Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
Now that you have mastered the limiting
reactant, it is time to move on to the last
calculation in this chapter, % yield.
Often when you are doing a chemical
reaction you get less product than you
would expect, based on your calculations.
For instance your calculations might say you
should have 5 g of product, but you only
isolate 3 g of product.
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
The % yield calculation is:
(actual yield/ theoretical yield) x100%
For this example, you predicted 5 g of
product, but only isolated 3 g of product
so the calculation would be:
(3g/5g) x 100% = 60%
One point to remember, you can calculate %
yield with either moles or grams,
grams it
doesn’t matter, you will get the same
answer.
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
Well that is too easy. Let’s make it a more
typical problem by rolling a % yield in
with a limiting reactant problem.
problem
Say you are running the reaction:
2 PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2,
and you obtain 20g of SO2 product, what
was your % yield for this reaction?
2 PbS(s) + 3O2(g)=2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2, and you
obtain
bt i 20
20g off SO2 product,
d t what
h t was your % yield
i ld
for this reaction?
The problem doesn’t say anything about
limiting reactant, but you actually have to
do that first
first, because that is how you get
your theoretical yield.
In our limiting reactant problems I said
earlier that you could use any product you
wanted. In this problem, since we will
need to use SO2 for our % yield
calculation, it makes sense to use that for
the limiting reactant calculation as well,
so you kill two birds with one calculation.
calculation
2 PbS(s) + 3O2(g)=2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2, and you
obtain
bt i 20
20g off SO2 product,
d t what
h t was your % yield
i ld
for this reaction?
PbS calculation:
1 mole PbS
2 mole SO2
100 g PbS ×
×
= .418 mole SO2
239.27 g PbS 2 mole PbS
O2 calculation:
1 mole O2 2 mole SO2
50 g O2 ×
×
= .44 mole SO2
32.00 g O2
3 mole O2
2 PbS(s) + 3O2(g)=2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2, and you
obtain
bt i 20
20g off SO2 product,
d t what
h t was your % yield
i ld
for this reaction?
PbS yields .418 mole, O2 yields .44 mole
so PbS is the limiting reagent and our final
theoretical yield is .418 moles of SO2
Just to prove that you get the same % yield
with moles or grams I will calculate it
both ways
2 PbS(s) + 3O2(g)=2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2, and you
obtain
bt i 20
20g off SO2 product,
d t what
h t was your % yield
i ld
for this reaction?
% yield with grams
Theoretical yield is .418
418 moles of SO2
.418 moles SO2 x (64.07g SO2/1 mole SO2)
= 26.8
26 8 g
% yield = (actual/theoretical) x 100%
= (20/26.8) x 100% = 75%
2 PbS(s) + 3O2(g)=2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2, and you
obtain
bt i 20
20g off SO2 product,
d t what
h t was your % yield
i ld
for this reaction?
% yield with moles
Theoretical yield is .418
418 moles of SO2
Actual yield is 20 g
20g SO2 x (1 mole SO2/64.07g SO2)
= .31 mole SO2
% yield = (actual/theoretical) x 100%
= (.31/.418) x 100% = 75%
2 PbS(s) + 3O2(g)=2PbO(s) + 2SO2(g)
You start with 100g of PbS and 50 g of O2, and you
obtain
bt i 20
20g off SO2 product,
d t what
h t was your % yield
i ld
for this reaction?
Since it doesn’t matter whether you calculate
% yield using moles or grams, choose the
one that
th t you like
lik the
th best,
b t or the
th one that
th t
is easier to do for a given problem.
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
How about a practice problem:
I am going to take a 750g sample of Fe2O3
and react it with 250g of C in the
following reaction:
Fe2O3(s) + 3C(s)=2Fe(s) + 3CO(g)
If I obtain 200g of Iron with this reaction,
what was my % yield?
(Answer on next slide,
slide work it yourself first)
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
Fe2O3:
1 mole Fe2 O3
2 mole Fe
750 g Fe2 O3 ×
×
= 9.4 mole Fe
159.7 g Fe2 O3 1 mole Fe2 O3
C:
1 mole C 2 mole Fe
250 g C ×
×
= 13.9 mole Fe
12.01g C
3C
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
750g of Fe2O3 limits the reaction to 9.4
moles
9 4 moles of Fe =
9.4
9.4 moles x (55.85g/1 mole) = 525g Fe
So your % yield is (200g/525g) x100% =
38%
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
One last comprehensive problem:
Ammonia, oxygen and methane reaction in
the following unbalanced reaction:
NH3(g) + O2(g) + CH4(g) = HCN(g) + H2O(g)
If I start the reaction with 10g each of
ammonia, oxygen and methane, and
obtain 10 g of HCN as a product, what
was the % yield for my reaction?
(Try it yourself and then check your answer
against the next few slides)
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
Step 1 Balance the equation:
NH3(g) + O2(g) + CH4(g)=HCN(g)
(g) HCN(g) + H2O(g)
Should be:
2NH3(g) + 3O2(g) + 2CH4(g)=2HCN(g) +
6H2O(g)
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
Step 2 Find the limiting reactant
1 mole NH 3 2 mole HCN
10 g NH 3 ×
×
= .59 mole HCN
17034 g NH 3 2 mole NH 3
1 mole O2 2 mole HCN
10 g O2 ×
×
= .21 mole HCN
32.00 g O2
3 mole O2
1 mole CH 4
2 mole HCN
10 g CH 4 ×
×
= .62 mole HCN
16.042 g CH 4 2 mole CH 4
Stoichiometric
Sto
c o et c Problems
ob e s IV:
% Yield
Step 3 Find % yield
O2 limits the reaction to .21
21 mole of HCN
Our Actual yield was 10 g of HCN and
10 g HCN x (1 mole/27.028g) = .37 mole
So our % yield
(.21/.37) x 100% = 57%