substance bold space bold in bold space bold grams bold space stretchy left parenthesis g stretchy right parenthesis over denominator bold molar bold space bold mass bold space stretchy left parenthesis g space mol to the power of negative 1 end exponent stretchy right parenthesis 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 equation
  • 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

Mass calculation using moles

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)
  2. 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
  3. 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
  4. 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
  5. Calculate the mass of magnesium oxide:
    • Mass = moles x Mr
    • Mass = 0.3086 mol x 40.3 g mol-1 = 12.44 g
    • Therefore, the mass of magnesium oxide produced is 12.44 g

Stoichiometric relationships

  • The stoichiometry of a reaction can be found if the exact amounts of reactants and products formed are known
  • The amounts can be found by using the following equation:

number of mol = Alternative text not available

  • The gas volumes can be used to deduce the stoichiometry of a reaction
    • E.g. in the combustion of 50 cm3 of propane reacting with 250 cm3 of oxygen, 150 cm3  of carbon dioxide is formed suggesting that the ratio of propane:oxygen:carbon dioxide is 1:5:3

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)

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    Previous:1.3.1 The Mole & the Avogadro ConstantNext:1.3.3 Reacting Volumes
    Stewart

    Author: Stewart

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Exam Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.