Atom economy and percentage yield

In a nutshell

The percentage yield and atom economy can be calculated for reactions. These two calculations show how much of the potential product is wasted during the reaction.

Equations

$Percentage \space yield \, = \, \dfrac{actual \space yield}{theoretical \space yield} \, \times \, 100$

$Atom \space economy \, = \, \dfrac{molecular \space mass \space of \space desired \space product}{sum \space of \space molecular \space mass \space of \space all \space products} \, \times \, 100$

Percentage yield

Definition

The percentage yield is the amount of product collected during a reaction. Percentage yield tells you how wasteful the process is and how much product is lost.

Percentage yield can be calculated by:

$Percentage \space yield \, = \, \dfrac{actual \space yield}{theoretical \space yield} \, \times \, 100$

The theoretical yield is the amount of product that should be made if nothing is lost during the reaction. Where as the actual yield is the amount of product collected once the reaction has completed.

Note: The theoretical yield can be calculated from masses by using a balanced equation to find molar ratios.

procedure

1.	Write the balanced equation for the reaction to find the molar ratios of both the reactants and products.
2.	Find the number of moles of the reactant by using the mass and the equation: $n \, = \dfrac{m}{M}$
3.	Use the moles of the reactant and molar ratios found from the balanced equation to find the number of moles of the desired product.
4.	Use the moles calculated to find the theoretical mass of the product using: $m \, = \, n \, \times \, M$
5.	Finally, use the actual mass gathered and the theoretical mass calculated to find the percentage yield: $Percentage \space yield \, = \, \dfrac{actual \space yield}{theoretical \space yield} \, \times \, 100$

Example

$2.56 \, g$ of sodium chloride is reacted with sulfuric acid. $2.45\, g$ of sodium sulfate is produced. What is the percentage yield?

First, write down the equation you will use to find the percentage yield:

$Percentage \space yield \, = \, \dfrac{actual \space yield}{theoretical \space yield} \, \times \, 100$

You need to find the theoretical yield.

Write out the balanced equation for the reaction:

$2NaCl(s) \, + \, H_2SO_4(aq) \, \rightarrow \, Na_2SO_4(s) \, + \, 2HCl(aq)$

Next, write down the equation you will use to find the number of moles of
$NaCl$ :

$n \, = \, \dfrac{m}{M}$

Find the molar mass for $NaCl$ :

$M \, (NaCl) \, = \, 23 \, + \, 35.5 = 58.5 \, g \, mol^{-1}$

Calculate the number of moles of $NaCl$ but substituting the mass and molar mass into the equation:

$n \, = \, \dfrac{2.56}{58.5} \, = \, 0.0438 \, mol$

Use the balanced equation to find a molar ratio between $NaCl$ and $Na_2SO_4$ :

The mole ratio from $NaOH$ to $Na_2SO_4$ is $2 : 1$ .

Use the mole ratio to find the number of moles of $Na_2SO_4$ :

$n \, = \, \dfrac{0.0438}{2} \, = \, 0.0219 \, mol$

Find the mass (theoretical yield) of $Na_2SO_4$ using the equation:

$m \, = \, n \, \times \, M$

Find the molar mass of $Na_2SO_4$ :

$M \, (Na_2SO_4) \, = \, (23 \, \times \,2) \, + 32.1 \, + \ (16 \, \times \, 4) \, = \,142.1 \, g \, mol^{-1}$

Substitute in the values for the molar mass and number of moles to find the mass of $Na_2SO_4$ :

$m \, = \, 0.0219 \, \times \, 142.1 \, = \, 3.11 \, g$

Substitute your values for actual yield (given in the question) and theoretical yield calculated to find the percentage yield:

$Percentage \space yield \, = \, \dfrac{2.45}{3.11} \, \times \, 100 \, = \, 78.8\%$

The percentage yield for this reaction is $\underline{78.8\%}$ .

Atom economy

Definition

Atom economy is a measure of the proportion of atoms in the reactants that are in the desired product. Atom economy tells you how wasteful the reaction itself is.

Atom economy can be calculated by:

$Atom \space economy \, = \, \dfrac{molecular \space mass \space of \space desired \space product}{sum \space of \space molecular \space mass \space of \space all \space products} \, \times \, 100$

Example

Find the atom economy for the sodium chloride product of the reaction below.

$Na_2S(s) \, + \, 2HCl(aq) \, \rightarrow \, 2NaCl(s) \, + \, H_2S(g)$

First, write down the equation you will use to find atom economy:

$Atom \space economy \, = \, \dfrac{molecular \space mass \space of \space desired \space product}{sum \space of \space molecular \space mass \space of \space all \space products} \, \times \, 100$

Find the molecular mass of all the products:

$M \, (2NaCl) \, = \, 2 \, \times \, (23 \, + \, 35.5) \, = \, 117 \, g \, mol^{-1} \newline M \, (H_2S) \, = \, 2 \, + \, 32.1 \, = \, 34.1 \, g \, mol^{-1} \newline Total \, = \, 117 \, + \, 34.1 \, = \, 151.1 \, g \, mol^{-1}$

Find the molecular mass for the desired product ( $NaCl$ ):

$M \, (2NaCl) \, = \, 2 \, \times \, (23 \, + \, 35.5) \, = \, 117 \, g \, mol^{-1}$

Substitute the molecular masses for the products and desired product to find the atom economy:

$Atom \space economy \, = \, \dfrac{117}{151.1} \, \times \, 100 \, = \, 77.4\%$

The atom economy is $\underline{77.4\%}$ .

When a reaction has one product the percentage atom economy will always be $100\%$ .

Example

Addition reactions will always form one product, leaving these reactions with an atom economy of $100\%$ .

Sustainability

Chemists will choose reactions with high atom economies to reduce wastage. If a reaction has a low atom economy more money is needed to separate and dispose of the waste products. It is also expensive to purchase reactants when they are not all being used to make the desired product.

Reactions that require high amounts of energy (high temperatures and pressures) are very costly. These conditions are not good for the environment. Raw materials sourced renewably are much better for the environment.