Percent Yield

One use of stoichiometry is predicting the amount of product we expect to make. This is very useful especially for industrial purposes. However, stoichiometry only predicts the theoretical maximum that can be made. In the lab incomplete reactions, impure chemicals, reactants lost from transferring vessels can all lead to less product being made than expected. This smaller amount is the actual yield.

Percent Yield

Percent yield is how close the actual yield is to the theoretical maximum. You may be familiar with percent yields because this is how test scores are often reported. An 80% yield on a test means you answered 80% of the questions correctly. In the lab we can "grade" experiments similarly. A good chemist may achieve 80% yield or more. Yields below 40% are considered poor and may require the experiment to be redone for the results to be considered accurate.

Example

Given the reaction: CaCO₃→CaO + CO₂

What is the percent yield if 60.0 grams of CaCO₃ is heated to give 15.0 grams of CaO?

First we must solve this as a mass-mass problem to find the theoretical yield.

1. Identify what information we are given and what we need to find.

given: 60.0 g of CaCO₃, find: theoretical grams of CaO (the 15.0 g of CaO is the actual)

2. Convert given into moles by dividing by molar mass.

60.0 g of CaCO₃ ÷ 100.1 g/mol H₂O = 0.599 mol CaCO₃

3. Create a conversion factor from the balanced equation.

We need 1 mole CaCO₃ for each 1 mole CaO

CaCO₃ /CaO

4. Multiply the given by the conversion factor.

0.599 mol CaCO₃ x CaO/CaCO₃ = 0.599 mol CaO

5. Convert moles to mass by multiplying by the molar mass.

Theoretical Yield → 0.599 mol CaO x 56.077 g/mol = 33.6 g CaO

Now we will take the answer from our mass-mass stoich and use it in our percent yield formula:

The theoretical yield is 33.6 g but the actual yield is 15.0 g CaO

Following the percent yield formula:

15g CaO / 33.6 CaO x 100 = 44.6%