# Is there a better way balancing reaction equations than systematic trial and error? [duplicate]

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I was introduced to the concept of balancing chemical equations in school and I did some practice examples, which were easy and took me about 2 minutes per question. I used the systematic trial and error method and I don't really like it.

I'm worried that my teacher will give us a more difficult question, so I'm looking for a method to solve chemical equations such as $$\ce{C3H8 + O2 -> CO2 + H2O}.$$

I have tried matrices, but I don't really understand them, as I am not in grade 12 yet. If you know matrices, can you explain how I can balance the chemical equation with aforementioned method (thoroughly)? I would also be grateful if you have any other useful and quick methods which work for pretty much any high school chemical equation balancing problem.

## marked as duplicate by ron, airhuff, Todd Minehardt, Wildcat, ButtonwoodMay 14 '17 at 20:42

• I really wouldn't bother learning matrices just for balancing at this early stage. You can just do these by observation, where you 'juggle' and make sure that you always have the same number of each element on either side. If you do insist on learning matrices just to do this (which I advise against), you can start here – surelyourejoking Oct 24 '15 at 4:41

When using matrices, you are really just solving a system of equations. You can use any technique you know to solve the system of equations that a chemical equation represents. Using matrices is only one technique that may be useful for complicated equations or if you have a calculator/MATLAB on hand. In the given example, it would be easier to solve it directly with guess and check.

# Writing the system of equations

In your example $$\ce{C3H8 + O2 -> CO2 + H2O},$$ write in some variables as coefficients: $$\ce{a C3H8 + b O2 -> c CO2 + d H2O}.$$ Now, we can write out a system of equations based on the stoichiometry of each element on either side of the equation. In other words, write equations that express the number of each element on the left hand side to the number of each element on the right hand side: \begin{align} 3a &= c\\ 8a &= 2d &\Longleftrightarrow&& 4a &= d\\ 2b &= 2c + d \end{align}

Finally, we have to choose a value for one variable to substitute in. This will be the fourth and final equation. We want integer coefficients here, so we will choose a substitution that gives us all integer coefficients. Otherwise, just multiply your coefficients at the end so that they are all integers. Here, I'll let $$a = 1$$.

# Solving the system

## Substitution

Substituting the first two equations into the third gives: \begin{align} 2b &= 6a + 4a = 10a &\Longleftrightarrow&& b &= 5a \end{align}

\begin{align} a &= 1\\ b &= 5\\ c &= 3\\ d &= 4 \end{align} And the balanced equation is $$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$

## Using matrices

To answer the question though, let's do it with matrices. I won't go through the whole explanation of solving the system but I'll show you how to build the matrices and get an answer. Remember, we are just representing the same system of equations from before, but with matrices now!

We need to build two matrices, $$A$$ and $$b$$. Reorganize the equations we wrote previously so that all of the variables are on the left hand side and a constant is on the right hand side. Also, notice I added in our equation let $$a = 1$$: \begin{align} 1a + 0b + 0c + 0d &= 1\\ 3a + 0b -1c + 0d &= 0\\ 0a + 2b -2c - 1d &= 0\\ 4a + 0b + 0c - 1d &= 0 \end{align}

These equations are written like this: \begin{align} A &= \begin{bmatrix} 1 & 0 & 0 & 0\\ 3 & 0 & -1 & 0\\ 0 & 2 & -2 & -1\\ 4 & 0 & 0 & -1\\ \end{bmatrix}\\ b &= \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix} \end{align}

Notice that $$b$$ is just a column of the constants on the right hand sides of the equations, in the same order as they were given. The order is important here!

There are a few ways to solve this manually or in a calculator/MATLAB. If you have a TI-8x calculator or similar, you could use its rref function (this transforms an augmented matrix into reduced row echelon form, I'm sure google could help you out) or you could multiply $$A^{-1}b = x$$. In MATLAB, just do: A\b. This gave me: $$x = \begin{bmatrix} 1 \\ 5 \\ 3 \\ 4 \end{bmatrix}$$ Each row in this final matrix $$x$$ is the value for $$a$$, $$b$$, $$c$$, $$d$$ in that order, giving the balanced chemical equation $$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$

• @aabmass How would I be able to solve this (matrices) manually? – Imagine Dragons Oct 25 '15 at 16:48
• @ImagineDragons Using Gauss-Jordan elimination would be simple for you. I can't explain in a comment, but I'm sure you can find great examples online. However, it boils down to the elimination method for solving a system of equations. You basically multiply/add/subtract rows with each other like you would for elimination. – aabmass Oct 27 '15 at 4:17

Always start balancing by equating the number of atoms of elements which appear only once. For example, in the equation you have mentioned, oxygen appears in both the compounds, whereas, hydrogen and carbon appear only once. So, balance $$\ce{H}\;(=8)$$ and $$\ce{C}\,(=3)$$. Now, we have $$\ce{C3H8 + O2 -> 3CO2 + 4H2O}.$$

The equation isn't balanced yet. Now we need to balance the only remaining element, $$\ce{O}$$.

We have $$\ce{2 O}$$ atoms on the left and $$10$$ on the right hand side, balancing these numbers we get: $$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$

Just follow these steps and you can balance almost all chemical equations.

We have the unbalanced equation: $$\ce{C3H8 + O2 -> CO2 + H2O}$$

You don't really need to solve a system of equations or matrices to balance such chemical equations.

First, observe which elements only occur once on each side of the equation:

We can see that $$\ce{C}$$ and $$\ce{H}$$ only occur once on each side, so let's begin by trying to balance these.

There are three $$\ce{C}$$ on the left and one on the right. So we will add $$3$$ as a coefficient to $$\ce{CO2}$$: \begin{align} \Rightarrow&& \ce{C3H8 + O2 &-> 3CO2 + H2O} \end{align}

There are eight $$\ce{H}$$ on the left and two on the right. So we will add $$4$$ as a coefficient to $$\ce{H2O}$$: \begin{align} \Rightarrow&& \ce{C3H8 + O2 &-> 3CO2 + 4H2O} \end{align}

It is useful to leave the $$\ce{O2}$$ on the left until the end because we can add any coefficient to it to balance with the right side.

Now, on the right side we have a total of ten $$\ce{O2}$$. So we will add $$5$$ as a coefficient to $$\ce{O2}$$: \begin{align} \Rightarrow&& \ce{C3H8 + 5O2 &-> 3CO2 + 4H2O} \end{align}

We now have a balanced chemical equation.