So building on the concepts of stoichiometry and limiting reagent
let's talk about percent yield problems
because it's basically a step up from those
and the final step on the topic.
And as always a percentage is just the part over the whole times 100.
So how are we to defining out percentage this time?
Well for percent yield
the percentage is going to be the actual experimental yield
divided by our calculated theoretical yield
times 100 as always.
So we've got to find the theoretical yield
the stoichiometry
and also check for limiting reagent.
So we might have to do the calculation twice to get theoretical yield.
Actual yield is usually a given although we can give all parts of the equation
but one and ask you to calculate it.
However the usual method especially in lab
is to measure your actual yield
that's given to you then calculate the theoretical yield
and calculate your percent yield.
So let's do an example.
For the following balanced equation 0.504 grams of hydrogen gas is produced.
What is the percent yield if four grams of sodium metal
and 13 grams of water are reacted to form hydrogen gas and sodium hydroxide.
This is a single replacement reaction it's already balance
if it weren't balanced we have to balance it immediately.
That's what we do with stoichiometric problems
and we immediately diagram the reaction for mass,
the number of moles corresponding to the given masses experimentally,
and then we calculate the molar mass from the periodic table
So what is given can already be put in.
We have 4 grams of sodium metal
and we have 13 grams of water
and we must calculate
the theoretical yield of hydrogen gas to compare it to
the actual yield of hydrogen gas.
Okay. We also have to calculate the molar masses from the periodic table.
I've done that ahead of time.
We don't care about sodium hydroxide because it's not involved in the problem.
What do we have to do? We can never compare grams to grams.
So we've got to convert to moles
and then use the mole ratio from the balanced equation
to convert to moles of hydrogen gas
and then eventually to grams of hydrogen gas.
And because we're given two set of information
we also have to calculate what the theoretical yield would be
If H2O is our limiting reagent.
Okay, so we're going to have to do this process twice
because we're in a limiting reagent situation where we do not know
what the limiting reagent is.
Let's do things one step at a time and find was a theoretical yield is
if sodium is limiting.
And so we say was the given? Well it's four grams
of sodium metal.
We want to get rid of sodium metal--
try to draw a straight line so we cancelled out on the bottom.
We want to go to moles of sodium metal.
Grams the mole is the molar mass so 22.99 grams for 1 mole.
Cancel the grams of sodium.
Now we want to get rid of moles of sodium
and get moles of hydrogen gas.
The balanced equation says one mole of hydrogen gas
two moles of sodium from the coefficients of the balanced equation.
Now rid of sodium we want moles of hydrogen to go away.
And we want to get grams--we can compare grams of one substance
to moles of that same substance--that's the molar mass.
And so 2.016 for one mole.
We can calculate what the theoretical yield if a sodium is limiting.
We get 0.1754 to three significant figures five grams of H2
as our theoretical yield.
But we also have to check for
what if water is the limiting reagent?
And so we've got to go through the roadmap again
and solve again.
So if water,
we start with 13 grams.
We want to get rid of water and get moles.
That's the molar mass so 18.02.
Want to get rid of moles water
and get moles of H2.
That's coefficients of the balanced equation.
One for H2.
Two for water again (same mole ratio by coincidence).
Grams, grams: moles, moles.
We want to get rid of moles of H2
and get grams H2 from the theoretical yield
And what do we have?
Well, grams H2 molar masses 2.016 just like before
for one mole.
Let's see what our theoretical yield is.
I get 0.727 grams.
of H2 to three significant figures.
So 0.727
is larger than .0175
so sodium is the limiting reagent.
That means the theoretical yield is 0.175.
And so if we finish the problem
and calculate our percent yield, we want actual yield
which is 0.504
divided by the theoretical yield 0.175 grams.
Percentage is unitless.
0.504 divided by 0.175
and we get times a hundred, 288% yield
which if we're paying attention
is impossible.
So we have to acknowledge that there is a problem with the experiment.
We can't get more yield
than is possible.
100% is maximum and is really unobtainable.
So getting a perfect reaction every atom out that you put in is very difficult to do.
We've got to be able to recognize
that when we get a percent yield greater than 100
that means we have a big problem with the experiment.
Either error on our part or inconsistent premise based on the experiment
So be careful about these in lab.
Most problems give you percentage under 100 percent.
However for the problems but they don't have to.
Remember whenever that happens it's a problem with the experiment.