Lesson 13:

Concepts conveyed | Materials | Procedure | Benefits | Resource

Concepts conveyed:

The purpose of this activity is to constructively teach students about stoichiometry and balanced chemical equations by using real-life situations.

Materials:

• 1 can of any diet soda (display purposes only)
• 1 worksheet per student (each containing the same problems to work and chemical equations to balance)
• 1 package of bread slices
• 1 package of cheese
• 1 package of deli meat slices
• 1 assembled sandwich (2 bread slices, 2 meat, 1 cheese, and 3 pickles)
• Wet-wipes and paper towels to clean hands when finished assembling the sandwiches

Procedure:

It is important to have the proper relative amounts, or “balance” of ingredients in food, paints, oils, cosmetics, medicines, plastics and other materials, methods of analysis of diseases, the water we drink, the air we breathe, etc. If the wrong amount of plasticizer were used in a particular plastic, its properties would not allow it to be used for its applications. What would that chocolate chip cookie taste like if ten times as much flour or no sugar were used?

This procedure provides two series of examples of balancing ingredients. These should be worked out by the instructor with the students. Afterward, the students are asked to solve the problems that follow, eventually applying what they know to real chemical reactions.

Work through the following two problems with the class. All of the problems should be on the worksheet that the students have already received at the beginning of the class period. Space should be left on those worksheets so that the students can work out the answers to the problems.

Part 1: Making Sandwiches

Joan’s famous sandwich consists of two slices of rye bread, two slices of pastrami, one slice of swiss cheese and three slices of dill pickle. Ask the students to write an equation for this recipe. One follows:

2 slices bread + 2 slices meat + 1 slice cheese + 3 slices pickle —> 1 sandwich

Note that the number in front of each item (also called a coefficient) represents the total number of each item that is needed in the recipe. Also note that there is a conservation of mass, in that the substances on the left are also on the right (in the form of a completed sandwich). To further simplify this equation, let us introduce the concept of subscripts to fully represent the components in the sandwich. In doing so, let us also represent B = bread, M = meat, C = cheese, and P = pickle, so that the sandwich is represented by B₂M₂CP₃. The above equation becomes:

2 B + 2 M + 1 C + 3 P —> 1 B₂M₂CP₃

Now it is more obvious that mass is conserved. This is starting to look like a chemical equation. We can drop coefficients of “1” if we assume that “no coefficient” means “1”. (If an item does not participate in the recipe, then we simply do not include it in the equation.) Thus, we get:

2 B + 2 M + C + 3 P —> B₂M₂CP₃ (1)

We can use the coefficients in the balanced equation (1) to relate the number of any item to the number of any other item in the recipe. For example, the ratio of bread slices to sandwiches is 2B to 1 B₂M₂CP₃. The ratio of meat slices to sandwiches is 2M to 1 B₂M₂CP₃ The ratio of cheese slices to sandwiches is 1 C to 1 B₂M₂CP₃. The ratio of pickles to sandwiches is 3 P to 1 B₂M₂CP₃. The ratio of bread slices to meat is 2 B to 2 M (or 1 to 1). The ratio of cheese slices to pickles is 1 C to 3 P, and so on!

When ingredients are in perfect ratio to each other
a. Determining the amount of product (number of sandwiches) based on the amount of ingredients.

Question: How many of Joan’s famous sandwiches can be produced when 8 slices of bread, 8 slices of meat, 4 slices of cheese, and 12 slices of pickles are available?

Answer:
When we multiply any of these numbers by the ratio of the number of sandwiches to that item based upon the coefficients in equation (1), the number is four sandwiches.

Considering bread: 8 B x (1 B₂M₂CP₃/ 2 B) = 4 B₂M₂CP₃
Considering meat: 8 M x (1 B₂M₂CP₃/ 2 M) = 4 B₂M₂CP₃
Considering cheese: 4 C x (1 B₂M₂CP₃/ 1 C) = 4 B₂M₂CP₃
Considering pickles: 12 P x (1 B₂M₂CP₃/ 3 P) = 4 B₂M₂CP₃

(Note how the conversion factor, that term in the parentheses, is always written so that the appropriate units cancel.)

Thus, the answer is that exactly four sandwiches can be prepared from these ingredients and there are no ingredients left over. If we wanted to calculate the number of ingredients left over, we would have to subtract the number required to make four sandwiches from the starting number of items:

Considering bread: 8 B – (4 B₂M₂CP₃ x (2 B/ 1 B₂M₂CP₃)) = 0 B
Considering meat: 8 M – (4 B₂M₂CP₃ x (2 M/ 1 B₂M₂CP₃)) = 0 M
Considering cheese: 4 C – (4 B₂M₂CP₃ x (1 C/ 1 B₂M₂CP₃)) = 0 C
Considering pickles: 12 P – (4 B₂M₂CP₃ x (3 P/ 1 B₂M₂CP₃)) = 0 P

This proves that there are no extra items left over when making 4 sandwiches from these amounts of these ingredients.

b. Determining the exact amount of ingredients (numbers) based on the amount of sandwiches wanted.

Question: How many of each ingredient is needed in order to make 10 of Joan’s famous sandwiches?

Answer: We can use the ratio of the number of items in the balanced equation (1) to help us here, too!

Considering bread: 10 B₂M₂CP₃ x (2 B/ 1 B₂M₂CP₃) = 20 B
Considering meat: 10 B₂M₂CP₃ x (2 M/ 1 B₂M₂CP₃) = 20 M
Considering cheese: 10 B₂M₂CP₃ x (1 C/ 1 B₂M₂CP₃) = 10 C
Considering pickles: 10 B₂M₂CP₃ x (3 P/ 1 B₂M₂CP₃) = 30 P

Thus, in order to construct 10 sandwiches, we need 20 slices of bread, 20 slices of meat, 10 slices of cheese, and 30 slices of pickles.

To make life easy, it would be nice if bread came in packages of 20 slices, meat in packages of 20 slices, cheese in packages of 10 slices and pickles in bottles of 30. Then, we would only need to buy one package of each ingredient to prepare 10 sandwiches. However, packages are not always available with the appropriate number of slices. See the next problem for more on this.

c. Consideration of packages of ingredients.

Question: How many packages of ingredients would be needed to make 10 sandwiches if bread came in bags of 20 slices, meat in packages of 10 slices, cheese in packages of 8 slices, and pickles in containers of 300?

Answer: We can use our answer in part b above to help us solve this question. Because of the work in part b, we know that 10 sandwiches require the use of 20 slices of bread, 20 slices of meat, 10 slices of cheese, and 30 slices of pickles. If a package does not contain enough ingredients, then we need to buy two or more packages. Calculations are shown below. We can use the conversion factors provided to us in this question.

Considering bread: 20 B x (1 B package / 20 B) = 1 B package
Considering meat: 20 M x (1 M package / 10 M) = 2 M packages
Considering cheese: 10 C x (1 C package / 8 C) = 1.25 C packages ~ 2 C packages
Considering pickles: 30 P x (1 P package / 300 P) = 0.1 P package ~ 1 P package

There are exactly the right number of bread slices in a package to make 10 sandwiches. We need exactly 2 packages of meat to make 10 sandwiches. Because one package of cheese slices contains only 8, we need 2 packages to have enough to make 10 sandwiches. There are so many pickles in a jar, that we won’t use all of them…only 30; so, we only need one jar of pickles to make 10 sandwiches.

When ingredients are not in perfect ratio to each other: Limiting ingredient problems.
a. Determining the amount of product (number of sandwiches) based on the amount of ingredients.

Question: How many of Joan’s famous sandwiches can be produced when 15 slices of bread, 10 slices of meat, 11 slices of cheese, and 24 slices of pickles are available? Are any ingredients left over? If so, how much?

Answer:
We can use the same strategy as part a above and multiply any of these numbers by the ratio of the number of sandwiches to that item based upon the equation (1).

Considering bread: 15 B x (1 B₂M₂CP₃/ 2 B) = 7.5 B₂M₂CP₃
Considering meat: 10 M x (1 B₂M₂CP₃/ 2 M) = 5 B₂M₂CP₃
Considering cheese: 11 C x (1 B₂M₂CP₃/ 1 C) = 11 B₂M₂CP₃
Considering pickles: 24 P x (1 B₂M₂CP₃/ 3 P) = 8 B₂M₂CP₃

We can only construct as many sandwiches as the limiting ingredient will allow. The ingredient that allows us to make the fewest sandwiches is the meat, which will only allow 5 sandwiches to be constructed. Thus, the meat is considered the “limiting” ingredient. All of the other ingredients are considered to be in “excess”. The maximum number of sandwiches that can be prepared is 5. When we do this, we will have the following amounts of ingredients left over:

Considering bread: 15 B – (5 B₂M₂CP₃ x (2 B/ 1 B₂M₂CP₃)) = 5 B
Considering meat: 10 M – (5 B₂M₂CP₃ x (2 M/ 1 B₂M₂CP₃)) = 0 M
Considering cheese: 11 C – (5 B₂M₂CP₃ x (1 C/ 1 B₂M₂CP₃)) = 6 C
Considering pickles: 24 P – (5 B₂M₂CP₃ x (3 P/ 1 B₂M₂CP₃)) = 9 P

We will have 5 slices of bread, 6 slices of cheese, and 9 slices of pickles left over.

b. Determining the amount of ingredients needed, based on the number of sandwiches wanted.

Question: How many of each ingredient is needed to make 242 sandwiches? How many packages are needed if there are 20 slices of bread, 10 slices of meat, 8 slices of cheese, and 300 slices of pickles in their corresponding packages?

Answer:
The first part of this answer is solved the same way as part b above. We can use the ratio of the number of items in the balanced equation (1) to help us.

Considering bread: 242 B₂M₂CP₃ x (2 B/ 1 B₂M₂CP₃) = 484 B
Considering meat: 242 B₂M₂CP₃ x (2 M/ 1 B₂M₂CP₃) = 484 M
Considering cheese: 242 B₂M₂CP₃ x (1 C/ 1 B₂M₂CP₃) = 242 C
Considering pickles: 242 B₂M₂CP₃ x (3 P/ 1 B₂M₂CP₃) = 726 P

Thus, in order to construct 242 sandwiches, we need 484 slices of bread, 484 slices of meat, 242 slices of cheese, and 726 slices of pickles.

The second part of this question is answered just like that answer in part c above.

Considering bread: 484 B x (1 B package / 20 B) = 24.2 B package ~ 25 B packages
Considering meat: 484 M x (1 M package / 10 M) = 48.4 M packages ~ 49 M packages
Considering cheese: 242 C x (1 C package / 8 C) = 30.25 C packages ~ 31 C packages
Considering pickles: 726 P x (1 P package / 300 P) = 2.42 P package ~ 3 P packages

We can’t split packages, so we must buy 25 packages of bread, 49 packages of meat, 31 packages of cheese, and 3 jars of pickles to construct 242 sandwiches.

Part 2: Making Soda

One recipe that makes enough diet cola for a case (24, 12-ounce cans) requires 240 ounces of carbonated water, 12 ounces of diet cola syrup, 8 ounces of artificial sweetener, and 24 ounces of phosphate stabilizers.

Write an equation for this.

240 oz. H₂O + 12 oz. syrup + 8 oz. sweetener + 24 oz. stabilizer—> 24 oz. diet colas

To simplify the coefficients of this equation, all coefficients can be divided by four:

60 oz. H₂O + 3 oz. syrup + 2 oz. sweetener + 6 oz. stabilizer—> 6 oz. diet colas (2)

Now the students should be able to apply what they learned in the sandwich examples to these problems. These problems are no longer categorized into ingredient-limiting or non-limiting, to encourage students to use the same procedure to solve all such problems.

Question: If you had 240 ounces of carbonated water, 15 ounces syrup, 10 ounces sweetener, and 17 ounces stabilizer, how much diet cola in ounces could you produce?

Answer:

Consider water: 240 oz. water x (6 oz. colas/60 oz. water) = 24 oz. colas
Consider syrup: 15 oz. syrup x (6 oz. colas/3 oz. syrup) = 30 oz. colas
Consider sweetner: 10 oz. sweetner x (6 oz. colas/ 2 oz. sweetner) = 30 oz. colas
Consider stabilizer: 17 oz. stabilizer x (6 oz. colas/ 6 oz. stabilizer) = 17 oz. colas

The ingredient that limits the cola to 17 oz. is the stabilizer. The amount of other ingredients that are left over after making 17 oz. of cola are:

Consider water: 240 oz – (17 oz. x (60 oz. water/6 oz. cola)) = 70 oz.
Consider syrup: 15 oz – (17 oz. x (3 oz. syrup/6 oz. cola)) = 6.5 oz.
Consider sweetner: 10 oz – (17 oz. x (2 oz. sweetner/6 oz. cola)) = 4.3 oz.
Consider stabilizer: 17 oz – (17 oz. x (6 oz. stabilizer/6 oz. cola)) = 0 oz.

After this example has been worked, allow students to work in pairs to answer the remaining problems on their own. Create additional problems as needed from the ones given.

Problems for Pairs of Students to Work on Their Own

• Question: A super-duper sandwich consists of 3 slices of bread, 4 slices of turkey, 2 slices of cheese, and 4 slices of tomato. If you want to prepare 50 sandwiches, how many packages of each ingredient do you need to obtain? In one package for a given ingredient, there are 25 slices of bread, 30 slice of meat, and 4 slices of cheese. One tomato typically provides 6 slices.

• Question: If 22 slices of bread, 30 slices of meat, 18 slices of cheese, and 32 slices of tomatoes are available, how many super-duper sandwiches can you make? How much of each ingredient will be leftover?

Answer: Use the balanced equation from the previous problem

• Question: If you have 237 ounces of carbonated water, 9 ounces syrup, 41 ounces sweetener, and 52 ounces stabilizer. How much diet cola could you produce in ounces? If you only want to make enough to fill the 12 ounce cans, how many diet cola cans could you produce? Use only whole numbers, no fractions or decimals. What amounts would you use and what would be left over if you limit yourself to making enough to fill whole cans? Use Equation (2) as your balanced equation.

Answer: Use balanced equation (2). 

• Question. The equations you have just developed to illustrate the recipes for cola or sandwiches are very similar to balanced chemical equations. Instead of food ingredients, the ingredients in chemical equations are atoms and molecules. Making chemical compounds, each of which has a certain formula, requires specific amounts of each starting chemical.

a. Write out and balance the following equation: Iron (Fe) reacts with oxygen in the air (O2) and produces rust, also known as ferric oxide (Fe2O3). Just as in the sandwich and soda examples, the composition of the compounds cannot be changed; thus, coefficients may only be placed in front of each formula, but subscripted numbers cannot be changed. Use integer coefficients for this example.

b. In a chemical reaction, the coefficients in front of the formulas represent numbers of molecules or numbers of moles (where 1 mole is 6.02 x 1023 molecules). If you have 10 moles of Fe and 6 moles of O2, how many moles of Fe2O3 are produced? How much of the reactants are left over?

c. The mass of one mole of Fe is 55.847 g/mol, that of O₂ is 31.9988 g/mol, and that of Fe₂O₃ is 159.692 g/mol. These are called the molecular weights (or molar masses) of these species. They are analogous to the “packages” in our sandwich and soda problems. For example, in a “package” of 55.847 grams, there is one mole of Fe. How many grams of Fe and of O₂ must be used in order to make 5.00 grams of Fe₂O₃?

Benefits:

• Instructors can use the activity as a means of assessing students grasp of information and of keeping record of class attendance.
• Students will be better prepared to balance and understand real chemical equations in lecture and in laboratory assignments.

Resource:

• Belcher, D. C. Science Scope. 1995, 19(3), 37-39.