Welcome to Percent Yield Made Easy brought to you by Ketzbook
This is the fourth tutorial in my stoichiometry series,
and in this video, we will learn how to distinguish and calculate
the three ways to quantify chemical reactions.
They are: theoretical yield, actual yield, and percent yield.
To make this easier, let’s start with an analogy.
Suppose that you have 6 slices of bread, a full jar of peanut butter,
and a full jar of jelly
How many peanut butter and jelly sandwiches could you theoretically make?
Because each sandwich uses two slices of bread, you could make three sandwiches, at most.
That is what we call the theoretical yield in chemistry.
It is the maximum amount of product you can make with what you have.
Suppose, however, that you weren’t that hungry.
So, instead of making three peanut butter and jelly sandwiches,
you only made two peanut butter and jelly sandwiches.
That is what we call the actual yield in chemistry.
It is the amount of product that is actually made and collected.
Chemists often like to compare the actual and theoretical yields,
and they do so by calculating the percent yield,
which is more or less a measure of reaction efficiency.
The percent yield tells you how well your reaction worked.
It is defined as the actual yield divided by the theoretical yield
all multiplied by 100 in order to convert it to a percentage.
We can calculate the percent yield of our peanut butter and jelly sandwiches
by dividing the actual yield of 2 sandwiches by the theoretical yield of 3 sandwiches.
After multiplying that by 100, we see that the percent yield of this experiment is 67%.
Okay, now, we’re ready for calculating the percent yield of a real chemical reaction.
Consider the decomposition of potassium chlorate.
Like other metal chlorates, it will decompose when heated,
and the products of its decomposition are potassium chloride and oxygen gas.
You can perform this reaction in the lab in order to make oxygen,
but you need to be careful because small amounts of impurities
or exposing potassium chlorate to organic compounds
such as plastic, grease, or rubber could result in a dangerous explosion.
In order to speed up this reaction, the catalyst manganese(IV) oxide is typically added.
Catalysts are often written below the reaction arrow.
Remember that a catalyst speeds up a reaction without being changed itself.
The manganese(IV) oxide is not really part of the reaction,
so we will ignore MnO2 for all of our stoichiometry calculations.
As it is written, the reaction is not balanced, so let’s go ahead and fix that.
There are 3 oxygens on the left side and 2 oxygens on the right side,
so we put a 2 in front of potassium chlorate and a 3 in front of O2
to give us 6 oxygens on both sides.
This messed up potassium and chlorine, which we can balance by writing a 2 in front
of potassium chloride.
Okay, now that the reaction is balanced, let’s go ahead and see what the question is.
In the lab, 10.0 g of potassium chlorate were carefully decomposed,
and 3.41 g of oxygen gas were collected.
What are the theoretical, actual, and percent yields of the reaction?
The theoretical yield is the amount of product that should be made
from the amount of reactant you are starting with.
So, this is a grams to grams stoichiometry problem going from the reactant to the product.
We are going to need the molar masses of the chemicals,
which we can solve by adding up the molar masses of the elements each one is made of.
Potassium chlorate has a molar mass of 122.5 grams per mole.
There are two products, which one should we solve for?
Because the question tells us the amount of O2 collected, we should solve for O2, not KCl.
The molar mass of O2 is 32 grams per mole.
For these percent yield questions, always start your stoichiometry calculation
by writing down the mass of the reactant, NOT the mass of the product.
In this case, our reactant is 10 grams of potassium chlorate.
Because this is a grams to grams stoichiometry problem,
we are going to need 3 conversion factor fractions.
As always, put “1 mol” on the top left and the bottom right.
This reminds us that the number on the bottom left is the mass of 1 mole
of potassium chlorate, 122.5 grams.
We also know that this is correct because grams on the top and bottom cancel each other out.
In the middle fraction we always write the chemicals with their coefficients
from the balanced reaction.
Because we are starting with potassium chlorate, write 2 KClO3 on the bottom.
Remember that the two is simply the coefficient from the balanced reaction.
Now which product goes on top?
Because the problem states the amount of O2 made from the reaction,
we write 3 O2 on the top.
Notice that potassium chlorate on the top and bottom now cancel each other out.
Lastly, on the top right goes the molar mass of O2, which is 32 grams per mole.
Moles on the top and bottom cancel out.
To solve this problem, we move from left to right
always multiplying by numbers on the top of fractions
and dividing by numbers on the bottom of fractions.
In your calculator type 10 divided by 122.5 times 3 divided by 2 times 32,
and the answer is 3.92 grams of oxygen.
This number is the theoretical yield.
It is the amount of product that we should make if everything worked perfectly.
It is the maximum amount of product that we can make
from the amount of reactant we started with.
What about the actual yield?
How much O2 did we actually make and collect?
3.41 grams.
Where did I get that number?
Unless you are doing a lab, you always find the actual yield in the question itself.
3.41 grams of O2 gas were collected.
That’s the actual yield.
The percent yield tells us how well the reaction worked,
and we calculate it by dividing the actual yield by the theoretical yield
and multiplying the result by 100 to turn it into a percent.
In your calculator type 3.41 divided by 3.92 times 100, which gives us 87%,
which is actually a pretty good yield as far as chemical reactions go.
Now, you might be wondering, why is the actual yield normally LESS than
the theoretical yield?
What’s going on?
A number of factors can contribute to this “loss” of product.
The reaction might be incomplete; that is, the reactant can’t or hasn’t
fully converted to product.
There may be side reactions that turn the reactants or products into something different.
Some of the product may get blown away or spilled.
Sometimes, some of the product sticks to the container and you can’t get it all out.
It is also possible to lose some of the product when trying to purify it.
These are all normal occurrences of things that can reduce your actual and percent yields.
But what if the actual yield is MORE than the theoretical yield?
That is a much bigger problem because it means that you made a mistake.
It is impossible to make more product than the theoretical yield.
Oftentimes this can seem to occur if the product is impure,
in which case you will need to purify it or dry it.
Alternatively, if the yield appears to be greater than a hundred percent,
this may indicate a calculation or measurement error
that you will need to go back and fix.
Thanks for watching.
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