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CU-103 / L13 · lecture

Theoretical, Actual, and Percentage Yield

~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable identifying the limiting reagent in a balanced chemical equation and calculating the mass of product theoretically possible from that reagent. Familiarity with molar mass calculations and unit conversions is assumed.

Why this matters

In the laboratory, reactions rarely proceed to perfect completion, making yield calculations essential for assessing efficiency and cost.

While stoichiometry provides the ideal mathematical relationship between reactants and products, real-world chemical processes are subject to side reactions, incomplete mixing, and physical losses during isolation. Understanding the distinction between the theoretical maximum and the actual amount obtained allows chemists to evaluate the practical success of a synthesis, optimise reaction conditions, and determine the economic viability of industrial processes. This lecture bridges the gap between idealized textbook calculations and the messy reality of experimental chemistry.

The Ideal vs. The Real

Imagine you are baking a cake. The recipe tells you that if you combine exactly two cups of flour, one cup of sugar, and three eggs, you will produce one standard-sized cake. This is your theoretical yield: the maximum amount of product possible if everything goes perfectly. In chemistry, this corresponds to the stoichiometric calculation based on the limiting reagent. However, in the real kitchen, you might spill some flour, the oven might not heat evenly, or you might scrape some batter off the bowl. Consequently, the cake you actually pull out of the oven is smaller than the recipe promised. This is the actual yield. The percentage yield is simply a measure of how close your real cake was to the perfect recipe cake. In chemistry, we define the theoretical yield as the amount of product calculated from the limiting reagent, assuming 100% efficiency. The actual yield is the measured mass of the pure product obtained in the laboratory. The percentage yield is the ratio of actual to theoretical yield, expressed as a percentage. It is crucial to understand that a percentage yield greater than 100% is physically impossible for a pure product, as it would imply you created matter from nothing. If you calculate a value above 100%, it indicates an error in measurement, impurities in the product, or incomplete drying. Conversely, a low percentage yield is common and often desirable to investigate, as it points to inefficiencies that can be improved to save resources and money.

Defining Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, assuming the reaction goes to completion and no side reactions occur. It is calculated using the stoichiometry of the balanced equation and the amount of the limiting reagent. Recall from the previous lecture that the limiting reagent is the reactant that is completely consumed first, thereby capping the amount of product that can be formed. To find the theoretical yield, you first determine the moles of the limiting reagent. Then, you use the mole ratio from the balanced equation to find the moles of product. Finally, you convert these moles to mass using the molar mass of the product. It is important to note that the theoretical yield is a calculated value, not a measured one. It represents an ideal scenario. For example, if you react 2 moles of hydrogen with 1 mole of oxygen to form water, the theoretical yield is 2 moles of water. If you start with 10 grams of hydrogen and excess oxygen, you calculate how many moles of hydrogen you have, determine that hydrogen is limiting (if oxygen is in excess), and then calculate the mass of water that 10 grams of hydrogen could theoretically produce. This value is your benchmark for success.

\[ n_{\text{limiting}} = \frac{m_{\text{reactant}}}{M_{\text{reactant}}} \]
\[ n_{\text{product}} = n_{\text{limiting}} \times \frac{\text{coeff. of product}}{\text{coeff. of limiting reagent}} \]
\[ m_{\text{theoretical}} = n_{\text{product}} \times M_{\text{product}} \]

Calculating Actual and Percentage Yield

The actual yield is the amount of product actually obtained from a reaction in the laboratory. It is always a measured value, determined by weighing the isolated product. The actual yield is almost always less than the theoretical yield due to various factors such as incomplete reactions, side reactions, loss of product during transfer or purification, and impurities. The percentage yield is a measure of the efficiency of the reaction. It is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. This value indicates how much of the theoretical maximum was actually achieved. A high percentage yield (close to 100%) suggests an efficient reaction with minimal loss. A low percentage yield indicates significant losses or inefficiencies. It is important to ensure that the actual yield is measured for the pure product only. If the product is wet or contains unreacted starting materials, the actual yield will be artificially high, leading to an incorrect percentage yield. Therefore, proper purification and drying are critical steps before weighing the product. The percentage yield is a dimensionless quantity, as the units of mass cancel out in the ratio.

\[ \%\,\text{yield} = \frac{m_{\text{actual}}}{m_{\text{theoretical}}} \times 100 \]

For instance, if the theoretical yield of a reaction is 50.0 \(\mathrm{g}\) and the actual yield obtained in the lab is 42.5 \(\mathrm{g}\), the percentage yield is \(\frac{42.5}{50.0} \times 100 = 85.0\%\). This means that 85% of the possible product was successfully isolated.

Factors Affecting Yield

Several factors can cause the actual yield to be lower than the theoretical yield. First, reactions may not go to completion. Some reactants may remain unreacted due to kinetic barriers or equilibrium limitations. Second, side reactions may occur, consuming reactants to form unwanted by-products. This reduces the amount of desired product formed. Third, physical losses can occur during the experimental procedure. For example, product may be lost during filtration, distillation, or transfer between containers. Fourth, the product may be impure, containing solvent or other substances that affect the mass measurement. If the product is not fully dried, the mass will be higher than the true mass of the product, leading to an overestimation of the actual yield. Conversely, if the product is not fully purified, the mass may include impurities, also leading to an overestimation. To maximise yield, chemists optimise reaction conditions such as temperature, pressure, and catalysts. They also use techniques to minimise physical losses, such as using efficient filtration methods and careful transfer techniques. Understanding these factors helps in troubleshooting low yields and improving reaction efficiency.

It is also worth noting that in industrial chemistry, yield is a critical economic factor. Even a small increase in percentage yield can lead to significant cost savings and reduced waste. Therefore, optimizing yield is a major goal in process development.

The result

◆ RESULT · T-015 · full treatment on its own page

The limiting reagent

For a balanced reaction \(aA+bB\to\text{products}\), the reactant whose available moles, divided by its own stoichiometric coefficient, is smallest (\(\min\{n_A/a,\ n_B/b\}\)) is the limiting reagent: it is fully consumed first, and the amount of product actually formed is capped by however much that reactant alone could produce, regardless of how much of the other reactant remains unreacted.

\[ \text{limiting} = \arg\min\left(\frac{n_A}{a},\ \frac{n_B}{b}\right), \qquad m_{\text{theoretical}}=r_{\text{limiting}}\cdot(\text{product coeff.})\cdot M_{\text{product}}, \qquad \%\,\text{yield}=\frac{m_{\text{actual}}}{m_{\text{theoretical}}}\times100 \]
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Tier 3 ยท watch out

A common misconception is that a percentage yield greater than 100% indicates a highly efficient reaction. In reality, a yield above 100% is a red flag indicating that the product is impure, likely containing solvent or unreacted starting materials, or that there was an error in the theoretical calculation. Since matter cannot be created, the actual mass of pure product cannot exceed the theoretical maximum. Always check for purity and ensure the theoretical yield is calculated correctly from the limiting reagent. C

Worked example

Problem. In a laboratory experiment, 12.0 \(\mathrm{g}\) of magnesium metal is reacted with excess hydrochloric acid to produce magnesium chloride and hydrogen gas. The balanced equation is \(\mathrm{Mg} + 2\mathrm{HCl} \to \mathrm{MgCl_2} + \mathrm{H_2}\). The actual yield of magnesium chloride obtained is 28.5 \(\mathrm{g}\). Calculate the theoretical yield of magnesium chloride and the percentage yield of the reaction. Use the following molar masses: \(\mathrm{Mg} = 24.31\ \mathrm{g\,mol^{-1}}\), \(\mathrm{MgCl_2} = 95.21\ \mathrm{g\,mol^{-1}}\).

  1. Identify the limiting reagent. Since hydrochloric acid is in excess, magnesium is the limiting reagent.
  2. Calculate the moles of magnesium used. \( n_{\mathrm{Mg}} = \frac{12.0\ \mathrm{g}}{24.31\ \mathrm{g\,mol^{-1}}} = 0.4936\ \mathrm{mol} \).
  3. Determine the moles of magnesium chloride produced using the stoichiometric ratio. From the equation, 1 mole of Mg produces 1 mole of \(\mathrm{MgCl_2}\). Therefore, \( n_{\mathrm{MgCl_2}} = 0.4936\ \mathrm{mol} \).
  4. Calculate the theoretical yield of magnesium chloride in grams. \( m_{\text{theoretical}} = 0.4936\ \mathrm{mol} \times 95.21\ \mathrm{g\,mol^{-1}} = 47.00\ \mathrm{g} \).
  5. Calculate the percentage yield. \( \%\,\text{yield} = \frac{28.5\ \mathrm{g}}{47.00\ \mathrm{g}} \times 100 = 60.6\% \).
\[ \boxed{\text{Theoretical yield} = 47.0\ \mathrm{g}, \quad \%\,\text{yield} = 60.6\%} \]

Units check. The theoretical yield is in grams, consistent with the input mass. The percentage yield is dimensionless, as expected for a ratio of masses.

CHECK YOURSELF

Three quick tests

  1. If the theoretical yield is 50.0 \(\mathrm{g}\) and the actual yield is 45.0 \(\mathrm{g}\), what is the percentage yield?
    Answer90.0%
  2. Why is the actual yield usually less than the theoretical yield?
    AnswerDue to side reactions, incomplete reactions, and physical losses during isolation.
  3. Can a percentage yield be greater than 100%?
    AnswerNo, for pure product. Values above 100% indicate impurities or errors.
UNIT

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NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Results cited: T-015 · Threads: structure