Calculation of Limiting Reagents and Yield in Reactions
We begin with high school chemistry–balance the reaction. For example,
| PBr_{3} |
CH_{3}CH_{2}CH_{2}OH | –––––> | CH_{3}CH_{2}CH_{2}Br |
| ether |
is an organic chemist's "summary reaction." It is not balanced. We get closer by writing:
CH_{3}CH_{2}CH_{2}OH + PBr_{3} –––––> CH_{3}CH_{2}CH_{2}Br + H_{3}PO_{3}
but, there are three Br atoms on the left and only one on the right. Also, there is only one O atom
on the left of the equation and three on the right. This leads us to the complete the balanced
reaction:
3 CH_{3}CH_{2}CH_{2}OH + PBr_{3} –––––> 3 CH_{3}CH_{2}CH_{2}Br + H_{3}PO_{3}
To make life easier, let's write this as:
a A + b B –––––> c C + d D
where a, b, c and d are called "stoichiometric factors" with the values of 3, 1, 3 and 1 respectively
in this particular case. That is, 3 moles of A combine with 1 mole of B giving 3 moles of C and 1
mole of D. Determine the limiting reagent from the balanced reaction by following these steps:
a. convert the measured quantity of A and B into millimoles. If the reactant is a solid which is
weighed, this only involves dividing the mass by the MW and multiplying by 1000. If the reactant
is a liquid whose volume was accurately measured, this involves multiplying the volume by the
liquid's density to give mass, then proceeding as above for a solid. In this way, determine the
number of millimoles of each reagent.
example: 1.5 grams of solid with MW = 75 g/mol
1.5 g ÷ 75 g/mol x 1000 = 20 mmol
example: 2.0 mL of a liquid with d = 0.800 g/mL and MW = 160 g/mol
2.0 mL x 0.800 g/mL = 1.6 g
1.6 g ÷ 160 g/mol x 1000 = 10 mmol
b. divide the number of millimoles of each reagent by the stoichiometric factor for that reagent. In
this case, we would divide the number of millimoles of CH_{3}CH_{2}CH_{2}OH by 3 and the number of
millimoles of PBr_{3} by 1. Call these results "reaction equivalents" (req).
example: 10 mL CH_{3}CH_{2}CH_{2}OH, density = 0.8035 g/mL, MW = 60.09 g/mol
10 mL x 0.8035 g/mL = 8.0 g
8.0 g ÷ 60.09 g/mol x 1000 = 130 mmol
130 mmol ÷ 3 = 44 req
3.0 mL PBr_{3}, density = 2.88 g/mL, MW = 270.67
3.0 mL x 2.88 g/mL = 8.6 g
8.6 g ÷ 270.76 g/mol x 1000 = 32 mmol
32 mmol ÷ 1 = 32 req
c. The reactant with the smallest number of reaction equivalents is the "limiting reagent." In this example in b)
we see that PBr_{3} is the limiting reagent.
d. To determine "expected yield" of product, multiply the reaction equivalents for the limiting
reagent by the stoichiometric factor of the product. This is the expected number of millimoles of
product. Multiply this result by the MW of the product to determine the expected mass of the
product. The answer will be in milligrams. You may wish to divide by 1000 to obtain the answer
in grams.
example: for reaction as in b) above, product of interest:
CH_{3}CH_{2}CH_{2}Br
32 req (limiting reagent) x 3 (stoichiometric factor) x 123 mg/mmol (MW of product)
= 11,808 mg
= 11.8 g expected yield
e. To determine the "percentage yield" of the product, divide the actual yield in grams by the
expected yield in grams and multiply by 100.
example: for reaction as in b) above, suppose the actual yield of
CH_{3}CH_{2}CH_{2}Br is 8.45 g
% yield = 8.45 (actual) ÷ 11.8 g (expected)
= 0.716
= 72% yield