Balancing Reactions

Balancing ReactionsLet’s start with a lab we’ve done recently: burning magnesium to form the oxide:

Mg (s) + O2 (g) MgO (s)

We started with pure magnesium, and oxygen must be O2 (because it’s a diatom),

and based on charges, the oxide must be a 1:1 ratio, but….

Balancing ReactionsLet’s start with a lab we’ve done recently: burning magnesium to form the oxide:

Mg (s) + O2 (g) MgO (s)

We started with pure magnesium, and oxygen must be O2 (because it’s a diatom),

and based on charges, the oxide must be a 1:1 ratio, but….

…where’d the other oxygen go?

Basic principle: matter can neither be created nor destroyed (though you can convert into a LOT of energy…E=mc2).

Since we didn’t observe a nuclear-sized explosion, that probably didn’t happen. The oxygen atom must be around somewhere.

Possible Explanation Number 1It just floated off on it’s own, and our reaction equation should be:

Mg (s) + O2 (g) MgO (s) + O (g)

There is one big problem with this suggestion: O really, really dislikes* being on it’s own.

*Yes, we’re anthropomorphizing the elements. And if you have a problem with that, oxygen will come be sarcastic at you.

Better Explanation Number 2It just found another magnesium atom, and reacted with that to make a second magnesium oxide:

2Mg (s) + O2 (g) 2MgO (s)

This is now a balanced reaction equation—for every atom that goes in, an equal number of that type come out.

¡CRITICALLY IMPORTANT! ¡READ THIS AS IF YOUR LIFE DEPENDED ON IT!

Balancing is done by changing coefficients (the number before each compound, aka how many of them there are).

You never ever ever never ever never (is that clear enough?) change the formula of a compound.

Pictorially

Mg O O+ Mg O O+

?????

Mg

O O+

Mg O

Mg Mg O

How to BalanceThere is no guaranteed formula for balancing successfully—it’s a trial and error process—but there are things you can do to make it easier:

1.Just fix one element at a time, even if it makes another one worse

1.If an element is in the equation as pure element, save it for last, because you can fix it without screwing up any of the others.

1.If an element shows up all over the place, just avoid it—if you get everything else balanced, it will probably work itself out.

1.If you have a group like nitrate or phosphate, you can probably* balance the group all together as if it were one element.

1.It can be useful to keep a running tally of the number of each element on each side, but is not necessary.

* This doesn’t apply if the group changes from one side to the other, but they usually don’t

An ExampleFeCl3 (aq) + AgNO3 (aq) Fe(NO3)3 (aq) + AgCl (s)

Step 1: I’ll pick chlorine to start with, for no particular reason:

An ExampleFeCl3 (aq) + AgNO3 (aq) Fe(NO3)3 (aq) + 3AgCl (s)

I’ll pick chlorine to start with, for no particular reason:

Start with three, end with one, so I need to change a coefficient.

But now silver is messed up—only one in, and three out.

An ExampleFeCl3 (aq) + 3AgNO3 (aq) Fe(NO3)3 (aq) + 3AgCl (s)

I’ll pick chlorine to start with, for no particular reason:

Start with three, end with one, so I need to change a coefficient.

But now silver is messed up—only one in, and three out.

So I add a coefficient to silver nitrate.

Conveniently, this also fixes everything else—three nitrogens and nine oxygens go in, and thee nitrogens and nine oxygens come out.

All elements are balanced.

A More Complicated ExampleC2H6 (g) + O2 (g) CO2 (g) + H2O (g)

I will avoid oxygen for two reasons: it’s in multiple products, and it’s pure as a reactant, so I can fix it without unbalancing anything else.

Starting with carbon for no particular reason:

A More Complicated ExampleC2H6 (g) + O2 (g) 2CO2 (g) + H2O (g)

I will avoid oxygen for two reasons: it’s in multiple products, and it’s pure as a reactant, so I can fix it without unbalancing anything else.

Starting with carbon for no particular reason.

Now hydrogen:

A More Complicated ExampleC2H6 (g) + O2 (g) 2CO2 (g) + 3H2O (g)

I will avoid oxygen for two reasons: it’s in multiple products, and it’s pure as a reactant, so I can fix it without unbalancing anything else.

Starting with carbon for no particular reason.

Now hydrogen.

Well, this is awkward: I have seven total oxygens in the products. How am I supposed to get seven oxygens in the starting materials from an O2?

Solution 1C2H6 (g) + 7/2 O2 (g) 2CO2 (g) + 3H2O (g)

This technically works, and you will see reactions balanced this way.

It’s a bit problematic from a realism point of view—how do you have half an O2?NOTE: O and ½O2 are not the same thing….it’s a little complicated.

Solution 22C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (g)

I just doubled everything from solution number 1.

If it was balanced before doubling, it’s balanced after doubling, but now with nice, integer values of O2.

If It Won’t Balance

Sometimes, you go around and around and something just won’t balance. When that happens, there are three possible reasons:

1. You tried to balance something incorrectly, and should just start over

2. You have the formula of something incorrect (are your charges balanced?)

3. The reaction is just impossible for some reason (probably number 2 on this list)

Summary• Matter can neither be created nor destroyed in a

chemical reaction, so the number of each element going in must be the same as coming out.

• Balance by adding coefficients (how many of something there are), not by altering formulas.

• There are some strategies to make you more efficient at it, but in the end it's a bit of trial-and-error.