# 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?

• 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

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.

• 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? – Sophie 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

One method I am particularly fond of is balancing the half reactions of the redox (reduction-oxidation) reaction that is occurring. In the following reaction:

$$\mathrm{Al} + \mathrm{O_2} \rightarrow \mathrm{Al_2 O_3}$$

it should be clear that Al is being oxidized and O is being reduced. Both Al and $\mathrm{O_2}$ are in their elemental states, and therefore both have an oxidation number of 0. Simply by looking at the periodic table, you can devise that the oxidation numbers of Al and O in $\mathrm{Al_2 O_3}$ are +3 and -2 respectively. Write the half reactions with the amount of each species present in the unbalanced reaction, along with the charge changes:

$$\mathrm{Al^0}\rightarrow 2\mathrm{Al^{+3}}$$

$$2\mathrm{O^0}\rightarrow 3\mathrm{O^{-2}}$$

Find the least common multiple of the coefficients of each half reaction reaction:

$$2(\mathrm{Al^0})\rightarrow 2\mathrm{Al^{+3}}$$

$$3(2\mathrm{O^0})\rightarrow 2(3\mathrm{O^{-2}})$$

Now consider the amount of electrons that must be lost by 2 atoms of Al to transform elemental Al into $\mathrm{Al^{+3}}$, as well as the amount of electrons that must be gained by 6 atoms of O to transform elemental $\mathrm{O_2}$ into $\mathrm{O^{-2}}$:

$$2(\mathrm{Al^0})\rightarrow 2\mathrm{Al^{+3}}+6e^-$$

$$3(2\mathrm{O^0})+12e^-\rightarrow 2(3\mathrm{O^{-2}})$$

The discrepancy between the two reactions is clear--the oxidation reaction must be multiplied through by 2 in order to balance the number of electrons lost by Al with the number of electrons gained by O:

$$2[2(\mathrm{Al^0})\rightarrow 2\mathrm{Al^{+3}}+6e^-]$$

This finally results in a balance in the number of electrons lost and gained, and thereby the coefficients of both the products and the reactants:

$$4(\mathrm{Al^0})\rightarrow 2(2\mathrm{Al^{+3}})+2(6e^-)$$

$$3(2\mathrm{O^0})+12e^-\rightarrow 2(3\mathrm{O^{-2}})$$

Now transcribe our coefficients back into the original equation and your balancing is done:

$$4\mathrm{Al} + 3\mathrm{O_2} \rightarrow 2\mathrm{Al_2 O_3}$$

Yes, although this process does seem extremely long, it is applicable to more complicated redox reactions. This includes reactions in which the the species participating in the redox reaction are not the only ones requiring balancing, and even reactions in which multiple species are oxidized/reduced at once. It is quite simple once you get the hang of it, and I hope this reduces the amount of time you spend balancing equations in the long run.

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 $A$.

$AN=0$

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

$(RREF(N^T))^T$

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.

• 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 '14 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 '14 at 19:03

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 $c=3$:

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

• Why copy an answer posted three years earlier? – Did May 2 '17 at 7:26
• @Did: Why address a user who hasn't been active since the posting of this answer, 4 years ago? And no, this answer doesn't look like a copied one to me (or maybe I'm blinded by my hatred for chemistry). – Alex M. May 2 '17 at 20:56
• @AlexM. ?? Why not? – Did May 2 '17 at 21:09
• @AlexM. "And no, this answer doesn't look like a copied one to me" Well, I am not sure if you are joking or not here, but merely looking at the posts should help you overcome your prejudice against chemistry and realize that this post indeed copies the previous ones (unsurprisingly, seeing the question)... – Did May 5 '17 at 6:22