# How to Balance Chemical Equations : An Easy Approach

How to balance chemical equations when you encounter one. But before that, we must understand what a chemical equation is and how it works.

A chemical equation is a written with a symbolic representation of a chemical reaction in form of various scientific symbols. The symbols are the elemental alphabet or alphabets representing the element. The reactant chemical(s) are always taken on the left-hand side and the product chemical(s) are written on the right-hand side.

The reactants and products are connected with an arrow leading from the left to the right which symbolizes the reaction. The law of conservation of mass always states that no matter can be created or destroyed in a chemical reaction, it can only be transformed from one form to the another. Therefore, the number of atoms, present in the reactants has to be equal to the number of atoms present in the product side.

Follow the given guidelines to carefully learn how to balance chemical equations every time.

**Steps to How to Balance Chemical Equations**

**Doing a Traditional Balance****Completing an Algebraic Balance**

**Step 1 : Doing a Traditional Balance**

**Write down your given equation. **

For an example, we will use: C3H8 + O2 –> H2O + CO2

The reaction takes place when propane (C3H8) is oxidized in the presence of oxygen to produce water and carbon dioxide.

**Write down the number of atoms per element. **

This step must be done for each side of the equation. Look at the subscripts written next to every atom to get the count of number of atoms in the equation. It is a good idea to connect it back to the original equation while writing it out, noting how every element appears in the equation.

In the given example, we have 3 oxygen atoms on the right side after get a total as a result from addition.

Left side: 3 carbon (C3), 8 hydrogen (H8) and 2 oxygen (O2).

Right side: 1 carbon (C), 2 hydrogen (H2) and 3 oxygen (O + O2).

**Save hydrogen and oxygen for last**

As hydrogen and oxygen are often on both sides, save them for last. Hydrogen and oxygen both are common in molecules. Therefore, it’s likely that you’ll may have them on both sides of the equation. It is best to balance them last. We are required to recount the atoms before balancing the hydrogen and oxygen. We are needed to utilize the coefficients of the to balance the another atoms in the equation.

**Start with single elements. **

If we have more than one element left to balance in the given equation, select the element which appears in only a single molecule of reactants and in only a single molecule of products in the equation. This means that we will be required to balance the carbon atoms first.

**Use a coefficient for balancing the single carbon atom. **

Put a coefficient to the single carbon atom on the right of the equation to balance it. This will definitely not going to change much of the equation but will help to balance the number of carbon atoms on the right to the left which happens to be 3 carbon atoms on the left of the equation.

C3H8 + O2 –> H2O + 3CO2

The coefficient 3 placed in front of the right side carbon atoms indicates 3 carbon atoms just as the subscript 3 on the left side shows 3 carbon atoms.

In a chemical equation, we can change coefficients, but we must never alter or change the number given as a subscript.

**Balance the hydrogen atoms next. **

Since we have balanced all atoms besides the hydrogen and oxygen, we can now check on the hydrogen atoms. We have 8 on the left side. So we are going to need 8 on the right side. Use a coefficient to get this thing right.

C3H8 + O2 –> 4H2O + 3CO2

On the right side, we now add 4 as the coefficient because the subscript showed that we already had 2 hydrogen atoms. When we multiply the coefficient with 4 to the subscript 2, we get 8. The other 6 atoms of oxygen come from 3CO2 i.e; 3×2=6 atoms of oxygen+ the other 4=10.

**Balance the oxygen atoms.**

Remember to account for the coefficients that we have been using to balance out the other atoms. Because we have added coefficients to the molecules on the right side of the equation, the number of oxygen atoms must have changed. We have 4 oxygen atoms in the water molecules and 6 oxygen atoms in the carbon dioxide molecules for the equation. That makes a total of 10 oxygen atoms. Add a coefficient of 5 to the oxygen molecule on the left side of the equation. We now have 10 oxygen atoms on each side.

C3H8 + 5O2 –> 4H2O + 3CO2. The carbon, hydrogen, and oxygen atoms are balanced. Our chemical equation is complete.

**Step 2 : Completing an Algebraic Balance**

The method, known as Bottomley’s method, is especially useful for all those more complex reactions. But it does take a bit longer to solve and balance the given chemical equation.

**Write down the given equation.**

For this example, we will use: PCl5 + H2O –> H3PO4 + HCl

**Assign a letter to each substance.**

aPCl5 + bH2O –> cH3PO4 + dHCl

**Check the number of each element.**

Go through the number of each element written on both sides, and try to set those equal to each other. aPCl5 + bH2O –> cH3PO4 + dHCl

On the left side, there are 2b atoms of hydrogen i.e.; 2 for every molecule of H2O, while on the right side, there are 3c+d atoms of hydrogen i.e.; 3 for every molecule of H3PO4 and 1 for every molecule of HCl). As the number of atoms of hydrogen has to remain equal on both sides to get a balanced chemical equation, 2b must be equal to 3c+d.

Do this for each element.

P: a=c

Cl: 5a=d

H: 2b=3c+d

**Solve this system of equations**

Solve the system of equations to get the numeric value for all the coefficients. As there are more variables than equations, there are multiple solutions. We must find the one where every variable is in its smallest, non-fractional form. Take one variable quickly and assign a value to it. Let’s take a=1 and start solving the system of equations to get the following values:

As P: a = c, we know that c = 1.

As Cl: 5a = d, we know that d = 5

As H: 2b = 3c + d, we can calculate b like this:

2b = 3(1) + 5

2b = 3 + 5

2b = 8

b=4

This shows us the values are as follows:

a = 1

b = 4

c = 1

d = 5

The whole balancing process, when done in an orderly manner, will definitely give us a balanced chemical equation.