Calculating Theoretical and Percent Yield

In chemistry, it is often important to know how efficient a reaction is. This is because when a reaction is carried out, the reactants may not always be present in the proportions written in the balanced equation. As a result, some of the reactants will be used, and some will be left over when the reaction is completed.

Theoretical, Actual, and Percents Yields

A reaction should theoretically produce as much of the product as the stoichiometric ratio of product to the limiting reagent suggests. This number can be calculated and is called the theoretical yield. However, the amount of product actually produced by the reaction will usually be less than the theoretical yield and is referred to as the actual yield. This is because often reactions have "side reactions" that compete for reactants and produce undesired products. To evaluate the efficiency of the reaction, chemists compare the theoretical and actual yields by calculating the percent yield of a reaction:

$\mbox{Percent yield} = \frac{\mbox{actual yield}}{\mbox{theoretical yield}} \times 100$

$% yield = (actual yield/theoretical yield) * 100$ To calculate percent yield using this equation, it is not necessary to use a particular unit of measurement (moles, mL, g etc.), but it is important that the two values being compared are consistent in units. The theoretical yield of a reaction is 100 percent, but this value becomes nearly impossible to achieve due to limitations.

To accurately calculate the yield, the equation needs to be balanced. Next, identify the limiting reagent. Then the theoretical yield of the product can be determined and, finally, compared to the actual yield. Then, percent yield can be calculated.

For example, consider the preparation of nitrobenzene (C₆H₅NO₂), starting with 15.6g of benzene (C₆H₆) in excess of nitric acid (HNO₃):

$C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O$

$15.6 \ g \ C_6H_6 \times \frac{1 \ mol \ C_6H_6}{78.1 \ g \ C_6H_6} \times \frac{1 \ mol \ C_6H_5NO_2}{1 \ mol \ C_6H_6} \times \frac{123.1 \ g \ C_6H_5NO_2}{1 \ mol \ C_6H_5NO_2}=24.6 \ g \ C_6H_5NO_2$

In theory, therefore, if all C₆H₆ were converted to product and isolated, 24.6 grams of product would be obtained (100 percent yield). If 18.0 grams were actually produced, the percent yield could be calculated:

percent yield = $\frac{18.0g}{24.6g}\times 100 $

percent yield = 73.2%