Deducing Stoichiometry (Edexcel GCSE Chemistry)

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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 Tip

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

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Caroline

Author: Caroline

Expertise: Physics Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.