7.4 Reaction Yields - Chemistry: Atoms First

Learning Objectives

By the end of this section, you will be able to:

Explain the concepts of theoretical yield and limiting reactants/reagents.
Derive the theoretical yield for a reaction under specified conditions.
Calculate the percent yield for a reaction.

The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.

Limiting Reactant

Consider another food analogy, making grilled cheese sandwiches (Figure 7.12):

1 slice of cheese + 2 slices of bread ⟶ 1 sandwich

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Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

This figure has three rows showing the ingredients needed to make a sandwich. The first row reads, “1 sandwich = 2 slices of bread + 1 slice of cheese.” Two slices of bread and one slice of cheese are shown. The second row reads, “Provided with: 28 slices of bread + 11 slices of cheese.” There are 28 slices of bread and 11 slices of cheese shown. The third row reads, “We can make: 11 sandwiches + 6 slices of bread left over.” 11 sandwiches are shown with six extra slices of bread.

Figure 7.12 Sandwich making can illustrate the concepts of limiting and excess reactants.

Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

H_{2} (g) + {Cl}_{2} (g) ⟶ 2HCl (g)

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The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H₂ and 2 moles of Cl₂. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.

An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield

mol HCl produced = 3 mol H_{2} \times \frac{2 mol HCl}{1 mol H_{2}} = 6 mol HCl

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Complete reaction of the provided chlorine would produce

mol HCl produced = 2 mol {Cl}_{2} \times \frac{2 mol HCl}{1 mol {Cl}_{2}} = 4 mol HCl

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The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 7.13).

The figure shows a space-filling molecular models reacting. There is a reaction arrow pointing to the right in the middle. To the left of the reaction arrow there are three molecules each consisting of two green spheres bonded together. There are also five molecules each consisting of two smaller, white spheres bonded together. Above these molecules is the label, “Before reaction,” and below these molecules is the label, “6 H subscript 2 and 4 C l subscript 2.” To the right of the reaction arrow, there are eight molecules each consisting of one green sphere bonded to a smaller white sphere. There are also two molecules each consisting of two white spheres bonded together. Above these molecules is the label, “After reaction,” and below these molecules is the label, “8 H C l and 2 H subscript 2.”

Figure 7.13 When H₂ and Cl₂ are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.

Link to Learning

View this interactive simulation illustrating the concepts of limiting and excess reactants.

Example 7.12

Identifying the Limiting Reactant

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

3Si (s) + 2 N_{2} (g) ⟶ {Si}_{3} N_{4} (s)

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Which is the limiting reactant when 2.00 g of Si and 1.50 g of N₂ react?

Solution

Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

mol Si = 2.00 g Si \times \frac{1 mol Si}{28.09 g Si} = 0.0712 mol Si

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mol N_{2} = 1.50 g N_{2} \times \frac{1 mol N_{2}}{28.02 g N_{2}} = 0.0535 mol N_{2}

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The provided Si:N₂ molar ratio is:

\frac{0.0712 mol Si}{0.0535 mol N_{2}} = \frac{1.33 mol Si}{1 mol N_{2}}

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The stoichiometric Si:N₂ ratio is:

\frac{3 mol Si}{2 mol N_{2}} = \frac{1.5 mol Si}{1 mol N_{2}}

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Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.

Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

mol {Si}_{3} N_{4} produced = 0.0712 mol Si \times \frac{1 mol {Si}_{3} N_{4}}{3 mol Si} = 0.0237 mol {Si}_{3} N_{4}

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while the 0.0535 moles of nitrogen would produce

mol {Si}_{3} N_{4} produced = 0.0535 mol N_{2} \times \frac{1 mol {Si}_{3} N_{4}}{2 mol N_{2}} = 0.0268 mol {Si}_{3} N_{4}

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Since silicon yields the lesser amount of product, it is the limiting reactant.

Check Your Learning

Which is the limiting reactant when 5.00 g of H₂ and 10.0 g of O₂ react and form water?

Answer:

O₂

Percent Yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

percent yield = \frac{actual yield}{theoretical yield} \times 100 %

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Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

Example 7.13

Calculation of Percent Yield

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

{CuSO}_{4} (a q) + Zn (s) ⟶ Cu (s) + {ZnSO}_{4} (a q)

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What is the percent yield?

Solution

The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:

1.274 g {CuSO}_{4} \times \frac{1 mol {CuSO}_{4}}{159.62 g {CuSO}_{4}} \times \frac{1 mol Cu}{1 mol {CuSO}_{4}} \times \frac{63.55 g Cu}{1 mol Cu} = 0.5072 g Cu

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Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be

percent yield = (\frac{actual yield}{theoretical yield}) \times 100

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\begin{matrix} percent yield = (\frac{0.392 g Cu}{0.5072 g Cu}) \times 100 \\ = 77.3 % \end{matrix}

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Check Your Learning

What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF₂Cl₂ from 32.9 g of CCl₄ and excess HF?

{CCl}_{4} + 2 HF ⟶ {CF}_{2} {Cl}_{2} + 2 HCl

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Answer:

48.3%

How Sciences Interconnect

Green Chemistry and Atom Economy

The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry” (see details at this website). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:

atom economy = \frac{mass of product}{mass of reactants} \times 100 %

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Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 7.14). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%. In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997.

This figure is labeled, “a,” and, “b.” Part a shows an open bottle of ibuprofen and a small pile of ibuprofen tablets beside it. Part b shows a reaction along with line structures. The first line structure looks like a diagonal line pointing down and to the right, then up and to the right and then down and to the right. At this point it connects to a hexagon with alternating double bonds. At the first trough there is a line that points straight down. From this structure, there is an arrow pointing downward. The arrow is labeled, “H F,” on the left and “( C H subscript 3 C O ) subscript 2 O,” on the right. The next line structure looks exactly like the first line structure, but it has a line angled down and to the right from the lower right point of the hexagon. This line is connected to another line which points straight down. Where these two lines meet, there is a double bond to an O atom. There is another arrow pointing downward, and it is labeled, “H subscript 2, Raney N i.” The next structure looks very similar to the second, previous structure, except in place of the double bonded O, there is a singly bonded O H group. There is a final reaction arrow pointing downward, and it is labeled, “C O, [ P d ].” The final structure is similar to the third, previous structure except in place of the O H group, there is another line that points down and to the right to an O H group. At these two lines, there is a double bonded O.

Figure 7.14 (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)