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In your opinion, what would be the simplest way to balance any size of 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...

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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). –  LordStryker Apr 16 at 14:45
    
@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 ? –  Trevör Anne Denise Apr 16 at 14:49
    
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. –  LordStryker Apr 16 at 14:51
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2 Answers 2

up vote 5 down vote accepted

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: C, H and O. This means that you need 3 equations to balance. First I write the chemical reaction as follows: $$a\;\ce{C12H26}+ b\;\ce{O2}\rightarrow c\;\ce{CO2}+d\;\ce{H2O}$$ Now I will write the 3 equalities for the 3 atomic species that we have: $$12a=c \tag{for C}$$ $$26a=2d \tag{for H}$$ $$2b=2c+d \tag{for O}$$ where 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 incorrect1, 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 $b=13.5$. So our balanced reaction becomes2: $$\;\ce{C12H26}+ 13.5\;\ce{O2}\rightarrow 12\;\ce{CO2}+13\;\ce{H2O}$$

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 $$2\;\ce{C12H26}+ 27\;\ce{O2}\rightarrow 24\;\ce{CO2}+26\;\ce{H2O}$$

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.

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Thanks a lot ! I'm not used at all to matrix but I understand much better now ! There is just something that I still don't understand : if 12a = c, why does 1a = 12c ? Shouldn't it be 1a = 1/12c ? –  Trevör Anne Denise Apr 16 at 18:08
    
Where do you see 1a=12c? I just mentioned that with a=1 I get c=12 which follows from 12a=c (plug in a=1 you get 12*1=c i.e. c=12) –  Michiel Apr 16 at 18:16
    
This procedure is clearly more relevant than my approach. (nice write-up) –  LordStryker Apr 16 at 19:33
    
@LordStryker thanks, I agree that it is more relevant for more complex reactions, but for simple ones imo it is just too much work and I would take your approach. But I think it is also a matter of experience, the more experienced you get with these kind of things, the more often you will just 'see' what the coefficients have to be. –  Michiel Apr 16 at 20:47
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Here is an example of the process I just used:

C$_{12}$H$_{26}$ + O$_2$ $\rightarrow$ CO$_2$ + H$_2$O

1.) Start with carbon (completely arbitrary choice). Put 12 on r.h.s.

C$_{12}$H$_{26}$ + O$_2$ $\rightarrow$ 12CO$_2$ + H$_2$O

2.) Carbons are balanced but oxygens are not. Therefore, move on to O. Since we added 12 to O$_2$ on r.h.s., do same for O$_2$ on l.h.s.

C$_{12}$H$_{26}$ + 12O$_2$ $\rightarrow$ 12CO$_2$ + H$_2$O

3.) Oxygen is still not balanced (24 vs. 25... ouch). What else can we do? Lets work with Hydrogen for now (completely arbitrary choice). We can return to oxygen later. Put 13 on r.h.s. to balance hydrogen.

C$_{12}$H$_{26}$ + 12O$_2$ $\rightarrow$ 12CO$_2$ + 13H$_2$O

4.) Oxygen still not balanced (even though everything else is). This is an even/odd situation. The oxygens on the r.h.s. must be multiplied by an even number. So lets multiply 13 by 2 to do this.

C$_{12}$H$_{26}$ + 12O$_2$ $\rightarrow$ 12CO$_2$ + 26H$_2$O

5.) Now our hydrogens are not balanced. Multiply l.h.s. by 2

2C$_{12}$H$_{26}$ + 12O$_2$ $\rightarrow$ 12CO$_2$ + 26H$_2$O

6.) Now our carbons are not balanced. Lets fix that.

2C$_{12}$H$_{26}$ + 12O$_2$ $\rightarrow$ 24CO$_2$ + 26H$_2$O

7.) Everything but oxygen is good. 24 on l.h.s. and 74 on r.h.s. We can get 74 on the left by changing our coefficient from 24 to 37 (since 37$\times$2$=$74).

2C$_{12}$H$_{26}$ + 37O$_2$ $\rightarrow$ 24CO$_2$ + 26H$_2$O

8.) 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).

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Thanks a lot for your answer ! –  Trevör Anne Denise Apr 16 at 18:09
    
Surely a solution based around common multiple recognition would be faster? –  Sam Apr 16 at 20:10
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