Ar & Mr (Cambridge (CIE) IGCSE Chemistry)
Revision Note
Written by: Caroline Carroll
Reviewed by: Stewart Hird
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Relative masses
Relative atomic mass
The symbol for relative atomic mass is Ar
The relative atomic mass for each element can be found in the Periodic Table, along with the atomic number
Relative atomic mass is shown on the atomic symbol
It is always larger than the atomic number (except for hydrogen, where they are the same)
Use the key on the Periodic Table to correctly identify the mass number
Key for chemical symbols on the Periodic Table
This key is given on the Periodic Table in exams and identifies the number that is the relative atomic mass.
Atoms are too small to accurately weigh but scientists needed a way to compare the masses of atoms
Carbon-12 is used as the standard atom and has a fixed mass of 12 units
The mass of all other atoms are compared against carbon-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
Relative molecular mass is the sum of the relative atomic masses of all the atoms in a molecule
The term relative formula mass is used when referring to the total mass of an ionic compound
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 Calculations Table
Substance | Atoms present | Calculation | Mr |
|---|---|---|---|
Hydrogen | 2 x H | (2 x 1) | 2 |
Water | (2 x H) + (1 x O) | (2 x 1) + (1 x 16) | 18 |
Potassium carbonate | (2 x K) + (1 x C) + (3 x O) | (2 x 39) + (1 x 12) + (3 x 16) | 138 |
Calcium hydroxide | (1 x Ca) + (2 x O) + (2 x H) | (1 x 40) + (2 x 16) + (2 x 1) | 74 |
Ammonium sulfate | (2 x N) + (8 x H) + (1 x S) + (4 x O) | (2 x 14) + (8 x 1) + (1 x 32) + (4 x 16) | 132 |
Worked Example
Calculate the relative formula mass of:
Sodium chloride, NaCl
Copper oxide, CuO
Magnesium nitrate, Mg(NO3)2
Answers:
Sodium chloride
NaCl = 23 + 35.5 = 58.5
Copper oxide
CuO = 63.5 + 16 = 79.5
Magnesium nitrate
Mg(NO3)2 = 24 + (14 x 1 x 2) + (16 x 3 x 2) = 148
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:
Reactants:
2 x 40 = 80 units of mass of calcium
2 x 16 = 32 units of mass of oxygen (O2 molecule, 16 + 16 = 32)
Products:
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
80 g of calcium will react with 32 g of oxygen to form 112 g of calcium oxide
80 tonnes of calcium will react with 32 tonnes of oxygen to form 112 tonnes of calcium oxide
So, 40 kg of calcium will react with 16 kg of oxygen to form 56 kg 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:
Using the balanced symbol equation:
Reactants:
12 + (4 x 1) = 16 units of mass of methane
2 x 16 = 32 units of mass of oxygen (O2 molecule, 16 + 16 = 32)
Products:
12 + (2 x 16) = 44 units of mass of carbon dioxide
2 x ((2 x 1) + 16) = 36 units of mass of water
CH4 + 2O2 → CO2 + 2H2O
16 + 64 → 44 + 36
So, 16 g of methane would react in excess oxygen to form 44 g of carbon dioxide
32 g of methane is double the amount of methane from the balanced symbol equation
So, this would produce double the amount of caron dioxide
2 x 44 = 88 g of carbon dioxide
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