Deducing Stoichiometry (Edexcel GCSE Chemistry)
Revision Note
Did this video help you?
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 Tips and Tricks
The molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.
You've read 0 of your 5 free revision notes this week
Sign up now. It’s free!
Did this page help you?