Determining Excess

Whenever a reaction occurs one of the reactants will be limiting. The only times this is not true is in decomposition (only one reactant) or if the chemist spends the time to create a perfect mixture with exactly the right amount of each reactant (called a stoichiometric mix). Even then, it’s almost certainly NOT a perfect mix at the molecular level.

What that means is that we always run out of one reactant before the other. The flip side of that concept is that some amount of one reactant is left at the end. This is called excess (not a terribly original name) and determining the amount of excess is the last type of stoichiometry problem that we need to understand.

Here is a sample problem:

When 2.05 grams of C₂H₆ react with 12.73 g of oxygen gas, how many grams of CO₂ are produced? What is the limiting reactant? What reactant is in excess? How much of that reactant remains after the reaction?

Here is the balanced reaction with the amounts written underneath:

Since we have amounts of two different reactants, we know that we will have to do two different stoichiometry problems:

Hopefully, it’s obvious that I set up this reaction to be “pretty.”

Looking at the two answers, it is clear that the correct (smaller) answer is 6.00 g CO2. We also know that C2H6 must be the limiting reactant, since it gave us the smaller answer. Logically, O2, must be the excess reactant since it is NOT limiting.

All of that means that at the end of the reaction, all of the C2H6 will  have been used but some of the O2 will remain unreacted. It is now our task to determine that amount.

There are multiple ways to solve this last problem. You do NOT have to know them all. Take a look at the set-ups below, pick one that makes sense to you and stick with that process. Once you understand them, you can probably even work out a few of the more obscure methods as well.

Method 1: Everything depends on the limiting reactant

The limiting reactant determines how far the reaction can proceed before it stops. It “decides” how much of each product will be created. It also “decides” how much of the other reactant will be used. So, we can do a stoichiometry problem that starts with the limiting reactant and solves for the amount of the other reactant required used. It will then be a simple matter of subtraction to determine the excess amount.

Here we go:

So, the reaction USED 7.64 g of O2. This is, of course, less than the amount we started with. So, we can determine the excess (the amount left over after the reaction) simply by subtracting:

Method 2: We know how much product was made

Since we know the amount of product produced, we can use that value to determine the amount of oxygen used in that process. This is not quite as “logical” as the first method, but it is just as straightforward and just as valid. Once again, it will then be a simple matter of subtraction to determine the excess amount.

Two things to notice here:

1. The answer here is VERY slightly different from the previous answer. That doesn’t make this wrong (or the other one wrong for that matter). It’s just the result of rounding an answer (using sig fig rules of course) and then using that answer in further calculations.

2. The set up for this calculation is exactly the same as one of the equations at the top of the page except upside-down and backwards.

Let’s finish up by subtracting:

Method 3: The Handler Maneuver

This last method is one that (to the best of my knowledge) you will not find anywhere else. I’d love to pretend that I invented it, but it was actually suggested by a high school sophomore in my Honors Chemistry class a number of years ago. I’ve named it after him (with his permission) and I’ve taught it ever since. It is an elegant solution (using that term in the way that mathematicians do to describe clear and creative methods rather than using “brute force”).

It is for this method that I designed a problem that gave “clean” answers. Mind you, the method works with any numbers, but it is easiest to see and understand this way. 

Here is the logic. We had enough O2 to make 10 g of CO2 but we only made 6 g. Therefore we only used 6/10ths of the oxygen provided.

or, more precisely with descriptive labels:

Then, of course, we'll need to subtract to determine the amount of oxygen remaining:

Other methods

There are a number of other ways to find the excess. Each of them involves a stoichiometry problem and a subtraction problem. None of them are as logical as the methods above.

For instance, you could subtract the two answers which would give you the amount of CO₂ that was NOT made in the reaction. You could then do a stoichiometry problem to find the amount of oxygen that was not used in the process of not making it. It sounds horrible in English, but the math is sound.

There are other variations on this theme, but first two methods above are the ones that your teacher is most likely to teach.