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This content explains the concepts of theoretical yield and percent yield in chemical reactions, detailing how to calculate the maximum possible product (theoretical yield) based on stoichiometry and then determining the efficiency of a reaction (percent yield) by comparing the actual obtained product to the theoretical maximum.
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hi everyone in this video we're
continuing to discuss reaction yields
our learning objectives are to explain
the concepts of theoretical yield and
limiting reagents we touched on limiting
reagents last time so here we're going
to focus on the theoretical yield we're
going to derive the theoretical yield
for reaction under specified conditions
and then we'll use it to calculate the
percent yield for a given
reaction so the amount of product that
can be produced by a reaction when we
calculate it based upon the reaction stochiometry
stochiometry
is called the theoretical yield of the
reaction right so assuming that we use
up all of our limiting reagent we
determine how much product that could
form that would be our theoretical yield
in reality the amount of product that we
obtain is What's called the actual yield
and it is usually less than the
theoretical yield for a number of
reasons right there could be competing
side reactions meaning that the
reactants could be doing some other
reaction that's not represent presented
uh in your chemical equation there could
be an incomplete reaction meaning that
not all of your limiting reagent is
actually used up for whatever reason uh
it could be difficult to recover the
product so for example the product that
you make would be mixed in with the
reagents right the reactants and if you
aren't able to separate them then you'll
only recover a portion of what you
actually make in terms of the
product so the extent to which reactions
theoretical yield is ad achieved is
commonly expressed as something called
the percent yield and this is calculated
by taking the ratio of the actual yield
over the theoretical yield and
multiplying by
100% so when we calculate the percent
yield it's important that the actual
yield and the theoretical yield have the
same units right so if they're both in
grams then grams would cancel and then
our percentage would simply be a
percent so for example if we have a
reaction in which we are able to recover
75 gam of product but based upon our
limiting reagent and how much product
that should have made right we would
have expected 100 grams so in this case
our actual yield is 75 grams our
theoretical yield is 100 gram if we want
to calculate the percent yield we would
take this ratio and then multiply by 100
and we would get
75% so this 75 % tells us that we were
able to to uh recover 75% of the
theoretical yield here we have that upon
1.2745 N2 gram of copper metal so this
is one of our products was obtained
according to the equation what is the %
yield so let's Orient ourselves to the
problem a little bit we can see that the
problem told us that we have copper
sulfate and excess zinc metal so this
excess tells me that the zinc is the
excess reagent and so the copper sulfate
must be our limiting reagent so the
amount of product that we could form it
needs to be based upon how much limiting
reagent we have we're also told the
amount of copper that is recovering in
the reaction so that's 0.392 G this is our
our actual
actual
yield right this is how much was
actually formed in the reaction so if we
want to calculate the percent yield
right we said in the previous slide that
our percent yield would be the actual
yield divided by the theoretical yield
and then multiplied by
100% so we already have the actual yield
which is given in the problem and we
need to calculate what the theoretical
yield would be so our theoretical
yield is going to be based upon using up
all of our limiting reagent the copper
sulfate so the problem told us that we
had 1. 1274 G of copper
sulfate just as we've done in previous
problems we know that our stochiometric
factor is in moles so first we're going
to need to convert GS of copper sulfate to
to
moles so 1 Mo of copper
copper and then finally we need our
theoretical yield to be in the same unit
as our actual right this was provided to
us in grams so we're going to need to
convert from moles of copper to grams of
copper so here again we're going to need
a molar mass I want moles of copper to
cancel so I'm going to put one mole of
copper in the denominator and then the
mass for that one mole is
63.5 5
G so moles of copper cancel and I'm left
with grams of copper and this should be
0.507 Gams of copper so this is my
theoretical yield we're trying to
calculate the percent yield so my percent
percent
yield is going to be the actual yield so
that was given in our problem as 0.392
0.392
G over the theoretical yield this is
what we calculated as 0.507
0.507
G now the grams will cancel so that's
what we want and we're multiplying this
by 100% to make it a
percentage and when you do this
77.3% this means that we were able to recover
recover
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