Theoretical Yield and Percent Yield - Chemistry

1. Fundamental Concepts

Theoretical Yield:

Definition: The maximum amount of product that can be produced from a given amount of reactant in ideal conditions.

Nature: It is a calculated value based on stoichiometry.

Actual Yield:

Definition: The measured amount of product obtained from a laboratory experiment.

Nature: It is a measured value (found by weighing). It is almost always less than the theoretical yield due to human error, side reactions, or loss during filtration/transfer.

Percent Yield:

Definition: A measure of the efficiency of a reaction, comparing the actual yield to the theoretical yield.

Formula: $$ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% $$

2. Key Concepts

Limiting Reactant Determines Theoretical Yield:

The theoretical yield is always calculated based on the limiting reactant (the substance that runs out first), not the excess reactant.

Unit Consistency:

Actual yield and theoretical yield must have the same units (e.g., both in grams or both in moles) when calculating the percentage.

Interpretation of Results:

< 100%: Normal. Indicates product loss or incomplete reaction.

= 100%: Ideal (rarely achieved).

> 100%: Impossible chemically. Usually indicates impurities (e.g., wet product, unreacted material) or calculation errors.

3. Examples

Easy

Question: The theoretical yield of a reaction is calculated to be 50.0 g. The actual yield obtained in the lab is 40.0 g. What is the percent yield?

Solution:

$$ \text{Percent Yield} = \left( \frac{40.0}{50.0} \right) \times 100\% = 80.0\% $$

Medium

Question: Given the balanced equation:

$$ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) $$

If 14.0 g of $N_2$ reacts with excess $H_2$ and produces 15.0 g of $NH_3$, what is the percent yield?

Solution:

1. Moles of $N_2$: $14.0\text{ g} / 28.0\text{ g/mol} = 0.50\text{ mol}$.

2. Mole Ratio: $1\text{ mol } N_2 \rightarrow 2\text{ mol } NH_3$.

Theoretical moles of $NH_3 = 0.50 \times 2 = 1.0\text{ mol}$.

3. Theoretical Yield: $1.0\text{ mol} \times 17.0\text{ g/mol} = 17.0\text{ g}$.

4. Percent Yield: $(15.0 / 17.0) \times 100\% \approx 88.2\%$.

Hard

Question: Given the balanced equation:

$$ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) $$

If 31.0 g of $P_4$ is mixed with 32.0 g of $O_2$, and the reaction produces 40.0 g of $P_4O_{10}$, what is the percent yield?

Solution:

1. Identify Limiting Reactant:

Moles $P_4 = 31.0\text{ g} / 124\text{ g/mol} = 0.25\text{ mol}$.

Moles $O_2 = 32.0\text{ g} / 32\text{ g/mol} = 1.0\text{ mol}$.

Ratio required: $1 P_4 : 5 O_2$.

$0.25\text{ mol } P_4$ needs $1.25\text{ mol } O_2$. We only have $1.0\text{ mol } O_2$.

$O_2$ is the Limiting Reactant.

2. Calculate Theoretical Yield (based on $O_2$):

Ratio $O_2 : P_4O_{10} = 5 : 1$.

Moles of $P_4O_{10} = 1.0 / 5 = 0.20\text{ mol}$.

Mass of $P_4O_{10} = 0.20\text{ mol} \times 284\text{ g/mol} = 56.8\text{ g}$.

3. Calculate Percent Yield: $(40.0 / 56.8) \times 100\% \approx 70.4\%$.

4. Problem-Solving Techniques

The "Stoichiometry Map":

Follow the conversion path: Grams (Reactant) $\rightarrow$ Moles (Reactant) $\rightarrow$ Moles (Product) $\rightarrow$ Grams (Theoretical Yield).

The Limiting Reactant Check:

If two reactant amounts are given, always determine the limiting reactant first. The theoretical yield is determined strictly by the limiting reactant.

Watch for Purity:

If the reactant is impure (e.g., "80% pure"), multiply the given mass by the purity percentage ($\times 0.80$) before starting your stoichiometry calculation.

Common Sense Verification:

If your final answer is greater than 100%, re-check your math immediately.

Ensure you did not mix up actual and theoretical yield in the numerator/denominator.