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Deducing Stoichiometry (Edexcel GCSE Chemistry)
Revision Note
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
- 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
- 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 + 3O2 ⟶ 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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