6.13 - Atom Economy
Atom economy of a reaction
- Industrial reactions often make more than one product. Any product other than the desired one is waste.
- Atom economy (or atom utilisation) measures the proportion of reactant mass that ends up as the desired product. It's given as a percentage and calculated from the balanced equation.
Formula for atom economy
Where:
- = Relative formula mass, using the numbers in the balanced equation
Using balancing numbers
- Both lines of the fraction use the balancing numbers: multiply each relative formula mass () by the balancing number in front of it in the equation.
- For example, 3H2 means multiplying the of H2 by 3, and 2Fe2O3 means multiplying the of Fe2O3 by 2.
Interpreting atom economy
- 100% atom economy - Every reactant atom ends up in the desired product, with no waste. A reaction with only one product always has an atom economy of 100%.
- Below 100% - Some of the reactant mass is lost as waste products.
- Atom economy is a theoretical value worked out from the balanced equation alone, not from practical measurements.
Worked example - Atom economy of a reaction
Copper is extracted from copper(II) oxide by heating it with carbon:
2CuO + C → 2Cu + CO2
Calculate the atom economy of this reaction for the production of copper. Give your answer to 2 significant figures.
Desired product = copper, Cu. Relative atomic masses (from the periodic table): C = 12, O = 16, Cu = 63.5
Step 1 - Total relative formula mass of the reactants
The equation uses 2CuO, so that balancing number is used here.
Mr(CuO) = 63.5 + 16 = 79.5
2 × Mr(CuO) = 2 × 79.5 = 159
Total Mr of reactants = 159 + 12 = 171
Step 2 - Relative formula mass of the desired product
The equation makes 2Cu, so the balancing number is used here. Carbon dioxide is the other product and plays no part in the top of the fraction.
2 × Ar(Cu) = 2 × 63.5 = 127
Step 3 - Substitution and calculation
Atom economy = (127 ÷ 171) × 100 = 74.269...%
Step 4 - Round the answer
74.269...% to 2 significant figures is 74%.
So about 74% of the mass of the copper(II) oxide and carbon ends up as copper. The other 26% leaves as carbon dioxide.
Why a high atom economy matters
Choosing a reaction with a high atom economy is important for both economic reasons and sustainable development.
Benefits of a high atom economy
- Conserves raw materials - A higher atom economy means less starting material is needed to make a given amount of product, reducing costs and preserving finite resources.
- Reduces waste - Less waste is produced, cutting disposal costs and reducing the risk of environmental pollution.
Improving industrial efficiency
- Use an alternative reaction with fewer unwanted products, such as an addition reaction where all reactant atoms become one product (100% atom economy).
- Find a use for by-products - If a by-product can be recycled or sold to another manufacturer, it becomes a useful product rather than waste.
Choosing a reaction pathway from data
- A manufacturer often has a choice of several reactions to make a desired product.
- Atom economy is one factor considered, but it's never the only one.
Five factors to consider
- Atom economy - The proportion of reactant mass converted into desired products.
- Percentage yield - The amount of product actually obtained compared with the maximum theoretical yield.
- Rate of reaction - How fast the reaction produces the desired product.
- Position of equilibrium - For reversible reactions, how much product is made at equilibrium and the cost of conditions (such as high temperature or pressure) needed to shift it.
- Usefulness of by-products - Whether waste products can be sold or used, or must be disposed of at a cost.
A route is chosen by balancing all these factors against each other. A pathway with a high atom economy might still be rejected if the yield is too low or the reaction is too slow to be commercially viable.
Worked example - Choosing a reaction pathway
Two reactions make the same product, and a manufacturer has to choose between them. This data is supplied for both:
| Route 1 | Route 2 | |
|---|---|---|
| Atom economy | 85% | 44% |
| Percentage yield | 40% | 95% |
| Rate | slow | fast |
| Reversible? | yes, and settles well short of the products unless a high pressure is used | no |
| Other product | waste, disposed of at a cost | sold to another manufacturer |
Which route should the manufacturer choose? Use the data to explain the choice.
Step 1 - Compare the atom economies
Route 1 turns 85% of its reactant mass into the product wanted; route 2 turns 44%. On atom economy alone, route 1 looks the better reaction.
Step 2 - Compare the percentage yields
Route 1 delivers 40% of the product its equation predicts; route 2 delivers 95% - almost everything.
Step 3 - Compare the rates
Route 1 is slow, so a plant running it makes less product each day. Route 2 is fast.
Step 4 - Compare the positions of equilibrium
Route 1 is reversible and settles well short of the products, holding its yield at 40%; shifting it further means paying for a high pressure. Route 2 is not reversible, so there's nothing to shift and nothing to pay for.
Step 5 - Compare the usefulness of the by-products
Route 1's other product is waste and must be paid for to be taken away. Route 2's is sold to another manufacturer, so the mass leaving as the other product earns money instead of costing it.
Step 6 - Interpretation
Route 1 wins on atom economy and loses on everything else the data gives. Route 2 turns less than half its reactant mass into the product wanted, but it delivers almost everything the equation predicts, delivers it quickly, and its other product is worth selling. On this data, route 2 is the one to choose.
No single figure settles the choice - the route picked is the one giving the best balance across all five for the particular product being made, which is why two different reactions making the same chemical can both be in industrial use.