Using matrices to balance chemical equations

• barryj
In summary: Hello,In summary, the conversation discusses a method for balancing chemical equations using matrices. The method involves assigning arbitrary coefficients to each element and then using matrices to solve for the coefficients. However, the conversation also touches on the limitations of this method, as it may not work for all reactions, and the need to balance charges in some cases. A specific example of a reaction is given and the conversation walks through the steps of balancing it using the matrix method.
barryj
TL;DR Summary
I need an equation to balance that has more terms than elements
I am working with matrices to balance chem equations. I have googled hard equations to balance and find that most, at least what I have found, have the number of elements to be less than the terms in the equation. for example
... FeCr2O4 + ... Na2CO3 + ... O2 -->... Na2CrO4 + ...Fe2O3 + 1CO has 5 elements and 6 terms.

So, I can assign arbitrary letters, ABCDE, as coefficients such as

(A)FeCr2O4 + (B)Na2CO3 + (C)O2 --> (D)Na2CrO4 + (E)Fe2O3 + 1CO

and then write equations like
A = 2E balancing Fe
2A = D balancing Cr
4A+B + 2C = D +3E + 1 balancing O
2B = 2D balancing Na
B = 1 balancing C

I can use matrices to solve for these cooeficients

In this case, and many others I have found, the number of elements are 1 less than the number of terms.

So my question is, is there an equation that I cannot balance using matrices in this manner?

If so, I would like to see one so I can see if I can solve it.

That's so called algebraic method. Sometimes you will need an additional equation for balancing charges.

Typically when the reaction can't be balanced this way it simply means it doesn't proceed. There are some interesting exceptions though, like the reaction between permanganate and hydrogen peroxide:

KMnO4 + H2O2 + H2SO4 -> MnSO4 + K2SO4 + O2 + H2O

Source of the problem here is that hydrogen peroxide decomposition produces water and oxygen as well - but the reaction as written takes place, and its stoichiometry is so solid there is a titration method based on it.

Compare https://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure

You are correct. I do not seem to be able to balance this equation using my algebraic method.
I have 6 variables and 5 equations. I need another eq as you say I think

Hi barryj,
Cheers,
Boris

Greenchemist said:
0=0 doesn't add much to the problem.

There is no 0.
Take the productsum of the ONs on the left side and this must be equal to the productsum of the ONs on the product side.

Greenchemist said:
There is no 0.
Take the productsum of the ONs on the left side and this must be equal to the productsum of the ONs on the product side.

Show how, I don't see how to make it a universal method.

Universal method calls for charge balance, which in this case doesn't add anything new, as none of these substances is charged.

True for the molecules, but the individual atoms also have Oxidation numbers. Changes in these represents the electrons that have moved.

Atoms don't "have" oxidation numbers - ON have no physical meaning. They are just an accounting device - quite useful, but artificial. We assign them using a set of arbitrary rules, that follow simplified logic of assigning electrons to atoms, but as these rules are simplified they are guaranteed to fail now and then. At the same time charge and mass conservation are universal rules, guaranteed to work always (at least in the "chemical" energy range). If they are not enough to balance the equation it most likely means stoichiometry is mechanism related and very difficult to predict.

As Borek pointed out, in the example, there are two competing reactions possible, one of them the disproportionation of H2O2. So you also have to specify the ratio of the two reactions. If you specify this ratio, you should have enough equations. In Boreks example, the disproportionation should not take place in the titration.

It might help to answer the original question if the correct products were given...
Try solving
... FeCr2O4 + ... Na2CO3 + ... O2 -->... Na2CrO4 + ...Fe2O3 + CO2

Let me walk you through a method to balance this. It's the matrix in disguise, sort of.
Start by looking for an element on each side of the equation which is also only present in one species on each side but is out of balance (not the same amount on each side). Avoid starting with hydrogen and oxygen initially, we usually leave them towards the end.So there is one Fe on the left and two on the right - we need 2FeCr2O4 on the left to balance the Fe atoms

2FeCr2O4 + ... Na2CO3 + ... O2 -->... Na2CrO4 + ...Fe2O3 + CO2So there are now four Cr atoms on the left, and only one on the right, both in one species on each side.
So we add a four in front of the Na2CrO4 on the RHS to balance that and get

2FeCr2O4 + ... Na2CO3 + ... O2 -->... 4Na2CrO4 + ...Fe2O3 + CO2But there are only two Na atoms on the left and 8 in total on the right, so we need 4Na2CO3

2FeCr2O4 + ... 4Na2CO3 + ... O2 -->... 4Na2CrO4 + ...Fe2O3 + CO2As the oxygen is in several species on each side, we will go for the carbon next.
There are four carbon atoms on the left, from the 4Na2CO3 , and only one on the right - we need four CO2

2FeCr2O4 + ... 4Na2CO3 + ... O2 -->... 4Na2CrO4 + ...Fe2O3 + 4CO2Finally, we will look at the oxygen, but remember that if we alter some of the things we have already changed, the rest of the equation goes wrong, and we end up going round in circles - this is a strong clue we started with the wrong element initially. And would have to start again with a different one.

On the left, we have eight oxygens from the 2FeCr2O4, 12 oxygens from the 4Na2CO3 and 2 from the O2. That makes 22 on the right.

On the right we have 16 oxygens from the 4Na2CrO4, three from the Fe2O3, and eight from the 4CO2. That makes 27 on the right.
This means we need five more oxygens on the left and we get that by using 3 1/2 O2, giving us a total of 27 on the right.
2FeCr2O4 + ... 4Na2CO3 + ... 3 1/2 O2 -->... 4Na2CrO4 + ...Fe2O3 + 4CO2

The half doesn't mean half of an O2 molecule, it means that's the ratio of O2 to the others - like adding 3 and a half buckets of one thing to every five of something else. To get rid of that awkward looking half, we multiply everything on each side by two !

4FeCr2O4 + 8Na2CO3 + 7O2 --> 8Na2CrO4 + 2Fe2O3 + 8CO2

And that's everything balanced out - check those atom counts for yourself.

This is a sneaky one to do, as there are oxygens everywhere and we have to choose carefully how we work through and then see the confusing 1/2, so get rid of it by doubling everything.

This is nearly the matrix method, but done without playing with mathematical equations and the matrix. If when using this compound by compound method, you find yourself changing things you thought were balanced, then changing others then others, you started with the wrong element. The matrix gets around this, but requires you working with maths.

When doing this, I usually wrote out the unbalanced starting point, then calculated the numbers to insert one by one and added them to the equation. But in a written exam, we expect you to show your working steps (so we know you didn't copy it by glancing at someone else's answer and can sometimes - not always - allow for you writing a number down wrongly having just calculated it correctly). And if I wrote down the unbalanced and converted it to the balanced in one go, no working out is visible. So you'd have to write out the equation at each step. This is tedious, but if you don't finish it, your working is visible and you will get some marks.

You would right out the unbalanced equation and then say "first balance the Fe" and write the first altered equation.
"Now balance the Cr" and write the second altered version and so on, no need to give anywhere near as much as I have written to explain things.

Hope this helps you understand how to do this.
And that you don't get more sneaky ones like this to try ;)

Last edited:
berkeman
DrJohn said:
It might help to answer the original question if the correct products were given...
While sometimes correcting the original question helps, it always helps to read the original question and answer what was asked, not what you want to answer. The OP clearly stated they know how to balance the equation by an algebraic method, so the explanation of how to do it by inspection is kind of pointless.

Plus, the original equation can be easily balanced without a need to "correct" it:

4FeCr2O4 + 8Na2CO3 + 3O2 → 8Na2CrO4 + 2Fe2O3 + 8CO

1. How do you use matrices to balance chemical equations?

Matrices can be used to balance chemical equations by setting up a system of linear equations. Each element in the equation is represented by a variable, and the coefficients are placed in a matrix. The matrix can then be solved using techniques such as Gaussian elimination to find the values of the variables, which correspond to the coefficients needed to balance the equation.

2. Why is it important to balance chemical equations?

Balancing chemical equations is important because it ensures that the law of conservation of mass is followed. This law states that matter cannot be created or destroyed in a chemical reaction, so the number of atoms of each element must be the same on both sides of the equation.

3. Can matrices be used to balance any chemical equation?

Yes, matrices can be used to balance any chemical equation. However, the process may be more complex for equations with multiple variables or equations involving redox reactions. In these cases, additional techniques such as oxidation number method or half-reaction method may also be needed.

4. What are the limitations of using matrices to balance chemical equations?

One limitation of using matrices is that it can be time-consuming and tedious for more complex equations. Additionally, it may not be suitable for equations with large coefficients or equations with fractional coefficients, as these can result in large matrices with many decimal values.

5. Are there any alternative methods for balancing chemical equations?

Yes, there are alternative methods for balancing chemical equations such as the trial and error method, which involves adjusting coefficients until the equation is balanced. There are also online tools and software programs available that can balance equations automatically using various methods.

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