Balance chemical equations without trial and error? In my AP chemistry class, I often have to balance chemical equations like the following:

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

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:

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

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?
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

$A (\mathrm{Al}) + B (\mathrm{O_2}) \rightarrow C (\mathrm{Al_2 O_3})$

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 $(A,B,C) = (1,\frac{3}{4},\frac{1}{2})$. 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.
Given that:

$$Al+O_2\to Al_2O_3,$$

introduce variable coeffecient i.e:

$$sAl+cO_2\to eAl_2O_3.$$

Comparing coeffecient on both sides


Comparing A: $1s=2e$;
Comparing O: $2c=3e$.


When $e=2$, $S=2\cdot2=4$. Again, $2c=3\cdot2\implies c=3$. Hence $e=2$,  $s=4$ and $c=3$:

$$4Al+3O2\longrightarrow 2Al_2O_3.$$