In your opinion, what would be the simplest way to balance any size of the chemical equation?

For example here is an equation: $$\ce{C12H26 + O2 -> CO2 + H2O}$$

I saw some ways by putting letters in front of each molecule, but I just don't get too much how to do this with this method! My teacher is always speaking about another method, but it simply doesn't feel logic/mathematics is enough for me.

  • $\begingroup$ This is a very broad question because chemical equations can be very long and complex (unlike the example you've provided). Given this concept, the simplest way to balance ANY chemical equation would be to use a computer program that is capable of doing so for any arbitrary equation (though I'm sure this isn't the type of answer you were looking for). $\endgroup$ Apr 16, 2014 at 14:45
  • $\begingroup$ @LordStryker Yes, sorry, I recognize that it's too broad ! Well, my question was before all about some pretty simple equations like the one I showed in my question but I would also like to create an algorithm to do that so in the future I will need a way to do that with any equation (that's why I was wondering if there was a pretty easy way to do that...). But about my example, how would you have done that ? $\endgroup$ Apr 16, 2014 at 14:49
  • $\begingroup$ Well, its easiest just to 'go big' then 'reduce'. So, start by putting a 12 in front of C (on the r.h.s. of the equation). Now scale up O on the l.h.s. to match that on the r.h.s. Continue 'scaling-up' then when everything is balanced, 'reduce' your coefficients. $\endgroup$ Apr 16, 2014 at 14:51

2 Answers 2


You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions.

Your chemical equation contains 3 atomic species: $\ce{C}$, $\ce{H}$ and $\ce{O}$. This means that you need 3 equations to balance. First I write the chemical reaction as follows: $$\ce{a C12H26 + b O2 -> c CO2 + d H2O}$$ Now I will write the 3 equalities for the 3 atomic species that we have: \begin{align} 12a &= c \tag{for C}\\ 26a &= 2d \tag{for H}\\ 2b &= 2c + d \tag{for O} \end{align} The numerical constants in front of $a$, $b$, $c$ and $d$ come from the number of atoms that are in the molecule for which they are included.

As you may have noticed, this system of equations is ill-defined, because we have 4 unknowns ($a$ to $d$) and only 3 equations. The reason is that we can pick any multiple of the equation without it becoming incorrect,1 just against convention. This means that we can simply set one of the unknowns to any value. Let's say $a = 1$ for now and revisit this choice after solving the equations.

With $a = 1$ it immediately follows that $c = 12$ and $d = 13$. From those two it then follows that $b = 18.5$. Our balanced reaction becomes:2 $$\ce{C12H26 + 18.5 O2 -> 12 CO2 + 13 H2O}$$

In principle this is already correct, but convention is to have only integer numbers in the equation (i.e. no decimal points) therefore we need to revisit our choice of $a = 1$ and pick it such that all numbers will become integers. In this case you can easily see that this will happen for $a = 2$ which then results in $$\ce{2C12H26 + 37 O2 -> 24 CO2 + 26 H2O}$$

If for some reason it is not easy to see which value for $a$ you would need then you can multiply by a large power of 10 to make all the decimals disappear and then check the greatest common divisor of the 4 numbers and divide by that to obtain the same equation.


  1. Compare $10=10$ to the multiplied version $30=30$, both are correct, but they are just different by a factor $3$.

  2. I'm not sure whether you are familiar with linear algebra, but if you are you will probably have noticed that the set of equations is a linear set so you could solve it through matrix manipulations which makes this method applicable to arbitrarily complex chemical reactions.

  • $\begingroup$ But still, this method would become quite tedious in, for example, complicated ionic equations. So in these cases, what do you suggest- should we go for the traditional half-reaction method, or is there a better alternative? $\endgroup$ Sep 28, 2018 at 12:05
  • $\begingroup$ If you know something like a little matlab or python with numpy or similar these linear equations should be very easy to solve $\endgroup$
    – Ian Bush
    Sep 28, 2018 at 13:30

Here is an example of the process I just used: $$\ce{C12H26 + O2 -> CO2 + H2O}$$

  1. Start with carbon (completely arbitrary choice). Put $12$ on the right hand side: $$\ce{C12H26 + O2 -> 12CO2 + H2O}$$

  2. Carbons are balanced but oxygens are not. Therefore, move on to $\ce{O}$. Since we added $12$ to $\ce{O2}$ on the right hand side, do same for $\ce{O2}$ on left hand side. $$\ce{C12H26 + 12O2 -> 12CO2 + H2O}$$

  3. Oxygen is still not balanced ($24$ vs. $25$... ouch). What else can we do? Let's work with hydrogen for now (completely arbitrary choice). We can return to oxygen later. Put $13$ on right hand side to balance hydrogen.

$$\ce{C12H26 + 12O2 -> 12CO2 + 13H2O}$$

  1. Oxygen still not balanced (even though everything else is). This is an even/odd situation. The oxygens on the right hand side must be multiplied by an even number. So lets multiply $13$ by $2$ to do this. $$\ce{C12H26 + 12O2 -> 12CO2 + 26H2O}$$

  2. Now our hydrogens are not balanced. Multiply left hand side by $2$: $$\ce{2C12H26 + 12O2 -> 12CO2 + 26H2O}$$

  3. Now our carbons are not balanced. Let's fix that. $\ce{2C12H26 + 12O2 -> 24CO2 + 26H2O}$$

  4. Everything but oxygen is good, $24$ on left hand side and $74$ on right hand side. We can get $74$ on the left by changing our coefficient from $24$ to $37$ (since $37\times2 = 74$). $$\ce{2C12H26 + 37O2 -> 24CO2 + 26H2O}$$

  5. Check to see if the coefficients can be reduced (divided evenly by the same number to get corresponding whole numbers). Since we have an odd number ($37$), we cannot. Therefore, the equation is balanced in Step 7.

This is an example of what I've just denoted as the 'Go-big then reduce' method (completely made this term up). I find it to be the easiest way to balance an equation. It may not be the most efficient but it will land you on the right answer (which, in a way, is efficient).

  • $\begingroup$ If the answer clarified that one should avoid starting with elements present in two places on the same side of the equation, or that you should start with one that exists only in one place on either side, and therefore it is best to start with C or H, but the choice between THEM is arbitrary then I would consider it to be an excellent answer and upvote it. I would like to upvote it but the use of the term "completely arbitrary" is imperfect and easy to fix. $\endgroup$ Dec 20, 2016 at 17:10

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