CIE IGCSE Chemistry

Revision Notes

3.1.4 Ar & Mr

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Relative Masses

Relative Atomic Mass

  • The symbol for the relative atomic mass is Ar
  • The relative atomic mass for each element can be found in the Periodic Table along with the atomic number
  • The relative atomic mass is shown underneath the atomic symbol and is larger than the atomic number (except for hydrogen where they are the same)
  • Atoms are too small to accurately weigh but scientists needed a way to compare the masses of atoms 
  • The carbon-12 is used as the standard atom and has a fixed mass of 12 units 
  • It is against this atom which the masses of all other atoms are compared 
  • Relative atomic mass (Ar) can therefore be defined as:
    • The average mass of the isotopes of an element compared to 1/12th of the mass of an atom of 12
  • The relative atomic mass of carbon is 12
    • The relative atomic mass of magnesium is 24 which means that magnesium is twice as heavy as carbon 
    • The relative atomic mass of hydrogen is 1 which means it has one twelfth the mass of one carbon-12 atom 

Relative molecular (formula) mass

  • The symbol for the relative molecular mass is Mr and it refers to the total mass of the molecule
  • To calculate the Mr of a substance, you have to add up the relative atomic masses of all the atoms present in the formula
  • Relative formula mass is used when referring to the total mass of an ionic compound

Relative Formula Mass Calculations Table

Substance, atoms present table, IGCSE & GCSE Chemistry revision notes

Reacting masses

  • The Law of Conservation of mass tells us that mass cannot be created or destroyed
  • In a chemical reaction, the total mass of reactants equals the total mass of the products
  • We can use this, along with relative atomic/formula masses to perform calculations to identify the quantities of reactants or products involved in a chemical reaction
  • Example:

2Ca + O2 → 2CaO

  • Relative atomic masses: Ca = 40; O = 16
  • Using the balanced symbol equation shows that 2 x 40 = 80 units of mass of calcium react with 2 x 16 = 32 units of mass of oxygen (O2 molecule, 16 + 16 = 32) to form 2 x (40 + 16) = 112 units of mass of CaO:

2Ca + O2 → 2CaO

80 + 32   =   112

  • The ratio of the mass of calcium and oxygen reacting will always be the same, regardless of the units
    • E.g. 80 g of calcium will react with 32 g of oxygen to form 112 g of calcium oxide
    • Or, 40 tonnes of calcium will react in excess oxygen to form 56 tonnes of calcium oxide

Worked example

Calculate the mass of carbon dioxide produced when 32 g of methane, CH4, reacts completely in excess oxygen:

CH4 + 2O2 → CO2 + 2H2O

Relative atomic masses, Ar: H = 1; C = 12; O = 16

Answer

    • In terms of relative mass the equation is:

                CH4      +       2O2        →         CO2     +        2H2O

              12 + (4 x 1) + 2 x (2 x 16) → 12 + (2 x 16) + 2 x (2 x 1 + 16)

               16       +          64        →         44        +         36

So 16 g of methane would react in excess oxygen to form 44 g of carbon dioxide

Therefore, 32 g of methane would produce 44 x 2 = 88 g of carbon dioxide

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