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The Law of Conservation of Mass

The law of conservation of mass

In order to balance a chemical equation, it is important to understand the law of conservation of mass.

Definition: The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the processes acting inside the system. Matter can change form, but cannot be created or destroyed.

For any chemical equation (in a closed system) the mass of the reactants must be equal to the mass of the products. In order to make sure that this is the case, the number of atoms of each element in the reactants must be equal to the number of atoms of those same elements in the products. An example is shown below:


Iron is a metal. When we represent it in a balanced chemical equation, we write only \(\text{Fe}\). Sulfur occurs as \(\text{S}_{8}\) but we write only the empirical formula: \(\text{S}\). We do this for all network structures. Writing formulae like this represents one unit of the compound or network structure.


\(\text{Fe} + \text{S} \rightarrow \text{FeS}\)

Mass of one atom of \(\text{Fe}\) is \(\text{55.8}\)

Mass of one atom of \(\text{S}\) is \(\text{32.1}\)

Mass of one atom of \(\text{FeS}\) is \(\text{87.9}\)

Mass of reactants is \(\text{87.9}\)

Mass of products is \(\text{87.9}\)

To calculate the mass of the molecules we use the relative atomic masses for iron and sulfur, as seen in the table above. You will notice that the mass of the reactants equals the mass of the product. A chemical equation that is balanced will always reflect the law of conservation of mass and the law of conservation of atoms.

Optional Activity: Balancing chemical equations


You will need: coloured balls (or marbles), prestik, a sheet of paper and coloured pens.

We will try to balance the following equation:

\[\text{Al} + \text{O}_{2} \rightarrow \text{Al}_{2}\text{O}_{3}\]

Take one ball of one colour. This represents a molecule of \(\text{Al}\). Take two balls of another colour and stick them together. This represents a molecule of \(\text{O}_{2}\). Place these molecules on your left. Now take two balls of the first colour and three balls of the second colour to form \(\text{Al}_{2}\text{O}_{3}\). Place this compound on your right. On a piece of paper draw coloured circles to represent the balls. Draw a line down the centre of the paper to represent the molecules on the left and on the right.

Count the number of balls on the left and the number on the right. Do you have the same number of each colour on both sides? If not, the equation is not balanced. How many balls of each colour will you have to add to each side to make the number of balls the same? How would you add these balls?

You should find that you need four balls of the first colour for \(\text{Al}\) and three pairs of balls of the second colour (i.e. six balls in total) for \()_{2}\) on the left side. On the right side you should find that you need \(\text{2}\) clusters of balls for \(\text{Al}_{2}\text{O}_{3}\). We say that the balanced equation is:

\[4\text{Al} + 3\text{O}_{2} \rightarrow 2\text{Al}_{2}\text{O}_{3}\]


Use jelly tots and toothpicks to build the following chemical equation. Make sure that your atoms are balanced. Use the same colour jelly tots for the same atoms.

\[\text{C} + \text{H}_{2}\text{O} \rightarrow \text{CO}_{2} + \text{CO} + \text{H}_{2}\]

Add compounds until the atoms are balanced. Write the equation down and use a coefficient to indicate how many compounds you used. For example if you had to use three water molecules then write \(3\text{H}_{2}\text{O}\)


Use ball and stick drawings to balance the atoms in the following reaction:

\[\text{NH}_{3} + \text{O}_{2} \rightarrow \text{NO} + \text{H}_{2}\text{O}\]

Use your drawings to write a balanced chemical equation for the reaction.


Lead \((\text{Pb})\), lead (IV) oxide \((\text{PbO}_{2})\) and sulfuric acid \((\text{H}_{2}\text{SO}_{4})\) are used in car batteries. The following reaction takes place: \(\text{Pb} + \text{PbO}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{PbSO}_{4} + \text{H}_{2}\text{O}\)

Cut out circles from four different colours of paper to represent each of the atoms. Build a few of the compounds (\(\text{Pb}\), \(\text{PbO}_{2}\), \(\text{H}_{2}\text{SO}_{4}\)). These are the reactants. Do not build the products. Rearrange the atoms so that the products are formed. Add more reactants if needed to balance the atoms (e.g. you will need two \(\text{H}_{2}\text{SO}_{4}\) molecules). Use what you have learnt to write a balanced equation for the reaction.

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