TRIAL BALANCING: STEPS, EXAMPLES AND EXERCISES - CHEMISTRY - 2025

The trial and error balancing consists of a trial and error method that seeks to ensure that the conservation of matter is fulfilled in a chemical equation for a given reaction; that is, to equalize the numbers of atoms of the reactants and products. Thus, they will not vanish or create atoms out of thin air.

Depending on the rush, this is usually an entertaining operation, reinforcing understandings about stoichiometric coefficients and subscripts. Although it may not seem like it, trial and error involves mastering many concepts, applied almost unconsciously to those who dabble in chemistry.

Balancing a chemical equation by trial and error is like trying to level a seesaw in the air. Image by michael maggiore from Pixabay

Thus, balancing resembles the effort that would be made to level a seesaw (or seesaw), such that neither end falls to one side while the other rises. A scale also illustrates this perfectly.

As you have experience, this balancing can even be done mentally, as long as the chemical equation is not too complicated. A bad swing completely ruins the interpretation of a reaction, so it is essential to do it calmly to avoid eating mistakes.

Steps

Write the unbalanced equation

Regardless of what balancing is required for, you should always start with the unbalanced equation at hand. Likewise, it is important to be clear about its elements. Assume the following chemical equation:

A + B → 3C + D

Where species A, B, C and D are molecular. This equation cannot be balanced because it tells us nothing about its atoms. The atoms are balanced, not the molecules.

Both A, B and D have a stoichiometric coefficient of 1, while C of 3. This means that 1 molecule or mole of A reacts with one molecule or mole of B, to produce 3 molecules or moles of C, and one molecule or mole of D. When we show the atoms, we introduce the stoichiometric subscripts.

Review stoichiometric coefficients and subscripts

Now suppose the following equation:

CH ₄ + O ₂ → CO ₂ + H ₂ O

Stoichiometric subscripts tell us how many atoms of each element make up a molecule, and they are recognized because they are the smallest numbers on the right side of an atom. For example, CH ₄ has one carbon atom (although the 1 is not listed) and four hydrogen atoms.

Balance the atoms in the smallest proportion first

According to the unbalanced equation above, carbon is the minor atom: it is part of a single reactant (CH ₄) and a single product (CO ₂). If you look at it, there is a C atom on both the reactants and products side.

Balance by oxygen atoms

CH ₄ + O ₂ → CO ₂ + H ₂ O

2 O 3 O

We cannot change the subscripts, but only the stoichiometric coefficients in order to balance an equation. There are more oxygens on the right side, so we try to add a coefficient to the O ₂:

CH ₄ + 2O ₂ → CO ₂ + H ₂ O

4 or 3

We do not want to affect the coefficient of CO ₂ because it would unbalance the C atoms. We then change the coefficient of H ₂ O:

CH ₄ + 2O ₂ → CO ₂ + 2H ₂ O

4 or 4

Lastly balance the hydrogen atoms

Once we have balanced the oxygen atoms, we finally balance the hydrogen atoms. Several times these are by themselves balanced at the end.

CH ₄ + 2O ₂ → CO ₂ + 2H ₂ O

4H 4H

And so the equation has been balanced by trial and error. The order of these steps is not always fulfilled.

Examples

Balanced equations are shown below to verify that the number of its atoms is equal on both sides of the arrow:

SO ₂ + 2H ₂ → S + 2H ₂ O

P ₄ + 6F ₂ → 4PF ₃

2HCl → H ₂ + Cl ₂

C + O ₂ → CO ₂

Exercises

Some proposed exercises will be solved below. In some of them it will be seen that sometimes it is convenient to break the order of the steps and balance the minority atom last.

Exercise 1

Balance by trial and error the following chemical equation:

SO ₃ → SO ₂ + O ₂

1S 1S

3 or 4

It is important to emphasize that the coefficients multiply the subscripts to give us the total number of atoms for an element. For example, 6N ₂ gives us a total of 12 N atoms.

The sulfur at the beginning is already balanced, so we continue with the oxygen:

3 O 4 O

We are forced to change the coefficient to SO ₃ to balance the oxygens on the left side:

2SO ₃ → SO ₂ + O ₂

6 O 4 O

2S S

Now we are interested in balancing the sulfur atoms first before the oxygen atoms:

2SO ₃ → 2SO ₂ + O ₂

2S 2S

6 OR 6O

Note that the oxygen atoms were left balanced by themselves in the end.

Exercises 2

Balance by trial and error the following chemical equation:

CH ₄ + H ₂ O → CO + H ₂

The carbons and oxygens are already balanced, not in the same way the hydrogens:

6H 2H

All we have to do is modify the coefficient of H ₂ to have more hydrogens on the right:

CH ₄ + H ₂ O → CO + 3H ₂

6H 6H

And the equation is totally balanced.

Exercise 3

Balance by trial and error the following chemical equation:

C ₂ H ₄ + O ₂ → CO ₂ + H ₂ O

We start balancing the carbon again:

C ₂ H ₄ + O ₂ → 2CO ₂ + H ₂ O

2C 2C

2O 5O

4H 2H

Note that this time it is easier to balance the hydrogens first than the oxygens:

C ₂ H ₄ + O ₂ → 2CO ₂ + 2H ₂ O

4H 4H

2O 6O

Now, we modify the coefficient of O ₂ to balance the oxygens:

C ₂ H ₄ + 3O ₂ → 2CO ₂ + 2H ₂ O

6O 6O

And the equation is already balanced.

Exercise 4

Finally, a challenging equation will be balanced by trial and error:

N ₂ + H ₂ O → NH ₃ + NO

Nitrogen and oxygen are already balanced, but not hydrogens:

2H 3H

Let's try changing the coefficient of H ₂ O and NH ₃:

N ₂ + 3H ₂ O → 2NH ₃ + NO

6H 6H

3O O

2N 3N

By trial and error we vary the coefficient of NO:

N ₂ + 3H ₂ O → 2NH ₃ + 3NO

6H 6H

3O 3O

2N 5N

And now the nitrogens are unbalanced. Here it is convenient to make a sudden change: quintuple the coefficient of N ₂:

5N ₂ + 3H ₂ O → 2NH ₃ + 3NO

10 N 5N

6H 6H

3O 3O

Thus, it remains for us to play with the coefficients of NH ₃ and NO in such a way that they add 10 nitrogens and balance the oxygen and hydrogen atoms at the same time. Let's try this score:

5N ₂ + 3H ₂ O → 5NH ₃ + 5NO

10 N 10 N

6 H 15H

3O 5O

However, the hydrogens look very unbalanced. Therefore, let's vary the coefficients again:

5N ₂ + 3H ₂ O → 4NH ₃ + 6NO

10 N 10N

6H 12H

3O 6O

Note that now the left side has twice the oxygen and hydrogen. At this point, it is enough to double the coefficient of H ₂ O:

5N ₂ + 6H ₂ O → 4NH ₃ + 6NO

10 N 10N

12H 12H

6O 6O

And the equation is finally balanced.

References

1. Whitten, Davis, Peck & Stanley. (2008). Chemistry (8th ed.). CENGAGE Learning.
2. Organic Chemistry. (sf). Balancing chemical equations by trial and error - Solved Exercises. Recovered from: quimica-organica.com
3. Nissa Garcia. (2020). Balanced Chemical Equation: Definition & Examples. Study. Recovered from: study.com
4. Helmenstine, Anne Marie, Ph.D. (October 21, 2019). How to Balance Chemical Equations. Recovered from: thoughtco.com
5. Study Guide. (May 11, 2019). Trial balancing of chemical reactions. Solved Exercises. Recovered from: quimicaencasa.com
6. University of Colorado Boulder. (2019). Balancing Chemical Equations. Recovered from: phet.colorado.edu