Percentage yield calculations

The theoretical yield of a reaction is the mass of product, obtained from our starting mass of reactant, as calculated from the equation for the reaction. However, the actual yield which is obtained from a practical reaction is nearly always less than the theoretical yield. This is because the reaction may not have been given sufficient time to be complete, or alternative side-reactions may have reduced the quantity of main product obtained. If it is an equilibrium, the reaction never reaches completion. We also lose some product during purification processes and in general transfer work.

The percentage yield indicates how close we have got to the maximum yield possible. The percentage yield may be limited by the success of the reaction itself, and a low percentage yield is not necessarily an indication of poor practical work. Practical Chemistry textbooks will usually indicate the yield attainable from a particular reaction. Of course, poor practical work will further reduce the percentage yield obtained.

Percentage yield is calculated using the following equation:

Example

In the dehydration of 15.0 g of cyclopentanol, 8.7 g of cyclopentene were obtained. What is the percentage yield of this reaction?

First write down the equation with the molar masses underneath:

This tells us that 86 grams of cyclopentanol should make 68 grams of cyclopentene. However, we only have 15.0 g of cyclopentanol. So we need to scale the quantities down in the same way that we might scale down a recipe if we wanted a smaller quantity. To scale 86 g down to 15.0 g first divide by itself (86) to give 1 and then multiply by the number you are looking for (15.0).This gives:

86/86 × 15.0 = 15.0 g

Carry out exactly the same maths on the product quantity to give the theoretical yield:

68/86 × 15.0 = 11.9 g (theoretical yield)

Now use the equation above to calculate the percentage yield:

Percentage yield = 8.7/11.9 × 100 = 73%

If you would like some practice at this technique you can try exercise 1 and exercise 2.

If we know the expected percentage yield for a reaction we can allow for this when designing our experiment. For example, if a reaction has a 50% yield, and we design our experiment according to the equation for the reaction, we shall only get half as much product as we want. The solution to this is to double the amount of reactants that we started with. In general, we need to use the following relationship:

Example

The formation of ethyl ethanoate from ethanol has a percentage yield of 29% using excess ethanoic acid. What mass of ethanol should be used to make 10 g of product?

We first need the equation and molar masses in order to work out the theoretical starting quantity:

You will see from the equation that this is a reversible reaction. This is the main reason for the low yield. The molar masses indicate the theoretical quantities required to give 88 g of product. In order to scale this down to 10 g we need to divide by 88 (to make 1 g) and then multiply by 10 (to give 10 g).

Theoretical quantity of ethanol = 46/88 × 10 = 5.23 g

Theoretical quantity of ethanoic acid = 60/88 × 10 = 6.82 g

We now need to use the above equation to take account of the poor percentage yield:

Actual quantity of ethanol = 5.23 × 100/29 = 18.0 g

Actual quantity of ethanoic acid = 6.82 × 100/29 = 23.5 g, but use excess

Now try exercise 3 and exercise 4.