Understanding Percent Yield Calculations

In chemistry, the percent yield is a measure of the efficiency of a reaction. It compares the actual yield (the amount of product obtained) to the theoretical yield (the amount expected based on stoichiometric calculations). Here, we present three diverse examples of percent yield calculations that illustrate how to apply this concept in real-world scenarios.

Example 1: Synthesizing Water from Hydrogen and Oxygen

In a laboratory experiment, a chemist aims to synthesize water (H₂O) by combining hydrogen gas (H₂) and oxygen gas (O₂). The balanced chemical equation for this reaction is:

2 H₂ + O₂ → 2 H₂O

The chemist starts with 4 grams of hydrogen and 32 grams of oxygen. The theoretical yield of water can be calculated as follows:

1. Calculate the moles of hydrogen and oxygen:

   • Molar mass of H₂ = 2 g/mol (1 g/mol × 2)
   • Molar mass of O₂ = 32 g/mol
   • Moles of H₂ = 4 g / 2 g/mol = 2 moles
   • Moles of O₂ = 32 g / 32 g/mol = 1 mole

2. Determine the limiting reactant:

   • From the balanced equation, 2 moles of H₂ react with 1 mole of O₂. Here, H₂ is in excess, and O₂ is the limiting reactant.

3. Calculate the theoretical yield of water:

   • According to the equation, 1 mole of O₂ produces 2 moles of H₂O.
   • Therefore, 1 mole of O₂ yields 36 grams of H₂O (2 moles × 18 g/mol, where 18 g/mol is the molar mass of water).
   • The theoretical yield is 36 grams.

Now, if the chemist actually produces only 30 grams of water:

4. Calculate the percent yield:

   • Percent Yield = (Actual Yield / Theoretical Yield) × 100
   • Percent Yield = (30 g / 36 g) × 100 = 83.33%

Notes: This example illustrates how to determine the limiting reactant and calculate both theoretical and actual yields in a reaction involving gases.

Example 2: Synthesis of Ammonium Chloride

A student is conducting an experiment to synthesize ammonium chloride (NH₄Cl) using ammonia (NH₃) and hydrochloric acid (HCl). The balanced equation is:

NH₃ + HCl → NH₄Cl

The student starts with 10 grams of ammonia and 10 grams of hydrochloric acid. To find the percent yield:

1. Calculate the moles of each reactant:

   • Molar mass of NH₃ = 17 g/mol
   • Molar mass of HCl = 36.5 g/mol
   • Moles of NH₃ = 10 g / 17 g/mol = 0.588 moles
   • Moles of HCl = 10 g / 36.5 g/mol = 0.274 moles

2. Identify the limiting reactant:

   • The balanced equation shows a 1:1 mole ratio, meaning HCl is the limiting reactant.

3. Calculate the theoretical yield of NH₄Cl:

   • 0.274 moles of HCl produces 0.274 moles of NH₄Cl.
   • Theoretical yield = 0.274 moles × 53.5 g/mol = 14.66 grams.

After completing the reaction, the student collects 12 grams of NH₄Cl:

4. Calculate the percent yield:

   • Percent Yield = (12 g / 14.66 g) × 100 = 81.83%

Notes: This example demonstrates the importance of understanding mole ratios and the limiting reactant in a straightforward synthesis reaction.

Example 3: Decomposition of Calcium Carbonate

In another experiment, a researcher is investigating the decomposition of calcium carbonate (CaCO₃) into calcium oxide (CaO) and carbon dioxide (CO₂). The reaction is represented as follows:

CaCO₃ → CaO + CO₂

The researcher begins with 50 grams of calcium carbonate. The theoretical yield of calcium oxide can be determined:

1. Calculate the moles of calcium carbonate:

   • Molar mass of CaCO₃ = 100 g/mol
   • Moles of CaCO₃ = 50 g / 100 g/mol = 0.5 moles.

2. Use the stoichiometry of the reaction:

   • The balanced equation indicates that 1 mole of CaCO₃ produces 1 mole of CaO. Thus, 0.5 moles of CaCO₃ will produce 0.5 moles of CaO.
   • Theoretical yield = 0.5 moles × 56 g/mol = 28 grams.

After the experiment, the researcher obtains 22 grams of CaO:

3. Calculate the percent yield:

   • Percent Yield = (22 g / 28 g) × 100 = 78.57%

Notes: This example highlights the decomposition reaction and how to accurately derive the theoretical yield and percent yield for solid products.

By examining these examples, you can see how percent yield calculations apply across various chemical reactions, enhancing your understanding of reaction efficiency.