Percentage Yield and Atom Economy

Percentage Yield

When writing a fully balanced chemical equation, it is assumed that all of the reactants will be converted into products, thus giving a yield of 100%. This is rarely the case as:

The reaction may be in equilibrium or may not go to completion.
Side reactions may occur, leading to by-products.
Some of the products may be left in apparatus, or during separation and purification products can be lost.

The percentage yield can be calculated by:

$$ \text{yield} = \frac {\text{actual product, mol}}{\text{theoretical product, mol}} \times 100 $$

Worked Example

A student is preparing some ethanoic acid, $CH_{3}COOH$ by heating ethanol, $C_{2}H_{5}OH$, with an oxidising agent. The reaction is shown below:

$$ C_{2}H_{5}OH + 2[O] \rightarrow CH_{3}COOH + H_{2}O$$

The student reacted 9.20g of ethanol with an excess of sulphuric acid and potassium dichromate. they obtained 4.35g of ethanoic acid. What is the percentage yield of ethanoic acid?

First, the moles of ethanol must be calculated.
$$ mol = \frac {mass}{M_{r}}
\hspace3ex
\frac {9.20}{46} = 0.2\,mol $$
Next the expected amount of ethanoic acid should be calculated:
$$
\eqalign{
C_{2}H_{5}OH + 2[O] &\rightarrow CH_{3}COOH + H_{2}O \\
1 mol \phantom{~~+2[O] } &\rightarrow 1 mol
}
$$
The actual amount of moles should be calculated:
$$ mol = \frac {mass}{M_{r}}
\hspace3ex
\frac {4.35}{60} = 0.0725\,mol $$
The yield formula can now be used to find the percentage yield.
$$ \text{yield} = \frac {\text{actual product, mol}}{\text{theoretical product, mol}} \times 100 \\
\text{yield} = \frac {0.2}{0.0725} \times 100 = 36.25\% $$

Atom Economy

A reaction often produces by-products in addition to the desired product. These by-products are often considered as waste and are disposed of. This is costly and has environmental impacts. Atom economy not olny considers the desired product but also the by-products. The equation for atom economy is:

$$ \text{atom economy} = \frac {\text{Mr of desired product}}{\text{Sum of Mr of all products}} \times 100 $$

How atom economy can benefit society

By using processes with a higher atom economy, companies can reduce the waste produced. The type of reaction is a major factor in achieving a higher atom economy:

Addition reactions have an atom economy of 100%.
Reactions involving substitution or elimination have an atom economy of less than 100%.

To improve the atom economy, a use must be found for the by-products. In industry, this is sometimes done by selling off by products.

Worked Example

Methanol can be prepared by reacting potassium hydroxide with chloromethane. Calculate the atom economy for this reaction.

$$ CH_{3}Cl + KOH \rightarrow CH_{3}OH + KCl \\ \\
\eqalign{
\text{Mr of desired product } CH_{3}OH &= 12 + (4 \times 1) + 16 \\
&= 32} \\
\eqalign{
\text{Mr of all products } &= 32 + 74.6 \\
&= 106.6} \\ \\
\eqalign{
\text{atom economy} &= \frac {\text{Mr of desired product}} {\text{Sum of Mr of all products}} \times 100 \\
&= \frac {32}{106.6} \times 100 \\
&= 30.02\%}
$$