Scaling reacting masses gives the theoretical yield

Definition

Theoretical yield

The maximum mass of product that could form from a given mass of reactant, calculated from the balanced equation.

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

A balanced equation shows the reacting masses of each substance when you replace each formula with its $M_{r}$ .
1. Mass is conserved, so these reacting masses always add up to the same total on both sides.
The reacting masses stay in the same ratio whatever mass of reactant is used.
To find the theoretical yield, multiply the mass of reactant by the mass ratio of product to reactant.
This gives the greatest mass of product possible, because it assumes all the reactant turns into product.

Example

Calculate the theoretical yield of calcium oxide when $g10.0\,\text{g}$ of calcium carbonate is heated: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
Work out the relative formula masses: $Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ and $Mr(CaO)=40.1+16.0=56.1M_r(\text{CaO}) = 40.1 + 16.0 = 56.1$ .
Write the reacting masses: $g100.1\,\text{g}$ of $CaCO3\text{CaCO}_3$ gives $g56.1\,\text{g}$ of $CaO\text{CaO}$ .
Scale to the mass used: $10.0 \times \dfrac{56.1}{100.1} = 5.60\,\text{g}$ of calcium oxide.

Multiply relative formula masses by the balancing numbers

A balancing number in front of a formula multiplies the whole formula, so it also multiplies that substance's reacting mass.
In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ $2 Mg + O_{2} \to 2 MgO$ , the reacting masses are $2×24.3=48.62 \times 24.3 = 48.6$ $2 \times 24.3 = 48.6$ for magnesium and $2×40.3=80.62 \times 40.3 = 80.6$ $2 \times 40.3 = 80.6$ for magnesium oxide.
1. So $g6.00\,\text{g}$ of magnesium gives a theoretical yield of $6.00 \times \dfrac{80.6}{48.6} = 9.95\,\text{g}$ of magnesium oxide.
Only the reactant you are given and the product you want are needed, so other substances in the equation can be ignored.
Round only at the end, and give the answer with a unit of mass, because relative formula masses themselves have no units.

Example

Calculate the theoretical yield of iron from $g80.0\,\text{g}$ of iron(III) oxide: $Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ ( $A_r$ : Fe = 55.8, O = 16.0).
Work out the reacting masses: $Fe2O3=(2×55.8)+(3×16.0)=159.6\text{Fe}_2\text{O}_3 = (2 \times 55.8) + (3 \times 16.0) = 159.6$ and $2Fe=2×55.8=111.62\text{Fe} = 2 \times 55.8 = 111.6$ .
Scale to the mass used: $80.0 \times \dfrac{111.6}{159.6} = 55.9\,\text{g}$ of iron.

Percentage yield compares actual yield with theoretical yield

Definition

Actual yield

The mass of product that is collected when a reaction is carried out.

Definition

Percentage yield

The actual yield of a product expressed as a percentage of its theoretical yield.

Percentage yield is calculated with this equation: $yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$
The actual yield is almost always less than the theoretical yield.
1. The reaction may be reversible or may not go to completion, so some reactant never turns into product.
2. Side reactions can turn some of the reactant into other, unwanted products.
3. Some product is lost when it is filtered, transferred between containers or purified.
A percentage yield above $100%100\%$ suggests the product was still wet or impure when it was weighed.

Example

The theoretical yield of calcium oxide is $g5.60\,\text{g}$ , but only $g4.48\,\text{g}$ is collected; calculate the percentage yield.
Substitute into the equation: $4.485.60×100\dfrac{4.48}{5.60} \times 100$ .
Percentage yield $80.0\%$ .

Atom economy shows how much reactant mass becomes product

Definition

Atom economy

The percentage of the total mass of reactants that ends up in the desired product, calculated from the balanced equation.

Definition

Desired product

The useful product that a reaction is carried out to make.

Atom economy is calculated from the balanced equation: $products×100\text{atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$
Multiply each $M_{r}$ by its balancing number before adding, as for reacting masses.
1. The total mass of the products equals the total mass of the reactants, so the bottom line can also be the sum of the reactants' $M_r$ values.
A reaction with only one product has an atom economy of $100%100\%$ , because every atom ends up in the desired product.
A low atom economy means much of the reactant mass becomes waste products, which wastes raw materials unless the by-products can be used or sold.

Example

Calculate the atom economy for making calcium oxide: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $M_r$ : CaO = 56.1, $CO2\text{CO}_2$ = 44.0).
Add the $M_r$ values of all the products: $56.1 + 44.0 = 100.1$ .
Atom economy $\dfrac{56.1}{100.1} \times 100 = 56.0\%$ .

Percentage yield and atom economy measure different things

Percentage yield depends on how well the reaction is carried out, so it can change from one attempt to the next.
Atom economy depends only on the balanced equation, so it is fixed for a given reaction.
A reaction can have a high atom economy but a low percentage yield.
1. Making ethanol from ethene and steam has an atom economy of $100%100\%$ , yet only a small fraction of the gases react each time they pass through the reactor.
Chemists want both values to be high, because together they mean less waste and more product from the same raw materials.

Exam technique

When a question asks for percentage yield, look for a measured mass of product, but when it asks for atom economy, use only the $M_r$ values from the balanced equation.
Show the reacting masses or $M_r$ values you used, then give the final answer with its unit or percentage sign.

Self review

What is the theoretical yield of a reaction?
What mass of calcium oxide could form from $g20.0\,\text{g}$ of calcium carbonate ( $M_r$ : $CaCO3\text{CaCO}_3$ = 100.1, CaO = 56.1)?
Give two reasons why the actual yield is usually less than the theoretical yield.
Why does a reaction with only one product have an atom economy of $100%100\%$ ?
Which of percentage yield and atom economy can be calculated without doing the experiment?

Scaling reacting masses gives the theoretical yield

Definition

Theoretical yield

The maximum mass of product that could form from a given mass of reactant, calculated from the balanced equation.

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

A balanced equation shows the reacting masses of each substance when you replace each formula with its $M_{r}$ .
1. Mass is conserved, so these reacting masses always add up to the same total on both sides.
The reacting masses stay in the same ratio whatever mass of reactant is used.
To find the theoretical yield, multiply the mass of reactant by the mass ratio of product to reactant.
This gives the greatest mass of product possible, because it assumes all the reactant turns into product.

Example

Calculate the theoretical yield of calcium oxide when $g10.0\,\text{g}$ of calcium carbonate is heated: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
Work out the relative formula masses: $Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ and $Mr(CaO)=40.1+16.0=56.1M_r(\text{CaO}) = 40.1 + 16.0 = 56.1$ .
Write the reacting masses: $g100.1\,\text{g}$ of $CaCO3\text{CaCO}_3$ gives $g56.1\,\text{g}$ of $CaO\text{CaO}$ .
Scale to the mass used: $10.0 \times \dfrac{56.1}{100.1} = 5.60\,\text{g}$ of calcium oxide.

Multiply relative formula masses by the balancing numbers

A balancing number in front of a formula multiplies the whole formula, so it also multiplies that substance's reacting mass.
In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ $2 Mg + O_{2} \to 2 MgO$ , the reacting masses are $2×24.3=48.62 \times 24.3 = 48.6$ $2 \times 24.3 = 48.6$ for magnesium and $2×40.3=80.62 \times 40.3 = 80.6$ $2 \times 40.3 = 80.6$ for magnesium oxide.
1. So $g6.00\,\text{g}$ of magnesium gives a theoretical yield of $6.00 \times \dfrac{80.6}{48.6} = 9.95\,\text{g}$ of magnesium oxide.
Only the reactant you are given and the product you want are needed, so other substances in the equation can be ignored.
Round only at the end, and give the answer with a unit of mass, because relative formula masses themselves have no units.

Example

Calculate the theoretical yield of iron from $g80.0\,\text{g}$ of iron(III) oxide: $Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ ( $A_r$ : Fe = 55.8, O = 16.0).
Work out the reacting masses: $Fe2O3=(2×55.8)+(3×16.0)=159.6\text{Fe}_2\text{O}_3 = (2 \times 55.8) + (3 \times 16.0) = 159.6$ and $2Fe=2×55.8=111.62\text{Fe} = 2 \times 55.8 = 111.6$ .
Scale to the mass used: $80.0 \times \dfrac{111.6}{159.6} = 55.9\,\text{g}$ of iron.

Percentage yield compares actual yield with theoretical yield

Definition

Actual yield

The mass of product that is collected when a reaction is carried out.

Definition

Percentage yield

The actual yield of a product expressed as a percentage of its theoretical yield.

Percentage yield is calculated with this equation: $yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$
The actual yield is almost always less than the theoretical yield.
1. The reaction may be reversible or may not go to completion, so some reactant never turns into product.
2. Side reactions can turn some of the reactant into other, unwanted products.
3. Some product is lost when it is filtered, transferred between containers or purified.
A percentage yield above $100%100\%$ suggests the product was still wet or impure when it was weighed.

Example

The theoretical yield of calcium oxide is $g5.60\,\text{g}$ , but only $g4.48\,\text{g}$ is collected; calculate the percentage yield.
Substitute into the equation: $4.485.60×100\dfrac{4.48}{5.60} \times 100$ .
Percentage yield $80.0\%$ .

Atom economy shows how much reactant mass becomes product

Definition

Atom economy

The percentage of the total mass of reactants that ends up in the desired product, calculated from the balanced equation.

Definition

Desired product

The useful product that a reaction is carried out to make.

Atom economy is calculated from the balanced equation: $products×100\text{atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$
Multiply each $M_{r}$ by its balancing number before adding, as for reacting masses.
1. The total mass of the products equals the total mass of the reactants, so the bottom line can also be the sum of the reactants' $M_r$ values.
A reaction with only one product has an atom economy of $100%100\%$ , because every atom ends up in the desired product.
A low atom economy means much of the reactant mass becomes waste products, which wastes raw materials unless the by-products can be used or sold.

Example

Calculate the atom economy for making calcium oxide: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $M_r$ : CaO = 56.1, $CO2\text{CO}_2$ = 44.0).
Add the $M_r$ values of all the products: $56.1 + 44.0 = 100.1$ .
Atom economy $\dfrac{56.1}{100.1} \times 100 = 56.0\%$ .

Percentage yield and atom economy measure different things

Percentage yield depends on how well the reaction is carried out, so it can change from one attempt to the next.
Atom economy depends only on the balanced equation, so it is fixed for a given reaction.
A reaction can have a high atom economy but a low percentage yield.
1. Making ethanol from ethene and steam has an atom economy of $100%100\%$ , yet only a small fraction of the gases react each time they pass through the reactor.
Chemists want both values to be high, because together they mean less waste and more product from the same raw materials.

Exam technique

When a question asks for percentage yield, look for a measured mass of product, but when it asks for atom economy, use only the $M_r$ values from the balanced equation.
Show the reacting masses or $M_r$ values you used, then give the final answer with its unit or percentage sign.

Self review

What is the theoretical yield of a reaction?
What mass of calcium oxide could form from $g20.0\,\text{g}$ of calcium carbonate ( $M_r$ : $CaCO3\text{CaCO}_3$ = 100.1, CaO = 56.1)?
Give two reasons why the actual yield is usually less than the theoretical yield.
Why does a reaction with only one product have an atom economy of $100%100\%$ ?
Which of percentage yield and atom economy can be calculated without doing the experiment?

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Multiply relative formula masses by the balancing numbers

Percentage yield compares actual yield with theoretical yield

Atom economy shows how much reactant mass becomes product

Percentage yield and atom economy measure different things

Review 5.1.2 Theoretical mass, percentage yield and atom economy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Multiply relative formula masses by the balancing numbers

Percentage yield compares actual yield with theoretical yield

Atom economy shows how much reactant mass becomes product

Percentage yield and atom economy measure different things

Review 5.1.2 Theoretical mass, percentage yield and atom economy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Multiply relative formula masses by the balancing numbers

Percentage yield compares actual yield with theoretical yield

Atom economy shows how much reactant mass becomes product

Percentage yield and atom economy measure different things

Review 5.1.2 Theoretical mass, percentage yield and atom economy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.2 Theoretical mass, percentage yield and atom economy Revision Guide

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Multiply relative formula masses by the balancing numbers

Percentage yield compares actual yield with theoretical yield

Atom economy shows how much reactant mass becomes product

Percentage yield and atom economy measure different things

Review 5.1.2 Theoretical mass, percentage yield and atom economy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.2 Theoretical mass, percentage yield and atom economy Revision Guide