The mole concept is fundamental to O Level Chemistry calculations. Master this and calculations become much easier. Struggling with stoichiometry? Our chemistry tuition Singapore program breaks down the mole concept into easy-to-follow steps.
What Is a Mole?
A mole is a unit for amount of substance.
One mole = 6.02 × 10²³ particles (Avogadro’s number)
This is the number of atoms in exactly 12g of carbon-12.
Molar Mass
Molar mass = mass of one mole of a substance
- Same numerical value as relative atomic/molecular mass
- – Unit: g/mol
Examples:
- H: 1 g/mol
- – O: 16 g/mol
- – H2O: 18 g/mol (2×1 + 16)
- – NaCl: 58.5 g/mol (23 + 35.5)
Key Formulas
Number of moles = Mass / Molar Mass
n = m/Mr
Number of moles = Volume (dm³) × Concentration (mol/dm³)
n = V × C
Number of moles (gas at rtp) = Volume (dm³) / 24
n = V/24
Number of particles = Number of moles × Avogadro’s number
= n × 6.02 × 10²³
Converting Units
Mass: g, kg (1 kg = 1000 g)
Volume: cm³, dm³ (1 dm³ = 1000 cm³)
Concentration: g/dm³, mol/dm³
To convert g/dm³ to mol/dm³:
Concentration (mol/dm³) = Concentration (g/dm³) / Mr
Calculating Molar Mass
Step 1: Write the formula
Step 2: Identify each element
Step 3: Multiply atomic mass by number of atoms
Step 4: Add all values
Example: H2SO4
H: 2 × 1 = 2
S: 1 × 32 = 32
O: 4 × 16 = 64
Total = 98 g/mol
Empirical Formula
The simplest whole number ratio of atoms in a compound.
Steps:
- Convert to moles
- 2. Find the simplest ratio
- 3. Write the formula
Example: A compound contains 40% calcium, 12% carbon, 48% oxygen by mass.
Ca: 40/40 = 1 mol
C: 12/12 = 1 mol
O: 48/16 = 3 mol
Ratio: 1:1:3
Empirical formula: CaCO3
Molecular Formula
The actual number of atoms in a molecule.
May be a multiple of the empirical formula.
Example: Empirical formula CH2O, molecular mass 180
Empirical formula mass = 12 + 2 + 16 = 30
Multiple = 180/30 = 6
Molecular formula = C6H12O6
Stoichiometry
Using balanced equations to calculate quantities.
Steps:
- Write balanced equation
- 2. Convert known quantities to moles
- 3. Use mole ratio from equation
- 4. Calculate unknown quantity
Example: What mass of CO2 is produced from 10g of CaCO3?
CaCO3 → CaO + CO2
Moles of CaCO3 = 10/100 = 0.1 mol
Mole ratio CaCO3:CO2 = 1:1
Moles of CO2 = 0.1 mol
Mass of CO2 = 0.1 × 44 = 4.4g
Limiting Reactants
The reactant that runs out first limits the reaction.
Steps:
- Calculate moles of each reactant
- 2. Use mole ratio to determine which is limiting
- 3. Use limiting reactant to calculate products
Example: 2g of H2 reacts with 20g of O2 to form H2O.
2H2 + O2 → 2H2O
Moles of H2 = 2/2 = 1 mol
Moles of O2 = 20/32 = 0.625 mol
Ratio needed: 2:1
For 1 mol H2, need 0.5 mol O2
We have 0.625 mol O2, so O2 is in excess
H2 is limiting
Moles of H2O = 1 mol (from 1 mol H2)
Mass of H2O = 1 × 18 = 18g
Concentration Calculations
Concentration = moles / volume (in dm³)
To prepare a solution:
- Calculate moles needed
- 2. Calculate mass needed
- 3. Dissolve in water
- 4. Transfer to volumetric flask
- 5. Make up to mark
Titration Calculations
Use results to find unknown concentration.
Example: 25.0 cm³ of NaOH requires 23.5 cm³ of 0.1 mol/dm³ HCl for neutralisation.
HCl + NaOH → NaCl + H2O
Moles of HCl = 0.0235 × 0.1 = 0.00235 mol
Mole ratio = 1:1
Moles of NaOH = 0.00235 mol
Concentration of NaOH = 0.00235/0.025 = 0.094 mol/dm³
Gas Calculations
At room temperature and pressure (rtp):
1 mole of gas = 24 dm³
Example: What volume of CO2 is produced from 10g of CaCO3?
CaCO3 → CaO + CO2
Moles of CaCO3 = 10/100 = 0.1 mol
Moles of CO2 = 0.1 mol
Volume of CO2 = 0.1 × 24 = 2.4 dm³
Percentage Yield
Percentage Yield = (Actual Yield / Theoretical Yield) × 100%
Example: Theoretical yield = 50g, actual yield = 42g
Percentage yield = (42/50) × 100% = 84%
Percentage Composition
% of element = (mass of element / total mass) × 100%
Example: % of Ca in CaCO3
= (40/100) × 100% = 40%
Common Mistakes
- Not converting units (cm³ to dm³)
- 2. Using wrong molar mass
- 3. Not balancing equations first
- 4. Ignoring mole ratios
- 5. Arithmetic errors
How Ace Scorers Helps
Our Chemistry programme includes:
- Mole concept mastery
- – Calculation practice
- – Step-by-step methods
- – Exam techniques
Contact us for Chemistry tuition.
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