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In my AP chem class, I often have to balance chemical equations like the following:

$$ \text{Al} + \text O_2 \to \text{Al}_2 \text O_3 $$

The goal is to make both side of the arrow have the same amount of atoms by adding compounds in the equation to each side.

A solution:

$$ 4 \text{Al} + 3 \text O_2 \to 2 \text{Al}_2 \text O_3 $$

When the subscripts become really large, or there are a lot of atoms involved, trial and error is impossible unless performed by a computer. What if some chemical equation can not be balanced? (Do such equations exist?) I tried one for a long time only to realize the problem was wrong.

My teacher said trial and error is the only way. Are there other methods?

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Are you being serious, or just seeding the site? –  Casebash Jul 24 '10 at 3:20
@Casebash: I would guess the latter, but why does it matter? This is a math question, albeit a simple one. –  Larry Wang Jul 24 '10 at 3:28
it's also probably one of the more useful math questions for high school chem! If I had a nickel for every chem student I've tutored whose teacher told them they just had to try coefficients til it worked... –  Jamie Banks Jul 24 '10 at 3:55

3 Answers 3

up vote 13 down vote accepted

Yes; it's possible to write a system of equations that can be solved to find the correct coefficients. Here's an example for the given formula.

We're trying to find coefficients A, B, and C such that

A (Al) + B (O_2) -> C (Al_2 O_3)

In order to do this, we can write an equation for each element based on how many atoms are on each side of the equation.

for Al:  A = 2C
for O:  2B = 3C

This is an uninteresting example, but these will always be linear equations in terms of the coefficients. Note that we have fewer equations than variables. This means that there's more than one way to correctly balance the equation (and there is, because any set of coefficients can be scaled by any factor). We just need to find one integral solution to these equations.

To solve, we can arbitrarily set one of the variables to 1 and we'll get a solution with (probably fractional) coefficients. If we add $A=1$, the solution is $(A,B,C) = (1,3/4,1/2)$. To get the smallest solution with integer coefficients, just multiply by the least common multiple of the denominators (4 in this case), giving us $(4,3,2)$.

If the set of equations has no solution where the coefficients are nonzero, then you know that the equation cannot be balanced.

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Ben: I think you left out the most important part: How to solve those two equations! –  Larry Wang Jul 24 '10 at 3:42
I agree with Kaester; it appears that your mathematical form, with two equations and three unknowns, reduces to guess and check just as well! –  Justin L. Jul 25 '10 at 0:23
Just edited; better now? –  Ben Alpert Jul 25 '10 at 1:00
thx, I changed the solution in the answer, idk what i was thinking xD –  Chao Xu Jul 25 '10 at 2:31

given that
A+O2---- >A203 introduce variable co effecient i.e sA+cO2---- >eA203 comparing coeffecient on both side comparing A: 1s=2e comparing O: 2c=3e when e=2 S=2(2)=4 again 2c=3(2) c=3 hence e=2, s=4 ,c=3 4A+3O2---- >2A203

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To solve this problem, create an ordered set of chemical elements $(Al,O)$ and an ordered set of chemical species $(Al,Al_2O_3,O_2)$.

The ordering of the chemical elements can be used to create a vector for each chemical species.

$Al = \begin{pmatrix}1\\0\end{pmatrix}$

$Al_2O_3 = \begin{pmatrix}2\\3\end{pmatrix}$

$O_2 = \begin{pmatrix}0\\2\end{pmatrix}$

The ordering of the chemical species (this ordering is arbitrary, but standard linear algebra algorithms will produce different results) can be used to create an element abundance matrix, $A$.

$A = \begin{pmatrix}1&2&0//0&3&2\end{pmatrix}

The coefficients of the set of linearly independent stoichiometric equations can be represented by the stoichiometric matrix, $N$, where N is a representation of the null space of $A$.


To put the stoichiometric matrix into 'canonical form' apply the operations:


The ordering of the species will preferentially make the stoichiometric coefficents of the species at the left side of matrix $A$ to have coefficents of $1$. This method will produce the maximal number of linearly independent stoichiometric equations. They can be added or multiplied and still represent a stoichiometric equation (they are linearly independent).

If you are going to try and calculate thermodynamic equilibrium with a mixture of gas and condensed species, I recommend ordering the gas species first in the list. This will try and isolate 1 gas per independent stoichiometric equation (if possible) and will facilite using the chemical equilibrium constant method $K_{eq}$ to calculate thermodynamic equilibrium.

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I'm not sure if you're aware or not, but the discussions of linear algebra here are (at least in US education) generally multiple grade levels above when the chemical equation balancing being talked about in the question is taught - they're often latter-year college material, whereas the OP's question is often in an 11th-grade curriculum, offered to students roughly 16 years old. This is a good approach in the abstract, but it's expressed well above the target educational level. –  Steven Stadnicki Feb 26 at 2:08
I understand that maybe some of the abstract concepts are to advanced for a high school student. But I solved systems of equations in high school math with augmented matrices to find solutions to homogeneous equations. If this is a place to learn, why dumb down answers like was suggested by the teacher to 'guess and check'. A curious person might not understand the full answer, but will often then look up what they dont understand. It appears to me that this person was curious enough to post the question. It is hard not to know something you dont even know exists! –  jpantina Feb 26 at 19:03

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