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).

Experimental yield is always less than theoretical

The experimental yield is the actual mass of product isolated from a paticular experiment.

The experimental 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 experimental 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 experimental yield is to the theoretical maximum.

To calculate it, use the formula:

$ \%\text{ yield}=\frac{\text{experimental yield}}{\text{theoretical yield}}\times100 $

For the example reaction:

$ \%\text{ yield}=\frac{5.94}{13.2}\times100=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 experimental yield

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

Calculate the experimental yield (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

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

Step 2: Substitution and correct evaluation

$\text{Experimental yield }=\frac{\text{77.7}}{100}\times34.0=26.4\text{ g}$

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

$\text{Percentage yield }=\frac{\text{experimental yield}}{\text{theoretical yield}}\times100$

Step 5: Substitution and correct evaluation

$\text{Percentage yield }=\frac{14.5}{18.0}\times100=80.6\%$