CIE IGCSE Chemistry

Revision Notes

4.2 The Mole Concept

Extended Only

The Mole & Avogadro’s Constant

The mole

  • This is the mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of 12C
  • The mole is the unit representing the amount of atoms, ions, or molecules
  • One mole is the amount of a substance that contains 6.02 x 1023 particles (Atoms, Molecules or Formulae) of a substance (6.02 x 1023 is known as the Avogadro Number)

Examples

  • 1 mole of Sodium (Na) contains 6.02 x 1023 Atoms of Sodium
  • 1 mole of Hydrogen (H2) contains 6.02 x 1023 Molecules of Hydrogen
  • 1 mole of Sodium Chloride (NaCl) contains  6.02 x 1023 Formula units of Sodium Chloride

Linking the mole and the atomic mass

  • One mole of any element is equal to the relative atomic mass of that element in grams
  • For example one mole of carbon, that is if you had 6.02 x 1023 atoms of carbon in your hand, it would have a mass of 12g
  • So one mole of helium atoms would have a mass of 4g, lithium 7g etc
  • For a compound we add up the relative atomic masses
  • So one mole of water would have a mass of 2 x 1 + 16 = 18g
  • Hydrogen which has an atomic mass of 1 is therefore equal to 1/12 the mass of a 12C atom
  • So one carbon atom has the same mass as 12 hydrogen atoms
Extended Only

The Mole & the Volume of Gases

Molar volume

  • This is the volume that one mole of any gas (be it molecular such as CO2 or monoatomic such as helium) will occupy
  • It’s value is 24dm3 or 24,000 cm3 at room temperature and pressure (r.t.p.)

 

Calculations involving gases

General equation:

Amount of gas (mol) = Volume of gas (dm3) ÷ 24

or

Amount of gas (mol) = Volume of gas (cm3) ÷ 24000

1. Calculating the volume of gas that a particular amount of moles occupies

Equation:

Volume of gas (dm3) = Amount of gas (mol) x 24

or

Volume of gas (cm3) = Amount of gas (mol) x 24000

Example:

extended_a table, IGCSE & GCSE Chemistry revision notes

 

2. Calculating the moles in a particular volume of gas

Equation:

Amount of gas (mol) = Volume of gas (dm3) ÷ 24

or

Amount of gas (mol) = Volume of gas (cm3) ÷ 24000

Example:

extended_b table, IGCSE & GCSE Chemistry revision notes

Extended Only

Calculating Reacting Masses, Solutions & Concentrations of Solutions in g/dm3 & mol/dm3

Calculating percentage composition, moles, mass and relative formula mass

 

Calculating w moles mass relative formula mass, IGCSE & GCSE Chemistry revision notesFormula triangle for moles, mass and formula mass

 

1. Calculating Moles

Equation:

Amount in Moles = Mass of Substance in grams ÷ Mr (or Ar)

Example:

extended_Calculating moles table, IGCSE & GCSE Chemistry revision notes

 

2. Calculating Mass

Equation:

Mass of substance (grams) = Moles x Mr (or Ar)

Example:

extended_Calculating mass table, IGCSE & GCSE Chemistry revision notes

 

3. Calculating Relative Formula Mass

Equation:

Mr (or Ar) = Mass of Substance in Grams ÷ Moles

Example:

10 moles of Carbon Dioxide has a Mass of 440 g. What is the Relative Formula Mass of Carbon Dioxide?

Relative Formula Mass = Mass ÷ Number of Moles
Relative Formula Mass = 440 ÷ 10 = 44

Relative Formula Mass of Carbon Dioxide = 44

 

4. Calculating Percentage Composition

  • The percentage composition is found by calculating the percentage by mass of each particular element in a compound

 Example:

Calculate the percentage of oxygen in CO2

 

Step 1 – Calculate the molar mass of the compound

Molar mass CO2  =  (2 x 16) + 12   =  44

Step 2 – Add the atomic masses of the element required as in the question (oxygen)

16 + 16 = 32

Step 3 – Calculate the percentage

% of oxygen in CO2 = 32/44 x 100 = 72.7%

 

Calculations of solutions: moles, concentration and volume

General Equation:

Concentration (mol / dm3)  =  Amount of substance (mol) ÷ Volume of solution (dm3)

This general equation is rearranged for the term as is asked in the question.

1. Calculating Moles

Equation:

Amount of Substance (mol) = Concentration x Volume of Solution (dm3)

Example:

Calculate the Moles of Solute Dissolved in 2 dm3 of a 0.1 mol / dm3 Solution

Concentration of Solution : 0.1 mol / dm3

Volume of Solution : 2 dm3

Moles of Solute   =   0.1   x   2   =   0.1 mol (the dm3 above and below the line cancel out)

Amount of Solute = 0.2 mol

 

2. Calculating Concentration

Equation:

Concentration (mol / dm3)  =  Amount of substance (mol) ÷ Volume of solution (dm3)

Example:

25.0 cm3 of 0.050 mol / dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration in mol / dm3 of the hydrochloric acid.

 

Step 1 – Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing by 1000 to convert cm3 to dm3

Amount of Na2CO3  =  (25.0 x 0.050) ÷ 1000  =  0.00125 mol

Step 2 – Calculate the amount, in moles, of hydrochloric acid reacted

1 mol of Na2CO3 reacts with 2 mol of HCl, so the Molar Ratio is 1 : 2

Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

Step 3 – Calculate the concentration, in mol / dm3 of the Hydrochloric Acid

1 dm3 = 1000 cm3

 Volume of HCl =  20 ÷ 1000  =   0.0200 dm3

Concentration HCl (mol / dm3) =  0.00250 ÷ 0.0200  =  0.125

Concentration of Hydrochloric Acid = 0.125 mol / dm3

 

3. Calculating Volume

Equation:

Volume (dm3)  =  Amount of substance (mol) ÷ Concentration (mol / dm3)

Example:

Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm3 that is required to react completely with 2.5g of calcium carbonate.

 

Step 1 – Calculate the amount, in moles, of calcium carbonate that reacts

Mr of CaCO3 is 100

Amount of CaCO3  =  (2.5 ÷ 100)  =  0.025 mol

Step 2 – Calculate the moles of hydrochloric acid required

CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

1 mol of CaCO3 requires 2 mol of HCl

So 0.025 mol of CaCO3 Requires 0.05 mol of HCl

Step 3 – Calculate the volume of HCl Required

Volume  =  (Amount of Substance(mol)  ÷  Concentration (mol / dm3)

=  0.05  ÷  1.0

=  0.05 dm3 (the moles cancel out above and below the line)

Volume of Hydrochloric Acid = 0.05 dm3

 

The limiting reactant and reacting masses

Limiting reactant

  • The limiting reactant is the reactant which is not present in excess in a reaction
  • It is always the first reactant to be used up which then causes the reaction to stop
  • In order to determine which reactant is the limiting reagent in a reaction, we have to consider the ratios of each reactant in the balanced equation

Example:

9.2g of sodium is reacted with 8.0g of sulfur to produce sodium sulfide, NaS. Which reactant is in excess and which is the limiting reactant?

 

Step 1 – Calculate the moles of each reactant

Moles = Mass ÷ Ar

Moles Na = 9.2/23 = 0.40

Moles S = 8.0/32 = 0.25

Step 2 – Write the balanced equation and determine the molar ratio

2Na + S → Na2S so the molar ratios is 2 : 1

Step 3 – Compare the moles. So to react completely 0.40 moles of Na require 0.20 moles of S and since there are 0.25 moles of S, then S is in excess. Na is therefore the limiting reactant.

 

Calculating reacting masses

  • Chemical equations can be used to calculate the moles or masses of reactants and products
  • Use information from the question to find the amount in moles of the substances being considered
  • Identify the ratio between the substances using the balanced chemical equation
  • Apply mole calculations to find answer

Example 1:

Calculate the Mass of Magnesium Oxide that can be made by completely burning 6 g of Magnesium in Oxygen

Magnesium (s) + Oxygen (g) Magnesium Oxide (s)

Symbol Equation:     

2Mg + O2 → 2MgO

Relative Formula Mass:    Magnesium : 24             Magnesium Oxide : 40

 

Step 1 – Calculate the moles of Magnesium Used in reaction

Moles = Mass ÷ Mr     Moles = 6 ÷ 24 = 0.25

Step 2 – Find the Ratio of Magnesium to Magnesium Oxide using the balanced Chemical Equation

extended_Calculating reacting masses_magnesium table, IGCSE & GCSE Chemistry revision notes

 

Step 3 – Find the Mass of Magnesium Oxide

Moles of Magnesium Oxide = 0.25

Mass = Moles x Mr     Mass = 0.25 x 40 = 10 g

Mass of Magnesium Oxide Produced = 10 g

 

Example 2:

Calculate the Mass, in Tonnes, of Aluminium that can be Produced from 51 Tonnes of Aluminium Oxide

Aluminium Oxide (s) Aluminium (s) + Oxygen (g)

Symbol Equation:

2Al2O3 → 4Al + 3O2

Ar and Mr:     Aluminium : 27    Oxygen : 16    Aluminium Oxide : 102

1 Tonne = 106 g

 

Step 1 – Calculate the moles of aluminium oxide used

Mass of Aluminium Oxide in Grams = 51 x 106 = 51,000,000  g

Moles = Mass ÷ Ar                             Moles = 51,000,000 ÷ 102 = 500,000

Step 2 – Find the ratio of aluminium oxide to aluminium using the balanced chemical equation

extended_Calculating reacting masses_aluminium table, IGCSE & GCSE Chemistry revision notes

 

Step 3 – Find the mass of aluminium

Moles of aluminium = 1,000,000

Mass in grams = Moles x Ar     Mass = 1,000,000 x 27 = 27,000,000

Mass in Tonnes = 27,000,000 ÷ 106 = 27 Tonnes

Mass of Aluminium Produced = 27 Tonnes

Extended Only

Using the Mole to Determine Empirical & Molecular Formulae

Empirical formula: gives the simplest whole number ratio of atoms of each element in the compound

  • Calculated from knowledge of the ratio of masses of each element in the compound

Example:

A compound that contains 10 g of Hydrogen and 80 g of Oxygen has an Empirical Formula of H2O. This can be shown by the following calculations:

Amount of Hydrogen Atoms = Mass in grams ÷ Ar of Hydrogen = (10 ÷ 1) = 10 moles

Amount of Oxygen Atoms = Mass in grams ÷ Ar of Oxygen = (80 ÷ 16) = 5 moles

The ratio of moles of hydrogen atoms to moles of oxygen atoms:

The Ratio of Moles of Hydrogen Atoms to Moles of Oxygen Atoms table, IGCSE & GCSE Chemistry revision notes

 

Since equal numbers of moles of atoms contain the same number of atoms, the ratio of hydrogen atoms to oxygen atoms is 2:1

Hence the empirical formula is H2O

 

Molecular formula: gives the exact numbers of atoms of each element present in the formula of the compound

  • Divide the relative formula mass of the molecular formula by the relative formula mass of the Empirical Formula
  • Multiply the number of each element present in the Empirical Formula by this number to find the Molecular Formula

Relationship between Empirical and Molecular Formula:

Relationship between Empirical _ Molecular Formula table, IGCSE & GCSE Chemistry revision notes

 

Example:

The Empirical Formula of X is C4H10S1 and the Relative Formula Mass of X is 180. What is the Molecular Formula of X?

Relative Formula Mass:       Carbon : 12      Hydrogen : 1      Sulfur : 32

 

Step 1 – Calculate Relative Formula Mass of Empirical Formula

(C x 4) + (H x 10) + (S x 1)    =   (12 x 4) + (1 x 10) + (32 x 1)   =   90

Step 2 – Divide Relative Formula Mass of X by Relative Formula Mass of Empirical

Formula

180 / 90 = 2

Step 3 – Multiply Each Number of Elements by 2

(C4 x 2) + (H10 x 2) + (S1 x 2)     =    (C8) + (H20) + (S2)

Molecular Formula of X = C8H20S2

Extended Only

Calculating Percentage Yield & Percentage Purity of the Product

Percentage yield

  • This is the calculation of the percentage yield obtained from the theoretical yield
  • In practice, you never get 100% yield in a chemical process for several reasons
  • These include some reactants being left behind in the equipment, the reaction may be reversible or product may also be lost during separation stages

Equation:

Percentage Yield = (Yield Obtained / Theoretical Yield) x 100

Example:

In an experiment to displace copper from copper sulfate, 6.5 g of Zinc was added to an excess of copper (II) sulfate solution. The copper was filtered off, washed and dried. The mass of copper obtained was 4.8 g. Calculate the percentage yield of copper.

Equation Of Reaction:    

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

 

Step 1: Calculate the Amount, in Moles of Zinc Reacted

Moles of Zinc = 6.5 ÷ 65 = 0.10 moles

Step 2: Calculate the Maximum Amount of Copper that could be formed from the

Molar ratio

Maximum Moles of Copper = 0.10 moles (Molar ratio is 1:1)

Step 3: Calculate the Maximum Mass of Copper that could be Formed

Maximum Mass of Copper = ( 0.10 x 64 ) = 6.4 g

Step 4: Calculate the Percentage of Yield of Copper

Percentage Yield = ( 4.8 ÷ 6.4 ) x 100 = 75%

Percentage Yield of Copper = 75%

 

Percentage purity

  • Often the product you are trying to fabricate may become contaminated with unwanted substances such as unreacted reactants, catalysts etc.

Equation:

Percentage Purity = (Mass of pure substance / Mass of impure substance) x 100

Example:

In an experiment 7.0g of impure calcium carbonate were heated to a very high temperature and 2.5g of carbon dioxide were formed. Calculate the percentage purity of the calcium carbonate.

Equation Of Reaction:

CaCO3 (s) → CaO(s) + CO2 (g)

Step 1: Calculate the relative formula masses

1 mole CaCO3 → 1 mole CO2

40+12+(3×16)      12+(2×16)

100 → 44

Step 2: Calculate the theoretical mass of calcium carbonate used if pure

From 2.5g CO2 we would expect 2.5/44 x 100 = 5.68g

Step 3: Calculate the percentage purity

(Mass of pure substance / mass of impure substance) x 100

= 5.68/7.0 x 100

= 81.1%

Author: Morgan

Morgan’s passion for the Periodic Table begun on his 10th birthday when he received his first Chemistry set. After studying the subject at university he went on to become a fully fledged Chemistry teacher, and now works in an international school in Madrid! In his spare time he helps create our fantastic resources to help you ace your exams.
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