Deducing Stoichiometry (Edexcel GCSE Chemistry)

Revision Note

Stewart Hird

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Writing & Balancing Equations

Nothing created - nothing destroyed

  • New substances are made during chemical reactions

    • However, the same atoms are always present before and after reaction

    • They have just joined up in different ways

    • Atoms cannot be created or destroyed, so if they exist in the reactants then they absolutely must be in the products!

  • Because of this the total mass of reactants is always equal to the total mass of products

  • This idea is known as the Law of Conservation of Mass

Conservation of Mass

  • The Law of Conservation of Mass enables us to balance chemical equations, since no atoms can be lost or created

  • You should be able to:

    • Write word equations for reactions outlined in these notes

    • Write formulae and balanced chemical equations for the reactions in these notes

Word Equations

  • These show the reactants and products of a chemical reaction using their full chemical names

  • The reactants are those substances on the left-hand side of the arrow and can be thought of as the chemical ingredients of the reaction

  • They react with each other and form new substances

  • The products are the new substances which are on the right-hand side of the arrow

  • The arrow (which is spoken as “goes to” or “produces”) implies the conversion of reactants into products

  • Reaction conditions or the name of a catalyst (a substance added to make a reaction go faster) can be written above the arrow

  • An example is the reaction of sodium hydroxide (a base) and hydrochloric acid produces sodium chloride (common table salt) and water:

sodium hydroxide + hydrochloric acid ⟶ sodium chloride + water

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

  • During chemical reactions as 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

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 numbers of atoms in each reactant and product

Atoms, Molecules & Stoichiometry Worked example - Balancing equations table, downloadable AS & A Level Chemistry revision notes
  • 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)

Ionic Equations

Higher Only

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, and not the spectator ions

Worked Example

1. Balance the following equation

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

2. Write down 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 numbers of atoms in each reactant and product. The equation is already balanced

Atoms, Molecules & Stoichiometry Worked example - Equations (balancing & ionic) table, downloadable AS & A Level Chemistry revision notes
  • 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)

Deducing Stoichiometry

Higher Only

  • Stoichiometry refers to the numbers in front of the reactants and products in an equation, which must be adjusted to make sure that the equation is balanced

  • These numbers are called coefficients (or multipliers) and if we know the masses of reactants and products, the balanced chemical equation for a given reaction can be found by determining the coefficients

  • First, convert the masses of each reactant and product in to moles by dividing by the molar masses using the periodic table

  • If the result yields uneven numbers, then multiply all of the numbers by the same number, to find the smallest whole number for the coefficient of each species

    • For example, if the resulting numbers initially were 1, 2 and 2.5, then you would multiply all of the numbers by 2, to give the whole numbers 2, 4 and 5

  • Then, use the molar ratio to write out the balanced equation

Worked Example

64 g of methanol, CH3OH, reacts with 96 g of oxygen gas to produce 88 g of carbon dioxide and 72 g of water. Deduce the balanced equation for the reaction.(C = 12, H = 1, O = 16).

Answer:

  • Calculate the molar masses of the substances in the equation

CH3OH = 32 g mol-1         O2 = 32 g mol-1        

CO2 =  44 g mol-1             H2O =  18 g mol-1  

  • Divide the masses present by the molar mass to obtain the number of moles

CH3OH = 64 g ÷ 32 g mol-1  =  2 mol 

O2 = 96 g ÷ 32 g mol-1          =  3 mol   

CO2 =  88 g ÷ 44 g mol-1      =  2 mol

H2O =  72 g ÷ 18 gmol-1       =  4 mol

  • The mole ratios are the same as the coefficients in the balanced equation

2CH3OH  +  3O⟶ 2CO2 + 4H2O

Examiner Tips and Tricks

The molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.

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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.