A5.8 Atom Economy

## How “green” is a chemical manufacturing process?

The answer to this question can be expressed as the chemical’s percentage atom economy. To evaluate the atom economy, this equation is used:
$\textsf{Atom economy} = {\frac{\textsf{molar mass of desired products}}{\textsf {total molar mass of all the reactants}}} \times \textsf{100}$

## Worked example 🖊

What is the atom economy of the manufacture of ethanol by: i. fermentation of glucose?
$\textsf{C}_\textsf{6}\textsf{H}_\textsf{12}\textsf O_\textsf6 \rightarrow \textsf{2C}_\textsf2\textsf H_\textsf5\textsf{OH + 2CO}_\textsf2$
ii. hydration of ethane?
$\textsf{C}_\textsf{2}\textsf H_\textsf4 \textsf{ + H}_\textsf2\textsf O \rightarrow \textsf{C}_\textsf2\textsf H_\textsf5\textsf{OH}$
Method i. molar mass of desired products
$=\textsf{2} \times [(\textsf{12} \times \textsf{2})+\textsf{5}+{\textsf{16}}+\textsf{1}]=\textsf{92}$
molar mass of reactants
$=[(\textsf{12} \times \textsf{6})+\textsf{12}+\textsf{(16}\times \textsf{6})]=\textsf{180}$
$\textsf{Atom economy} = {\frac{\textsf{molar mass of desired products}}{\textsf{total molar mass of all the reactants}}} \times \textsf{100}=\textsf{50\%}$
(of course, in practice this reaction is even less efficient as the reaction usually stops before all the glucose has been changed into ethanol!) Method ii. Atom economy = 100% as there is only one product.