# How To Write A Chemical Equation

Show/Hide Sub-topics (Stoichiometry | O Level)

A chemical equation gives you the essential information about a chemical reaction, specifically:

• The nature of the reactants (chemicals reacting together) and products (chemicals produced)
• The quantities of reactants and products
• The physical state of the reactants and products (E.g. solid, liquid, gas, aqueous)

### Step by step guide to writing a chemical equation for a chemical reaction

• Write down a word equation for the chemical reaction
• Write down the unbalanced equation using correct chemical symbols for all the reactants and products
• Balance the equation by inspection. (Ensure that the same number of each type of atom appears on both sides of the equation)
• Write the state symbol after each chemical symbol

### Tips for balancing the equation

• Typically, you balance H (hydrogen) atoms first, followed by O (oxygen) atoms, then any other atoms. OR you can start by balancing the most complicated molecule.
• Never change the chemical symbol of a substance. (Do not change the subscripts in the chemical symbols of reactants or products)
• A number in front of a formula multiplies every symbol that follows it
• Remember that atoms cannot be created or destroyed in a chemical reaction (i.e. do not ‘eat’ any chemical symbols)

### Examples:

Formation of iron(III) chloride from the reacting iron with chlorine gas

• Iron + chlorine $\rightarrow$ iron(III) chloride
• Before balancing: $\text{Fe} \, + \, \text{Cl}_{2} \, \rightarrow \, \text{FeCl}_{3}$
• Balance number of Cl atoms: There are 2 Cl atoms on the left and 3 Cl atoms on the right. The simplest way to balance Cl atoms is to multiply 3 on the left and 2 on the right to make a total of 6 Cl atoms.
• We obtain: $\text{Fe} \, + \, 3\text{Cl}_{2} \, \rightarrow \, 2\text{FeCl}_{3}$.
• Notice that the number of Fe atoms are not the same for both side: 1 on the left and 2 on the right. Hence, we will just have to multiply 2 on the left.
• We have: $2\text{Fe} \, + \, 3\text{Cl}_{2} \, \rightarrow \, 2\text{FeCl}_{3}$
• Not forgetting to include the state symbols: $2\text{Fe} (s) \, + \, 3\text{Cl}_{2}(g) \, \rightarrow \, 2\text{FeCl}_{3}(s)$

Formation of carbon dioxide and water from methane and oxygen. (OR known as Combustion of methane in oxygen)

• methane + oxygen $\rightarrow$ carbon dioxide + water
• Convert the above to chemical symbols: $\text{CH}_{4} \, + \, \text{O}_{2} \, \rightarrow \, \text{CO}_{2} \, + \, \text{H}_{2}\text{O}$
• Balance H first: There are 4 H atoms on the left and 2 on the right. Hence, we will have to multiply the chemical symbol containing H on the left by 2. (I.e. multiply $\text{H}_{2}\text{O}$ by 2)
• We have: $\text{CH}_{4} \, + \, \text{O}_{2} \, \rightarrow \, \text{CO}_{2} \, + \, 2\text{H}_{2}\text{O}$
• Balance O: There are 2 O atoms on the left and 4 O atoms on the right. We will have to multiply the $\text{O}_{2}$ on the left by 2.
• We have: $\text{CH}_{4} \, + \, 2\text{O}_{2} \, \rightarrow \, \text{CO}_{2} \, + \, 2\text{H}_{2}\text{O}$
• The carbon atoms are balanced so we do not have to multiply them.
• Including the state symbol: $\text{CH}_{4} (g) \, + \, 2\text{O}_{2}(g) \, \rightarrow \, \text{CO}_{2}(g) \, + \, 2\text{H}_{2}\text{O}(l)$

Balance: $\text{Pb}_{3}\text{O}_{4} \, + \, \text{HNO}_{3} \, \rightarrow \, \text{PbO}_{2} + \text{Pb}(\text{NO}_{3})_{2} \, + \, \text{H}_{2}\text{O}$

• Notice that there are 3 Pb atoms on the left and 2 Pb atoms on the right. There are several ways to go about balancing the Pb atoms – we can multiply $\text{PbO}_{2}$ by 2 or $\text{Pb}(\text{NO}_{3})_{2}$ by 2 or multiply both sides of the equation. Any one of the ways is correct. But some ‘ideal’ way will be shorter than the others.
• We shall look at multiplying $\text{Pb}(\text{NO}_{3})_{2}$ by 2. (shortest method in this case)
• We have: $\text{Pb}_{3}\text{O}_{4} \, + \, \text{HNO}_{3} \, \rightarrow \, \text{PbO}_{2} + 2\text{Pb}(\text{NO}_{3})_{2} \, + \, \text{H}_{2}\text{O}$
• We have 1 N atom on the left and 4 N atoms on the right. We will have to multiply the $\text{HNO}_{3}$ by 4
• We have: $\text{Pb}_{3}\text{O}_{4} \, + \, 4\text{HNO}_{3} \, \rightarrow \, \text{PbO}_{2} + 2\text{Pb}(\text{NO}_{3})_{2} \, + \, \text{H}_{2}\text{O}$
• Balance H atom: $\text{Pb}_{3}\text{O}_{4} \, + \, 4\text{HNO}_{3} \, \rightarrow \, \text{PbO}_{2} + 2\text{Pb}(\text{NO}_{3})_{2} \, + \, 2\text{H}_{2}\text{O}$
• Notice that the number of O atoms are balanced. We’re done!
• Including state symbol: $\text{Pb}_{3}\text{O}_{4}(s) \, + \, 4\text{HNO}_{3}(aq) \, \rightarrow \, \text{PbO}_{2}(s) + 2\text{Pb}(\text{NO}_{3})_{2}(aq) \, + \, 2\text{H}_{2}\text{O}(l)$

Important: Always express your equation as the simplest form.

• For example:  $2\text{Pb}_{3}\text{O}_{4}(s) \, + \, 8\text{HNO}_{3}(aq) \, \rightarrow \, 2\text{PbO}_{2}(s) + 4\text{Pb}(\text{NO}_{3})_{2}(aq) \, + \, 4\text{H}_{2}\text{O}(l)$
• The above equation is not correct as it is not in its simplest form. (You can divide throughout by 2)