Percentage Yield and Atom Economy

This section explains percentage yield and atom economy as part quantitative chemistry. 

Percentage Yield

In any chemical reaction, the percentage yield refers to the amount of product actually obtained from a reaction, compared to the maximum possible amount that could have been obtained under ideal conditions. Often, the actual yield is less than the theoretical yield due to factors such as incomplete reactions, side reactions, or losses during the process (e.g., evaporation or filtering).

The formula for calculating percentage yield is:

$$\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100$$ 

Where:

• Actual yield is the amount of product you get from the reaction, usually measured in grams.
• Theoretical yield is the maximum amount of product that could be formed from the given quantities of reactants, calculated based on the limiting reagent.

Example:

Suppose a reaction produces 30 grams of a product, but the theoretical yield (calculated from the stoichiometry of the reaction) was 40 grams. To calculate the percentage yield:

$$\text{Percentage Yield} = \left( \frac{30 \, \text{g}}{40 \, \text{g}} \right) \times 100 = 75\%$$ 

So, the percentage yield of the reaction is 75%.

Atom Economy

Atom economy is a measure of the efficiency of a chemical reaction in terms of how well the atoms in the reactants are utilised to form the desired product. A reaction with high atom economy produces a greater proportion of useful product compared to waste products. This is important not only for the economics of the reaction but also for reducing environmental impact by minimising waste.

The formula for atom economy is:

$$\text{Atom Economy} = \left( \frac{\text{Relative Formula Mass of Desired Product}}{\text{Sum of Relative Formula Masses of All Reactants}} \right) \times 100$$ 

Where:

• Relative formula mass of desired product refers to the molecular mass of the desired product (or main product).
• Sum of relative formula masses of all reactants is the total mass of all the reactants used in the reaction.

Example:

Consider the reaction for the production of water (H₂O) from hydrogen and oxygen:

$$\text{2H₂ (g)} + \text{O₂ (g)} \rightarrow \text{2H₂O (l)}$$

To calculate the atom economy:

1. Relative formula mass of desired product (H₂O):

• H = 1, O = 16.
• Mr of H₂O = (2 × 1) + (1 × 16) = 18 g/mol.

2. Sum of relative formula masses of all reactants:

• 2 moles of H₂ = (2 × 2) = 4 g/mol.
• 1 mole of O₂ = 2 × 16 = 32 g/mol.
• Total = 4 + 32 = 36 g/mol.

3. Atom economy:

$$\text{Atom Economy} = \left( \frac{18 \, \text{g/mol}}{36 \, \text{g/mol}} \right) \times 100 = 50\%$$ 

Thus, the atom economy of this reaction is 50%, meaning half of the reactant atoms are used in the desired product, and the rest are part of waste or by-products.

Examples

Example 1: Percentage Yield

In a reaction to produce nitrogen dioxide (NO₂), the theoretical yield is 20 grams, but only 16 grams of NO₂ are actually produced.

To calculate the percentage yield:

$$\text{Percentage Yield} = \left( \frac{16 \, \text{g}}{20 \, \text{g}} \right) \times 100 = 80\%$$ 

So, the percentage yield is 80%, indicating that the reaction was relatively efficient, but some product was lost or not fully formed.

Example 2: Atom Economy of a Reaction

Consider the following reaction between ethene (C₂H₄) and oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O):

$$\text{C₂H₄ (g)} + 3\text{O₂ (g)} \rightarrow 2\text{CO₂ (g)} + 2\text{H₂O (l)}$$ 

To calculate the atom economy:

1. Relative formula mass of desired products:

• CO₂ = 12 + (2 × 16) = 44 g/mol.
• H₂O = 2 + 16 = 18 g/mol.
• Total mass of desired products = (2 × 44) + (2 × 18) = 88 + 36 = 124 g/mol.

2. Sum of relative formula masses of all reactants:

• C₂H₄ = (2 × 12) + (4 × 1) = 28 g/mol.
• 3 moles of O₂ = (3 × 32) = 96 g/mol.
• Total = 28 + 96 = 124 g/mol.

3. Atom economy:

$$\text{Atom Economy} = \left( \frac{124 \, \text{g/mol}}{124 \, \text{g/mol}} \right) \times 100 = 100\%$$ 

This reaction has 100% atom economy, meaning that all the atoms in the reactants are used to form the desired products, with no waste or by-products.

Percentage yield and atom economy are key concepts in quantitative chemistry. The percentage yield helps evaluate how effectively a reaction produces the expected amount of product, while atom economy gives a measure of the efficiency of a reaction in terms of how well the reactants are converted into useful products, reducing waste. These concepts are essential in assessing the practical efficiency and environmental impact of chemical processes.