Percentage Yield in Quantitative Chemistry

Understanding Chemical Yield

In chemical reactions, atoms are neither created nor destroyed, but they are rearranged to form new substances. According to the conservation of mass, the mass of the products should theoretically equal the mass of the reactants used, assuming all reactants are converted into the desired product. However, in practical laboratory or industrial settings, the mass of product actually obtained is almost always less than the mass predicted by calculation.

The amount of product collected is known as the yield. To quantify the efficiency of a process, we calculate the percentage yield by comparing the actual yield to the theoretical maximum, which is the predicted yield.

Why is the Yield Rarely 100 Percent?

Even when no atoms are lost in a fundamental sense, several practical factors prevent a perfect 100 percent yield:

1. Incomplete reactions: If a reaction is reversible, it will reach a state of dynamic equilibrium where both the forward and reverse reactions occur at the same rate. In such cases, the reactants never fully convert into products.
2. Product loss during separation: When a product is separated from the reaction mixture through processes such as filtration, evaporation, or distillation, small amounts are inevitably left behind on filter paper, in glassware, or lost during transfer.
3. Side reactions: Reactants may participate in unexpected chemical pathways, forming minor unwanted byproducts instead of the primary product intended.

Calculating Percentage Yield

To determine the percentage yield, you must first calculate the predicted yield using the balanced chemical equation and the relative formula masses of the substances involved.

Worked Example: Determining Percentage Yield

Consider the oxidation of ethanol to form ethanoic acid:

$\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

If $23.0\text{ g}$ of ethanol is reacted and $25.0\text{ g}$ of ethanoic acid is formed, what is the percentage yield? ($A_r\text{ values: H } = 1.0;\text{ C } = 12.0;\text{ O } = 16.0$)

1. Calculate the relative molar masses ($M_r$):

   • $M_r(\text{CH}_3\text{CH}_2\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0$
   • $M_r(\text{CH}_3\text{COOH}) = (2 \times 12.0) + (4 \times 1.0) + (2 \times 16.0) = 60.0$

2. Calculate the predicted yield:

   • The equation shows $1\text{ mol}$ of ethanol forms $1\text{ mol}$ of ethanoic acid.
   • $46.0\text{ g}$ of ethanol is predicted to form $60.0\text{ g}$ of ethanoic acid.
   • Therefore, $23.0\text{ g}$ of ethanol is predicted to form $\frac{23.0}{46.0} \times 60.0 = 30.0\text{ g}$ of ethanoic acid.

3. Apply the percentage yield formula:

   • $\text{Percentage yield} = \frac{\text{actual yield (25.0 g)}}{\text{predicted yield (30.0 g)}} \times 100 = 83.3\%$

Using Percentage Yield in Other Calculations

The percentage yield can also be used to find the actual mass produced or the mass of reactant required for a specific outcome.

Worked Example: Finding Actual Yield from Percentage Yield

Ammonia is produced via the reaction: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})$. Under specific conditions, the yield is $15\%$. What mass of ammonia is formed from $2.8\text{ g}$ of nitrogen? ($A_r\text{ values: H } = 1.0;\text{ N } = 14.0$)

1. Calculate molar masses: $M_r(\text{N}_2) = 28.0$ and $M_r(\text{NH}_3) = 17.0$.
2. Calculate predicted yield: $1\text{ mol}$ of $\text{N}_2$ ($28.0\text{ g}$) forms $2\text{ mol}$ of $\text{NH}_3$ ($34.0\text{ g}$). So, $2.8\text{ g}$ of $\text{N}_2$ should form $\frac{2.8}{28.0} \times 34.0 = 3.4\text{ g}$ of $\text{NH}_3$.
3. Calculate actual yield: $\frac{15}{100} \times 3.4 = 0.51\text{ g}$ of ammonia.

Worked Example: Finding Reactant Mass Required

Ethyl ethanoate is formed by: $\text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$. If the yield is $50\%$, what mass of ethanol is needed to produce $4.4\text{ g}$ of ethyl ethanoate?

1. Find the required predicted yield: Since the actual yield is $50\%$ of the predicted, the predicted yield must be $4.4 \times \frac{100}{50} = 8.8\text{ g}$.
2. Determine $M_r$ values: $M_r(\text{ester}) = 88.0$ and $M_r(\text{ethanol}) = 46.0$.
3. Calculate reactant mass: $1\text{ mol}$ of ester ($88.0\text{ g}$) requires $1\text{ mol}$ of ethanol ($46.0\text{ g}$). Therefore, $8.8\text{ g}$ of ester requires $46.0 \times \frac{8.8}{88.0} = 4.6\text{ g}$ of ethanol.