Percentage yield and atom economy

In a nutshell

Percentage yield calculates the conversion of theoretical yield to actual yield. Atom economy is the measure of atoms in reactants compared to atoms in useful products. 

Equations

​$$\frac{mass\ of\ products}{max\ theoretical\ mass\ of\ products}\ \times 100\ = \% \ yield$$​

​$$\frac{M_r\ of\ desired\ products}{M_r\ of\ reactants}\ \times\ 100\ =\ \% \ atom\ economy$$​​

Percentage yield

Theoretical yield is if $$100\%$$ of the reactants react to form the products. In practice, yield is never $$100\%$$​ and therefore ways of calculating yields are required.

There are several ways in which a product can be lost, therefore lowering the yield. Loss in transfer is one way, a solution might not be completely emptied or residue can be left behind. Certain chemicals can evaporate more easily than others so the reaction may not have gone to completion.

Calculating percentage yield

​$$\frac{mass\ of\ products}{max\ theoretical\ mass\ of\ products}\ \times 100\ = \% \ yield$$​

$$100\%$$​ means that only the product was made and $$0\%$$​ means that no product was formed. 

Procedure

Example

The theoretical maximum yield of a reaction results in $$2.0\ g$$​ of $$CuSO_4$$ product. After the reaction and transfer, $$1.2\ g$$ of $$CuSO_4$$ is collected.

Mass of products achieved:

$$1.2\ g\ CuSO_4\ (real)$$

​​

Theoretical maximum mass of product:

​$$2.0\ g\ CuSO_4\ (theoretical)$$​​

Insert the given values into the equation for percentage yield:

$$\frac{mass\ of\ products}{max\ theoretical\ mass\ of\ products}\ \times 100\ = \% \ yield$$​​

​$$\frac{1.2\ g\ (real)}{2.0\ g\ (theoretical)}\ =\ 0.6$$

​

Multiply by $$100$$​ to convert the fraction to a percentage:

​$$0.6 \ \times\ 100\ =\ \underline{60\%}$$​​

The percentage yield achieved for this reaction was $$\underline{60\%}$$​.

Atom economy 

Definition

Atom economy compares the total amount of atoms used in the reactants to the amount of atoms present in the desired product. Whilst no atoms are lost in a chemical reaction they may end up as wasted a product. 

Calculating atom economy

​$$\frac{M_r\ of\ desired\ products}{M_r\ of\ reactants}\ \times\ 100\ =\ \% \ atom\ economy$$

​​

Procedure

Example 

The electrolysis of water is seen as a route to produce eco-friendly $$H_2$$ for industrial processes. $$O_2$$ is a waste product. Calculate the atom economy for:

$$2H_2O\ \rightarrow\ 2H_2\ +\ O_2$$ 

Calculate the molecular mass of the reactants and the desired product:

$$M_r(H_2O)\ = \ (1\ \times\ 2)\ + (16\ \times\ 1) = 18 \\ M_r(H_2) = (1 \times2)\ =\ 2$$

Multiply the molecular mass of the reactants and the desired product by the stoichiometry:

$$M_r(2H_2O)\ =\ 18\ (M_r)\ \times \ 2\ =\ 36 \\M_r(2H_2)\ =\ 2\ (M_r)\ \times 2\ =\ 4$$​​

Note: Stoichiometry means the ratio of each substance in the balanced equation.

Insert the calculated values into the equation for percentage atom economy:

$$\frac{M_r\ of\ desired\ products}{M_r\ of\ reactants}\ \times\ 100\ =\ \% \ atom\ economy$$​​

​$$\frac{4\ }{36\ }\ =\ 0.111$$

Multiply by $$100$$​ to convert the fraction to a percentage:

​$$0.111\ \times\ 100 \ = \ 11.1\%$$

The percentage atom economy for the $$H_2$$ production is $$\underline{11.1\%}$$.