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Balanced Equations & Associated Calculations (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

Balancing Equations

  • A symbol equation is a shorthand way of describing a chemical reaction using chemical symbols to show the number and type of each atom in the reactants and products

  • A word equation is a longer way of describing a chemical reaction using only words to show the reactants and products

Balancing equations

  • During chemical reactions, atoms cannot be created or destroyed

  • The number of each atom on each side of the reaction must therefore be the same

    • E.g. the reaction needs to be balanced

  • When balancing equations remember:

    • Not to change any of the formulae

    • To put the numbers used to balance the equation in front of the formulae

    • To balance firstly the carbon, then the hydrogen and finally the oxygen in combustion reactions of organic compounds

  • When balancing equations follow the following the steps:

    • Write the formulae of the reactants and products

    • Count the numbers of atoms in each reactant and product

    • Balance the atoms one at a time until all the atoms are balanced

    • Use appropriate state symbols in the equation

  • The physical state of reactants and products in a chemical reaction is specified by using state symbols

    • (s) solid

    • (l) liquid

    • (g) gas

    • (aq) aqueous

Ionic equations

  • In aqueous solutions, ionic compounds dissociate into their ions

  • Many chemical reactions in aqueous solutions involve ionic compounds, however, only some of the ions in solution take part in the reactions

  • The ions that do not take part in the reaction are called spectator ions

  • An ionic equation shows only the ions or other particles taking part in a reaction, without showing the spectator ions

Worked Example

Balance the following equation:

magnesium + oxygen → magnesium oxide

Answer:

  • Step 1: Write out the symbol equation showing reactants and products

Mg + O2 → MgO

  • Step 2: Count the number of atoms in each reactant and product

 

Mg

O

Reactants

1

2

Products

1

1

  • Step 3: Balance the atoms one at a time until all the atoms are balanced

2Mg + O2 → 2MgO

This is now showing that 2 moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide

  • Step 4: Use appropriate state symbols in the fully balanced equation

2Mg (s) + O2 (g) → 2MgO (s)

Worked Example

1. Balance the following equation:

zinc + copper(II) sulfate → zinc(II) sulfate + copper

2. Write the ionic equation for the above reaction.

Answer 1:

  • Step 1: To balance the equation, write out the symbol equation showing reactants and products

Zn  + CuSO4  → ZnSO4 + Cu

  • Step 2: Count the number of atoms in each reactant and product. The equation is already balanced

 

Zn

Cu

S

O

Reactants

1

1

1

4

Products

1

1

1

4

  • Step 3: Use appropriate state symbols in the equation

Zn (s)  + CuSO4 (aq)  → ZnSO4 (aq) + Cu (s)

Answer 2:

  • Step 1:  The full chemical equation for the reaction is

Zn (s)  + CuSO4 (aq)  → ZnSO4 (aq) + Cu (s)

  • Step 2:  Break down reactants into their respective ions

Zn (s)  + Cu2+ SO42- (aq)  → Zn2+SO42- (aq) + Cu (s) 

  • Step 3:  Cancel the spectator ions on both sides to give the ionic equation

Zn (s)  + Cu2+SO42- (aq)  → Zn2+SO42- (aq) + Cu (s)

Zn (s)  + Cu2+(aq)  → Zn2+ (aq) + Cu (s)

Calculations from Balanced Equations

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 balanced 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

Volumes of gases

  • At room temperature and pressure, one mole of any gas has a volume of 24.0 dm3 

    • Room temperature is 20 oC

    • Room pressure is 100 000 Pa 

Examiner Tips and Tricks

If the information given does not state room temperature and pressure, or give the specific values, then you should consider using the ideal gas equation PV = nRT

  • Using the following equations, the molar gas volume, 24.0 dm3, can be used to find:

    • The volume of a given mass or number of moles of gas

    • The mass or number of moles of a given volume of gas

volume of gas (dm3) = amount of gas (mol) x 24.0

amount of gas (mol) = fraction numerator bold volume bold space bold of bold space bold gas bold space stretchy left parenthesis dm cubed stretchy right parenthesis over denominator bold 24 bold. bold 0 end fraction

Worked Example

Calculating the volume of gas

Complete the table to calculate the volume that the gases occupy:

Gas

Amount of gas
(mol)

Volume of gas
(dm3)

Hydrogen

3.0

Carbon dioxide

0.25

Oxygen

5.4

Ammonia

0.02

Answers:

Gas

Amount of gas
(mol)

Volume of gas
(dm3)

Hydrogen

3.0

3.0 x 24.0 = 72.0

Carbon dioxide

0.25

0.25 x 24.0 = 6.0

Oxygen

5.4

5.4 x 24.0 = 129.6

Ammonia

0.02

0.02 x 24.0 = 0.48

Worked Example

Calculating the number of moles of gas

Complete the table to calculate the number of moles of gas:

Gas

Amount of gas
(mol)

Volume of gas
(dm3)

Methane

225.6

Carbon monoxide

7.2

Sulfur dioxide

960

Answers:

Gas

Amount of gas
(mol)

Volume of gas
(dm3)

Methane

fraction numerator bold 225 bold. bold 6 over denominator bold 24 bold. bold 0 end fraction = 9.4

225.6

Carbon monoxide

fraction numerator bold 7 bold. bold 2 over denominator bold 24 bold. bold 0 end fraction = 0.30

7.2

Sulfur dioxide

fraction numerator bold 960 over denominator bold 24 bold. bold 0 end fraction = 40

960

Percentage yield

  • In a lot of reactions, not all reactants react to form products which can be due to several factors:

    • Other reactions take place simultaneously

    • The reaction does not go to completion

    • Reactants or products are lost to the atmosphere

  • The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

percentage yield = fraction numerator bold actual bold space bold yield over denominator bold predicted bold space bold or bold space bold theoretical bold space bold yield end fraction bold cross times bold 100

  • Where actual yield is the number of moles or mass of product obtained experimentally

  • The predicted yield is the number of moles or mass obtained by calculation

Worked Example

Calculate % yield using moles

In an experiment to displace copper from copper(II) sulfate, 6.54 g of zinc was added to an excess of copper(II) sulfate solution.

The copper was filtered off, washed and dried.

The mass of copper obtained was 4.80 g.

Calculate the percentage yield of copper.

Answer:

  1. Write the balanced symbol equation:

    • Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

  2.  Calculate the number of moles of zinc:

    • n(Zn) = fraction numerator 6.54 space straight g over denominator 65.4 space straight g space mol to the power of negative 1 end exponent end fraction = 0.10 moles

  3. Deduce the number of moles of copper, using the balanced chemical equation:

    • 1 mole of zinc forms 1 mole of copper

      • The ratio is 1 : 1

    • Therefore, n(Cu) = 0.10 moles

  4. Calculate the maximum mass (theoretical yield) of copper:

    • Mass = mol x Mr

    • Mass = 0.10 mol x 63.5 g mol-1

    • Mass = 6.35 g

  5. Calculate the percentage yield of copper:

    • Percentage yield = fraction numerator 4.80 space straight g over denominator 6.35 space straight g end fraction x 100 = 75.6 %

Percentage atom economy

  • The atom economy of a reaction shows how many of the atoms used in the reaction become the desired product

    • The rest of the atoms or mass is wasted

  • It is found directly from the balanced equation by calculating the Mr of the desired product

percentage atom economy = fraction numerator bold molecular bold space bold mass bold space bold of bold space bold desired bold space bold product over denominator bold sum bold space bold of bold space bold molecular bold space bold masses bold space bold of bold space bold all bold space bold reactants end fraction bold cross times bold 100

  • In addition reactions, the atom economy will always be 100%, because all of the atoms are used to make the desired product

    • Whenever there is only one product, the atom economy will always be 100%

    • For example, in the reaction between ethene and bromine:

CH2=CH2 + Br2 → CH2BrCH2Br

  • The atom economy could also be calculated using mass, instead or Mr

  • In this case, you would divide the mass of the desired product formed by the total mass of all reactants, and then multiply by 100

  • Efficient processes have high atom economies and are important to sustainable development

    • They use fewer resources

    • Create less waste

Worked Example

Qualitative atom economy

Ethanol can be produced by various reactions, such as:

Hydration of ethene: 

C2H4 + H2O → C2H5OH

Substitution of bromoethane:

C2H5Br + NaOH → C2H5OH + NaBr

Explain which reaction has a higher atom economy.

Answer:

  • The hydration of ethene has a higher atom economy

    • The atom economy is 100 %

  • This is because all of the reactants are converted into products

  • Whereas the substitution of bromoethane produces NaBr as a waste product

Worked Example

Quantitative atom economy

The blast furnace uses carbon monoxide to reduce iron(III) oxide to iron.

Fe2O3 + 3CO → 2Fe + 3CO2

Calculate the atom economy for this reaction, assuming that iron is the desired product.

Answer:

  1. Write the equation:

    Atom space economy space equals space fraction numerator molecular space mass space of space desired space product over denominator sum space of space molecular space masses space of space ALL space reactants end fraction space cross times space 100

  2. Calculate the relevant atomic / molecular masses:

    • Fe2O3 = (2 x 55.8) + (3 x 16.0) = 159.6

    • CO = (1 x 12.0) + (1 x 16.0) = 28.0

    • Fe = 55.8

    • The Mr of CO2 is not required as it is not the desired product

  3. Substitute values and evaluate:

    • Atom economy = fraction numerator 2 cross times 55.8 over denominator 159.6 plus left parenthesis 3 cross times 28.0 right parenthesis end fraction cross times 100

    • Atom economy = 45.8 %

Concentrations and volumes for reactions in solutions

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

    • The solute is the substance that dissolves in a solvent to form a solution

    • The solvent is often water

bold concentration bold space stretchy left parenthesis mol space dm to the power of negative 3 end exponent stretchy right parenthesis bold equals fraction numerator bold number bold space bold of bold space bold moles bold space bold of bold space bold solute stretchy left parenthesis mol stretchy right parenthesis over denominator bold volume bold space bold of bold space bold solution bold space stretchy left parenthesis dm cubed stretchy right parenthesis end fraction

  • A concentrated solution is a solution that has a high concentration of solute

  • A dilute solution is a solution with a low concentration of solute

  • When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:

    • Change mass in grams to moles

    • Change cm3 to dm3 

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

    • Rearrange the concentration equation

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

  • 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 1.0 mol dm-3 hydrochloric acid required to completely react with 2.5 g of calcium carbonate.

Answer:

  1. Write the balanced symbol equation

    • CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

  2. Calculate the amount, in moles, of calcium carbonate:

    • 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

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

    • 1 mol of CaCO3 requires 2 mol of HCl

    • So 0.025 mol of CaCO3 requires 0.05 mol of HCl

  4. Calculate the volume of HCl required:

    • Volume (HCl) = fraction numerator amount space open parentheses mol close parentheses over denominator concentration space open parentheses mol space dm to the power of negative 3 end exponent close parentheses end fraction

    • Volume (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 = 0.05 dm3

    • So, the volume of hydrochloric acid required is 0.05 dm3 

Worked Example

Calculating concentration from volume

25.0 cm3 of 0.050 mol dm-3 sodium carbonate solution 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  →  Na2Cl2  +  H2O  +  CO2

  2. Calculate the amount, in moles, of sodium carbonate reacted

    • n(Na2CO3) = 0.025 dm3 x 0.050 mol dm-3

    • n(Na2CO3) = 0.00125 mol

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

    • 1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2

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

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

    • [HCl] = fraction numerator amount space open parentheses mol close parentheses over denominator volume space open parentheses dm cubed close parentheses end fraction

    • [HCl] = fraction numerator 0.00250 over denominator 0.0200 end fraction = 0.125 mol dm-3

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    Author
    Reviewer
    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    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.

    Stewart Hird

    Author: Stewart Hird

    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 Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.