Balancing a Chemical Equation

Balancing a chemical equation ensures that the total number of atoms on each side of the equation is balanced by applying the Law of Conservation of Matter.

Certain rules must be observed for an equation to be properly balanced.

1. Molecules should never be altered: it is impossible to add an atom to a molecule or to modify subscripts of a molecule.

2. Molecules should never be added or removed.

3. Writing the coefficient 1 is not required, as it is implied.

4. Coefficients must be whole numbers. If fractions are needed to balance a reaction, the entire reaction must be multiplied by the same factor so as to have integer coefficients for all molecules.

5. The coefficients must be as small as possible.

6. After balancing the equation, it is always necessary to check whether the equation is correctly balanced by counting the atoms.

When using the most complex molecule method, the balancing of equations should start with the most complex molecule, which is the one with the most atoms, and save the simpler molecules for the end.

What is the balanced equation for the combustion of methane?

|CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O|

In this equation, the most complex molecule is the methane molecule  |(CH_{4})|. We must therefore start with this molecule to carry out the balance of atoms.

The number of carbon atoms is the same on the reactant and product side.
However, the number of hydrogen atoms is different: there are four atoms in the methane molecule, but only two atoms in the water molecule. Therefore, the molecule |H_{2}O| must be multiplied by 2.

The most complex molecules are, subsequently, |CO_{2}| and |H_{2}O|. The atoms of these molecules must therefore be balanced.

The carbon atoms are properly balanced. However, there are only two oxygen atoms on the reactant side, while there are four on the product side (two in the molecule |CO_{2}| and two in the molecule |H_{2}O| ). A coefficient of 2 must therefore be added in front of the molecule |O_{2}| .

The balanced equation is therefore given by this formula:

|CH_{4} + \color {red} {2} \space O_{2} ​\rightarrow CO_{2} + \color {red}{2}\space H_{2}O| 

What is the balanced equation for water synthesis?

|H_{2} + O_{2} \rightarrow H_{2}O|

In this equation, the most complex molecule is the water molecule. Therefore, to balance the atoms, the first step is to start with this molecule.

The number of hydrogen atoms is the same on the reactant side and on the product side.
However, the number of oxygen atoms is different: there is only one atom in the more complex molecule. Therefore the |O_{2}| molecule must be multiplied by |1/2| .

Since it is impossible to have fractional coefficients, all the coefficients must be multiplied by 2 to obtain whole coefficients.

|2 \times(H_{2} + \color{blue} {1/2} \space O_{2} ​\rightarrow H_{2}O)|

The balanced equation is therefore given by the following formula:

|\color {red}{2} \space H_{2} + \color {red}{1} \space O_{2} ​\rightarrow \color {red}{2} \space H_{2}O|

In the table method, each atom is balanced one after the other. Hydrogen |(H)| and oxygen |(O)| atoms are balanced at the end, in that order. All the other atoms are balanced at the beginning, without any specific order.

What is the balanced equation for ammonia synthesis?

|N_{2} + H_{2} \rightarrow NH_{3}|

Nitrogen |(N)| and hydrogen atoms must be balanced. Since hydrogen is more complex to balance, the first atom to balance will be nitrogen.

Nitrogen atoms are not balanced on either side of the equation. Two nitrogen atoms are present on the left side of the equation, while there is only one atom on the right side. The right side of the equation must then be multiplied by 2.

There are two hydrogen atoms on the left side of the reaction, while there are six on the right side. The |H_{2}| molecule must therefore be multiplied by 3 to balance the hydrogen atoms.

The balanced equation is therefore given by the following formula:

|N_{2} + \color {blue}{3}\space H_{2} ​\rightarrow \color {red}{2}\space NH_{3}|

What is the balanced equation for octane combustion?

|C_{8}H_{18} + O_{2} \rightarrow CO_{2} + H_{2}O|

Carbon |(C)| , hydrogen |(H)| , and oxygen |(O)| atoms must be balanced, in that order.

There are 8 carbon atoms in the reactants and only 1 in the products. A coefficient of 8 must therefore be placed in front of the |CO_{2}| molecule to balance the carbon atoms.

There are 18 hydrogen atoms in the reactants and 2 hydrogen atoms in the products. The molecule of |H_{2}O| must then be multiplied by 9 to obtain as many hydrogen atoms on each side of the equation.

To balance the oxygen atoms, the |O_{2}| molecule must be multiplied by a coefficient. Since there are two oxygen atoms in the reactants and 25 oxygen atoms in the reactants (16 from the |CO_{2}| molecule and 9 atoms from the |H_{2}O| molecule), the |O_{2}| molecule must thus be multiplied by |25/2| .

Since it is impossible to have fractional coefficients, all coefficients must be multiplied by 2 to obtain integer coefficients.

|2 \times (C_{8}H_{18} + \color {green}{25/2}\space O_{2} ​\rightarrow \color {red}{8} \space CO_{2} + \color {blue}{9} \space H_{2}O)|

So, the balanced equation is:

|2 \space C_{8}H_{18} + \color {green}{25}\space O_{2} ​\rightarrow \color {red}{16} \space CO_{2} + \color {blue}{18} \space H_{2}O|

In the algebraic method, the coefficients of the chemical equation are replaced by algebraic variables. Equations are then created for each atom.

Balance the following equation:

|Fe_{2}O_{3} + C \rightarrow Fe + CO|

Initially, algebraic variables must be substituted for the coefficients of each molecule in the chemical equation.

|\color {red}{a} \space Fe_{2}O_{3} +\color {red}{b} \space C \rightarrow \color {red}{c} \space Fe + \color {red}{d} \space CO|

In the equation to be balanced, atoms of iron, oxygen, and carbon are present. An equation must therefore be created for each atom, taking into account the coefficients and the number of atoms in each of the molecules.

For the iron atom: |2a = c|
For the oxygen atom: |3a = d|
For the carbon atom: |b = d|

To solve the equation, a variable must be substituted by an arbitrary value. In the example below, the value of |a| will be 1, as this variable allows to deduce the values ​​of |c| and |d| .

Since |a = 1|, the equation of the iron atom becomes |2 \times 1 = c|, or |c = 2|.
Since |a = 1|, the equation of the oxygen atom becomes |3 \times 1 = d|, or |d = 3|.
Since |d = 3|, the equation of the carbon atom becomes |b = 3|.

Thus, the balanced equation is:

|\color {red}{1} \space Fe_{2}O_{3} +\color {red}{3} \space C \rightarrow \color {red}{2} \space Fe + \color {red}{3} \space CO|

Balance the following equation.

|C_{6}H_{5}COOH + O_{2} \rightarrow CO_{2} + H_{2}O|

Initially, algebraic variables must be substituted for the coefficients of each molecule in the chemical equation.

|\color {red}{a} \space C_{6}H_{5}COOH +\color {red}{b} \space O_{2} \rightarrow \color {red}{c} \space CO_{2} + \color {red}{d} \space H_{2}O|

In the equation to be balanced, atoms of carbon, hydrogen, and oxygen are present. An equation must therefore be created for each atom, taking into account the coefficients and the number of atoms in each of the molecules.

For the carbon atom: |7a = c|
For the hydrogen atom: |6a = 2d|
For the oxygen atom: |2a + 2b = 2c + d|

To solve the equation, a variable must be substituted with an arbitrary value. In the example below, the value of |a| will be 1, as this variable allows to deduce the values of |c| and |d|.

Since |a = 1|, the equation of the carbon atom becomes |7 \times 1 = c| , or |c = 7|.
Since |a = 1|, the equation of the hydrogen atom becomes |6 \times 1 = 2d| , or |d = 3|.
Since |a = 1|, |c = 7| and |d = 3|, the equation of the oxygen atom becomes:

|2 + 2b = 2 \times 7 + 3|
|2 + 2b = 14 + 3|
|2b = 15|
|b = 15/2|

Since there is a fractional coefficient, all coefficients must be multiplied to have only integer coefficients.

|a = 1 \color {red}{\times 2} = 2|
|b = 15/2 \color {red}{\times 2} = 15|
|c = 7 \color {red}{\times 2} = 14|
|d = 3 \color {red}{\times 2} = 6|

Thus, the balanced equation is:

|\color {red}{2} \space C_{6}H_{5}COOH +\color {red}{15} \space O_{2} \rightarrow \color {red}{14} \space CO_{2} + \color {red}{6} \space H_{2}O|