Answer: % yield
=
actual yield
theoretical yield
⋅
100
%
So, let's say you want to do an experiment in the lab. You want to measure how much water is produced when 12.0 g of glucose (
C
6
H
12
O
6
) is burned with enough oxygen.
C
6
H
12
O
6
+
6
O
2
→
6
C
O
2
+
6
H
2
O
Since you have a
1
:
6
mole ratio between glucose and water, you can determine how much water you would get by
12.0
g glucose
⋅
1 mole glucose
180.0 g
⋅
6 moles of water
1 mole glucose
⋅
18.0 g
1 mole water
=
7.20
g
This represents your theoretical yield. If the percent yield is 100%, the actual yield will be equal to the theoretical yield. However, after you do the experiment you discover that only 6.50 g of water were produced.
Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%. This is confirmed by
% yield
=
6.50 g
7.20 g
⋅
100
%
=
90.3
%
You can backtrack from here and find out how much glucose reacted
65.0 g of water
⋅
1 mole
18.0 g
⋅
1 mole glucose
6 moles water
⋅
180.0 g
1 mole glucose
=
10.8
g
Step-by-step explanation: