Percentage yield

A balanced chemical equation tells what the theoretical or ideal yield of the reaction should be, assuming perfect completion of the reaction. In reality, the actual yield may be less than the theoretical yield. Reasons for this include:

1.

There may be competing reactions which hinder the process under consideration.

2.

External conditions may not be perfectly maintained.

3.

Reactants may not be pure.

The ratio of the actual yield to the theoretical yield is a quantity known as the fractional yield:

$${\rm fractional\ yield} = {{\rm actual\ yield} \over {\rm theoretical\ yield}}$$

and the percentage yield is obtained from the fractional yield according to

$${\rm percentage\ yield} = 100\%\times ({\rm fractional\ yield})$$

Example: Pure zinc can be produced in the following two-step process:

$$2{\rm ZnS} + 3{\rm O}_2 \longrightarrow 2{\rm ZnO} + 2 {\rm SO}_2$$

$${\rm ZnO} + {\rm CO} \longrightarrow {\rm Zn} + {\rm CO}_2$$

Suppose 5.32 kg of ZnS yield 3.30 kg of pure zinc. What is the percentage yield?

Solution: Convert the mass of ZnS to moles:

$${\rm moles\ ZnS} = {5320 {\rm\ g} \over 97.46 {\rm\ g/mol}} = 54.6 {\rm\ moles\ ZnS}$$

How do we figure out how many moles of Zn are produced in the two-step reaction? We need to obtain the overall reaction, which is just a ``sum'' of the individual steps. We can obtain the equation of the overall reaction by adding the equations for the individual steps. However, this addition must be carried out in a particular way. Specifically, note that the species ZnO is produced in the first step and consumed in the second step. Thus, it does not appear among the final products of the process, and for this reason, it is called an intermediate. The equations for the individual steps must be added in such a way that all intermediates cancel out. Since adding reaction equations works like the adding of algebraic equations, this means that the coefficients of intermediates must be the same in the individual steps. In the above process the coefficients of ZnO are different in each step. But, if the first step is divided by 2, so that the reaction equation reads:

$${\rm ZnS} + {3 \over 2}{\rm O}_2 \longrightarrow {\rm ZnO} + {\rm SO}_2$$

then ZnO has the same coefficient in the first step as is does in the second and the equations can now be added:

$${\rm ZnS} + {3 \over 2}{\rm O}_2 \longrightarrow {\rm ZnO} + {\rm SO}_2$$

$${\rm ZnO} + {\rm CO} \longrightarrow {\rm Zn} + {\rm CO}_2$$

$$--------- ---------$$

$${\rm ZnS} + {3 \over 2}{\rm O}_2 + {\rm ZnO} + {\rm CO} \longrightarrow {\rm ZnO} + {\rm SO}_2 + {\rm Zn} + {\rm CO}_2$$

$${\rm ZnS} + {3 \over 2}{\rm O}_2 + {\rm CO} \longrightarrow {\rm SO}_2 + {\rm Zn} + {\rm CO}_2$$

where ZnO has been ``cancelled'' algebraically on both sides. From the overall reaction, we see that

1 mole of ZnS yields 1 mole of Zn

or

54.6 moles of ZnS yields 54.6 moles of Zn

Thus, the mass of Zn produced is

$${\rm mass\ of\ Zn} = (54.6 {\rm\ mol})\times(65.39 {\rm\ g/mol}) = 3570 {\rm\ g} = 3.57{\rm\ kg}$$

This is the theoretical yield predicted by the reaction equations. The actual yield is only 3.30 kg, so that the fractional yield is

$${\rm fractional\ yield} = {3.3{\rm\ kg} \over 3.57{\rm\ kg}} = 0.924 = 92.4\%$$

In multi-step processes, it is important that each step have as high a percentage yield as possible. The reason for this is that the overall percentage yield will be a product of the percentage yields of each step. As an example, suppose there is a 10-step process with a 50% yield for each step:

$${\rm Step\ 1}\;\;\;\;\;\;\;\;\;\;{\rm percentage\ yield} = 50\%$$

$${\rm Step\ 2}\;\;\;\;\;\;\;\;\;\;{\rm percentage\ yield} = 50\%$$

$$\cdots$$

$${\rm Step\ 10}\;\;\;\;\;\;\;\;\;\;{\rm percentage\ yield} = 50\%$$

The overall percentage yield would be the product of the individual percentages, or

$$(0.5)^{10} = 0.001 = 0.1\%$$

Even if the percentage yield of each step were 90%, the overall yield would be $(0.9)^{10} = 35\%$, which is better, but still somewhat low. If the percentage yield of each step were 95%, then the overall yield would be $(0.95)^{10} = 60\%$, which might make the process feasible.