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

Revision Note

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⁺SO₄^2- (aq) → Zn²⁺SO₄^2- (aq) + Cu (s)

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

Zn (s) + Cu²⁺SO₄^2- (aq) → Zn²⁺SO₄^2- (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:

Write the balanced chemical equation:
- 2Mg (s) + O₂(g) → 2MgO (s)
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
Calculate the moles of magnesium used in the reaction:
- n(Mg) = $\frac{7.5 g}{24.3 g {mol}^{- 1}}$ = 0.3086 moles
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
Calculate the mass of magnesium oxide:
- Mass = moles x M_r
- 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) = $\frac{volume of gas ({dm}^{3})}{24.0}$

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	$\frac{225.6}{24.0}$ = 9.4	225.6
Carbon monoxide	$\frac{7.2}{24.0}$ = 0.30	7.2
Sulfur dioxide	$\frac{960}{24.0}$ = 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 = $\frac{actual yield}{predicted or theoretical yield} \times 100$

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:

Write the balanced symbol equation:
- Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)
Calculate the number of moles of zinc:
- n(Zn) = $\frac{6.54 g}{65.4 g {mol}^{- 1}}$ = 0.10 moles
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
Calculate the maximum mass (theoretical yield) of copper:
- Mass = mol x M_r
- Mass = 0.10 mol x 63.5 g mol^-1
- Mass = 6.35 g
Calculate the percentage yield of copper:
- Percentage yield = $\frac{4.80 g}{6.35 g}$ 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 M_r of the desired product

percentage atom economy = $\frac{molecular mass of desired product}{sum of molecular masses of all reactants} \times 100$

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 M_r
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:

Write the equation:
$Atom economy = \frac{molecular mass of desired product}{sum of molecular masses of ALL reactants} \times 100$
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 M_r of CO₂ is not required as it is not the desired product
Substitute values and evaluate:
- Atom economy = $\frac{2 \times 55.8}{159.6 + (3 \times 28.0)} \times 100$
- 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 dm³of solution
- The solute is the substance that dissolves in a solvent to form a solution
- The solvent is often water

$concentration (mol {dm}^{- 3}) = \frac{number of moles of solute (mol)}{volume of solution ({dm}^{3})}$

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^-3the following points need to be considered:
- Change mass in grams to moles
- Change cm³to 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:

Write the balanced symbol equation
- CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Calculate the amount, in moles, of calcium carbonate:
- n(CaCO₃) = $\frac{2.5 g}{100 g {mol}^{- 1}}$ = 0.025 mol
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
Calculate the volume of HCl required:
- Volume (HCl) = $\frac{amount (mol)}{concentration (mol {dm}^{- 3})}$
- Volume (HCl) = $\frac{0.05 mol}{1.0 mol {dm}^{- 3}}$ = 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:

Write the balanced symbol equation:
- Na₂CO₃ + 2HCl → Na₂Cl₂ + H₂O + CO₂
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
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
Calculate the concentration, in mol dm^-3 of hydrochloric acid:
- [HCl] = $\frac{amount (mol)}{volume ({dm}^{3})}$
- [HCl] = $\frac{0.00250}{0.0200}$ = 0.125 mol dm^-3

Last updated: 16 April 2024

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

Revision Note

Balancing Equations

Balancing equations

Ionic equations

Worked Example

Worked Example

Calculations from Balanced Equations

Reacting masses

Worked Example

Volumes of gases

Examiner Tips and Tricks

Worked Example

Worked Example

Percentage yield

Worked Example

Percentage atom economy

Worked Example

Worked Example

Concentrations and volumes for reactions in solutions

Worked Example

Worked Example

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Author: Richard Boole

Author: Stewart Hird