Percent Yield

10.3 Percent Yield

Learning Objective

To calculate the quantities of compounds produced or consumed in a chemical reaction.

Percent Yield

You have learned that when reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yieldThe maximum amount of product that can be formed from the reactants in a chemical reaction, which theoretically is the amount of product that would be obtained if the reaction occurred perfectly and the method of purifying the product were 100% efficient., the amount you would obtain if the reaction occurred perfectly and your method of purifying the product were 100% efficient.

In reality, you almost always obtain less product than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not go all the way to completion, thus resulting in a mixture of products and reactants. This last possibility is a common occurrence. So the actual yieldThe measured mass of products actually obtained from a reaction. The actual yield is always less than the theoretical yield., the measured mass of products obtained from a reaction, is always less than the theoretical yield (often much less). The percent yieldThe ratio of the actual yield of a reaction to the theoretical yield multiplied by 100 to give a percentage. of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage:

Equation 3.23

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

The method used to calculate the percent yield of a reaction is illustrated in the following examples.

Example 10.3-1

A worker reacts 30.5 g of Zn with nitric acid and evaporates the remaining water to obtain 65.2 g of Zn(NO₃)₂. What are the theoretical yield, the actual yield, and the percent yield?

Zn(s) + 2HNO₃(aq) → Zn(NO₃)₂(aq) + H₂(g)

Solution

A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g/mol) and Zn(NO₃)₂ (189.41 g/mol). In three steps, the mass-mass calculation is

30.5 g Zn \times \frac{1 mol Zn}{65.39 g Zn} \times \frac{1 {mol Zn(NO}_{3})_{2}}{1 mol Zn} \times \frac{189.41 {g Zn(NO}_{3})_{2}}{1 {mol Zn(NO}_{3})_{2}} = 88.3 {g Zn(NO}_{3})_{2}

Thus, the theoretical yield is 88.3 g of Zn(NO₃)₂. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO₃)₂. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:

\frac{65.2 {g Zn(NO}_{3})_{2}}{88.3 {Zn(NO}_{3})_{2}} \times 100 % = 73.8 %

The worker achieved almost three-fourths of the possible yield.

Test Yourself

A synthesis produced 2.05 g of NH₃ from 16.5 g of N₂. What is the theoretical yield and the percent yield?

N₂(g) + 3H₂(g) → 2NH₃(g)

Answer

theoretical yield = 20.1 g; percent yield = 10.2%

Example 10.3-2

Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H₂SO₄ (indicated above the arrow) by the reaction

\underset{p -aminobenzoic acid}{C_{7} H_{7} {NO}_{2}} + \underset{2-diethylaminoethanol}{C_{6} H_{15} NO} \overset{H_{2} {SO}_{4}}{\to} \underset{procaine}{C_{13} H_{20} N_{2} O_{2}} + H_{2} O

If we carried out this reaction using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and we isolated 15.7 g of procaine, what was the percent yield?

The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water.

Given: masses of reactants and product

Asked for: percent yield

Strategy:

A Write the balanced chemical equation.

B Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.

C Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.

Solution:

A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.

B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C₇H₇NO₂), 137.14 g/mol; 2-diethylaminoethanol (C₆H₁₅NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:

\begin{array}{l} moles p -aminobenzoicacid = 10 .0 g \times \frac{1 mol}{137 .14 g} = 0 .0729 mol p -aminobenzoic acid \\ moles 2-diethylaminoethanol = 10 .0 g \times \frac{1 mol}{117 .19 g} = 0 .0853 mol 2-diethylaminoethanol \end{array}

The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C₁₃H₂₀N₂O₂) to calculate its molar mass, which is 236.31 g/mol.

theoretical yield of procaine = 0 .0729 mol \times \frac{236 .31 g}{1 mol} = 17 .2 g

C The actual yield was only 15.7 g of procaine, so the percent yield was

percent yield = \frac{15 .7 g}{17 .2 g} \times 100 = 91.3 %

(If the product were pure and dry, this yield would indicate that we have very good lab technique!)

Exercise

Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:

PbO(s) + C(s) → Pb(l) + CO(g)

If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?

Crystalline galena from a mine in Elizabeth, IL (a) and a sample of lead shot smashed with a hammer (b). Pure lead is soft enough to be shaped easily with a hammer, unlike the brittle mineral galena, the main ore of lead. Image Credit: Highland Community College, Freeport, IL

Answer: 89.2%

Percent yield can range from 0% to 100%.In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments! A 100% yield means that everything worked perfectly, and you obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows how unlikely a 100% yield is. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.

Summary

The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

Key Takeaways

Theoretical yield is what you calculate the yield will be using the balanced chemical reaction.
Actual yield is what you actually get in a chemical reaction.
Percent yield is a comparison of the actual yield with the theoretical yield.

Problems

What is the difference between the theoretical yield and the actual yield?
What is the difference between the actual yield and the percent yield?
A worker isolates 2.675 g of SiF₄ after reacting 2.339 g of SiO₂ with HF. What are the theoretical yield and the actual yield?
SiO₂(s) + 4HF(g) → SiF₄(g) + 2H₂O(l)
A worker synthesizes aspirin, C₉H₈O₄, according to this chemical equation. If 12.66 g of C₇H₆O₃ are reacted and 12.03 g of aspirin are isolated, what are the theoretical yield and the actual yield?
C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + HC₂H₃O₂
A chemist decomposes 1.006 g of NaHCO₃ and obtains 0.0334 g of Na₂CO₃. What are the theoretical yield and the actual yield?
2NaHCO₃(s) → Na₂CO₃(s) + H₂O(ℓ) + CO₂(g)
A chemist combusts a 3.009 g sample of C₅H₁₂ and obtains 3.774 g of H₂O. What are the theoretical yield and the actual yield?
C₅H₁₂(l) + 8O₂(g) → 5CO₂ + 6H₂O(ℓ)
What is the percent yield in Exercise 3?
What is the percent yield in Exercise 4?
What is the percent yield in Exercise 5?
What is the percent yield in Exercise 6?
Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.)
1. ${KClO}_{3} (s) \overset{Δ}{\to} KCl (s) + O_{2} (g);$ 2.14 g of KClO₃ produces 0.67 g of O₂
2. Cu(s) + H₂SO₄(aq) → CuSO₄(aq) + SO₂(g) + H₂O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide
3. AgC₂H₃O₂(aq) + Na₃PO₄(aq) → Ag₃PO₄(s) + NaC₂H₃O₂(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate
Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction?
A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction?
Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting.
A + 3B → 2C
Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol.
1. Write a balanced chemical equation for this reaction.
2. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide.
Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction:
NaClO₂(s) + Cl₂(g) → ClO₂(aq) + NaCl(aq)
1. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO₂?
2. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide.
The reaction of propane gas (CH₃CH₂CH₃) with chlorine gas (Cl₂) produces two monochloride products: CH₃CH₂CH₂Cl and CH₃CHClCH₃. The first is obtained in a 43% yield and the second in a 57% yield.
1. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion?
2. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane?
3. Use the actual percent yield to calculate how many grams of each product would actually be obtained.
Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions:
$2 {PaI}_{5} (s) \overset{Δ}{\to} 2 {Pa(s) + 5I}_{2} (s)$
1. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized?
2. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction?
3. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI₅ were used in the preparation?
Aniline (C₆H₅NH₂) can be produced from chlorobenzene (C₆H₅Cl) via the following reaction:
C₆H₅Cl(l) + 2NH₃(g) → C₆H₅NH₂(l) + NH₄Cl(s)
Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia.
1. Which is the limiting reactant?
2. Which reactant is present in excess?
3. What is the theoretical yield of ammonium chloride in grams?
4. If 4.78 g of NH₄Cl was recovered, what was the percent yield?
5. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess.
A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following:
1. 9.36 × 10²⁴ formula units of NaCl
2. 8.5 × 10⁴ mol of Br₂
3. 3.7 × 10⁸ g of NaCl

Answers

Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.
theoretical yield = 4.052 g; actual yield = 2.675 g
theoretical yield = 0.635 g; actual yield = 0.0334 g
66.02%
5.26%
1. 80%
2. 30%
3. 35.7%
45%.
1. CO + 2H₂ → CH₃OH
2. 58.28%
1. 2.24 g Cl₂
2. 4.95 g
3. 2.13 g CH₃CH₂CH₂Cl plus 2.82 g CH₃CHClCH₃
1. chlorobenzene
2. ammonia
3. 8.74 g ammonium chloride.
4. 55%
5. Theoretical yield (NH₄Cl) = $\frac{mass of chlorobenzene (g) \times 0 .92 \times 53 .49 g/mol}{112 .55 g/mol}$