Atom Economy

What is atom economy and why do we care?

In industrial and research chemistry, most reactions produce side-products alongside the products we are interested in. These side-products can cause a number of issues: they are often mixed in with our product, so are expensive and time-consuming to separate they could be harmful or toxic they could be harmful to the environment. If we have the choice between a reaction that produces more side products or a reaction that produces fewer side products, it is usually best to choose the reaction that produces fewer side products.

Atom economy is how we measure how "much" of the reaction leads to products. It is measured as a percentage, with 100% atom economy meaning there are no side products at all (i.e. only our product is made).

Worked example

Let`s work through an example (AQA AS 2022 Paper 2, question 6.1):

"The equation for a process used to manufacture CCl₃F is

SbF₃Br₂ + CCl₄ -> CCl₃F + SbF₂Br₂Cl

Calculate the percentage atom economy for the production of CCl₃F in this reaction. Give your answer to 3 significant figures."

Step 1: Calculate the mass of the product

We know from the question that CCl₃F is the product, and SbF2Br2Cl is a side-product. So, using masses from our insert, we find the mass of the product:

mass_Product = mass_C + (3 × mass_Cl) + mass_F

massProduct = 12 + (3 × 35.5) + 90

massProduct = 137.5

Step 2: Calculate the masses of all the reactants

Our reactants are SbF3Br2 CCl4, and the masses are again found in the insert:

mass_Reactants = mass_Sb + (3 × mass_F) + (2 × mass_Br) + mass_C + (4 × mass_Cl)

massReactants = 121.8 + (3 × 19) + (2 × 79.9) + 12 + (4 × 35.5)

massReactants = 492.6

Step 3: Calculate atom economy

atom economy = (mass_Product ÷ mass_Reactants) × 100

Simply divide the answer from step 1 by the answer from step 2, and multiply by 100.

As a sanity check, the answer must always be less than 100%. If your answer is more than 100%, you have probably divided the wrong way around.

Let`s finish off the example:
atom economy = (mass_Product ÷ mass_Reactants) × 100
atom economy = (137.5 ÷ 492.6) × 100
atom economy = 27.9 %
(make sure you pay attention to the number of significant figures asked in the question!)


A more complicated example(AQA AS 2017 Paper 1, question 6.4)
"Copper can also be produced by the reaction of carbon with copper(II) oxide according to the equation
2CuO + C → 2Cu + CO₂
Calculate the percentage atom economy for the production of copper by this process."

Step 1: Calculate the mass of the product
Notice that there are 2 moles of product formed. This must be included in the mass of the product:
massProduct = 2 × mass_Cumass_Product = 2 × 63.5
mass_Product = 127


Step 2: Calculate the masses of all the reactants
Notice that there are 2 moles of CuO in the reactants. This must also be included in the masses of the reactants:
mass_Reactants = (2 × mass_Cu) + (2 × mass_O) + mass_C
mass_Reactants = (2 × 63.5) + (2 × 16.0) + 12.0mass_Reactants = 171


Step 3: Calculate the atom economy
atom economy = (mass_Product ÷ mass_Reactants) × 100
atom economy = (127 ÷ 171) × 100
atom economy = 74.3 %