4.3 Theoretical Yield, Actual Yield, and Percent Yield | Stoichiometry | Chemistry 11

This section explores the relationship between the calculated amount of product from a chemical reaction and the amount actually obtained in a laboratory setting.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced from the given amounts of reactants, as calculated from the balanced chemical equation. This calculation assumes that the reaction goes to completion and that the limiting reactant is completely consumed without any losses or side reactions.

The theoretical yield is determined using stoichiometric calculations based on the balanced chemical equation and the amount of the limiting reactant.

Actual Yield

The actual yield is the quantity of product that is actually produced and measured experimentally in a chemical reaction. In practice, the actual yield is almost always less than the theoretical yield.

Reasons for the difference:

Side Reactions: Reactants may undergo alternative reactions, forming unintended products.
Incomplete Reaction: The reaction may not proceed to completion; it might reach equilibrium or be stopped prematurely.
Product Loss: Some product may be lost during separation and purification processes (e.g., remaining dissolved in a solution, adhering to glassware).

Percent Yield

The percent yield is a measure of the efficiency of a chemical reaction. It is calculated as the ratio of the actual yield to the theoretical yield, multiplied by 100 to be expressed as a percentage.

The formula is:

$Percent Yield = \frac{Actual Yield}{Theoretical Yield} \times 100%$

Worked Examples

Example 4.12: Calculating Percent Yield of Ozone

Problem: Calculate the percent yield of Ozone ( $O_{3}$ ) that could be produced by 10 g of Oxygen ( $O_{2}$ ). The actual yield of $O_{3}$ obtained experimentally was 1.05 g.

Solution:

1. Write the balanced chemical equation: $3 O_{2} \to 2 O_{3}$

2. List the given values and molar masses:

Mass of $O_{2}$ = 10 g
Actual yield of $O_{3}$ = 1.05 g
Molar mass of $O_{2} = 2 \times 16.00 = 32.00 g/mol$
Molar mass of $O_{3} = 3 \times 16.00 = 48.00 g/mol$

3. Calculate the moles of the reactant ( $O_{2}$ ): $Moles of O_{2} = \frac{10 g}{32 g/mol} = 0.3125 mol$

4. Calculate the theoretical moles of $O_{3}$ using stoichiometry: $Moles of O_{3} = 0.3125 mol \times \frac{2 mol O _{3}}{3 mol O _{2}} \approx 0.2083 mol$

5. Calculate the theoretical yield (mass) of $O_{3}$ : $Theoretical Yield = 0.2083 mol \times 48 g/mol = 10 g$

6. Calculate the percent yield: $Percent Yield = \frac{1.05 g}{10 g} \times 100% = 10.5%$

Concept Assessment Exercises

Exercise 4.6 (a): Ferric Chloride from Limiting Reactant

Problem: Determine the amount of Ferric Chloride ( $FeC l_{3}$ ) produced by the reaction of $KMn O_{4}$ , 10 moles of $FeC l_{2}$ , and 22 moles of HCl.

Reaction: $KMn O_{4} + 5 FeC l_{2} + 8 HCl \to MnC l_{2} + KCl + 5 FeC l_{3} + 4 H_{2} O$

Solution:

Step 1 — Find moles of product each reactant can produce:

From $FeC l_{2}$ (10 mol): $10 mol FeC l_{2} \times \frac{5 mol FeC l _{3}}{5 mol FeC l _{2}} = 10 mol FeC l_{3}$

From HCl (22 mol): $22 mol HCl \times \frac{5 mol FeC l _{3}}{8 mol HCl} = 13.75 mol FeC l_{3}$

Step 2 — Identify the limiting reactant: $FeC l_{2}$ produces less product (10 mol vs 13.75 mol), so $FeC l_{2}$ is the limiting reactant.

Step 3 — Theoretical yield of $FeC l_{3}$ : $Theoretical Yield = 10 mol FeC l_{3} \times 162.2 g/mol = 1622 g$

Exercise 4.6 (b): Percent Yield of Baking Soda

Problem: Baking soda ( $NaHC O_{3}$ ) is commercially prepared by passing ammonia ( $N H_{3}$ ) and carbon dioxide ( $C O_{2}$ ) through a saturated NaCl solution. If reacting 20 g $N H_{3}$ and 30 g $C O_{2}$ produced 40 g of baking soda, what is the percentage yield? (Assume NaCl and $H_{2} O$ are in excess.)

Reaction: $NaCl + H_{2} O + C O_{2} + N H_{3} \to NaHC O_{3} + N H_{4} Cl$

Molar masses: $N H_{3} = 17 g/mol$ , $C O_{2} = 44 g/mol$ , $NaHC O_{3} = 84 g/mol$

Step 1 — Convert masses to moles: $Moles of N H_{3} = \frac{20 g}{17 g/mol} = 1.176 mol$ $Moles of C O_{2} = \frac{30 g}{44 g/mol} = 0.682 mol$

Step 2 — Identify the limiting reactant (1:1 ratio): $C O_{2}$ (0.682 mol) $<$ $N H_{3}$ (1.176 mol), so $C O_{2}$ is the limiting reactant.

Step 3 — Calculate theoretical yield of $NaHC O_{3}$ : $Theoretical Yield = 0.682 mol \times 84 g/mol = 57.3 g$

Step 4 — Calculate percent yield: $Percent Yield = \frac{40 g}{57.3 g} \times 100% \approx 69.8%$

This section explores the relationship between the calculated amount of product from a chemical reaction and the amount actually obtained in a laboratory setting.

Theoretical Yield

The theoretical yield is determined using stoichiometric calculations based on the balanced chemical equation and the amount of the limiting reactant.

Actual Yield

Reasons for the difference:

Side Reactions: Reactants may undergo alternative reactions, forming unintended products.
Incomplete Reaction: The reaction may not proceed to completion; it might reach equilibrium or be stopped prematurely.
Product Loss: Some product may be lost during separation and purification processes (e.g., remaining dissolved in a solution, adhering to glassware).

Percent Yield

The formula is:

$Percent Yield = \frac{Actual Yield}{Theoretical Yield} \times 100%$

Worked Examples

Example 4.12: Calculating Percent Yield of Ozone

Problem: Calculate the percent yield of Ozone ( $O_{3}$ ) that could be produced by 10 g of Oxygen ( $O_{2}$ ). The actual yield of $O_{3}$ obtained experimentally was 1.05 g.

Solution:

1. Write the balanced chemical equation: $3 O_{2} \to 2 O_{3}$

2. List the given values and molar masses:

Mass of $O_{2}$ = 10 g
Actual yield of $O_{3}$ = 1.05 g
Molar mass of $O_{2} = 2 \times 16.00 = 32.00 g/mol$
Molar mass of $O_{3} = 3 \times 16.00 = 48.00 g/mol$

3. Calculate the moles of the reactant ( $O_{2}$ ): $Moles of O_{2} = \frac{10 g}{32 g/mol} = 0.3125 mol$

4. Calculate the theoretical moles of $O_{3}$ using stoichiometry: $Moles of O_{3} = 0.3125 mol \times \frac{2 mol O _{3}}{3 mol O _{2}} \approx 0.2083 mol$

5. Calculate the theoretical yield (mass) of $O_{3}$ : $Theoretical Yield = 0.2083 mol \times 48 g/mol = 10 g$

6. Calculate the percent yield: $Percent Yield = \frac{1.05 g}{10 g} \times 100% = 10.5%$

Concept Assessment Exercises

Exercise 4.6 (a): Ferric Chloride from Limiting Reactant

Problem: Determine the amount of Ferric Chloride ( $FeC l_{3}$ ) produced by the reaction of $KMn O_{4}$ , 10 moles of $FeC l_{2}$ , and 22 moles of HCl.

Reaction: $KMn O_{4} + 5 FeC l_{2} + 8 HCl \to MnC l_{2} + KCl + 5 FeC l_{3} + 4 H_{2} O$

Solution:

Step 1 — Find moles of product each reactant can produce:

From $FeC l_{2}$ (10 mol): $10 mol FeC l_{2} \times \frac{5 mol FeC l _{3}}{5 mol FeC l _{2}} = 10 mol FeC l_{3}$

From HCl (22 mol): $22 mol HCl \times \frac{5 mol FeC l _{3}}{8 mol HCl} = 13.75 mol FeC l_{3}$

Step 2 — Identify the limiting reactant: $FeC l_{2}$ produces less product (10 mol vs 13.75 mol), so $FeC l_{2}$ is the limiting reactant.

Step 3 — Theoretical yield of $FeC l_{3}$ : $Theoretical Yield = 10 mol FeC l_{3} \times 162.2 g/mol = 1622 g$

Exercise 4.6 (b): Percent Yield of Baking Soda

Reaction: $NaCl + H_{2} O + C O_{2} + N H_{3} \to NaHC O_{3} + N H_{4} Cl$

Molar masses: $N H_{3} = 17 g/mol$ , $C O_{2} = 44 g/mol$ , $NaHC O_{3} = 84 g/mol$

Step 1 — Convert masses to moles: $Moles of N H_{3} = \frac{20 g}{17 g/mol} = 1.176 mol$ $Moles of C O_{2} = \frac{30 g}{44 g/mol} = 0.682 mol$

Step 2 — Identify the limiting reactant (1:1 ratio): $C O_{2}$ (0.682 mol) $<$ $N H_{3}$ (1.176 mol), so $C O_{2}$ is the limiting reactant.

Step 3 — Calculate theoretical yield of $NaHC O_{3}$ : $Theoretical Yield = 0.682 mol \times 84 g/mol = 57.3 g$

Step 4 — Calculate percent yield: $Percent Yield = \frac{40 g}{57.3 g} \times 100% \approx 69.8%$