Theoretical yield is the maximum possible

The theoretical yield of a chemical reaction is the maximum mass of product that could be produced, assuming the reaction goes to completion with no loss of product.

To calculate the theoretical yield, follow these steps:

1. Use a balanced chemical equation to find the mole ratio between reactants and products.
2. Calculate the moles of the limiting reagent present.
3. Determine the mass of product that these moles of limiting reagent could produce, using the mole ratio.

For example, in the reaction:

C₂H₅OH + [O] ➔ CH₃CHO + H₂O

The C₂H₅OH : CH₃CHO mole ratio is 1 : 1.

So if 0.300 mol (13.8 g) of C₂H₅OH is present and [O] is in excess, the theoretical yield of CH₃CHO is 0.300 mol (13.2 g).

Actual yield is always less than theoretical

The actual yield is the mass of product actually produced and collected from a reaction.

The actual yield is always lower than the theoretical yield because of several factors:

• Some starting material may not react completely.
• Products can be lost during workup procedures, such as filtering or transferring between containers.
• Side reactions may occur, reducing the yield of the desired product.

For the example reaction, if only 5.94 g of CH₃CHO was collected, this is the actual yield, which is less than the theoretical yield of 13.2 g.

Calculating percentage yield

Percentage yield is a measure of how efficient a reaction is, indicating how close the actual yield is to the theoretical maximum.

To calculate it, use the formula:

$\%\text{ yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$

For the example reaction:

$\%\text{ yield}=\frac{5.94}{13.2}\times 100=45.0\%$

The percentage yield offers valuable insights:

• A yield above 90% is seen as efficient and high-yielding.
• Lower yields suggest that the process needs optimisation.

Worked example 1 - Calculating actual yield

The balanced chemical equation for the synthesis of ammonia (NH_3_) through the Haber process is: N₂ + 3H₂ ➔ 2NH₃

Calculate the actual mass (in g) of ammonia produced given a theoretical yield of ammonia of 34.0 g and a percentage yield of 77.7%.

Step 1: Rearrange equation

$\text{Actual yield }=\frac{\text{percentage yield}}{100}\times \text{ theoretical yield}$

Step 2: Substitution and correct evaluation

$\text{Actual yield }=\frac{\text{77.7}}{100}\times 34.0=26.4\text{ g}$

Worked example 2 - Calculating theoretical yield and percentage yield

The balanced chemical equation for the synthesis of water from hydrogen and oxygen gas is: 2H₂ + O₂ ➔ 2H₂O

Calculate the percentage yield of this reaction if 2.00 g of hydrogen reacts with excess oxygen to produce 14.5 g of water.

Step 1: Calculate number of moles of H₂

$\text{n }=\frac{\text{m}}{{\text{M}}_{\text{r }}}=\frac{2.00}{2.0}=1.00\text{ mol}$

Step 2: Calculate number of moles of H₂O

H₂ : H₂O mole ratio = 2:2 = 1:1

Moles of H₂O = 1.00 mol

Step 3: Calculate theoretical yield

m = n x M_r = 1.00 x 18.0 = 18.0 g

Step 4: Equation

$\text{Percentage yield }=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$

Step 5: Substitution and correct evaluation

$\text{Percentage yield }=\frac{14.5}{18.0}\times 100=80.6\%$

Atom economy measures reaction efficiency

The atom economy of a reaction indicates how efficient the chemical reaction is at incorporating the reactants into the desired product. Specifically, atom economy measures the percentage of reactant atoms that end up in the final desired product, rather than in by-products.

A reaction with 100% atom economy converts all reactant atoms into the desired product. Reactions with lower atom economies generate more waste.

Calculating atom economy

Atom economy is calculated using the following formula:

$\%\text{ atom economy}=\frac{{\text{M}}_{\text{r}}\text{ of desired product}}{\mathrm{\Sigma }{\text{M}}_{\text{r}}\text{ of all products}}\times 100$

This formula essentially compares the mass of the desired product to the total mass of all reactants. The higher the percentage, the more efficient the reaction is at converting reactants to the desired product.

Some key points about atom economy:

• It is based on the balanced chemical equation, so make sure to use the correct mole ratios.
• Atom economy gives no indication of the percentage yield or losses during processing.

Worked example 3 - Calculating atom economy for the production of methanol

Calculate the atom economy for the production of methanol (CH₃OH) from chloromethane (CH₃Cl) and potassium hydroxide (KOH). The balanced equation is:

CH₃Cl + KOH ➔ CH₃OH + KCl

Step 1: Calculate M_r_ of desired product

M_r_ of CH₃OH = 12.0 + 4(1.0) + 16.0 = 32.0

Step 2: Calculate ΣM_r_ of all products

M_r_ of KCl = 39.1 + 35.5 = 74.6

ΣM_r = 32.0 + 74.6 = 106.6

Step 3: Equation

$\%\text{ atom economy}=\frac{{\text{M}}_{\text{r}}\text{ of desired product}}{\mathrm{\Sigma }{\text{M}}_{\text{r}}\text{ of all products}}\times 100$

Step 4: Substitution and correct evaluation

$\%\text{ atom economy}=\frac{32.0}{106.6}\times 100=30.0\%$

This calculation shows that only 30.0% of the reactant atoms are incorporated into the desired product, indicating a low atom economy for this substitution reaction.

Worked example 4 - Calculating atom economy for the fermentation of glucose

Calculate the atom economy for the production of ethanol (C₂H₅OH) from the fermentation of glucose (C₆H₁₂O₆). The balanced chemical equation is:

C₆H₁₂O₆ ➔ 2C₂H₅OH + 2CO₂

Step 1: Calculate M_r of desired product

M_r_ of C₂H₅OH = 2(12.0) + 6(1.0) + 16.0 = 46.0

Step 2: Multiply by reaction stoichiometry

Since the equation produces 2 moles of C₂H₅OH the total M_r_ for C₂H₅OH = 2 × 46.0 = 92.0

Step 2: Calculate ΣM_r of all products

M_r_ of CO₂ = 12.0 + 2(16.0) = 44.0

Since the equation produces 2 moles of CO₂ the total M_r_ for 2CO₂ = 2 × 44.0 = 88.0

ΣM_r = 92.0 + 88.0 = 180.0

Step 4: Equation

$\%\text{ atom economy}=\frac{{\text{M}}_{\text{r}}\text{ of desired product}}{\mathrm{\Sigma }{\text{M}}_{\text{r}}\text{ of all products}}\times 100$

Step 5: Substitution and correct evaluation

$\%\text{ atom economy}=\frac{92.0}{180.0}\times 100=51.1\%$

This calculation shows that 51.1% of the reactant atoms are incorporated into the desired product, indicating a moderate atom economy for this reaction.

Atom economy in different reaction types

Addition reactions

In an addition reaction, two reactants combine to form a single product. For example, ethene and hydrogen undergo an addition reaction:

C₂H₄ + H₂ ➔ C₂H₆

Since all reactant atoms end up in the desired product ethane, addition reactions always have 100% atom economy.

Substitution reactions

In a substitution reaction, some atoms are substituted between reactants, generating at least two products - the desired product and an unavoidable by-product. For example, chloromethane reacts with potassium hydroxide in a substitution reaction:

CH₃Cl + KOH ➔ CH₃OH + KCl

Here the desired product is methanol, but potassium chloride is also produced as waste. The atom economy will be less than 100%.