Percent Yield Calculations
Percent yield calculations are one of the most commonly used calculations in organic chemistry
lab. Unfortunately the calculations are also one of the most common areas where students
make mistakes. Understanding how to routinely and accurately carry out percent yield
calculations will bolster understanding of the chemistry behind the calculation and
simultaneously help improve grades.
The amount of product which can form from each starting material must be calculated. This
defines what is the limiting reagent and allows the percent yield to be determined.
The calculations necessary to determine the Percent Yield calculations can be broken down into
3 simple steps.
Necessary Steps
1. Determine the amount of product (usually in grams) which can form from EACH starting
material. A separate calculation must be done for each starting material. Catalysts and solvents
are not stoichiometric reagents and cannot be converted into product.
2. The reactant which will produce the smallest amount of the product is the limiting reagent.
The limiting reagent is the starting material which will be depleted first. The amount of product
formed from the limiting reagent is the theoretical maximum amount of product which can form.
3. Determine the percent yield of the reaction by dividing actual grams of product formed by the
theoretical maximum amount of product possible X 100%.
Dimensional analysis is very helpful in arriving at the correct answer. Include all units and
identifiers. Often when a student is unsure how to proceed while solving a problem, a close
examination of the units will direct the correct next multiplication or division.
Reminders
* Density is used to convert from volume to mass.
* Molecular weight is used to convert from mass to moles.
* Reaction Stoichiometry is used to convert from moles of reactants to moles product.
* If a solution as opposed to a neat liquid is used, then the volume (in liters) of solution used can
be multiplied by the molarity (M = moles/L) of the solution to determine moles.
* Keep track of all units.
* Be very aware of the stoichiometry of the reaction.
* All data provided may not have to be used to determine answer.
Revised August 23, 2012
Practice problem 1. One material measured as a solid (g) one material as a neat liquid (mL).
2.897 g of maleic anhydride (MA) is reacted with 2.56 mL of cyclohexadiene (CHD) to produce 3.979
grams of the bicyclic product. What is the limiting reagent and what is the percent yield?
name
density (g/ml)
mol. Wt.
(g/mol)
volume (mL)
mass(g)
cyclohexadiene
(CHD)
0.8405
80.15
maleic anhydride
(MA)
1.314
98.06
5,6-anhydride [2.2.2]bicyclo-2octene
2.56
-
2.897
3.979
178.20
1) Amount of product possible from each starting material
For maleic anhydride
2.897 _ g _ MA( g ) 1 _ mole _ MA(mol) 1 _ mol _ Pr oduct 178.20 _ grams _ product
x
x
x
 5.265 _ grams _ Pr oduct
98.06 grams _ MA( g )
1 _ mol _ MA
1 _ mol _ Pr oduct
Step 1
Step 2
Step 3
For cyclohexadiene
0.845 gCHD 1 _ moles _ CHD 1 _ moles _ product 178.20 grams _ product
2.45 _ mL _ CHDx
x
x
x
 4.78 _ g _ Pr oduct
1 _ mL _ CHD 80.15 _ g _ CHD
1 _ mole _ CHD
1 _ mole _ product
Step 1
Step 2
Step 3
2) Theoretical Maximum and Limiting Reagent
Since 4.78 is smaller than 5.265, cyclohexadiene is the limiting reagent and 4.78 g is the theoretical
maximum amount of product which can form.
3) Percent Yield. Using knowledge of actual amount of product formed, calculate the % yield.
Actual _ Mass _ Pr oduct _ Formed ( g )
x100%  % yield _ Pr oduct
Theoretica l _ Maximum _ Mass _ Pr oduct ( g )
3.979 _ g _ Pr oduct
x100%  83.2 _ % _ yield _ Pr oduct
4.78 _ g _ product
Practice Problem 2. One material measured as a solution (M) one as a mass (g).
4.24 mL of 2-butanol is reacted with 10.2 mL of 6.0 M HCl to form 2.8798 grams of 2-chlorobutane. What
is the limiting reagent and what is the percent yield.
name
density
(g/ml)
2-butanol
(BuOH)
0.81
Hydrochloric Acid
(6.0 M)
1.314
2-chlorobutane
Water
1.00
Revised August 23, 2012
mol. Wt.
(g/mol)
volume
(mL)
mass(g)
74.1
36.5
92.6
4.24
10.2
-
-
-
2.8798
18.00
Amount of product possible from each starting material
4.24 _ ml _ BuOH
0.810 _ g _ BuOH 1 _ mole _ BuOH 1 _ mol _ product 92.6 _ g _ Pr od
x
x
x
 4.28 grams Pr oduct
1 _ ml _ BuOH
74.1 _ g _ BuOH 1 _ mol _ BuOH
1 _ mol _ Pr od
10.2 _ ml _ HCl
1L _ HCl
6.0 _ mol _ HCl 1 _ mol _ product 92.6 _ g _ Pr od
x
x
x
x
 5.7 _ grams Pr oduct
1000 _ ml _ HCl
1 _ L _ HCl
1 _ mol _ HCl
1 _ mol _ Pr od
x
2) Theoretical Maximum and Limiting Reagent
Since 4.28 is less than 5.7, 2-butanol is the limiting reagent and the maximum theoretical amount of
product which can form is 4.28 g.
Percent Yield. Using knowledge of actual amount of product formed, calculate the % yield.
Actual _ Mass _ Pr oduct _ Formed ( g )
x100%  % yield _ Pr oduct
Theoretica l _ Maximum _ Mass _ Pr oduct ( g )
2.8798 _ g _ Pr oduct
x100%  67.3 _ % _ yield _ 2  ChloroBu tan e
4.28 _ g _ product
Practice Problem 3. Non 1 :1 stoichiometry. Both materials measured as a neat liquid (mL).
7.3 mL of methylbenzoate is reacted with 10.0 mL of bromobenzene in an excess of magnesium
shavings and an acid work-up to produce triphenylmethanol. What is the limiting reagent, the theoretical
maximum amount of product which can form, and the percent yield? (An excess of magnesium is used.)
name
methyl benzoate
Bromobenzene
(MB)
(BrB)
density (g/ml)
1.09
1.50
mol. Wt. (g/mol)
136.1
157.02
volume (mL)
7.3
10.0
mass(g)
Amount of product possible from each starting material
triphenylmethanol
260.3
10.655
7.3 _ ml _ MB 1.09 _ g _ MB 1 _ mole _ MB 1 _ mol _ Pr od 260.5 _ g _ Pr od
x
x
x
x
 15 _ g _ Pr oduct
1 _ ml _ MB 136.1 _ g _ MB 1 _ mol _ MB
1 _ mol _ prod
10.0 _ ml _ BrB
x
1.50 _ g _ BrB
1 _ mol _ BrBe 1 _ mol _ Pr od 260.5 _ g _ Pr od
x
x
x
 12.5 _ g _ Pr oduct
1.00 _ mL _ BrB 157.02 _ g _ BrB 2 _ mol _ BrB
1 _ mol _ Pr od
2) Theoretical Maximum and Limiting Reagent
Because 12.5 is smaller than 15, bromobenzene is the limiting reagent and the theoretical maximum
amount of product which can form is 12.5 grams.
Percent Yield. Using knowledge of actual amount of product formed, calculate the % yield.
Actual _ Mass _ Pr oduct _ Formed ( g )
10.655 _ g _ Pr oduct
x100% 
x100%  85.2 _ % _ yield
Theoretica l _ Maximum _ Mass _ Pr oduct ( g )
12.5 _ g _ product
Revised August 23, 2012