How can I find the percentage yield?

Jan 2, 2015

Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%.

"% 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.

${C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2} \to 6 C {O}_{2} + 6 {H}_{2} O$

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") = 7.20g

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") = 10.8g

So not all the glucose reacted, which means that oxygen was not sufficient for the reaction - it acted as a limiting reagent.

As a conclusion, percent yield problems always have one reactant act as a limiting reagent, thus causing a difference between what is calculated and what is actually obtained. A percent yield that exceeds 100% is never possible, under any circumstances, and means that errors were made in the calculations.