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Reaction Yields Simplified Revision Notes

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1.2.4 Reaction Yields

In chemical reactions, the amount of product obtained is often less than the amount theoretically predicted. Reaction yields quantify how efficient a reaction is in terms of producing the desired product.

Theoretical Yield

infoNote

The theoretical yield is the amount of product that should be produced if the reaction goes to completion without any losses. It is based on the stoichiometric ratios from the balanced chemical equation.

Steps to Calculate Theoretical Yield

  1. Balance the chemical equation.
  2. Calculate moles of reactants using the given mass or volume.
  3. Use the stoichiometric ratio from the balanced equation to find the moles of the product.
  4. Convert moles of the product to mass (if needed) using the molar mass.

Actual Yield

infoNote

The actual yield is the true amount of product obtained from the reaction in practice. It is usually less than the theoretical yield due to various reasons, such as:

  • Mechanical loss (e.g., transferring materials between containers).
  • Loss during separation techniques (e.g., filtration).
  • Side reactions that form unwanted products.
  • The reaction may not go to completion.

Percentage Yield

The percentage yield compares the actual yield to the theoretical yield, indicating the reaction's efficiency. It is calculated using the following formula:

Percentage Yield=Actual YieldTheoretical Yield×100\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100

A high percentage yield indicates a more efficient reaction.


Worked Example

infoNote

Example: Calculating the Yield for the Conversion of Magnesium to Magnesium Oxide Question: Consider the reaction where magnesium (Mg) is burned to form magnesium oxide (MgO):

2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO

Step 1: Calculate the Theoretical Yield

Assume 5.00 g of magnesium is used. To calculate the theoretical yield of magnesium oxide:

  1. Calculate the moles of magnesium:
Moles of Mg=Mass of MgAr of Mg\text{Moles of Mg} = \frac{\text{Mass of Mg}}{A_r \text{ of Mg}} =5.00 g24.31 g/mol=0.206 mol= \frac{5.00 \, \text{g}}{24.31 \, \text{g/mol}} = 0.206 \, \text{mol}
  1. From the balanced equation, the ratio of magnesium to magnesium oxide is 1:1.
  2. Therefore, the moles of magnesium oxide will also be 0.206 mol.
  3. Calculate the mass of magnesium oxide:
Mass of MgO=Moles of MgO×Mr of MgO\text{Mass of MgO} = \text{Moles of MgO} \times M_r \text{ of MgO} =0.206×40.31=8.30 g= 0.206 \times 40.31 = 8.30 \, \text{g}

So, the theoretical yield of magnesium oxide is 8.30 g.


Step 2: Calculate the Percentage Yield

Assume the actual yield of magnesium oxide is 7.85 g. The percentage yield is calculated as:

Percentage Yield=7.85 g8.30 g×100=94.6%\text{Percentage Yield} = \frac{7.85 \, \text{g}}{8.30 \, \text{g}} \times 100 = 94.6\%

This means that 94.6% of the theoretical yield was obtained in the reaction.

Common Reasons for Yield Losses

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