In chemistry, percent yield is a comparison of actual yield to theoretical yield, expressed as a percentage. Here is a look at the percent yield formula, how to calculate it, and why it may be less than or greater than 100%.
The percent yield formula is actual yield divided by theoretical yield in moles, multiplied by 100%:
Percent Yield = Actual Yield/Theoretical Yield x 100%
It does not matter whether you express actual and theoretical yield in grams or moles, as long as you use the same units for both values.
Calculating percent yield requires two values: the actual yield and the theoretical yield. Yield depends on the mole ratio between reactants and products. The actual yield is the amount of product obtained from a reaction or experiment. You weigh the product and then convert the mass (usually in grams) to moles.
The theoretical yield comes from stoichiometry. In other words, it comes from the mole ratio between reactants and products in the balanced equation for the chemical reaction. Once you have the balanced equation, the next step is finding the limiting reactant. The limited reactant is the reactant that limits the amount of product because it’s consumed before the other reactant runs out. In a decomposition reaction, there may only be one reactant, which makes it the limiting reactant. In other reactions, you compare the molar masses and mole ratios. Next, use the number of moles of limiting reactant and mole ratio and calculate the theoretical yield. Finally, calculate theoretical yield.
First, here is a simple example of the percent yield calculation in action:
The decomposition of magnesium carbonate forms 15 grams of magnesium oxide in an experiment. The theoretical yield is 19 grams. What is the percent yield of magnesium oxide?
Here, you know the actual yield (15 grams) and the theoretical yield (19 grams), so just plug the values into the formula:
Percent Yield = Actual Yield/Theoretical Yield x 100%
Percent Yield = 15 g/19 g x 100%
Percent Yield = 79%
Find the percent yield of a reaction when you obtain 4.88 g of AlCl3(s) from a reaction between 2.80 g Al(s) and 4.15 g Cl2(g).
First, write out the balanced equation for the reaction:
Next, find the limiting reactant. Start with the molar masses of the reactants and products:
2.80 g Al x (1 mol Al/26.98 g Al) = 0.104 mol Al
4.15 g Cl2 x (1 mol Cl2/70.90 g Cl2) = 0.0585 mol Cl2
Compare the mole ratio to the actual number of moles present in the reaction. From the balanced equation, you see 2 moles of Al reacts with 3 moles of Cl2.
Mole ratio: moles Al/moles Cl2 = 2/3 = 0.6667
Actual ratio moles Al/moles Cl2= 0.104/0.0585 = 1.78
The actual ratio is larger than the mole ratio, so there is excess Al and Cl2 is the limiting reactant. (If the actual ratio is smaller than the mole ratio, it means there is excess Cl2 and Al is the limiting reactant.)
Use the actual number of moles of Cl2 and the mole ratio and find the maximum amount of AlCl3.
0.00585 mol Cl2 x (2 mol AlCl3/3 mol Cl2) = 0.00390 mol AlCl3
Convert the number of moles of product into grams so the units of actual and theoretical yield are the same. Get this from the molar mass.
0.00390 mol AlCl3 x (133.33 g AlCl3/1 mol AlCl3) = 5.20 g AlCl3
Finally, calculate percent yield. Actual yield is 4.88 g of AlCl3 (given in the problem) and theoretical yield is 5.20 g AlCl3.
Percent Yield = Actual Yield/Theoretical Yield x 100%
Percent Yield = 4.88 g AlCl3 / 5.20 g AlCl3 x 100%
Percent Yield = 93.8%
Percent yield is always less than 100% (often by a lot), yet it’s possible to calculate a value greater than 100%.
There are a few reasons why percent yield always falls short.
And yet, sometimes you get more product than predicted. Occasionally, an impurity contributes to product formation. But, usually there’s less product than the theoretical yield. Yet, if you collect an impure product, the mass exceeds the theoretical yield. The most common situation is weighing product that isn’t completely dry. Part of the mass is solvent, so it appears you got more product than predicted.