Word and symbol equations

In a nutshell

Chemical reactions can be represented by word and symbol equations. Word equations show the reactants and products involved in a reaction. Symbol equations show the substances involved but they can provide more information, such as how the atoms are rearranged and proportions of each substance.

Chemical reaction

A chemical reaction can be represented by a word equation which uses the full name of reactants and products involved. Word questions have the following layout:

$$Reactant(s) \rightarrow Product(s)$$​

There is always an arrow between the reactants and products. The different products/reactants are separated by a $$+$$​ sign.

Example

Calcium reacts with chlorine to form calcium chloride. This can be represented with the following word equation:

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​$$calcium + chlorine \rightarrow calcium \space chloride$$​

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Example

Magnesium can react with carbon dioxide to form magnesium oxide and carbon. The word equation for this reaction is:

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​$$magnesium + carbon \space dioxide \rightarrow magnesium \space oxide + carbon$$​

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Symbol equations

Symbol equations show the same information as a word equation using chemical formulas and symbols.

Atoms cannot be created or destroyed in a chemical reaction, therefore there must be the same number of atoms of each element on both sides of the symbol equation.

Balanced symbol equations are able to show the ratio of each substance involved in the reaction. 

It is also easier to see how the arrangement of atoms changes in a chemical reaction with symbol equations.

Example 

Nitrogen reacts with hydrogen to form ammonia. For every molecule of nitrogen, three molecules of hydrogen are needed to react with. 

​$$N_2 + 3H_2 \rightarrow 2NH_3$$

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Balancing an equation

Equations must have the same number of each atom on both sides of the arrow, given that atoms are not created or destroyed in a chemical reaction. Equations are balanced by putting numbers in front of the chemical formulas.

Example

Pentane $$(C_5H_{12})$$​ reacts with oxygen to produce carbon dioxide and water.

$$C_5H_{12} + O_2 \rightarrow CO_2 + H_2O$$

There are five carbon atoms in butane, but only one carbon on the right-hand side. Therefore a five is placed in front of $$CO_2$$:​

​​$$C_5H_{12} + O_2 \rightarrow 5CO_2 + H_2O$$

There are now $$12$$​ hydrogen atoms on the left-hand side but only two on the right-hand side. Therefore a six is placed in front of $$H_2O$$:

​​$$C_5H_{12} + O_2 \rightarrow 5CO_2 + 6H_2O$$

There are now only two oxygen atoms on the left-hand side and $$16$$​ oxygen atoms in total on the right-hand side. Therefore an eight is placed in front of $$O_2$$:​

​​​$$C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O$$​

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The equation is now balanced.

Example

Benzoic acid reacts with oxygen to produce carbon dioxide and water.

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$$C_7H_6O_2+ O_2 \rightarrow CO_2 + H_2O$$

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There are seven carbon atoms on the left-hand side but only one on the right-hand side. Therefore a seven is placed in front of $$CO_2$$​​

​$$C_7H_6O_2+ O_2 \rightarrow 7CO_2 + H_2O$$​

There are six hydrogen atoms on the left-hand side but only two on the right-hand side. Therefore a three is placed in front of $$H_2O$$:

$$C_7H_6O_2+ O_2 \rightarrow 7CO_2 + 3H_2O$$

There are four oxygen atoms on the left-hand side and $$17$$​ oxygens on the right-hand side. To get $$17$$​ oxygen atoms on the left-hand side $$O_2$$​ can be multiplied by seven point five:

$$C_7H_6O_2+ 7.5O_2 \rightarrow 7CO_2 + 3H_2O$$

In balanced equations, fractions are often avoided. Therefore, everything in the equation is multiplied by two to get whole numbers:
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$$2C_7H_6O_2+ 15O_2 \rightarrow 14CO_2 + 6H_2O$$​

The equation is now balanced.