Stoichiometry

Everyday Stoichiometry

How much equipment do you need for an experiment?

You are in charge of setting out the lab equipment for a chemistry experiment. If you have twenty students in the lab (and they will be working in teams of two) and the experiment calls for three beakers and two test tubes, how much glassware do you need to set out? Figuring this out involves a type of balanced equation and the sort of calculations you would do for a chemical reaction.

Everyday Stoichiometry

You have learned about chemical equations and the techniques used in order to balance them. Chemists use balanced equations to allow them to manipulate chemical reactions in a quantitative manner. Before we look at a chemical reaction, let’s consider the equation for the ideal ham sandwich.

Our ham sandwich is composed of 2 slices of ham (H), a slice of cheese (C), a slice of tomato (T), 5 pickles (P), and 2 slices of bread (B). The equation for our sandwich is shown below:

Now let us suppose that you are having some friends over and need to make five ham sandwiches. How much of each sandwich ingredient do you need? You would take the number of each ingredient required for one sandwich (its coefficient in the above equation) and multiply by five. Using ham and cheese as examples and using a conversion factor, we can write:

The conversion factors contain the coefficient of each specific ingredient as the numerator and the formula of on sandwich as the denominator. The result is what you would expect. In order to make five ham sandwiches, you would need 10 slices of ham and 5 slices of cheese.

This type of calculation demonstrates the use of stoichiometry. Stoichiometry is the calculation of the amount of substances in a chemical reaction from the balanced equation. The sample problem below is another stoichiometry problem involving ingredients of the ideal ham sandwich.

Sample Problem: Ham Sandwich Stoichiometry

Kim looks in the refrigerator and finds that she has 8 slices of ham. In order to make as many sandwiches as possible, how many pickles does she need? Use the equation above.

Step 1: List the known quantities and plan the problem.

Known

• have 8 ham slices (H)
• 2 H = 5 P (conversion factor)

Unknown

• How many pickles (P) needed?

The coefficients for the two reactants (ingredients) are used to make a conversion factor between ham slices and pickles.

Step 2: Solve.

Since 5 pickles combine with 2 ham slices in each sandwich, 20 pickles are needed to fully combine with 8 ham slices.

Step 3: Think about your result.

The 8 ham slices will make 4 ham sandwiches. With 5 pickles per sandwich, the 20 pickles are used in the 4 sandwiches.

Summary

• An example of everyday stoichiometry is given.

Practice

Questions

Use the link below to answer the following questions:

http://www.chem4kids.com/files/react_stoichio.html

1. What does stoichiometry help you figure out?
2. What are all reactions dependent upon?
3. If I have ten hydrogen molecules and three oxygen molecules, how many molecules of water can I make?
4. What will be left over and how much?

Review

Questions

1. I don’t like pickles. What would my ideal ham sandwich be?
2. How does that change the equation?
3. Will this change affect the amounts of other materials?

Mole Ratios

What does this porch need?

You want to add some sections to the porch seen above. Before you go to the hardware store to buy lumber, you need to determine the unit composition (the material between two large uprights). You count how many posts, how many boards, how many rails – then you decide how many sections you want to add before you calculate the amount of building material needed for your porch expansion.

Mole Ratios

Stoichiometry problems can be characterized by two things: (1) the information given in the problem, and (2) the information that is to be solved for, referred to as the unknown . The given and the unknown may both be reactants, both be products, or one may be a reactant while the other is a product. The amounts of the substances can be expressed in moles. However, in a laboratory situation, it is common to determine the amount of a substance by finding its mass in grams. The amount of a gaseous substance may be expressed by its volume. In this concept, we will focus on the type of problem where both the given and the unknown quantities are expressed in moles.

Chemical equations express the amounts of reactants and products in a reaction. The coefficients of a balanced equation can represent either the number of molecules or the number of moles of each substance. The production of ammonia (NH ₃ ) from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber.

The balanced equation can be analyzed in several ways, as shown in the Figure below .

We see that 1 molecule of nitrogen reacts with 3 molecules of nitrogen to form 2 molecules of ammonia. This is the smallest possible relative amounts of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia.

The most useful quantity for counting particles is the mole. So if each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation.

Finally, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. 1 mol of nitrogen has a mass of 28.02 g, while 3 mol of hydrogen has a mass of 6.06 g, and 2 mol of ammonia has a mass of 34.08 g.

Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved.

A mole ratio is a conversion factor that relates the amounts in moles of any two substances in a chemical reaction. The numbers in a conversion factor come from the coefficients of the balanced chemical equation. The following six mole ratios can be written for the ammonia forming reaction above.

In a mole ratio problem, the given substance, expressed in moles, is written first. The appropriate conversion factor is chosen in order to convert from moles of the given substance to moles of the unknown.

Sample Problem: Mole Ratio

How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?

Step 1: List the known quantities and plan the problem.

Known

• given: H ₂ = 4.20 mol

Unknown

• mol of NH ₃

The conversion is from mol H ₂ → NH ₃ . The problem states that there is an excess of nitrogen, so we do not need to be concerned with any mole ratio involving N ₂ . Choose the conversion factor that has the NH ₃ in the numerator and the H ₂ in the denominator.

Step 2: Solve.

The reaction of 4.20 mol of hydrogen with excess nitrogen produces 2.80 mol of ammonia.

Step 3: Think about your result.

The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.

Summary

• Mole ratios allow comparison of the amounts of any two materials in a balanced equation.
• Calculations can be made to predict how much product can be obtained from a given number of moles of reactant.

Practice

Do problems 1-4 at the link below:

http://myweb.astate.edu/mdraganj/Moles1.html

Review

Questions

1. If a reactant is in excess, why do we not worry about the mole ratios involving that reactant?
2. What is the mole ratio of H to N in the ammonia molecule?
3. The formula for ethanol is CH ₃ CH ₂ OH. What is the mole ratio of H to C in this molecule?

Mass-Mole and Mole-Mass Stoichiometry

Need nails?

When you are doing a large construction project, you have a good idea of how many nails you will need (lots!). When you go to the hardware store, you don’t want to sit there and count out several hundred nails. You can buy nails by weight, so you determine how many nails are in a pound, calculate how many pounds you need, and you’re on your way to begin building.

While the mole ratio is ever-present in all stoichiometry calculations, amounts of substances in the laboratory are most often measured by mass. Therefore, we need to use mole-mass calculations in combination with mole ratios to solve several different types of mass-based stoichiometry problems.

Mass to Moles Problems

In this type of problem, the mass of one substance is given, usually in grams. From this, you are to determine the amount in moles of another substance that will either react with or be produced from the given substance.

The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation.

Sample Problem: Mass-Mole Stoichiometry

Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?

Step 1: List the known quantities and plan the problem.

Known

• given: 75.0 g Sn
• molar mass of Sn = 118.69 g/mol
• 1 mol Sn = 2 mol HF (mole ratio)

Unknown

• mol HF

Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.

g Sn → mol Sn → mol HF

Step 2: Solve.

Step 3: Think about your result.

The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.

Moles to Mass Problems

In this type of problem, the amount of one substance is given in moles. From this, you are to determine the mass of another substance that will either react with or be produced from the given substance.

The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into mass in grams by use of the molar mass of that substance from the periodic table.

Sample Problem: Mole-Mass Stoichiometry

Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor.

What mass of oxygen gas is consumed in a reaction that produces 4.60 mol SO ₂ ?

Step 1: List the known quantities and plan the problem.

Known

• given: 4.60 mol SO ₂
• 2 mol SO ₂ = 3 mol O ₂ (mole ratio)
• molar mass of O ₂ = 32.00 g/mol

Unknown

• mass O ₂ = ? g

Use the mole ratio to convert from mol SO ₂ to mol O ₂ . Then convert mol O ₂ to grams. This will be done in a single two-step calculation.

mol SO ₂ → mol O ₂ → g O ₂

Step 2: Solve.

Step 3: Think about your result.

According to the mole ratio, 6.90 mol O ₂ is produced with a mass of 221 g. The answer has three significant figures because the given number of moles has three significant figures.

Summary

• Calculations involving conversions of mass to moles and moles to mass are described.

Practice

Work problems 11-20 at the link below:

http://myweb.astate.edu/mdraganj/Moles1.html

Review

Questions

1. In the first problem, what would happen if you multiply grams Sn by 118.69 grams/mole Sn?
2. Why is a balanced equation needed?
3. Does the physical form of the material matter for these calculations?

Mass-Mass Stoichiometry

How many walnuts are needed to equal 250 grams?

I want to send 250 grams of shelled walnuts to a friend (don’t ask why – just go with the question). How many walnuts in shells do I need to buy? To figure this out, I need to know how much the shell of a walnut weighs (about 40% of the total weight of the unshelled walnut). I can then calculate the mass of walnuts that will give me 250 grams of shelled walnuts and then determine how many walnuts I need to buy.

Mass to Mass Problems

Mass-mass calculations are the most practical of all mass-based stoichiometry problems. Moles cannot be measured directly, while the mass of any substance can generally be easily measured in the lab. This type of problem is three steps and is a combination of the two previous types.

The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Finally, the moles of the unknown are converted to mass by use of its molar mass.

Sample Problem: Mass-Mass Stoichiometry

Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.

In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each of the products formed.

Step 1: List the known quantities and plan the problem.

Known

• given: 45.7 g NH ₄ NO ₃
• 1 mol NH ₄ NO ₃ = 1 mol N ₂ O = 2 mol H ₂ O (mole ratios)
• molar mass of NH ₄ NO ₃ = 80.06 g/mol
• molar mass of N ₂ O = 44.02 g/mol
• molar mass of H ₂ O = 18.02 g/mol

Unknown

• mass N ₂ O = ? g
• mass H ₂ O = ? g

Perform two separate three-step mass-mass calculations as shown below.

Step 2: Solve.

Step 3: Think about your result.

The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.

Summary

• Mass-mass calculations involve converting the mass of a reactant to moles of reactant, then using mole ratios to determine moles of product which can then be converted to mass of product.

Practice

Read the material at the link below, then do the mass-mass problems at the link found at the bottom of the page:

http://www.chemteam.info/Stoichiometry/Mass-Mass.html

Review

Questions

1. If matter is neither created nor destroyed, why can’t we just go directly from grams of reactant to grams of product?
2. Why is it important to get the subscripts correct in the formulas?
3. Why do the coefficients need to be correct?

Volume-Volume Stoichiometry

How much propane is left in the tank?

As the weather gets warmer, more and more people want to cook out on the back deck or back yard. Many folks still use charcoal for grilling because of the added flavor. But increasing numbers of back yard cooks like to use a propane grill. The gas burns clean, the grill is ready to go as soon as the flame is lit – but how do you know how much propane is left in the tank? You can buy gauges at hardware stores that measure gas pressure and tell you how much is left in the tank.

Volume-Volume Stoichiometry

Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of gas particles. Further, one mole of any gas at standard temperature and pressure (0°C and 1 atm) occupies a volume of 22.4 L. These characteristics make stoichiometry problems involving gases at STP very straightforward. Consider the reaction of nitrogen and oxygen cases to form nitrogen dioxide.

Because of Avogadro’s work, we know that the mole ratios between substances in a gas-phase reaction are also volume ratios. The six possible volume ratios for the above equation are:

1. 
2. 
3. 

The volume ratios above can easily be used when the volume of one gas in a reaction is known and you need to determine the volume of another gas that will either react with or be produced from the first gas. The pressure and temperature conditions of both gases need to be the same.

Sample Problem: Volume-Volume Stoichiometry

The combustion of propane gas produces carbon dioxide and water vapor.

What volume of oxygen is required to completely combust 0.650 L of propane? What volume of carbon dioxide is produced in the reaction?

Step 1: List the known quantities and plan the problem.

Known

• given: 0.650 L C ₃ H ₈
• 1 volume C ₃ H ₈ = 5 volumes O ₂
• 1 volume C ₃ H ₈ = 3 volumes CO ₂

Unknown

• volume O ₂ = ? L
• volume CO ₂ = ? L

Two separate calculations can be done using the volume ratios.

Step 2: Solve.

Step 3: Think about your result.

Because the coefficients of the O ₂ and the CO ₂ are larger than that of the C ₃ H ₈ , the volumes for those two gases are greater. Note that total volume is not necessarily conserved in a reaction because moles are not necessarily conserved. In this reaction, 6 total volumes of reactants become 7 total volumes of products.

Summary

• Calculations of volume-volume ratios are based on Avogadro’s hypothesis.
• Pressures and temperatures of the gases involved need to be the same.

Practice

Read the material and work the Example One practice problems at the link below:

Stoichiometry

Review

Questions

1. What is Avogadaro’s hypothesis?
2. How much volume is occupied by one mole of a gas at STP?
3. In the sample problem above, assume we combust 1.3 L of propane. How much CO ₂ will be produced?

Mass-Volume and Volume-Mass Stoichiometry

How much azide is needed to fill an air bag?

Cars and many other vehicles have air bags in them. In case of a collision, a reaction is triggered so that the rapid decomposition of sodium azide produces nitrogen gas, filling the air bag. If too little sodium azide is used, the air bag will not fill completely and will not protect the person in the vehicle. Too much sodium azide could cause the formation of more gas that the bag can safely handle. If the bag breaks from the excess gas pressure, all protection is lost.

Mass to Volume and Volume to Mass Problems

Chemical reactions frequently involve both solid substances whose mass can be measured as well as gases for which measuring the volume is more appropriate. Stoichiometry problems of this type are called either mass-volume or volume-mass problems.

Because both types of problems involve a conversion from either moles of gas to volume or vice-versa, we can use the molar volume of 22.4 L/mol provided that the conditions for the reaction are STP.

Sample Problem: Mass-Volume Stoichiometry

Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas.

Determine the volume of hydrogen gas produced at STP when a 2.00 g piece of aluminum completely reacts.

Step 1: List the known quantities and plan the problem.

Known

• given: 2.00 g Al
• molar mass Al = 26.98 g/mol
• 2 mol Al = 3 mol H ₂

Unknown

• volume H ₂ = ?

The grams of aluminum will first be converted to moles. Then the mole ratio will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.

Step 2: Solve.

Step 3: Think about your result.

The volume result is in liters. For much smaller amounts, it may be convenient to convert to milliliters. The answer here has three significant figures. Because the molar volume is a measured quantity of 22.4 L/mol, three is the maximum number of significant figures for this type of problem.

Sample Problem: Volume-Mass Stoichiometry

Calcium oxide is used to remove sulfur dioxide generated in coal-burning power plants according to the following reaction.

What mass of calcium oxide is required to react completely with 1.4 × 10 ³ L of sulfur dioxide?

Step 1: List the known quantities and plan the problem.

Known

• given: 1.4 × 10 ³  L = SO ₂
• 2 mol SO ₂ = 2 mol CaO
• molar mass CaO = 56.08 g/mol

Unknown

• mass CaO = ? g

The volume of SO ₂ will be converted to moles, followed by the mole ratio, and finally a conversion of moles of CaO to grams.

Step 2: Solve.

Step 3: Think about your result.

The resultant mass could be reported as 3.5 kg, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.

Summary

• Calculations are described for determining the amount of gas formed in a reaction.
• Calculations are described for determining amounts of a material needed to react with a gas.

Practice

Answer the questions at the link below:

http://www.docbrown.info/page04/4_73calcs/MVGmcTEST.htm

Review

Answers

1. What are the conditions for all gases in these calculations?
2. How can you tell if all the ratios were set up correctly?
3. Why was 2 mol CaO/2mol SO ₂ included in the second example if it did not affect the final number?

Limiting Reactant

Don’t you hate running out of cooking ingredients?

Cooking is a great example of everyday chemistry. In order to correctly follow a recipe, a cook needs to make sure that he has plenty of all the necessary ingredients in order to make his dish. Let us suppose that you are deciding to make some pancakes for a large group of people. The recipe on the box indicates that the following ingredients are needed for each batch of pancakes:

1 cup of pancake mix

 cup milk

1 egg

1 tablespoon vegetable oil

Now you check the pantry and the refrigerator and see that you have the following ingredients available:

2 boxes of pancake mix (8 cups)

Half gallon of milk (4 cups)

2 eggs

Full bottle of vegetable oil (about 3 cups)

The question that you must ask is: How many batches of pancakes can I make? The answer is two. Even though you have enough pancake mix, milk, and oil to make many more batches of pancakes, you are limited by the fact that you only have two eggs. As soon as you have made two batches of pancakes, you will be out of eggs and your “reaction” will be complete.

Limiting Reactant

For a chemist, the balanced chemical equation is the recipe that must be followed. As you have seen earlier, the Haber process is a reaction in which nitrogen gas is combined with hydrogen gas to form ammonia. The balanced equation is shown below.

We know that the coefficients of the balanced equation tell us the mole ratio that is required for this reaction to occur. One mole of N ₂ will react with three moles of H ₂ to form two moles of NH ₃ .

Now let us suppose that a chemist were to react three moles of N ₂ with six moles of H ₂ (see Figure below ).

So what happened in this reaction? The chemist started with 3 moles of N ₂ . You may think of this as being 3 times as much as the “recipe” (the balanced equation) requires since the coefficient for the N ₂ is a 1. However, the 6 moles of H ₂ that the chemist started with is only two times as much as the “recipe” requires, since the coefficient for the H ₂ is a 3 and 3 × 2 = 6. So the hydrogen gas will be completely used up while there will be 1 mole of nitrogen gas left over after the reaction is complete. Finally, the reaction will produce 4 moles of NH ₃ because that is also two times as much as shown in the balanced equation. The overall reaction that occurred in words:

All the amounts are doubled from the original balanced equation.

The limiting reactant (or limiting reagent) is the reactant that determines the amount of product that can be formed in a chemical reaction. The reaction proceeds until the limiting reactant is completely used up. In our example above, the H ₂ is the limiting reactant. The excess reactant (or excess reagent) is the reactant that is initially present in a greater amount than will eventually be reacted. In other words, there is always excess reactant left over after the reaction is complete. In the above example, the N ₂ is the excess reactant.

Summary

• The amount of limiting reactant determines how much product will be formed in a chemical reaction.

Practice

Questions

Watch the video at the link below and answer the following questions:

http://www.sophia.org/limiting-reactant-definition/limiting-reactant-definition–2-tutorial

1. What reaction is occurring?
2. How is the reaction measured?
3. What do the balloons tell us?

Review

Questions

1. In the Haber reaction illustrated above, how do we know that hydrogen is the limiting reactant?
2. What if hydrogen were left over?
3. Which material would be limiting if no hydrogen or nitrogen were left over?

Determining the Limiting Reactant

Who’s coming for dinner?

You have ten people that show up for a dinner party. One of the guest brings twenty brownies for dessert. The decision about serving dessert is easy: two brownies are placed on every plate. If someone wants more brownies, they will have to wait until they go to the store. There are only enough brownies for everyone to have two.

Determining the Limiting Reactant

In the real world, amounts of reactants and products are typically measured by mass or by volume. It is first necessary to convert the given quantities of each reactant to moles in order to identify the limiting reactant.

Sample Problem: Determining the Limiting Reactant

Silver metal reacts with sulfur to form silver sulfide according to the following balanced equation:

What is the limiting reactant when 50.0 g Ag is reacted with 10.0 g S?

Step 1: List the known quantities and plan the problem.

Known

• given: 50.0 g Ag
• given: 10.0 g S

Unknown

• limiting reactant

Use the atomic masses of Ag and S to determine the number of moles of each present. Then, use the balanced equation to calculate the number of moles of sulfur that would be needed to react with the number of moles of silver present. Compare this result to the actual number of moles of sulfur present.

Step 2: Solve.

First, calculate the number of moles of Ag and S present:

Second, find the moles of S that would be required to react with all of the given Ag:

The amount of S actually present is 0.312 moles. The amount of S that is required to fully react with all of the Ag is 0.232 moles. Since there is more sulfur present than what is required to react, the sulfur is the excess reactant. Therefore, silver is the limiting reactant.

Step 3: Think about your result.

The balanced equation indicates that the necessary mole ratio of Ag to S is 2:1. Since there were not twice as many moles of Ag present in the original amounts, that makes silver the limiting reactant.

There is a very important point to consider about the preceding problem. Even though the mass of silver present in the reaction (50.0 g) was greater than the mass of sulfur (10.0 g), silver was the limiting reactant. This is because chemists must always convert to molar quantities and consider the mole ratio from the balanced chemical equation.

There is one other thing that we would like to be able to determine in a limiting reactant problem – the quantity of the excess reactant that will be left over after the reaction is complete. We will go back to the sample problem above to answer this question.

Sample Problem: Determining the Amount of Excess Reactant Left Over

What is the mass of excess reactant remaining when 50.0 g Ag reacts with 10.0 g S?

Step 1: List the known quantities and plan the problem.

Known

• Excess reactant = 0.312 mol S (from sample problem 12.9)
• Amount of excess reactant needed = 0.232 mol S (from sample problem 12.9)

Unknown

• Mass of excess reactant remaining after the reaction = ? g

Subtract the amount (in moles) of the excess reactant that will react from the amount that is originally present. Convert moles to grams.

Step 2: Solve.

There are 2.57 g of sulfur remaining when the reaction is complete.

Step 3: Think about your result.

There were 10.0 g of sulfur present before the reaction began. If 2.57 g of sulfur remains after the reaction, then 7.43 g S reacted.

This is the amount of sulfur that reacted. The problem is internally consistent.

Summary

• Determining the limiting reactant requires that all mass quantities first be converted to moles to evaluate the equation.

Practice

Carry out the calculations on the problem set at the link below:

Limiting Reagents

Review

Questions

1. Why do all mass values need to be converted to moles before determining the limiting reactant?
2. If we used 0.7 moles Ag, would it still be the limiting reactant?
3. If we ran the reaction using the original amounts of Ag and S and had 5.22 grams S left over, what might we assume about the reaction?

Theoretical Yield and Percent Yield

Can we save some money?

The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.

Percent Yield

Chemical reactions in the real world don’t always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.

To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield , the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.

Typically, percent yields are understandably less than 100% because of the reasons indicated earlier. However, percent yields greater than 100% are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction.

Sample Problem: Calculating the Theoretical Yield and the Percent Yield

Potassium chlorate decomposes upon slight heating in the presence of a catalyst according to the reaction below:

In a certain experiment, 40.0 g KClO ₃ is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed and the oxygen gas is collected and its mass is found to be 14.9 g. What is the percent yield for the reaction?

Part 12.11A : First, we will calculate the theoretical yield based on the stoichiometry.

Step 1: List the known quantities and plan the problem.

Known

• given: mass of KClO ₃ = 40.0 g
• molar mass KClO ₃ = 122.55 g/mol
• molar mass O ₂ = 32.00 g/mol

Unknown

• theoretical yield O ₂ = ? g

Apply stoichiometry to convert from the mass of a reactant to the mass of a product:

Step 2: Solve.

The theoretical yield of O ₂ is 15.7 g.

Step 3: Think about your result.

The mass of oxygen gas must be less than the 40.0 g of potassium chlorate that was decomposed.

Part 12.11B :  Now we use the actual yield and the theoretical yield to calculate the percent yield.

Step 1: List the known quantities and plan the problem.

Known

• Actual yield = 14.9 g
• Theoretical yield = 15.7 g (from Part 12.11A)

Unknown

• Percent yield = ? %

Use the percent yield equation above.

Step 2: Solve.

Step 3: Think about your result.

Since the actual yield is slightly less than the theoretical yield, the percent yield is just under 100%.

Summary

• Theoretical yield is calculated based on the stoichiometry of the chemical equation.
• The actual yield is experimentally determined.
• The percent yield is determined by calculating the ratio of actual yield/theoretical yield.

Practice

Work the problems found on the link below:

http://science.widener.edu/svb/tutorial/percentyieldcsn7.html

Review

Questions

1. What do we need in order to calculate theoretical yield?
2. If I spill some of the product before I weigh it, how will that affect the actual yield?
3. How will spilling some of the product affect the percent yield?
4. I make a product and weigh it before it is dry. How will that affect the actual yield?