﻿As much as we’d like it if things always went according to plan, the world unfortunately
doesn’t work that way.
It’s pretty much impossible to get things perfect all the time, no matter how hard we try.
*Will making mistakes*
Chemistry’s no different — we’re never really doing experiments in perfect conditions.
Sometimes, the measurements we take aren’t exact.
The amount of a substance we weigh can only tell us how much is there to a certain degree.
Plus, we’re humans, and we’re bound to make mistakes once in a while.
Even cross-contamination and disobedient cats might be some reasons a reaction won’t produce
exactly what we expect it to. When this happens, we need to
calculate which substances are the limiting or excess reactants.
That will help us find the theoretical and actual yields, and calculate percent yields to compare
what we do produce to what we planned to. Hi, I’m Will Comar, and welcome back to Study
Hall Chemistry, presented by Arizona State University and Crash Course.
Let’s dive in!
At this point in the series, we already know the general difference
between theoretical and actual yield. As a refresher, theoretical yield is the amount of
product we’re expecting to get out of a reaction, and actual yield is how much we really get.
But we still need to know how to calculate theoretical yield,
because in the examples we’ve done so far, those numbers have already been given to us.
So let’s say that we’ve decided to start a mountain bike business.
Each bike requires two tires and one lightweight frame.
Written as a reaction, we’ll say 2 tires plus one frame produces one bike.
So we order all our parts, and the wholesaler sends us something
funky: nine tires and seven frames. We’ll need to know how many bikes we can
theoretically make with what we’ve got. Let’s walk through this with
some help from TestTube. We’ll start by writing out our reaction
and plugging in our givens: nine tires and seven frames becomes some mystery number of bikes.
We’ll need to find out how many bikes we could make for each of our reactant parts individually
by using the ratio between each part and a completed bike.
In our original reaction, we make 1 bike for every 2 tires.
So, if we have nine tires and it takes two to build one bike,
we can make four and a half bikes.
But it’d be pretty difficult to sell someone half a bike (unless they’re into unicycles…
no judgement here), so we can really only make four whole bikes.
Next, let’s calculate how many bikes we can make with our frames.
It takes one frame to make one bike, so with seven frames we can make seven bikes. Great!
But hold up. We’re limited by how many tires we’ve got — a bike
won’t work without those, and we don’t have enough to build seven bikes, we can only make four.
That’s why we need to find out how many bikes we can make with each part individually —
it tells us what our limit is. So in this case, our tires are our
limiting reactant, the piece that’s used up first, and ends the reaction.
And, since we now have extra frames left over, our frames become our excess reactant.
This also means that four bikes, the amount we can feasibly make before running out of materials,
is our theoretical yield. Now that we’ve calculated our theoretical yield,
let’s take this a little bit further. Maybe when we actually build the bikes, two of
the tires have punctures and can’t be used. That means our actual yield would only be
three bikes. We can represent the relationship
between what we were expecting and what we really produced by calculating the percent yield.
This is done by dividing the actual yield by the theoretical yield and multiplying by 100%.
That way, we can turn the ratio between what we really produced and what we could
produce into a percentage, making it easier to see how efficient the reaction was.
So our percent yield of the bikes we could actually build would be
3 over 4, times 100%, which would be 75%. In other words, we made 75% of the product we
were hoping to, and we should probably order parts from a different company next time.
Yields and limitations play a big role in chemical reactions.
Sometimes, when we measure or transfer liquids or powders between containers,
some of the substance gets left behind. There’s no way for us to ensure we got
all of it into the reaction. Plus, any scale we use to weigh materials can only
measure to a certain level of exactness. If it measures 10.3 grams, that could be
10.30, 10.28, or 10.34. With that in mind, we shouldn’t expect
to get 100% yield in any reaction. But like we saw with the bike example,
we can calculate what we should get. And lucky for us, we can find the theoretical
yield of more complex reactions using stoichiometry.
Let’s try out an example. Potassium dioxide, sometimes called potassium
superoxide, or KO2, is a solid additive that’s combined with water to generate oxygen.
It’s commonly used in medical breathing masks.
So if a mask setup uses 0.283 moles of KO2 combined with 0.166 moles of water,
let’s find out how much oxygen we can theoretically produce in terms of moles.
We’d need to take the same approach as our bike problem:
use stoichiometry to find how much oxygen each of our reactants produces individually,
and then compare those amounts to find whichever is smallest.
So, let’s set up our first of the two calculations. We have 0.283 moles of KO2,
and we need a result in moles of oxygen. A mole ratio will get us from moles of
KO2 to moles of O2. From our balanced equation,
our coefficients tell us we have three moles of oxygen produced for every four moles of KO2,
so we can use this for our mole ratio. When we write out our calculation,
we’ll need to cancel moles of KO2, so we should put our four moles on the bottom
of our ratio, with 3 moles of O2 on the top. When we calculate across and cancel units,
we end up with 0.212 moles of oxygen produced by KO2.
Now, we need to do the exact same calculation setup for 0.166 moles of water.
At this point, we know the drill. We need a mole ratio, this time between
oxygen and water, with water on the bottom so it cancels out.
When we multiply and cancel units, we get 0.249 moles of oxygen produced by H2O.
Now it’s time to compare our results. The smaller number, 0.212 moles of oxygen,
is the theoretical yield — it’s the most we can make before we run out of KO2,
our limiting reactant here. Of course, we still should measure
the amount of oxygen actually produced to make sure it’s enough to give to a patient who needs
a mask with a percent yield above 92.0%. If our reading shows we’re producing 0.199 moles
of O2, let’s calculate our percent yield. If we remember the equation we talked about a
couple minutes ago, all we need to do is divide our actual yield by theoretical
yield and multiply by 100%. In other words, 0.199 moles of oxygen
over 0.212 moles of oxygen, times 100%. We get 93.9% yield as our result, and since
that’s above the hospital’s standard, the mask’s producing enough oxygen for what we need.
Unfortunately, not every chemistry scenario gives us our numbers in terms of moles.
In those cases, we’ll need to first convert from mass, usually in grams, to moles,
and then calculate our theoretical yield. Let’s roll with it and see what that
calculation would look like. If we worked for an environmental
regulations group examining the efficiency of fuels, we might need to find out the mass
of carbon dioxide emitted during the combustion of acetylene fuel.
If the reaction involves 15.2 grams of acetylene and 16.1 grams of oxygen, we can use these
givens to solve for an answer. First we’ll need to figure out which reactant is our limiting reactant.
Beginning with 15.2 grams of acetylene, we’ll use three conversion factors to find out how much carbon dioxide
each reactant can produce on its own. Since we need so many conversions, it might
be helpful to write out the calculation and ratios with just units first,
then fill in the numbers afterward. We’ll do that unit-first approach here to show
clearly how we get from step to step. Alright, conversion factor time.
We’ll need a molar mass to first get us from grams to moles of acetylene.
Then, we need a mole ratio to get us from moles of acetylene to moles of carbon dioxide,
and finally, a molar mass to get us from moles of carbon dioxide back to grams.
From the periodic table, we can find the molar masses we need.
From the reaction coefficients, we can also tell that it takes 2 moles of acetylene to make
4 moles of CO2 gas, so we can use this for our mole ratio.
Since we’ve set up our structure, let’s populate our calculation with these numbers,
based on how the units are arranged. When we multiply across, our grams of acetylene cancel,
and so do our moles of acetylene and moles of CO2. We’re left with our answer,
51.4 grams of CO2 produced by acetylene. Moving on, we’ll do this same setup for our other
reactant, 16.1 grams of diatomic oxygen. We still need three conversion factors to navigate
us to the units we want. This time though, we’ll use the molar mass of oxygen.
We’ll also need a new mole ratio that gets us from moles of O2 to moles of CO2.
From our reaction, it takes 5 moles of O2 to produce 4 moles of CO2, so we’ll use
that relationship as our mole ratio. Once we plug in our known numbers, we can
cancel out grams of oxygen, moles of oxygen, and moles of carbon dioxide, and our result is 17.7
grams of CO2 produced by oxygen. When we compare our results from each
of our reactants, there’s a pretty drastic difference.
Oxygen produces fewer grams of CO2 than acetylene does, so oxygen is our limiting reactant.
And, since we’d have a lot of leftover acetylene, that would be our excess reactant.
This means oxygen will run out first, so our theoretical yield is 17.7 grams.
And, if our actual yield was revealed to be 17.0 grams of carbon dioxide,
we could calculate our percent yield. We’d write 17.0 grams over our theoretical
yield, 17.7 grams, and multiply by 100%. We now know that this particular reaction yields
96.0% of the amount we expected. That's not too bad! Limiting and excess reactions are relevant to
everything from bike shops to energy to medicine.
In fact, the African American chemist Charles Richard Drew even used the concepts of percent
yield and limiting reactants in his research on blood chemistry.
In the first half of the 20th century, he figured out that blood plasma, or blood with cells
removed, lasts a lot longer in storage than normal blood and can still be used in transfusions.
He increased the yields of blood that could be donated to save lives,
providing blood donations for those who needed it most during the second World War.
There’s opportunities in nearly any career path to solve problems using
percent yields and limiting reactants. They’re also important for things like energy
efficiency, acid neutralization, and pollution prevention.
So at this point, we’ve covered how to calculate theoretical yield
and identify limiting and excess reactants, but we’ve still got a bit more to learn.
Next time, we’ll continue our discussion by learning how to calculate how much is left
over in reactions. See y’all there!
Thanks for watching Study Hall: Chemistry, which is produced by Arizona State University
and the Crash Course team at Complexly. If you liked this video and want to keep
learning with us here in Study Hall, be sure to subscribe.
You can learn more about ASU and the videos produced by Crash Course in the links in the description.
See you all next time!