Percent Yield: Videos & Practice Problems

Stoichiometry helps determine the theoretical amount of product from given reactants, while percent yield measures the actual success of a chemical reaction. Percent yield is calculated using the formula: $\mathrm{Percent}\mathrm{Yield}=\frac{\mathrm{Actual}}{Yield}/\frac{\mathrm{Theoretical}}{Yield}100$. Yields are categorized as excellent (≥90%), very good (≥80%), good (≥70%), or poor (<40%). Actual yield is always less than theoretical yield due to inefficiencies in reactions.

Percent Yield

Stoichiometry is a crucial concept in chemistry that allows us to predict the amount of product that can be generated from a given quantity of reactants. However, the true measure of success in a chemical reaction is determined by the percent yield, which reflects the efficiency of the reaction. A higher percent yield indicates a more successful synthesis of the desired product, similar to aiming for high scores on an exam.

Percent yield can be categorized into different levels of efficiency: an excellent yield is defined as 90% or greater, a very good yield ranges from 80% to 89%, a good yield falls between 70% and 79%, and a poor yield is anything below 40%. Understanding these classifications helps in evaluating the effectiveness of a chemical process.

The formula for calculating percent yield is essential to remember, as it is often not provided on formula sheets. The formula is expressed as:

\[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\]

In this formula, the actual yield refers to the amount of pure product obtained from the experiment, while the theoretical yield is the amount predicted through stoichiometric calculations. It is important to note that the units of actual yield and theoretical yield must match; for instance, if the actual yield is measured in grams, the theoretical yield must also be converted to grams from moles if necessary.

No chemical reaction can achieve a 100% yield due to various factors such as loss of material during handling, incomplete reactions, or external influences. Consequently, the actual yield is always less than the theoretical yield. By analyzing percent yield alongside actual and theoretical yields, chemists can gain valuable insights into the efficiency and effectiveness of their chemical reactions.

Percent Yield Example 1

In a balanced chemical reaction involving benzene (C₆H₆) and oxygen, the stoichiometry plays a crucial role in determining the theoretical yield of a product, in this case, water (H₂O). The reaction can be summarized as:

2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

To calculate the percent yield of water produced from a 2.6 gram sample of benzene that yields 1.25 grams of water, we first need to establish the theoretical yield using stoichiometry. The percent yield formula is:

Percent Yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)

Here, the actual yield of water is given as 1.25 grams. To find the theoretical yield, we start by converting the mass of benzene to moles. The molar mass of benzene is calculated as follows:

Molar Mass of C₆H₆ = (6 × 12.011 g/mol) + (6 × 1.008 g/mol) = 78.108 g/mol

Next, we convert the grams of benzene to moles:

Moles of C₆H₆ = \(\frac{2.6 \text{ g}}{78.108 \text{ g/mol}} \approx 0.0333 \text{ moles}\)

Using the stoichiometric coefficients from the balanced equation, we find the moles of water produced. The ratio of moles of C₆H₆ to moles of H₂O is 2:6, or 1:3. Therefore:

Moles of H₂O = 0.0333 moles C₆H₆ × \(\frac{6 \text{ moles H}_2\text{O}}{2 \text{ moles C}_6\text{H}_6}\) = 0.0999 moles H₂O

Now, we convert moles of water to grams using the molar mass of water, which is:

Molar Mass of H₂O = (2 × 1.008 g/mol) + (16.00 g/mol) = 18.016 g/mol

Theoretical Yield of H₂O = 0.0999 moles × 18.016 g/mol ≈ 1.799 grams

Now that we have both the actual yield (1.25 grams) and the theoretical yield (1.799 grams), we can calculate the percent yield:

Percent Yield = \(\frac{1.25 \text{ g}}{1.799 \text{ g}} \times 100 \approx 69.48\%\)

This calculation illustrates the application of stoichiometry in determining the efficiency of a chemical reaction, highlighting the importance of understanding both actual and theoretical yields in evaluating reaction performance.

What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO₄?  
Zn (s) + CuSO₄ (aq) → Cu (s) + ZnSO₄ (aq)

• A. 63.1 %
• B. 90.7 %
• C. 79.2 %
• D. 67.1 %
• E. 50.0 %

Answer: (C)

Ammonia, NH₃, reacts with hypochlorite ion, OCl^–, to produce hydrazine, N₂H₄. How many grams of hydrazine are produced from 115.0 g NH₃ if the reaction has a 81.5% yield?  
2 NH₃ + OCl^– → N₂H­₄ + Cl^– + H₂O

In this problem, we are tasked with determining the grams of hydrazine (N₂H₄) produced from 115 grams of ammonia (NH₃) given an 81.5% yield. To solve this, we first need to understand the relationship between theoretical yield and actual yield, which is expressed through the percent yield formula:

Percent Yield = (Actual Yield / Theoretical Yield) × 100

Given that the percent yield is 81.5%, we can rearrange the formula to find the actual yield (x):

x = (Percent Yield × Theoretical Yield) / 100

To find the theoretical yield, we will use stoichiometry. We start by converting grams of ammonia to moles. The molar mass of ammonia is 17.034 g/mol, so:

Number of moles of NH₃ = 115 g / 17.034 g/mol ≈ 6.75 moles

Next, we need to determine the moles of hydrazine produced. The balanced chemical reaction indicates that 2 moles of ammonia produce 1 mole of hydrazine. Therefore, we can set up the mole ratio:

For every 2 moles of NH₃, 1 mole of N₂H₄ is produced. Thus:

Moles of N₂H₄ = (6.75 moles NH₃) × (1 mole N₂H₄ / 2 moles NH₃) = 3.375 moles N₂H₄

Now, we convert moles of hydrazine to grams. The molar mass of hydrazine is 32.052 g/mol:

Grams of N₂H₄ = 3.375 moles × 32.052 g/mol ≈ 108.0 g

This value represents the theoretical yield of hydrazine. Now, we can substitute this into our earlier equation to find the actual yield:

81.5 = (x / 108.0) × 100

Solving for x gives:

x = (81.5 / 100) × 108.0 ≈ 88.2 g

Thus, the actual yield of hydrazine produced from 115 grams of ammonia, considering the 81.5% yield, is approximately 88.2 grams. This process illustrates the importance of stoichiometry in determining yields in chemical reactions, as well as the application of the percent yield formula to relate theoretical and actual outcomes.

The reduction of iron (III) oxide creates the following reaction: 
Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g) 
If the above reaction only went to 75% completion, how many moles of Fe₂O₃ were require to produce 0.850 moles of Fe? 

• A. 0.963 mol
• B. 2.27 mol
• C. 0.425 mol
• D. 0.567 mol
• E. 1.70 mol

Answer: (D)

Here’s what students ask on this topic:

What is the formula for calculating percent yield?

The formula for calculating percent yield is:

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

In this formula, the actual yield is the amount of product obtained from the experiment, while the theoretical yield is the amount of product predicted by stoichiometric calculations. The result is then multiplied by 100 to convert it to a percentage.

Why is percent yield important in chemistry?

Percent yield is important in chemistry because it measures the efficiency of a chemical reaction. It indicates how successful a scientist was in producing the desired product. A higher percent yield means a more efficient reaction, which is crucial for both academic research and industrial applications. It helps in evaluating the practicality of a reaction and in optimizing conditions to maximize product yield. Additionally, it provides insight into potential losses and inefficiencies in the experimental process.

What are the categories of percent yield values?

Percent yield values are categorized as follows:

• Excellent: ≥90%
• Very Good: ≥80%
• Good: ≥70%
• Poor: <40%

These categories help in assessing the efficiency of a chemical reaction. An excellent yield indicates a highly efficient reaction, while a poor yield suggests significant inefficiencies or losses during the process.

Why is it impossible to achieve a 100% percent yield in a chemical reaction?

It is impossible to achieve a 100% percent yield in a chemical reaction because no reaction is perfectly efficient. Various factors contribute to this, such as incomplete reactions, side reactions, loss of product during transfer, and measurement inaccuracies. Additionally, external factors like temperature, pressure, and human error can also affect the yield. Therefore, the actual yield is always less than the theoretical yield, making a 100% yield unattainable.

How do you convert theoretical yield units to match actual yield units?

To convert theoretical yield units to match actual yield units, you need to ensure both yields are in the same unit of measurement. For example, if the actual yield is given in grams and the theoretical yield is in moles, you must convert the theoretical yield from moles to grams. This is done using the molar mass of the product:

$\text{Theoretical Yield (g)}=\text{Theoretical Yield (mol)}\times \text{Molar Mass (g/mol)}$

This ensures consistency in units, allowing for accurate calculation of percent yield.