Atom economy

What you'll learn

What atom economy means and why chemists care about it.
How to calculate atom economy from a balanced symbol equation.
How atom economy is different from percentage yield.
How to compare reaction pathways, especially for Higher Tier questions.

The big idea: where do the atoms go?

In a chemical reaction, atoms are rearranged. They are not created or destroyed. But not every atom from the starting materials necessarily ends up in the product you actually want.

Some atoms may end up in by-products, which are extra substances made alongside the desired product. If those by-products are useless, they may become chemical waste.

Definition

Atom economy

Atom economy, also called atom utilisation, is a measure of the amount of starting materials that end up as useful products.

A high atom economy means a large proportion of the reactant atoms become the desired product.

Schematic comparing 100% atom economy with lower atom economy

Why atom economy matters

Atom economy matters for two linked reasons: sustainability and cost.

Definition

Sustainable development

Sustainable development means meeting present needs without preventing future generations from meeting their own needs. In chemistry, this often means using resources efficiently and reducing waste.

A reaction with high atom economy is usually better because:

less waste is produced
fewer raw materials are wasted
less energy and money may be needed to separate and dispose of by-products
fewer harmful chemicals may be released into the environment

Key Idea

High atom economy is efficient

Industrial chemists often prefer reactions with high atom economy because more of the starting material becomes useful product, so the process is usually cheaper and produces less waste.

The prerequisites

Before calculating atom economy, you need to be comfortable with two ideas: balanced equations and relative formula mass.

Balanced symbol equations

A reactant is a starting material in a reaction. A product is a substance made in a reaction.

A balanced symbol equation shows the correct chemical formulae and the correct numbers of each substance, so that the number of atoms of each element is the same on both sides.

For example:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

The big number in front of a formula is called a coefficient. In this equation, the coefficient 2 in front of O₂ means there are two lots of O₂ molecules.

Tip

Coefficients count whole formulae

If the equation says 2H₂O, you have two complete H₂O formula units. So you count two oxygen atoms and four hydrogen atoms in total.

Relative formula mass

Definition

Relative formula mass

Relative formula mass, written as $M_r$ , is the total of the relative atomic masses of all the atoms in a formula. It has no unit.

You use the relative atomic masses from the periodic table. For GCSE calculations, you are normally given the values you need or can read them from the periodic table.

For example, for ethanol, C₂H₅OH, you can think of the formula as C₂H₆O:

Mr(C2H5OH)=(2×12)+(6×1)+16=46M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12) + (6 \times 1) + 16 = 46

Example

Calculating relative formula mass

Calculate the relative formula mass of calcium carbonate, CaCO₃. Use Ca = 40, C = 12, O = 16.

Identify the atoms in the formula: CaCO₃ contains one calcium atom, one carbon atom and three oxygen atoms.
Multiply each relative atomic mass by the number of atoms of that element:
$\times 40) + (1 \times 12) + (3 \times 16)$
Add the values:
$Mr(CaCO3)=40+12+48=100M_r(\text{CaCO}_3) = 40 + 12 + 48 = 100$

The atom economy formula

For GCSE Chemistry, percentage atom economy is calculated from the balanced equation:

equation×100\text{percentage atom economy} = \frac{\text{relative formula mass of desired product from equation}} {\text{sum of relative formula masses of all reactants from equation}} \times 100

The desired product is the product you are trying to make. The denominator uses all the reactants, not all the products.

Common Mistake

Using the products in the denominator

Do not put the total mass of all products on the bottom of the fraction. For atom economy, the denominator is the total relative formula mass of all reactants from the balanced equation.

How to calculate atom economy

A reliable method is:

Write or use the balanced equation.
Identify the desired product.
Calculate the total $M_r$ of the desired product, remembering any coefficient in front.
Calculate the total $M_r$ of all reactants, remembering coefficients.
Substitute into the atom economy formula and multiply by 100.

Example: fermentation of glucose

Fermentation can produce ethanol from glucose:

C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)

The desired product is ethanol, C₂H₅OH. Carbon dioxide is a by-product.

Example

Calculating atom economy from a balanced equation

Calculate the atom economy for making ethanol by fermentation. Use C = 12, H = 1, O = 16.

Calculate the $M_r$ of the desired product, ethanol:
$Mr(C2H5OH)=(2×12)+(6×1)+16=46M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12) + (6 \times 1) + 16 = 46$
Use the coefficient in the balanced equation. The equation makes 2C₂H₅OH, so the total desired product mass from the equation is:
$\times 46 = 92$
Calculate the total $M_r$ of the reactant glucose:
$Mr(C6H12O6)=(6×12)+(12×1)+(6×16)=180M_r(\text{C}_6\text{H}_{12}\text{O}_6) = (6 \times 12) + (12 \times 1) + (6 \times 16) = 180$
Substitute into the atom economy formula:
$economy=92180×100=51.1%\text{percentage atom economy} = \frac{92}{180} \times 100 = 51.1\%$

So the atom economy is 51.1%.

Tip

A quick sense check

Atom economy cannot be more than 100%. If your answer is bigger than 100%, you have probably used the wrong denominator or forgotten a coefficient.

Reactions with 100% atom economy

Some reactions have 100% atom economy because all reactant atoms end up in the desired product.

For example, making ethanol by hydration of ethene:

C₂H₄(g) + H₂O(g) → C₂H₅OH(g)

There is only one product, and it is the desired product, so all reactant atoms are used to make ethanol.

Example

Recognising 100% atom economy

Show that hydration of ethene has 100% atom economy. Use C = 12, H = 1, O = 16.

Calculate the total $M_r$ of the reactants:
$Mr(C2H4)+Mr(H2O)=28+18=46M_r(\text{C}_2\text{H}_4) + M_r(\text{H}_2\text{O}) = 28 + 18 = 46$
Calculate the $M_r$ of the desired product, ethanol:
$Mr(C2H5OH)=46M_r(\text{C}_2\text{H}_5\text{OH}) = 46$
Substitute into the formula:
$economy=4646×100=100%\text{percentage atom economy} = \frac{46}{46} \times 100 = 100\%$

Atom economy is not the same as yield

This is a very common source of confusion.

Definition

Percentage yield

Percentage yield compares the actual mass of product made with the maximum theoretical mass that could be made.

Atom economy is about the balanced equation: where the atoms could go if the reaction happens as written.

Percentage yield is about the real experiment or industrial process: how much product is actually obtained after losses, incomplete reaction and purification.

Key Idea

Atom economy versus yield

A reaction can have 100% atom economy but still have a low percentage yield if the reaction does not go to completion or product is lost during separation.

Choosing a reaction pathway

A reaction pathway is a route used to make a particular product. In industry, chemists may have several possible routes and must decide which is best.

For Higher Tier, you may be asked to explain why a particular pathway is chosen using information such as:

atom economy
percentage yield
rate of reaction
equilibrium position
usefulness of by-products

If a reaction is fast, product can be made quickly. If an equilibrium reaction lies more towards the products, more product is present at equilibrium. If by-products are useful and can be sold or reused, that can make a pathway more attractive even if the atom economy is not perfect.

Example

Choosing a pathway to make ethanol

A company is choosing between two routes to make ethanol.

Route A, fermentation of glucose, has atom economy 51.1%, is slow, and produces carbon dioxide as a by-product.
Route B, hydration of ethene, has atom economy 100%, is fast with a catalyst, and produces no significant by-product.

Explain which route is likely to be preferred for efficient large-scale production, using the data.

Compare atom economy: Route B is better because 100% of the reactant atoms can become ethanol, while Route A sends some atoms into carbon dioxide.
Compare rate: Route B is faster with a catalyst, so ethanol can be produced more quickly in an industrial process.
Consider by-products: Route A makes carbon dioxide, which may need collecting or releasing, while Route B produces no significant by-product.
Make a justified decision: Route B is likely to be preferred for efficient large-scale production because it has higher atom economy, faster rate and less waste.

Common Mistake

The best route depends on the data

Do not automatically choose the highest atom economy if the question gives other important information. A route with lower atom economy might still be chosen if it uses cheaper renewable raw materials, has a much higher yield, or makes useful by-products.

Rearranging the formula

Sometimes you may be given the atom economy and asked to find a missing mass from the equation. This uses the same formula, just rearranged.

Starting formula:

mass×100\text{atom economy} = \frac{\text{desired product mass}} {\text{total reactant mass}} \times 100

To find the desired product mass:

mass100\text{desired product mass} = \frac{\text{atom economy} \times \text{total reactant mass}}{100}

Example

Finding the desired product mass from atom economy

A reaction has an atom economy of 75.0%. The sum of the relative formula masses of the reactants is 120. Find the relative formula mass of the desired product from the equation.

Choose the rearranged formula because the desired product mass is unknown:
$mass100\text{desired product mass} = \frac{\text{atom economy} \times \text{total reactant mass}}{100}$
Substitute the values:
$mass=75.0×120100\text{desired product mass} = \frac{75.0 \times 120}{100}$
Calculate:
$mass=90.0\text{desired product mass} = 90.0$

So the relative formula mass of the desired product from the equation is 90.0.

Exam technique

In the exam

Start from the balanced equation and include any coefficients in front of formulae.
Put the desired product from the equation on the top of the fraction, and the total of all reactants on the bottom.
For comparison questions, use the data given: atom economy, yield, rate, equilibrium position and whether by-products are useful.

Self review

Check yourself

Why does a reaction with only one product often have 100% atom economy?
In the atom economy formula, why do coefficients in the balanced equation matter?
How is atom economy different from percentage yield?

Atom economy

What you'll learn

What atom economy means and why chemists care about it.
How to calculate atom economy from a balanced symbol equation.
How atom economy is different from percentage yield.
How to compare reaction pathways, especially for Higher Tier questions.

The big idea: where do the atoms go?

In a chemical reaction, atoms are rearranged. They are not created or destroyed. But not every atom from the starting materials necessarily ends up in the product you actually want.

Some atoms may end up in by-products, which are extra substances made alongside the desired product. If those by-products are useless, they may become chemical waste.

Definition

Atom economy

Atom economy, also called atom utilisation, is a measure of the amount of starting materials that end up as useful products.

A high atom economy means a large proportion of the reactant atoms become the desired product.

Schematic comparing 100% atom economy with lower atom economy

Why atom economy matters

Atom economy matters for two linked reasons: sustainability and cost.

Definition

Sustainable development

A reaction with high atom economy is usually better because:

less waste is produced
fewer raw materials are wasted
less energy and money may be needed to separate and dispose of by-products
fewer harmful chemicals may be released into the environment

Key Idea

High atom economy is efficient

Industrial chemists often prefer reactions with high atom economy because more of the starting material becomes useful product, so the process is usually cheaper and produces less waste.

The prerequisites

Before calculating atom economy, you need to be comfortable with two ideas: balanced equations and relative formula mass.

Balanced symbol equations

A reactant is a starting material in a reaction. A product is a substance made in a reaction.

A balanced symbol equation shows the correct chemical formulae and the correct numbers of each substance, so that the number of atoms of each element is the same on both sides.

For example:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

The big number in front of a formula is called a coefficient. In this equation, the coefficient 2 in front of O₂ means there are two lots of O₂ molecules.

Tip

Coefficients count whole formulae

If the equation says 2H₂O, you have two complete H₂O formula units. So you count two oxygen atoms and four hydrogen atoms in total.

Relative formula mass

Definition

Relative formula mass

Relative formula mass, written as $M_r$ , is the total of the relative atomic masses of all the atoms in a formula. It has no unit.

You use the relative atomic masses from the periodic table. For GCSE calculations, you are normally given the values you need or can read them from the periodic table.

For example, for ethanol, C₂H₅OH, you can think of the formula as C₂H₆O:

Mr(C2H5OH)=(2×12)+(6×1)+16=46M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12) + (6 \times 1) + 16 = 46

Example

Calculating relative formula mass

Calculate the relative formula mass of calcium carbonate, CaCO₃. Use Ca = 40, C = 12, O = 16.

Identify the atoms in the formula: CaCO₃ contains one calcium atom, one carbon atom and three oxygen atoms.
Multiply each relative atomic mass by the number of atoms of that element:
$\times 40) + (1 \times 12) + (3 \times 16)$
Add the values:
$Mr(CaCO3)=40+12+48=100M_r(\text{CaCO}_3) = 40 + 12 + 48 = 100$

The atom economy formula

For GCSE Chemistry, percentage atom economy is calculated from the balanced equation:

equation×100\text{percentage atom economy} = \frac{\text{relative formula mass of desired product from equation}} {\text{sum of relative formula masses of all reactants from equation}} \times 100

The desired product is the product you are trying to make. The denominator uses all the reactants, not all the products.

Common Mistake

Using the products in the denominator

Do not put the total mass of all products on the bottom of the fraction. For atom economy, the denominator is the total relative formula mass of all reactants from the balanced equation.

How to calculate atom economy

A reliable method is:

Write or use the balanced equation.
Identify the desired product.
Calculate the total $M_r$ of the desired product, remembering any coefficient in front.
Calculate the total $M_r$ of all reactants, remembering coefficients.
Substitute into the atom economy formula and multiply by 100.

Example: fermentation of glucose

Fermentation can produce ethanol from glucose:

C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)

The desired product is ethanol, C₂H₅OH. Carbon dioxide is a by-product.

Example

Calculating atom economy from a balanced equation

Calculate the atom economy for making ethanol by fermentation. Use C = 12, H = 1, O = 16.

Calculate the $M_r$ of the desired product, ethanol:
$Mr(C2H5OH)=(2×12)+(6×1)+16=46M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12) + (6 \times 1) + 16 = 46$
Use the coefficient in the balanced equation. The equation makes 2C₂H₅OH, so the total desired product mass from the equation is:
$\times 46 = 92$
Calculate the total $M_r$ of the reactant glucose:
$Mr(C6H12O6)=(6×12)+(12×1)+(6×16)=180M_r(\text{C}_6\text{H}_{12}\text{O}_6) = (6 \times 12) + (12 \times 1) + (6 \times 16) = 180$
Substitute into the atom economy formula:
$economy=92180×100=51.1%\text{percentage atom economy} = \frac{92}{180} \times 100 = 51.1\%$

So the atom economy is 51.1%.

Tip

A quick sense check

Atom economy cannot be more than 100%. If your answer is bigger than 100%, you have probably used the wrong denominator or forgotten a coefficient.

Reactions with 100% atom economy

Some reactions have 100% atom economy because all reactant atoms end up in the desired product.

For example, making ethanol by hydration of ethene:

C₂H₄(g) + H₂O(g) → C₂H₅OH(g)

There is only one product, and it is the desired product, so all reactant atoms are used to make ethanol.

Example

Recognising 100% atom economy

Show that hydration of ethene has 100% atom economy. Use C = 12, H = 1, O = 16.

Calculate the total $M_r$ of the reactants:
$Mr(C2H4)+Mr(H2O)=28+18=46M_r(\text{C}_2\text{H}_4) + M_r(\text{H}_2\text{O}) = 28 + 18 = 46$
Calculate the $M_r$ of the desired product, ethanol:
$Mr(C2H5OH)=46M_r(\text{C}_2\text{H}_5\text{OH}) = 46$
Substitute into the formula:
$economy=4646×100=100%\text{percentage atom economy} = \frac{46}{46} \times 100 = 100\%$

Atom economy is not the same as yield

This is a very common source of confusion.

Definition

Percentage yield

Percentage yield compares the actual mass of product made with the maximum theoretical mass that could be made.

Atom economy is about the balanced equation: where the atoms could go if the reaction happens as written.

Percentage yield is about the real experiment or industrial process: how much product is actually obtained after losses, incomplete reaction and purification.

Key Idea

Atom economy versus yield

A reaction can have 100% atom economy but still have a low percentage yield if the reaction does not go to completion or product is lost during separation.

Choosing a reaction pathway

A reaction pathway is a route used to make a particular product. In industry, chemists may have several possible routes and must decide which is best.

For Higher Tier, you may be asked to explain why a particular pathway is chosen using information such as:

atom economy
percentage yield
rate of reaction
equilibrium position
usefulness of by-products

Example

Choosing a pathway to make ethanol

A company is choosing between two routes to make ethanol.

Explain which route is likely to be preferred for efficient large-scale production, using the data.

Compare atom economy: Route B is better because 100% of the reactant atoms can become ethanol, while Route A sends some atoms into carbon dioxide.
Compare rate: Route B is faster with a catalyst, so ethanol can be produced more quickly in an industrial process.
Consider by-products: Route A makes carbon dioxide, which may need collecting or releasing, while Route B produces no significant by-product.
Make a justified decision: Route B is likely to be preferred for efficient large-scale production because it has higher atom economy, faster rate and less waste.

Common Mistake

The best route depends on the data

Rearranging the formula

Sometimes you may be given the atom economy and asked to find a missing mass from the equation. This uses the same formula, just rearranged.

Starting formula:

mass×100\text{atom economy} = \frac{\text{desired product mass}} {\text{total reactant mass}} \times 100

To find the desired product mass:

mass100\text{desired product mass} = \frac{\text{atom economy} \times \text{total reactant mass}}{100}

Example

Finding the desired product mass from atom economy

A reaction has an atom economy of 75.0%. The sum of the relative formula masses of the reactants is 120. Find the relative formula mass of the desired product from the equation.

Choose the rearranged formula because the desired product mass is unknown:
$mass100\text{desired product mass} = \frac{\text{atom economy} \times \text{total reactant mass}}{100}$
Substitute the values:
$mass=75.0×120100\text{desired product mass} = \frac{75.0 \times 120}{100}$
Calculate:
$mass=90.0\text{desired product mass} = 90.0$

So the relative formula mass of the desired product from the equation is 90.0.

Exam technique

In the exam

Start from the balanced equation and include any coefficients in front of formulae.
Put the desired product from the equation on the top of the fraction, and the total of all reactants on the bottom.
For comparison questions, use the data given: atom economy, yield, rate, equilibrium position and whether by-products are useful.

Self review

Check yourself

Why does a reaction with only one product often have 100% atom economy?
In the atom economy formula, why do coefficients in the balanced equation matter?
How is atom economy different from percentage yield?

Atom economy

What you'll learn

The big idea: where do the atoms go?

Why atom economy matters

The prerequisites

Balanced symbol equations

Relative formula mass

The atom economy formula

How to calculate atom economy

Example: fermentation of glucose

Reactions with 100% atom economy

Atom economy is not the same as yield

Choosing a reaction pathway

Rearranging the formula

You've reached the end

Atom economy

What you'll learn

The big idea: where do the atoms go?

Why atom economy matters

The prerequisites

Balanced symbol equations

Relative formula mass

The atom economy formula

How to calculate atom economy

Example: fermentation of glucose

Reactions with 100% atom economy

Atom economy is not the same as yield

Choosing a reaction pathway

Rearranging the formula

You've reached the end

Atom economy

What you'll learn

The big idea: where do the atoms go?

Why atom economy matters

The prerequisites

Balanced symbol equations

Relative formula mass

The atom economy formula

How to calculate atom economy

Example: fermentation of glucose

Reactions with 100% atom economy

Atom economy is not the same as yield

Choosing a reaction pathway

Rearranging the formula

You've reached the end

Atom economy Revision Guide

Atom economy

What you'll learn

The big idea: where do the atoms go?

Why atom economy matters

The prerequisites

Balanced symbol equations

Relative formula mass

The atom economy formula

How to calculate atom economy

Example: fermentation of glucose

Reactions with 100% atom economy

Atom economy is not the same as yield

Choosing a reaction pathway

Rearranging the formula

You've reached the end

Atom economy Revision Guide