# Balancing chemical equations using the algebraic method

*In the last of my series on balancing chemical equations, we look at the algebraic method, which is useful for balancing the hardest equations*

Whilst the combination of algebra and balancing chemical equations might sound horrifying, it’s not as bad as it sounds. The algebra part is pretty straightforward, nothing more difficult than the simultaneous equations you’d encounter at GCSE-level maths, if that.

That being said, it is a mathematical approach that is **only suitable for mathematically-minded A Level or IB students**. If you learn it, it can be a very fast way to balance difficult equations that are hard to balance by inspection, but there is **no formal requirement to learn this method**, nor should you ever need to use it to balance a chemical equation on an A Level or IB Diploma exam.

There are two algebraic methods, the normal algebraic method and a simplified version, which is the best way to crack the toughest equations. Let’s look at the algebraic method first.

## Algebraic method for balancing chemical equations

The strategy for balancing chemical equations algebraically is as follows:

- Write a different letter coefficient in front of each compound in the equation
- Write algebraic expressions or rules for each element that equate its atoms on the LHS and RHS
- Substitute and simplify to obtain a rule that equates only two letter coefficients that you can solve
- Substitute the values into the other rules to obtain the balancing coefficients

That makes absolutely no sense I’m sure without an example, so here’s an equation to balance using this strategy:

_KMnO_{4} + _HCl → _MnCl_{2} + _KCl + _Cl_{2} + _H_{2}O

First thing we do is give each compound a letter coefficient:

*a*KMnO_{4} + *b*HCl → *c*MnCl_{2} + *d*KCl + *e*Cl_{2} + *f*H_{2}O

Next, applying the Conservation of Mass, which tell us that the total number of atoms of each element must be the same on both sides, write algebraic rules for each element.

K: *a* = *d*

Mn: *a* = *c*

O: 4*a* = *f*

H: *b* = 2*f*

Cl: *b* = 2*c* + *d* + 2*e*

To explain the logic behind this using Cl as an example, we know that the number of chlorine atoms must be the same on both sides of the equation. On the reactant side, we would have a total of *b* chlorine atoms. On the product side, MnCl_{2} contains two chlorine atoms, so if its coefficient is *c*, it must contain *2c* chlorine atoms, whilst KCl contains *d* chlorine atoms and so on, adding up the total number of chlorine atoms on the RHS.

There are too many unknowns here, but we can substitute the rules for K and Mn into the rule for Cl to get rid of *c* and *d*:

*b* = 2*a* + *a* + 2*e*

*b* = 3*a* + 2*e*

We can also get rid of *b* using the rule for H:

2*f* = 3*a* + 2*e*

And finally, get rid of *f* using the rule for O:

2(4*a*) = 3*a* + 2*e*

8*a* = 3*a* + 2*e*

5*a* = 3*a*, hence *a* = 5 and *b* = 3

Having now found two coefficients, substituting into the rules for Cl and O and using the fact that *a* = *c* = *d* = 2 solves for *b* and *f*:

*b* = 2*c* + *d* + 2*e*

*b* = 3*a* + 2*e*

*b* = 3 x 2 + 2 x 5

∴ *b* = 16

4*a* = *f*

∴ *f* = 8

2KMnO_{4} + 16HCl → 2MnCl_{2} + 2KCl + 5Cl_{2} + 8H_{2}O

In the unlikely event this showed up on an exam paper as a balancing question, you would ordinarily use the redox balancing method but I think you’d agree that this method is faster.

Note that when you use the algebraic method, you may end up performing different substitutions in order to eliminate the unknowns. This is perfectly fine, there is no right or wrong approach.

## Simple algebraic method for balancing chemical equations

Now, if in the above example you were thinking “*hang on, if a = c = d, why introduce c and d at all? Why not simplify the whole thing?*” you would have been right and this is where the simplified algebraic method comes in. This method does exactly that – it uses logic to reduce the number of unknowns you need to solve.

It’s the best method I know of for balancing extremely difficult equations and is a very fast method for balancing hard redox equations too, once you get the hang of it. The downside if there is one is that this method does need a little bit of intuition. By that, I mean some equations need you to apply certain balancing by inspection principles in order to simplify the algebraic expressions you end up needing to solve.

The procedure for using the simple algebraic method is below, but to be honest it will read like gobbledegook and the only way to make sense of it is to work through a lot of practise questions, which thankfully I have provided.

- Identify the element(s) appearing only once on the LHS and RHS of the equation
- If the element(s) is already balanced, write the same letter coefficient in front of its compounds
- If the element(s) is not balanced, write the letter coefficient in front of the compound containing the greater number of its atoms, then balance the element on the other side of the equation using the same letter coefficient (applying Conservation of Mass)

- Place letter coefficients in front of the remaining compounds.
- Minimise the number of letters needed by applying the Conservation of Mass and logic, so that coefficients representing an element on one side are expressed in terms of existing coefficients representing the element on the other side. The idea is to reduce the number of unknowns.
- Aim to use
**at least two**letter coefficients. - Write an algebraic rule for the remaining elements that equates them on each side of the equation, substitute and simplify to obtain solutions for each letter coefficient.
- Use the Conservation of Charge principle wherever possible as this can significantly reduce the number of unknowns (an example is given below)

We’ll use the same equation as above to show how this works:

_KMnO_{4} + _HCl → _MnCl_{2} + _KCl + _Cl_{2} + _H_{2}O

This time, we’ll start by noticing that oxygen appears only once on each side. Applying the rules above, we’ll use the same letter coefficient in front of KMnO_{4} and H_{2}O, putting *a* in front of KMnO_{4} and *4a* in front of H_{2}O, since logically the number in front of H_{2}O must be four times the number in front of KMnO_{4} to balance oxygen:

*a*KMnO_{4} + _HCl → _MnCl_{2} + _KCl + _Cl_{2} + *4a*H_{2}O

Potassium and manganese also appear only once on each side, as does hydrogen. Using what I called the ‘forced coefficients’ rule in this blog on balancing chemical equations, we know that the coefficient in front of KCl and MnCl_{2} must also be *a*:

*a*KMnO_{4} + _HCl → *a*MnCl_{2} + *a*KCl + _Cl_{2} + *4a*H_{2}O

For hydrogen, the coefficient in front of HCl would have to be twice the coefficient in front of H_{2}O, which is *8a*:

*a*KMnO_{4} + *8a*HCl → *a*MnCl_{2} + *a*KCl + _Cl_{2} + *4a*H_{2}O

That just leaves chlorine, which we’ll give the coefficient *b*:

*a*KMnO_{4} + *8a*HCl → *a*MnCl_{2} + *a*KCl + *b*Cl_{2} + *4a*H_{2}O

It’s now possible to write an algebraic rule for chlorine:

8*a* = 2*a* + *a* + 2*b*

5*a* = 2*b*, giving *a* = 2 and *b* = 5

2KMnO_{4} + 16HCl → 2MnCl_{2} + 2KCl + 5Cl_{2} + 8H_{2}O

Here’s another example, this time one where we can use the Conservation of Charge principle to solve it easily:

_IO_{3} + _I^{–} + _H^{+} → _I_{2} + _H_{2}O

The first step is to note that oxygen appears once on each side of the equation, so following the above guidelines, we’ll start by putting the letter coefficient *a* in front of IO_{3} and *3a* in front of H_{2}O:

*a*IO_{3} + _I^{–} + _H^{+} → _I_{2} + *3a*H_{2}O

Hydrogen also appears once on each side of the equation, and if there are *3a* in H_{2}O, there must be *6a* H^{+} ions on the LHS:

*a*IO_{3} + _I^{–} + *6a*H^{+} → _I_{2} + *3a*H_{2}O

That leaves iodine. If we put *b* in front of I_{2} on the RHS, then the total number of atoms of iodine on the RHS is *2b* and the total number of iodine atoms on the LHS (applying the Conservation of Mass) must be *2b -a* (*a* being the number of atoms present in IO_{3}):

*a*IO_{3} + *2b-a*I^{–} + *6a*H^{+} → *b*I_{2} + *3a*H_{2}O

This is where Conservation of Charge comes in. The RHS has no net charge, which means the LHS also must have no net charge. That means that the coefficient in front of I^{–} and H^{+} must be equal, so we can write an expression to represent that:

*2b-a* = *6a*

*2b* = *7a*, hence *a* = 2 and *b* = 7

Substituting these values gives the balanced equation:

**2**IO_{3} + **12**I^{–} + **12**H^{+} → **7**I_{2} + **6**H_{2}O

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