Revision NotesExam QuestionsPast Papers
A LevelChemistryOCRRevision Notes2. Foundations in ChemistryAmount of SubstanceReaction Calculations

Reaction Calculations (OCR A Level Chemistry A): Revision Note

Exam code: H432

Richard Boole

Written by: Richard Boole

Reviewed by: Philippa Platt

Updated on

Mass calculations

  • The number of moles of a substance (n) can be calculated using the equation:

n = fraction numerator m a s s space open parentheses m close parentheses over denominator m o l a r space m a s s space open parentheses M close parentheses end fraction

  • It is important to identify the type of particle being measured:

    • For example, one mole of CaF2 contains:

      • One mole of CaF2 formula units

    • But it also contains:

      • One mole of Ca2+ ions

      • Two moles of F- ions

Reacting mass calculations

  • Reactant masses help determine how much of each substance is needed or produced in a reaction

  • To calculate reacting masses, you need:

    • A balanced chemical equation

    • The mass of at least one substance

    • The molar mass of that substance

Worked Example

Mass calculation using moles

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:

  1. The balanced symbol equation is:

2Mg (s) + O2 (g) → 2MgO (s)

  1. The relative masses are:

Mg = 24.3, O2 = 32.0, MgO = 40.3

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

n(Mg) = begin mathsize 16px style space fraction numerator 6.0 space g over denominator 24.3 space g space m o l to the power of negative 1 end exponent end fraction end style = 0.25 moles

  1. Find the ratio of magnesium to magnesium oxide using the balanced chemical equation:

Mg

MgO

Moles

2

2

Ratio

1

1

Change in moles

-0.25

+0.25

  • Therefore, 0.25 mol of MgO is formed

  1. Calculate the mass of magnesium oxide:

Mass = mol x Mr

Mass = 0.25 mol x 40.3 g mol-1 

Mass = 10.1 g

  • Therefore, the mass of magnesium oxide produced is 10.1 g

Stoichiometric ratios

  • Stoichiometry describes the mole ratios between reactants and products in a balanced equation

  • These ratios can be used to calculate unknown masses, moles, or volumes

Example using gas volumes:

  • 50 cm3 of propane reacts with 250 cm3 of oxygen and produces 150 cm3 of carbon dioxide

  • This gives a ratio of 1 propane : 5 oxygen : 3 carbon dioxide

    • Therefore, the balanced chemical equation is:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + _H2O (l)

  • To complete the equation, the number of hydrogen atoms from propane must be balanced:

    • Propane contains 8 hydrogen atoms

    • These form 4 water molecules (4 × H2O)

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

  • This example shows how known gas volumes help deduce mole ratios

    • But, full balancing may also require using atomic counts

Volume calculations

Concentration and volume

  • Concentration is the amount of solute dissolved in a solvent to make 1 dm3 of solution

    • Solute = the substance being dissolved

    • Solvent = the liquid doing the dissolving (often water)

  • A concentrated solution has a high amount of solute

    • A dilute solution has a low amount

  • Concentration, moles and volume are linked by the equation:

concentration (mol dm-3) = fraction numerator number space of space moles space left parenthesis mol right parenthesis over denominator volume space of space solution space left parenthesis dm cubed right parenthesis end fraction

  • To calculate the mass of a substance present in a solution of known concentration and volume:

  1. Rearrange the concentration equation

number of moles (mol) = concentration (mol dm-3) x volume (dm3)

  1. Multiply the moles of solute by its molar mass

mass of solute (g) = number of moles (mol) x molar mass (g mol-1)

Worked Example

Calculating volume from concentration

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm-3 that is required to react completely with 2.5 g of calcium carbonate

Answer

  1. Write the balanced symbol equation:

CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

  1. Calculate the amount, in moles, of calcium carbonate that reacts:

n(CaCO3) = fraction numerator 2.5 space straight g over denominator 100 space straight g space mol to the power of negative 1 end exponent end fraction = 0.025 mol

  1. Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry:

    • 1 mol of CaCO3 requires 2 mol of HCl

      • The molar ratio is 1 : 2

    • So, 0.025 mol of CaCO3 requires 0.05 mol of HCl

  2. Calculate the volume of HCl required:

Volume of HCl (dm3) = fraction numerator amount space left parenthesis mol right parenthesis over denominator concentration space left parenthesis mol space dm to the power of negative 3 end exponent right parenthesis end fraction

Volume of HCl = fraction numerator 0.05 space mol over denominator 1.0 space mol space dm to the power of negative 3 end exponent end fraction

Volume of HCl = 0.05 dm3

Worked Example

Neutralisation calculation

25.0 cm3 of 0.050 mol dm–3 sodium carbonate was completely neutralised by 20.0 cm3  of dilute hydrochloric acid. 

Calculate the concentration in mol dm–3 of the hydrochloric acid.  

Answer:

  1. Write the balanced symbol equation

Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

  1. Calculate the amount, in moles, of sodium carbonate:

    • Rearrange the equation for amount of substance (mol)

    • Divide the volume by 1000 to convert cm3 to dm3

amount (Na2CO3) = 0.025 dm3 x 0.050 mol dm-3 = 0.00125 mol

  1. Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry:

    • 1 mol of Na2CO3 reacts with 2 mol of HCl

      • The molar ratio is 1 : 2

    • So, 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

  2. Calculate the concentration, in mol dm-3, of hydrochloric acid:

concentration of HCl (mol dm-3) = fraction numerator amount space left parenthesis mol right parenthesis over denominator volume space left parenthesis dm cubed right parenthesis end fraction

concentration of HCl = fraction numerator 0.00250 over denominator 0.0200 end fraction

concentration of HCl = 0.125 mol dm-3

Gas volumes and molar volume

  • Avogadro’s hypothesis states that:

‘equal gas volumes contain equal numbers of molecules’

  • At room temperature and pressure ( (20 oC, 1 atm), one mole of any gas occupies 24.0 dm3

  • Gas volumes and molar volumes are linked by the equation:

volume of gas (dm3) = amount of gas (mol) x 24 dm3 mol-1

Worked Example

Calculation volume of gas using excess & limiting reagents

  1. Calculate the volume the following gases occupy:

    1. Hydrogen (3 mol)

    2. Carbon dioxide (0.25 mol)

    3. Oxygen (5.4 mol)

    4. Ammonia (0.02 mol)

  2. Calculate the moles in the following volumes of gases:

    1. Methane (225.6 dm3)

    2. Carbon monoxide (7.2 dm3)

    3. Sulfur dioxide (960 dm3)

Answers:

  • Using the equation:

volume of gas (dm3) = amount of gas (mol) x 24 dm3 mol-1

  • So, the volume that each gas occupies is:

Gas

Amount of gas (mol)

Volume of gas (dm3)

a. Hydrogen

3

3 x 24.0 = 72.0

b. Carbon dioxide

0.25

0.25 x 24.0 = 6.0

c. Oxygen

5.4

5.4 x 24.0 = 129.6

d. Ammonia

0.02

0.02 x 24.0 = 0.48

  1. Rearranging volume = amount of gas x 24 gives:

amount of gas (mol) = fraction numerator volume space of space gas space left parenthesis dm cubed right parenthesis over denominator 24 space dm cubed space mol to the power of negative 1 end exponent end fraction

  • So, the number of moles of each gas is:

Gas

Volume of gas (dm3)

Amount of gas (mol)

a. Methane

225.6

fraction numerator 225.6 over denominator 24.0 end fraction = 9.4

b. Carbon monoxide

7.2

fraction numerator 7.2 over denominator 24.0 end fraction = 9.4

c. Sulfur dioxide

960

fraction numerator 960 over denominator 24.0 end fraction = 40

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself
    Previous:Determining FormulaeNext:The Ideal Gas Equation
    Author
    Reviewer
    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.

    Philippa Platt

    Reviewer: Philippa Platt

    Expertise: Chemistry Content Creator

    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