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+ 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:
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) = = 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 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) =
Worked Example
Calculating the volume of gas
Complete the table to calculate the volume that the gases occupy:
Gas | Amount of gas | Volume of gas |
---|---|---|
Hydrogen | 3.0 | |
Carbon dioxide | 0.25 | |
Oxygen | 5.4 | |
Ammonia | 0.02 |
Answers:
Gas | Amount of gas | Volume of gas |
---|---|---|
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 | Volume of gas |
---|---|---|
Methane | 225.6 | |
Carbon monoxide | 7.2 | |
Sulfur dioxide | 960 |
Answers:
Gas | Amount of gas | Volume of gas |
---|---|---|
Methane | = 9.4 | 225.6 |
Carbon monoxide | = 0.30 | 7.2 |
Sulfur dioxide | = 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 =
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) = = 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 Mr
Mass = 0.10 mol x 63.5 g mol-1
Mass = 6.35 g
Calculate the percentage yield of copper:
Percentage yield = 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 =
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:
Write the equation:
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
Substitute values and evaluate:
Atom economy =
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
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:
Write the balanced symbol equation
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Calculate the amount, in moles, of calcium carbonate:
n(CaCO3) = = 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) =
Volume (HCl) = = 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] =
[HCl] = = 0.125 mol dm-3
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