4.2 Writing and Balancing Chemical Equations
Learning Objectives
- Write chemical equations using chemical formulas.
- Balance chemical equations.
Writing and balancing chemical equations is a fundamental aspect of chemistry, as it allows us to describe and understand chemical reactions in a concise and quantitative manner. Chemical equations consist of reactants, which are the substances that participate in a reaction, and products, which are the substances formed as a result of the reaction. Balancing the equation ensures that the number of atoms of each element is conserved in accordance with the law of conservation of mass.
Here’s a step-by-step description of how to write and balance chemical equations:
Write the balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia gas.
Step 1: Write the reactants and products of the chemical equation using the correct symbols and formulas.
- Reactants are written on the left side, and products are written on the right side. Use a plus sign to separate each reactant or product from others, as shown below:
[latex]\ce{N}_{2}+\ce{H}_{2}\ce{\qquad}\ce{NH}_{3}[/latex]
- Place an arrow pointing towards the products between reactants and products, as shown below:
[latex]\ce{N}_{2}+\ce{H}_{2}\ce{\rightarrow}\ce{NH}_{3}[/latex]
Step 2: Balance the chemical equation using suitable coefficients.
In a balanced chemical equation, an equal number of atoms of each element involved must be present on both sides (i.e. left and right sides of the equation).
- To balance a chemical equation, begin by selecting a specific atom to adjust first. It’s often advantageous to commence with the compound with the most significant number of atoms. Within that compound, focus on the element with the highest atom count. In this context, let’s initiate the balancing process with the hydrogen found in ammonia. Ammonia consists of three hydrogen atoms, while the left side has only two hydrogen atoms. Six hydrogen atoms are necessary for each side to balance, as this number allows both two and three to be evenly divided. Consequently, add the coefficient “2” in front of ammonia and “3” in front of molecular hydrogen to ensure a total of six hydrogen atoms on both sides of the equation. The process is shown below:
\[\begin{align*}
2\ce{NH}_{3} & \qquad2\times3=6\\
3\ce{H}_{2} & \qquad3\times2=6\\
\ce{N}_{2}+3\ce{H}_{2} & \rightarrow2\ce{NH}_{3}
\end{align*}\]
- Next, balance the rest of the atoms in the selected compound. In this example, the next atom will be nitrogen. The right side has two nitrogen atoms, and the left side also has two nitrogen atoms. Therefore, nitrogen is already balanced.
- Balance the remaining elements in the equation, if any. The selected equation only contains two types of elements. Therefore, no more atoms are left to be balanced.
Step 3: Do a final check to ensure all types of atoms in the equation are balanced.
At this point, the equation is balanced for both atoms present. Two nitrogens on each side and six hydrogens on each side.
[latex]\ce{N}_{2}+3\ce{H}_{2}\rightarrow2\ce{NH}_{3}[/latex]
Table 4.2.1 summarises the total number of atoms for each element involved on each side of the equation:
Table 4.2.1 Number of atoms on each side of the equation.
| Left side | Right side | |
|---|---|---|
| [latex]\ce{N}[/latex] | [latex]2[/latex] | [latex]2[/latex] |
| [latex]\ce{H}[/latex] | [latex]3\times2=6[/latex] | [latex]2\times3=6[/latex] |
Step 4: Check whether the coefficients are in their lowest possible whole numbers.
The ratio of [latex]1:3:2[/latex] is the lowest possible in whole numbers, as displayed in the following equation.
[latex]\ce{1N}_{2} + \ce{3H}_{2} {\rightarrow} \ce{2NH}_{3}[/latex]
Balance the following chemical equation:
[latex]\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{CO}_{2}+\ce{H}_{2}\ce{O}[/latex]
Step 1: Write the reactants and products of the chemical equation using the correct symbols and formulas.
As the chemical equation is already given, this step is not applicable:
[latex]\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{CO}_{2}+\ce{H}_{2}\ce{O}[/latex]
Step 2: Balance the chemical equation using suitable coefficients.
- Choose one atom to balance first. Generally, it helps to start with the compound that is composed of the greatest number of atoms. Focus first on the element of that compound which has the highest number of atoms.
In this case, start with hydrogen. The left side has six hydrogens, and the right side has two hydrogens. Placing three in front of water can give six hydrogen atoms on the right side as well. As can be seen in this equation:
[latex]\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{CO}_{2}+\ce{3H}_{2}\ce{O}[/latex]
- Next, balance the carbon atoms. The left side contains two carbon atoms, and the right side contains one carbon atom. Using two as the coefficient for carbon dioxide makes two carbons on each side. Thus: \[
\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{2CO}_{2}+\ce{3H}_{2}\ce{O}
\] - Finally, balance the oxygen atoms. The left side has three oxygen atoms, and the right side has seven oxygen atoms. Placing three in front of molecular oxygen on the left side gives seven oxygen atoms on each side. as can be seen in the following equation:\[
\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{3O}_{2}\rightarrow\ce{2CO}_{2}+\ce{3H}_{2}\ce{O}
\]
Step 3: Do a final check to ensure all types of atoms in the equation are balanced.
The following Table 4.2.2 summarises the number of atoms of each element present on both sides of the equation.
Table 4.2.2 Number of atoms on each side of the equation.
| Left side | Right side | |
|---|---|---|
| [latex]\ce{H}[/latex] | [latex]6[/latex] | [latex]3\times2=6[/latex] |
| [latex]\ce{C}[/latex] | [latex]2[/latex] | [latex]2[/latex] |
| [latex]\ce{O}[/latex] | [latex]1+\left(3\times2\right)=7[/latex] | [latex]\left(2\times2\right)+3=7[/latex] |
Step 4: Check whether the coefficients are in their lowest possible whole numbers.
The ratio of [latex]1:3:2:3[/latex] is the smallest possible whole number for the given equation. Therefore:
\[
\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{3O}_{2}\rightarrow\ce{2CO}_{2}+\ce{3H}_{2}\ce{O}
\]
Examples
Write and balance the chemical equation for the decomposition of hydrogen peroxide in the presence of light to water and oxygen gas.
Step 1: Write the reactants and products of the chemical equation using the correct symbols and formulas.
Reactants: Hydrogen peroxide
Products: Water and oxygen gas
Write the reactants on the left side and products on the right side of the equation. Then, place an arrow pointing towards the products as shown by the following equation. \[
\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\ce{O}_{2}
\]
Step 2: Balance the chemical equation using suitable coefficients.
\[
\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\ce{O}_{2}
\]
Table 4.2.3 Number of atoms on each side of the equation.
| Left side | Right side | |
|---|---|---|
| [latex]\ce{H}[/latex] | [latex]2[/latex] | [latex]2[/latex] |
| [latex]\ce{O}[/latex] | [latex]2[/latex] | [latex]3[/latex] |
According to the table 4.2.3, hydrogen is already balanced. There are two oxygen atoms on the left side and three oxygen atoms on the right side. The number of oxygen atoms can be balanced by placing [latex]\frac{1}{2}[/latex] in front of molecular oxygen. This gives two oxygen atoms on each side. Thus:
[latex]\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\frac{1}{2}\ce{O}_{2}[/latex]
Step 3: Do a final check to ensure all types of atoms in the equation are balanced.
At this point, the equation is balanced for both atoms present: two hydrogens on both sides and two oxygens on both sides, as shown in table 4.2.4.
Table 4.2.4 Number of atoms on each side of the equation.
| Left side | Right side | |
|---|---|---|
| [latex]\ce{H}[/latex] | [latex]2[/latex] | [latex]2[/latex] |
| [latex]\ce{O}[/latex] | [latex]2[/latex] | [latex]2[/latex] |
Step 4: Check whether the coefficients are in their lowest possible whole numbers.
The balanced chemical equation obtained in step 3 is as follows:
[latex]\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\frac{1}{2}\ce{O}_{2}[/latex]
Except for the coefficient of oxygen on the right side of the equation, the rest have whole-number coefficients. To make all coefficients in their smallest whole number, multiply all coefficients by two as has been done here:
[latex]\ce{2H}_{2}\ce{O}_{2}\rightarrow\ce{2H}_{2}\ce{O}+\ce{O}_{2}[/latex]
Watch the following simulation about balancing chemical equations.
Key Takeaways
- When writing chemical equations, place the reactants on the left side of the arrow and the products on the right side of the arrow.
- When balancing chemical equations, it’s often helpful to commence with the compound that has the most significant number of atoms.
- You can only change coefficients, not subscripts, in chemical formulas to balance the equation.
