Calculation of Limiting Reagents and Yield in Reactions

We begin with high school chemistry–balance the reaction. For example,

 PBr3 CH3CH2CH2OH –––––> CH3CH2CH2Br ether

is an organic chemist's "summary reaction." It is not balanced. We get closer by writing:

CH3CH2CH2OH + PBr3 –––––> CH3CH2CH2Br + H3PO3

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 CH3CH2CH2OH + PBr3 –––––> 3 CH3CH2CH2Br + H3PO3

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 CH3CH2CH2OH by 3 and the number of millimoles of PBr3 by 1. Call these results "reaction equivalents" (req).

example: 10 mL CH3CH2CH2OH, 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 PBr3, 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 PBr3 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: CH3CH2CH2Br

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 CH3CH2CH2Br is 8.45 g

% yield = 8.45 (actual) ÷ 11.8 g (expected)
= 0.716
= 72% yield