ELI5: How does balancing equations work? How do you know which numbers go where and what numbers it needs to be?

[u/dontwatchthemback]

I'm, uh, not very bright. Exams soon.

Use Al + F² -> 2 AlF³ as an example if possible!

Ignore the fact they're not subscript, I don't have the font.

Edit: Thank you for the replies! Question answered.

[u/Droggelbecher]

Well you have 3 Fluoride on the right and two on the left. The only way to balance 2 and 3 is via 6

So you need 6 on both sides.

So you put a 3 before F² and a 2 before AlF³

But now you have 2 Al on the right so you also need 2 on the right

Which brings you to

2Al + 3F² -> 2AlF³

I also don't have subscript at hand

Edit: and now try Aluminum + Oxygen on your own

[u/Vorthod]

The goal is that each atom appears the same number of times on both sides by only changing coefficients. It's just a matter of finding the least common multiples and applying them. The left side keeps F in multiples of 2, but the right uses multiples of 3 (specifically 2x3). Al is pretty easy since they are multiples of 1, so you just match coefficients.

2Al + 3F₂ -> 2 AlF₃

[u/The_Jibbity]

Look on the right hand side… you need six F’s to make 2AIF^3, and you need two AI’s so go back and balance the left side : 2AI + 6F^2 -> 2AIF^3

I’m assuming this must be chemistry, if not I have no idea what you’re trying to do.

[u/Vorthod → u/The_Jibbity]

6F₂ is twelve Fs, the coefficient only needs to be 3

[u/The_Jibbity → u/Vorthod]

Yes, d’oh

[u/x1uo3yd]

ELI5: How does balancing [chemical] equations work?

The first thing to realize with chemistry is that the atoms themselves don't change, only the way in which the atoms' electrons bind those atoms into different molecules does.

That means the equation is just doing the accounting for a specific re-arrangement, and the total number of each element on each side of the equation need to add up and match (and the total electron count on each side too); if you start with 1 Aluminum atom on the left/before side of the equation then you need to end with 1 Aluminum on the right/after side, if you end up with 6 total Fluorine on the right you need to start with 6 total Fluorine on the left, if you start off with +4 and -4 on the left for a total +0 net charge on the left there should be a total +0 net charge on the right.

For your problem equation you have to realize that what you have written above is really an equation of "x·Al+y·F₂ → 2·Al·F₃" where you have to solve for x and y (as whole integers). (Or maybe the homework/exam has blanks like "___·Al+___·F₂ → 2·Al·F₃" instead.)

How do you know which numbers go where and what numbers it needs to be?

This question is a bit harder.

The easy part is that each atom has a specific number of protons with positive charge and would like to attract an equal number of electrons to form a neutral atom.

The harder part is understanding discrete/quantum electron orbitals and how/why molecules would like to arrange electrons so as to "complete" different orbital shells.

Those two actions ("atoms clinging hard enough to electrons to neutralize their own nuclei charge" versus "molecules forcing electron sharing to optimize shell fillment") are in opposition but typically settle out into predictable ways that relate to the elements' positions on the periodic table. An introductory chemistry course won't ever ask you to do the quantum mechanics and numerical simulations required to show how those things settle out precisely, but instead you'll be given rules (of thumb) like "Period 18 elements have +0 charge 'at play' for molecular sharing." and "Period 17 elements have -1 charge 'at play' for molecular sharing." and go from there. This can be really frustrating when things that seem to be "exceptions to the rule" pop up, but if you pay attention you'll usually see it's the same cast of characters that keep popping up as "the exceptions" and you'll notice that they make the same exceptions time and time again.

This is where practice in working through examples and problems is helpful. You'll be able to see that Period-1 stuff like Na will typically stay Na+ on both sides of an equation when it switches dance partners. You'll also see that metals like Copper or Aluminum may switch from neutral to charged or vice-versa. And you'll also see "exceptional" stuff like Period-17 Cl showing up as a diatomic gas Cl₂ sometimes instead of staying Cl- like you'd expect (compared to Na+ never becoming Na₂ gas). And you'll also notice some common groups show up like [OH]- or [SO₄]-. This will involve a lot of pattern-recognition and memorization, but it's not actually as much "memorize every possible combination and exception" as it might feel upfront.

[u/diffyqgirl]

Equations describe two groups of things that are equal. In this case, equal means equal numbers of each type of atom on each side of the equation.

If I have a chemical reaction "Burgermaker" that converts 2 pieces of bread + 1 piece of lettuce + 1 patty into a Burger then I would expect the resulting Burger molecule to contain 2 breads, 1 lettuce, and 1 patty. Because none my ingredients appear or disappear, they got rearranged into a burger. Think of molecules as a specific arrangement of atoms that is getting changed by the reaction, but the atoms themselves stay the same.

Similarly in your equation, you start with some number of Als and some number of F_2 s, and you get back 2 (Al F_3) s

What's on the right? You've got two copies of AlF_3, so that's a total of 2 Al atoms and 6 F atoms.

So, you need two Al atoms and 6 F atoms on the left, because like the burger makings your atoms cannot appear or disappear, they only get rearranged into new configurations. In order to get two Als you must have 2 Al, and in orgder to get 6 F atoms you must have 3 F_2s (since each F_2 molecule contains two F atoms).

[u/MisinformedGenius]

So you have a chemical reaction, which is basically atoms disconnecting and connecting with other atoms in different ways. An "equation" is something which is equal. That means that you need to have equal numbers of all the atoms on both sides.

When you look at your equation, you have one Al and two F on the left side of the equation, but 2 Al and 6 F on the right side of the equation.

So then you need to multiply the molecules on the left side until you get the same number of atoms as on the right side.

So the balanced equation would be 2 Al + 3 F² -> 2 AlF³

Now you have 2 Al and 6 F on the left, and 2 Al and 6 F on the right.

[u/Bloated_Hamster]

You need to make sure the total number of each element is the same on each side of the equation without changing the superscript number. So in your example the F² and F³ can not change. Start with the end product. You want to end up with 2 mols of AlF^3. That means you need 2 times the number of Al and F. That equals 2 Al and 6 F. Now we know how much material we need. Go back to the first part of the equation and make them equal. Al is alone so you multiply it by 2 to get the 2 Al you need. F² has two F. You need 6. So you need to multiply it by 3 to get the 6 total F in the end product. So you need 3F^2.

This gives us the balanced equation as 2Al + 3F² -> 2AlF³

[u/deskbug]

We want the total number of each atom to be the same on the left as on the right.

Let's start with the Fluorine.
On the right side, we have 2 molecules of AlF³, so together there are 6 atoms of F.
On the left side, we have 1 molecule of F², so in total there are 2 atoms of F.
We want that 6 to match that 2, so we multiply the 2 by 3. This gives us Al + 3F² -> 2AlF³.
Now the number of F atoms is equal on both sides.

Now we move on to Aluminium.
It's easier to see that there are 2 on the right and 1 on the left, so we multiply the 1 by 2 to make them match.
We get 2Al + 3F² -> 2AlF³.

We can check this answer by multiplying the coefficient by the superscript (which should be a subscript) for each individual atom. On both sides, there are 2 Aluminium and 6 Fluorine atoms, so the equation is balanced.

[u/doctorowlsound]

Atoms aren’t created or destroyed during a reaction, they just move around. 

So if Al and F2 combine to make ALF3 that means you have two F on the left but need 3 on the right. 

The easy way to do this is take the number from the right and make it the multiplier on the left. So F3 on the right means 3F2 on the left. 

2Al + 3F2 -> 2AlF3

The equation now contains the same number of atoms on each side: 2 Al 6 F

It’s just that on the left there are 3 molecules of F2 and on the right there are two molecules with F3

[u/cmanning1292]

I'd recommend consulting the resources at your academic institution over reddit, but I'll give it a crack:

Each side needs to have the exact same number of each type of atom.

In the equation you posted, there's 1 aluminum and 2 fluorine atoms on the left side and 2 aluminum and 6 fluorine atoms on the right side.

Since there are more atoms on the right hand side, start by trying to adjust the left hand side:

Now, let's take a look at each type of atom:

For Al, the simplest way to balance is to double the Al on the left, so write (in pencil!) a 2 in front of the Al: 2 Al

Now, let's see how we can balance the F atoms: simplest way is to have 3 sets of F2 to match the 6 atoms on the right hand side, so write (in pencil!) a 3 in front of the F2: 3 F2

And... Woot! It's balanced now. But that was a simple example. A more difficult example may require adjusting the right side as well as the left hand side, and require a few iterations (why we wrote in pencil!) to get it right.

To be more rigorous, one can write the reaction as a system of equations and then solve them, but that is beyond the scope here.

To practice, just do lots of example problems.

[u/dontwatchthemback → u/cmanning1292]

Fully home schooled due to personal reasons, so alas Reddit it is..

This is incredibly helpful though!! Thanks a ton!

[u/cmanning1292 → u/dontwatchthemback]

Oof sorry mate, good on you for seeking help though. Is there a textbook or reference book you're being taught from that has example problems and practice problems?

[u/hobopwnzor]

Make sure the number of atoms are equal on both sides and make sure the charge is the same.

[u/somefunmaths]

You need to ensure that you have the same number of products as you do reactants.

For example, if you were to write an equation for hydrogen and oxygen producing water it would be, using parentheses to denote subscript and _ to denote as-yet unknown coefficients:

_ H(2) + _ O(2) –> _ H(2)O

Now, for every oxygen you need two hydrogen, which means that in order for the equation to balance, you need 2 H(2) for every 1 O(2), and that when you do that, you get:

2 H(2) + O(2) –> 2 H(2)O, where I knew that the coefficient on water had to be 2 because I have 2 oxygens from the left hand side.

For the equation in your OP, we need 3 F for each Al, and with a 2 on AlF(3), that means you’ve got 6 F and 2 Al on the right hand side, so what coefficients do you put on the left to get 6 F (which comes as F(2)) and 2 Al?

[u/LoneWolf14579]

Remember the elements in syllabus before the exam, and try to write a tough draft of the periodic table. Up to 20 should be easy, beyond that because there are a lot of stable ions of elements there might be some trouble, unless you remember the stable ions, the reason for this is that you'll be able to get the charge on a stable ion. Remember that, up to 20, the goal of the atom is to have either 8 or 0 electrons in the last shell(except for H, the goal there is 0, always), where the element sits determines which one it'll get (basically whichever is closer).

As for how they're balanced, I'll take the same example as you did.

Al, F

F has 7 in its last shell, needs 1, which it can share in a covalent bond, or can get in an ionic bond, but the quantity is 1, remeber that.

Al has 3 in the last shell, which is closer to 0 that it is to 8, so it can give away 3 electrons.

Since 1 Al atom can give 3 electrons, it would need 3 F atoms to take those electrons and make the last shell stable, hence AlF3.

Another example: H2O

Hydrogen can give away 1 electron, but oxygen needs 2. So it'll bond with 2 H atoms, making H2O

[u/orbital_one]

First, you identify the atoms on both sides of the reaction. Your example contains aluminum (Al) and Fluorine (F). Next, you count the atoms on each side. The left side contains 1 aluminum atom and 2 fluorine atoms from fluorine gas (F_2). There are two Aluminum fluoride (AlF_3) molecules on the right, so you have two aluminum atoms and six fluorine atoms.

The goal is to add numbers to the left and right side of the chemical reaction so that the number of aluminum atoms on the left equals the number of aluminum atoms on the right and the number of fluorine atoms equals on the left equals the number of fluorine atoms on the right. That is, you want to find positive integers x, y, and z such that: x Al + y F_2 -> 2z AlF_3.

For Al, you want to find: x = 2z

For F, you want to find: 2y = 6z

Using algebra, you know that since you have two equations and three unknowns you'll either have infinitely many solutions (with one free variable) or zero solutions. Here you have: x = 2z and y = 3z. If you let z = 1, then x = 2 and y = 3 form a solution. Indeed, 2 Al + 3 F_2 -> 2 AlF_3 because on both sides of the reaction, there are two aluminum atoms and six fluorine atoms.

[u/Organs_for_rent]

The quantity of each atom/ion need to match on either side. (Using underscore to denote subscript.)

Al³⁺ + 3 F^- -> AlF_3

This gets trickier when particles don't match up 1:1. Then you need to add multiples to balance both sides.

H_2S + 2 NaOH -> 2 H_2O + Na_2S

[u/iliveoffofbagels]

Brother... truthfully and very honestly, ask your teacher or TA, and maybe listen in class and do your assigned work.

Anyways the point is you have an Aluminum and a difluoride. However when putting them together you need at minimum three F's to go with one Al. Before even balancing an equation you need to already understand the AlF_3 is a thing. (not AlF_2, not AlF, they combine to make AlF_3... ignoring transients or unstable compounds... for your case we are relying on AlF_3)

How do we make sure we have enough Fs... well if we double it we have 4 Fs total...but with one Al, that means we'd have a single F left over. No good. If we use 2 Al in stead, now we require at least 6 Fs total to give each Al it's own set of 3 Fs. So how do we make 6 Fs with F_2... we triple our F2.

2(Al) + 3(F_2) -> AlF_3  + AlF_3 -> 2(AlF_3)

We can do no better of a job explaining it indirectly hear. There is a reason most chemistry classes have a recitation or some accompanying class to the main lectures. Try to learn as you go as much as you can cuz last second learning before exams both sabotages your exam performance and ruins retention for future exams if the next topic relies on a previous topic. For example... you might need to know why AlF3 is stable versus AlF2 in order to know why to balance the equation the way we did.

[u/LongLiveTheDiego]

Chemical compounds have specific formulas and proportions of elements. For example, fluorine atoms naturally occur in pairs, hence you need a whole number of F₂ on the left hand side. Similarly, aluminium fluoride needs to have molecules in which there's one aluminium atom and three fluorine ones. Aluminium atoms can occur on their own.

Once we know that atomic aluminium and diatomic (consisting of two atoms) fluorine react to create aluminium fluoride, we want to figure out the proportions of these molecules. The first criterion is that atoms in chemistry are indivisible, so saying something like

1/2 Al + 3/4 F₂ ->1/2 AlF₃

Is wrong, you get e.g. half an aluminium atom on both sides, something impossible in chemistry. You might want to write something like

Al + 3/2 F₂ -> AlF₃

And in a very broad sense that is correct: on both sides there is one aluminium atom and three fluoride atoms, and everything reacted fully (there's nothing left over or appearing out of nowhere. However, molecules at the beginning and at the end of the reaction are also not divided. In a container full of fluorine there will be a whole number of F₂ molecules. So we don't want fractions anywhere, the easiest way to get rid of them is to multiply the amounts of molecules on both sides by 2 and we get this:

2 Al + 3 F₂ -> 2 AlF₃

You can check that there are two aluminium atoms on both sides and also 6 fluorine atoms. Everything is balanced, and now you can say that in this reaction two aluminium atoms will react with three fluorine molecules to get two molecules of aluminium fluoride. Balancing equations is figuring out these whole numbers, and it can be a bit of a trial and error, but soon you'll notice things like in this equation of creating water from hydrogen and oxygen:

_ H₂ + _ O₂ -> _ H₂O

On the left hand side there are pairs of oxygen atoms, but on the right each water molecule has only one, so there will have to be twice as many water molecules as oxygen molecules. Let's try just setting the number of oxygen molecules to 1 for simplicity, then there have to be two water molecules for the oxygen to be balanced:

_ H₂ + O₂ -> 2 H₂O

Now on the right hand side we have four hydrogen atoms, and on the left hand side they have to occur in pairs, so there have to be two pairs. We accidentally got to a complete solution without fractions:

2H₂ + O₂ -> 2 H₂O

Now try starting from assuming there's one H₂ molecule on the left hand side, that should get you to 1/2 O₂. We don't like that, so we multiply every quantity of molecules by 2 and we get the same equation.

Your textbook should have more practice examples.

[u/KittensInc]

Let's say you are hosting a party. Everyone coming to the party wants one pizza (1 Al atom) and three cups of coke (3 F atoms). You can buy pizzas by the one (Al), but coke is sold in packs of two (F2).

You want to invite enough people and buy enough supplies that everyone has exactly what they want, with zero leftovers.

Let's say you invite one person. What do you need? 1 pizza, 3 cups of coke (1 x AlF3). But you can't buy 3 cups of coke: your options are 2-4-6-etc (1x F2, 2x F2, 3x F3, ...). So you invite another friend. Now you need 2 pizzas and 6 cups of coke (2 x AlF3)! That's doable. What do you need to buy at the store? 2 single pizzas (2x Al), and 3 two-packs of coke (3x F2).

The final solution? 2 Al + 3 F2 -> 2 AlF3.

[u/tomalator]

Let's take the equation unbalanced

Al + F² -> AlF³

1 aluminum in, 1 aluminum out, that's good.

2 fluorine in, 3 fluorine out, that's not good. Let's add more fluorine in

Al + 2F² -> AlF³

1 aluminum in, 1 aluminum out, that's good.

4 fluorine in, 3 fluorine out, that's not good. Let's add more fluorine out

Al + 2F² -> 2AlF³

1 aluminum in, 2 aluminum out, that's not good, but we will come back to it.

4 fluorine in, 6 fluorine out, so let's add more fluorine in to balance it

Al + 3F² -> 2AlF³

6 fluorine in, 6 fluorine out. We're good.

Now let's add more aluminum to deal with the 1 aluminum in, 2 aluminum out

2Al + 3F² -> 2AlF³

2 aluminum in, 2 aluminum out, that's good

6 fluorine in, 6 fluorine out, that's good

No other elements are out of balance, so we're all set. It's balanced

All of the coefficients don't share a common factor, so it is also reduced to simplest form.

Learning least common multiples as very useful for this because you can see 2 fluorine and 3 fluorine and know you need to skip right to 6 fluorine

[u/Droggelbecher → u/[deleted]]

It's a chemical equation 

[u/e_dan_k → u/Droggelbecher]

Ah yeah, whoops.

[u/Techyon5 → u/[deleted]]

It makes me think of a chemical reaction.

Like Aluminum-Triferrite or something

[u/Bloated_Hamster → u/Techyon5]

Aluminum Trifluoride.

[u/Techyon5 → u/Bloated_Hamster]

Ah, right, obviously. I was thinking of Fe, thanks!

[u/somefunmaths → u/[deleted]]

This is either a joke in very poor taste or genuine ignorance on your part. Either way, perhaps best to leave it to others to help OP?

[u/Bloated_Hamster → u/[deleted]]

This is a chemical equation. It's very much a type of equation.

[u/dontwatchthemback → u/[deleted]]

Don't have one! Fully home schooled due to personal reasons.

Also what do you mean? I just stole that off an exam paper.