How Do You Calculate The Percent Yield

The percent yield is a critical concept in chemistry that allows us to evaluate the efficiency of a chemical reaction. It essentially tells us how much of the desired product we actually obtained compared to the maximum amount we could have theoretically obtained. Understanding how to calculate percent yield is crucial for chemists, researchers, and students alike, as it helps in optimizing reactions, minimizing waste, and accurately assessing experimental results.

Understanding the Basics: Theoretical Yield vs. Actual Yield

Before diving into the calculation, it's essential to understand two key terms:

• Theoretical Yield: This is the maximum amount of product that can be produced from a given amount of reactants, assuming the reaction goes to completion and no product is lost. It's a calculated value based on stoichiometry.
• Actual Yield: This is the amount of product that is actually obtained from a chemical reaction in the laboratory. It is an experimental value, always less than or equal to the theoretical yield. The actual yield can be affected by various factors, such as incomplete reactions, side reactions, and loss of product during purification.

The percent yield is then calculated by comparing the actual yield to the theoretical yield, expressed as a percentage.

The Percent Yield Formula

The formula for calculating percent yield is straightforward:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Where:

• Actual Yield is the experimentally obtained mass (or moles) of the product.
• Theoretical Yield is the calculated maximum possible mass (or moles) of the product.

Step-by-Step Guide to Calculating Percent Yield

Let's break down the process of calculating percent yield into a series of manageable steps with examples:

Step 1: Write the Balanced Chemical Equation

The first and most crucial step is to write a balanced chemical equation for the reaction. A balanced equation ensures that the number of atoms of each element is the same on both sides of the equation, adhering to the law of conservation of mass.

Example:

Consider the reaction between hydrogen gas (H₂) and nitrogen gas (N₂) to produce ammonia (NH₃). The unbalanced equation is:

H₂ + N₂ → NH₃

Balancing this equation gives us:

3H₂ + N₂ → 2NH₃

Step 2: Determine the Molar Masses of Reactants and Products

Next, you need to determine the molar masses of the reactants and the desired product(s). You can find these values on the periodic table.

Example (Continuing from above):

• H₂: 2.02 g/mol (2 x 1.01 g/mol)
• N₂: 28.02 g/mol (2 x 14.01 g/mol)
• NH₃: 17.03 g/mol (14.01 g/mol + 3 x 1.01 g/mol)

Step 3: Identify the Limiting Reactant

The limiting reactant is the reactant that is completely consumed in a chemical reaction. It determines the maximum amount of product that can be formed. To identify the limiting reactant, you need to:

1. Convert the mass of each reactant to moles using its molar mass.
2. Divide the number of moles of each reactant by its stoichiometric coefficient in the balanced equation.
3. The reactant with the smallest value is the limiting reactant.

Example:

Suppose we react 6.06 g of H₂ with 28.02 g of N₂.

1. Moles of H₂ = 6.06 g / 2.02 g/mol = 3.00 mol
2. Moles of N₂ = 28.02 g / 28.02 g/mol = 1.00 mol

Now, divide by the stoichiometric coefficients:

• H₂: 3.00 mol / 3 = 1.00
• N₂: 1.00 mol / 1 = 1.00

In this case, both reactants have the same value. This means neither is in excess, and they will both be completely consumed. We can choose either as the basis for our calculation. Let's choose N₂.

Step 4: Calculate the Theoretical Yield

Using the stoichiometry of the balanced equation and the moles of the limiting reactant, calculate the theoretical yield of the product in moles. Then, convert the moles of product to grams using its molar mass.

Example:

From the balanced equation (3H₂ + N₂ → 2NH₃), we know that 1 mole of N₂ produces 2 moles of NH₃. Since we have 1.00 mole of N₂ (our limiting reactant), the theoretical yield of NH₃ is:

Moles of NH₃ = 1.00 mol N₂ x (2 mol NH₃ / 1 mol N₂) = 2.00 mol NH₃

Now, convert moles of NH₃ to grams:

Theoretical Yield of NH₃ = 2.00 mol x 17.03 g/mol = 34.06 g

Step 5: Determine the Actual Yield

The actual yield is the amount of product you actually obtain from the experiment. This value will be given to you or you will have to measure it experimentally.

Example:

Let's say you performed the reaction and obtained 30.0 g of NH₃.

Step 6: Calculate the Percent Yield

Finally, use the percent yield formula:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Example:

Percent Yield of NH₃ = (30.0 g / 34.06 g) x 100% = 88.08%

Therefore, the percent yield for this reaction is 88.08%.

Example Problems with Detailed Solutions

Let's work through a few more examples to solidify your understanding:

Problem 1:

When 10.0 g of ethanol (C₂H₅OH) are reacted with excess acetic acid (CH₃COOH), 14.8 g of ethyl acetate (CH₃COOC₂H₅) are obtained. Calculate the percent yield.

C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O

Solution:

1. Balanced Equation: The equation is already balanced.
2. Molar Masses:
   • C₂H₅OH: 46.07 g/mol
   • CH₃COOC₂H₅: 88.11 g/mol
3. Limiting Reactant: Since acetic acid is in excess, ethanol is the limiting reactant.
4. Theoretical Yield:
   • Moles of C₂H₅OH = 10.0 g / 46.07 g/mol = 0.217 mol
   • From the balanced equation, 1 mole of C₂H₅OH produces 1 mole of CH₃COOC₂H₅.
   • Moles of CH₃COOC₂H₅ (theoretical) = 0.217 mol
   • Theoretical Yield of CH₃COOC₂H₅ = 0.217 mol x 88.11 g/mol = 19.12 g
5. Actual Yield: 14.8 g
6. Percent Yield:
   • Percent Yield = (14.8 g / 19.12 g) x 100% = 77.41%

Problem 2:

In a reaction, 5.0 g of copper (II) oxide (CuO) was reacted with excess hydrogen gas (H₂) to produce copper (Cu) and water (H₂O). 3.9 g of copper was collected. What is the percentage yield of copper?

CuO + H₂ → Cu + H₂O

Solution:

1. Balanced Equation: The equation is already balanced.
2. Molar Masses:
   • CuO: 79.55 g/mol
   • Cu: 63.55 g/mol
3. Limiting Reactant: Hydrogen gas is in excess, so copper (II) oxide is the limiting reactant.
4. Theoretical Yield:
   • Moles of CuO = 5.0 g / 79.55 g/mol = 0.0629 mol
   • From the balanced equation, 1 mole of CuO produces 1 mole of Cu.
   • Moles of Cu (theoretical) = 0.0629 mol
   • Theoretical Yield of Cu = 0.0629 mol x 63.55 g/mol = 4.00 g
5. Actual Yield: 3.9 g
6. Percent Yield:
   • Percent Yield = (3.9 g / 4.00 g) x 100% = 97.5%

Problem 3:

If 25.0 g of salicylic acid (C₇H₆O₃) reacts with excess acetic anhydride to produce 27.0 g of aspirin (C₉H₈O₄), what is the percent yield?

C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH

Solution:

1. Balanced Equation: The equation is already balanced.
2. Molar Masses:
   • C₇H₆O₃: 138.12 g/mol
   • C₉H₈O₄: 180.16 g/mol
3. Limiting Reactant: Acetic anhydride is in excess, so salicylic acid is the limiting reactant.
4. Theoretical Yield:
   • Moles of C₇H₆O₃ = 25.0 g / 138.12 g/mol = 0.181 mol
   • From the balanced equation, 1 mole of C₇H₆O₃ produces 1 mole of C₉H₈O₄.
   • Moles of C₉H₈O₄ (theoretical) = 0.181 mol
   • Theoretical Yield of C₉H₈O₄ = 0.181 mol x 180.16 g/mol = 32.61 g
5. Actual Yield: 27.0 g
6. Percent Yield:
   • Percent Yield = (27.0 g / 32.61 g) x 100% = 82.80%

Factors Affecting Percent Yield

Several factors can influence the percent yield of a reaction, leading to values less than 100%. Understanding these factors can help in optimizing reactions and improving yields:

• Incomplete Reactions: Not all reactions proceed to completion. Equilibrium reactions, in particular, reach a state where the forward and reverse reactions occur at the same rate, resulting in a mixture of reactants and products.
• Side Reactions: Reactants may participate in unintended side reactions, forming byproducts and reducing the amount of desired product.
• Loss of Product During Purification: During the isolation and purification of the product (e.g., filtration, recrystallization, distillation), some of the product may be lost.
• Experimental Error: Inaccurate measurements of reactants or products can lead to errors in the calculation of percent yield.
• Impurities in Reactants: Impure reactants may reduce the effective amount of the limiting reactant, leading to a lower yield.
• Handling Losses: Transferring liquids or solids between containers always results in some loss of material.

Significance of Percent Yield

The percent yield is a crucial parameter in chemistry for several reasons:

• Evaluating Reaction Efficiency: It provides a quantitative measure of how efficiently a reaction converts reactants into the desired product.
• Optimizing Reaction Conditions: By analyzing the percent yield under different reaction conditions (e.g., temperature, pressure, catalyst), chemists can optimize the reaction to maximize product formation.
• Assessing Experimental Technique: A low percent yield may indicate problems with the experimental technique, such as incomplete reactions, product loss, or inaccurate measurements.
• Cost Analysis: In industrial chemical processes, percent yield is a critical factor in determining the cost-effectiveness of a reaction. Higher yields translate to less waste and lower production costs.
• Research and Development: In research settings, percent yield helps evaluate the success of new synthetic methods and identify areas for improvement.

Strategies to Improve Percent Yield

While achieving a 100% yield is often impossible, there are strategies to improve the percent yield of a reaction:

• Ensure the Reaction Goes to Completion: Optimize reaction conditions (e.g., temperature, reaction time, catalyst concentration) to drive the reaction towards completion.
• Minimize Side Reactions: Choose reactants and reaction conditions that minimize the formation of unwanted byproducts.
• Optimize Product Isolation and Purification: Use efficient techniques for isolating and purifying the product to minimize losses.
• Use High-Purity Reactants: Start with high-purity reactants to ensure that the limiting reactant is present in the correct amount.
• Careful Handling: Employ careful handling techniques to avoid loss of material during transfers and manipulations.
• Use a Catalyst: Catalysts can speed up a reaction without being consumed, potentially leading to higher yields in a reasonable time frame.
• Remove Products as They Form: In some reactions, removing the product as it forms can shift the equilibrium towards product formation (Le Chatelier's principle).

Common Mistakes to Avoid

Calculating percent yield correctly is essential for accurate results. Here are some common mistakes to avoid:

• Using an Unbalanced Equation: Always ensure that the chemical equation is balanced before performing any calculations.
• Incorrectly Identifying the Limiting Reactant: Failing to correctly identify the limiting reactant will lead to an incorrect theoretical yield.
• Using Incorrect Molar Masses: Double-check the molar masses of reactants and products to avoid errors in calculations.
• Confusing Actual and Theoretical Yield: Make sure you are using the correct values for actual yield (experimental) and theoretical yield (calculated).
• Not Converting Units: Ensure that all quantities are expressed in consistent units (e.g., grams for mass, moles for amount of substance).
• Rounding Errors: Avoid rounding intermediate values too early in the calculation, as this can lead to significant errors in the final result.
• Ignoring Significant Figures: Pay attention to significant figures throughout the calculation and report the percent yield with the appropriate number of significant figures.

Percent Yield vs. Other Yield Measurements

While percent yield is the most common way to express the efficiency of a reaction, other measures exist. It's important to understand the differences:

• Quantitative Yield: This term is sometimes used when the percent yield is very close to 100%, implying that the reaction is essentially complete and there is minimal loss of product.
• Isolated Yield: This refers to the amount of pure product obtained after all purification steps. It's often lower than the crude yield (the amount of product obtained directly from the reaction mixture) due to losses during purification.
• Atom Economy: This is a theoretical measure of the proportion of reactant atoms that become incorporated into the desired product. It's a concept used in green chemistry to design reactions that minimize waste. Atom economy doesn't consider the actual yield obtained in the lab.

Conclusion

Calculating percent yield is a fundamental skill in chemistry that allows us to quantify the efficiency of chemical reactions. By understanding the concepts of theoretical yield, actual yield, and limiting reactants, and by following the step-by-step guide outlined in this article, you can accurately calculate percent yield and interpret its significance. Remember to pay attention to factors that can affect percent yield and implement strategies to improve it. Mastering this skill is essential for success in the laboratory and for advancing your understanding of chemical processes.