Law of Conservation of Mass Calculations
The Law of Conservation of Mass is defined and explained using examples of reacting mass
calculations using the law are fully explained with worked out examples using the balanced
symbol equation. The method involves reacting masses deduced from the balanced symbol
equation.
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When elements and compounds react to form new products, mass cannot be lost or gained.
"The Law of Conservation of Mass" definition states that mass cannot be created or
destroyed, but changed into different forms.
So, in a chemical change, the total mass of reactants must equal the total mass of
products.
By using this law, together with atomic and formula masses, you can calculate the quantities
of reactants and products involved in a reaction and the simplest formula of a compound
NOTE: (1) the symbol equation must be correctly balanced to get the right answer! (2) There are good
reasons why, when doing a real chemical preparation-reaction to make a substance you will not get
100% of what you theoretically calculate.
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Example 3.1: Magnesium + Oxygen ==> Magnesium oxide
o 2Mg + O2 ==> 2MgO (atomic masses required: Mg=24 and O=16)
o think of the ==> as an = sign, so the mass changes in the reaction are:
o (2 x 24) + (2 x 16) = 2 x (24 + 16)
o 48 + 32 = 2 x 40 and so 80 mass units of reactants = or produces 80 mass units of
products (you can work with any mass units such as g, kg or tonne (1 tonne = 1000
kg)
Example 3.2: iron + sulphur ==> iron sulphide (see the diagram at the top of the page!)
o Fe + S ==> FeS (atomic masses: Fe = 56, S = 32)
o If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left
unreacted? (assuming all the sulphur reacted)
o From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS.
o This means 59 - 56 = 3g Fe unreacted.
Example 3.3: When limestone (calcium carbonate) is strongly heated, it undergoes thermal
decomposition to form lime (calcium oxide) and carbon dioxide gas.
o CaCO3 ==> CaO + CO2 (relative atomic masses: Ca = 40, C = 12 and O = 16)
o Calculate the mass of calcium oxide and the mass of carbon dioxide formed by
decomposing 50 tonnes of calcium carbonate.
o (40 + 12 + 3x16) ==> (40 + 16) + (12 + 2x16)
o 100 ==> 56 + 44, scaling down by a factor of two, 50 ==> 28 + 22
o so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of
carbon dioxide gas.