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## Group Activity: Stoichiometry

### Background

The study of chemistry is concerned with understanding how and why chemicals react with one another. A critical tool in building this knowledge is called stoichiometry. The word originates in Greek roots for elements and measurement. This is appropriate because stoichiometry is a mathematical approach to understanding the proportions in which elements (and compounds) react in balanced chemical equations. These proportions are based on the whole-number ratio of molecules (or formula units) which react with one another. For example, consider the combustion of butane in oxygen to make carbon dioxide and water:

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

In this balanced chemical equation two molecules of butane (C4H10) are shown to require 13 molecules of oxygen (O2) to react to make carbon dioxide (CO2) and water (H2O). Based on this observation it’s straightforward to conclude that 4 molecules of butane would require 26 molecules of oxygen to react. And 8 molecules of butane require 52 molecules of oxygen and so on. This mathematical relationship continues to hold no matter how many molecules of butane you start with. Consider the case when 6.022 × 1023 molecules of butane react with oxygen. In that case,

```                             13 molecules O2
6.022 × 1023 molecules C4H10 × ---------------- = 3.914 × 1024 molecules O2
2 molecules C4H10
```

You may recall that 6.022 × 1023 is the number of items in one mole (1 mol). By this reasoning it is just as true to say that two moles of butane react with thirteen moles of oxygen as to say that two molecules of butane reacts with thirteen molecules of oxygen. So the equation states that two moles of butane react with thirteen moles of oxygen to make eight moles of carbon dioxide and ten moles of water.

In fact, expressing the mathematical proportions of compounds and elements in chemical equations in terms of moles is far more useful to chemists than expressing them in terms of molecules. A mole of a substance is a measurable amount of it: one mole of butane has a mass of 58.122 g. Remember, molar mass is calculated by adding up all the atomic mass in the chemical formula: C4H10 has a molar mass of 4 × 12.011 g/mol + 10 × 1.00794 g/mol = 58.122 g/mol. Single molecules of butane are far too small to measure or work with. One molecule has a mass of 9.65 × 10-23 grams.

### Molar Ratios

Mathematical proportions from chemical equations are expressed as molar ratios. A molar ratio is the relative number of formula units of two chemicals in a balanced chemical equation. The full set of possible molar ratios from the chemical equation for the burning of butane is as follows:

 2 mol C4H10 = 13 mol O2 2 mol C4H10 = 8 mol CO2 2 mol C4H10 = 10 mol H2O 13 mol O2 = 8 mol CO2 13 mol O2 = 10 mol H2O 8 mol CO2 = 10 mol H2O ```any ratio can be expressed as a fraction: 2 mol C4H10 13 mol O2 ----------- or ----------- 13 mol O2 2 mol C4H10 ```

### Stoichiometry Problems

#### Using Moles

When the ratios are expressed as fractions (see above, at right) they can be used as conversion factors of a sort. If you remember your dimensional analysis this will be familiar. Molar ratios are really proportions that can be used to find out how many moles of any of the other chemicals correspond to a given amount of the first. For example, 2 mol of butane corresponds to 13 mol of oxygen; how many moles of oxygen correspond to 4 mol of butane? Twenty-six:

```             13 mol O2
4 mol C4H10 × ---------- = 26 mol O2
2 mol C4H10
```

To the well-informed this means that to burn four moles of butane you need twenty-six moles of oxygen gas. How about if you have a number like 3.19 moles of butane? It works even then:

```                8 mol CO2
3.19 mol C4H10 × ---------- = 12.8 mol CO2
2 mol C4H10
```

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In this calculation, as in all of these examples, you use the molar ratio to convert moles of one substance to moles of another substance. In the example immediately above we found that burning 3.19 mol C4H10 produces 12.8 mol CO2.

Now let’s look at a few examples in the way they might be phrased in a set of chemistry problems or on a quiz.

##### Example 1

How many moles of carbon dioxide result from the burning of 8.9 moles of butane?

```                8 mol CO2
8.9 mol C4H10 × ---------- = 35.6 mol CO2  (because 8.9 × 8 ÷ 2 = 35.6)
2 mol C4H10
```

##### Example 2

If 42 mol of carbon dioxide were produced then how many moles of oxygen were used up in making it?

```               13 mol O2
42 mol CO2 × ---------- = 68 mol O2  (because 42 × 13 ÷ 8 = 68.25)
8 mol CO2
```

#### An Analogy

It is often helpful to relate new ideas to familiar ones. In this case a useful analogy is to the construction of a simple cheese sandwich. To make it more appetizing you may imagine grilling the sandwich in butter but that part does not play a part in our analogy.

One simple way to make a cheese sandwich is with two pieces of bread and a single slice of cheese. To make the chemical analogy clear, this recipe can be written as a sort of equation:

The equation expresses a set of proportions: exactly two slices of bread are needed for each one slice of cheese. Another proportion implied in the recipe (or the equation) is that exactly one slice of cheese is needed for each sandwich. These proportions can be written as fractions:

``` 2(Bread)          1(Cheese)         2(Bread)
-----------  and  ------------ and -------------
1(Cheese)         1(Sandwich)      1(Sandwich)

```

This is a complicated way of talking about something simple, but since the complicated way of thinking is what you are meant to learn, this is helpful. For example, it is probably obvious that if you want to make 10 sandwiches you need 20 slices of bread. But this can be calculated, with a little labor, as follows:

```                2(Bread)
1(Sandwich)
```

These ratios can scale up to any amount so that if instead of 10, you had 100 sandwiches to make then you would need 200 slices of bread. Similarly, if you have a mole of sandwiches to make (which is an insane amount to think about, but still) the ratios still work. One mole of sandwiches would require 2 moles of slices of bread. This way of thinking and calculating is exactly what stoichiometry is all about, with the added complication that you are calculating the number of molecules, which are too small to see or to count directly, rather than the number of slices of bread or cheese, which are easy to count for normal situations. But if molecules are too small to count directly, then how can stoichometry be used in any practical way?

#### Using Masses

To make stoichiometry practical it is necessary to be able to relate the mass of a reactant or product to the mass of the other reactants and products. It’s not possible to measure individual molecules in the lab or to measure moles directly. But moles are still useful because theycan be measured indirectly in the lab. Every chemical has a characteristic called its molar mass. The molar mass of a chemical is the sum of the atomic masses of the atoms in its chemical formula. You have probably already learned about molar mass and how to calculate it. As a reminder, here is an example calculation:

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##### Example 3
```Molar Mass of Fe2O3 (iron(III) oxide, aka rust)
Fe: 55.845 g/mol and O: 15.9994 g/mol
2 × 55.845 g/mol + 3 × 15.9994 g/mol = 159.69 g/mol          159.69 g
The molar mass of Fe2O3 is ----------
1 mol
```

So one mole of iron(III) oxide has a mass of 159.69 g. Butane has a molar mass of 58.122 g/mol. The molar mass of a chemical is the proportion between the mass of a chemical and the number of moles of that chemical. Since moles are just a way of counting things in a (very big) group, this is a way of counting atoms and molecules by weighing the chemical. It is far easier to measure a chemical by mass than by counting atoms and molecules.

Chemical equations express the proportions that exist between the reactants and products in a chemical reaction. These proportions are based on numbers of atoms and molecules and not on mass. Mass is a quantity easy to measure in a lab and once the mass is known, it is a simple matter to convert the mass to the number of moles, which is equivalent to the number of molecules. Since chemical equations relate the number of atoms and molecules it is critical to convert masses to moles before using a chemical equation to determine how much of a reactant is needed to react with a given amount of some chemical. The following example shows how to use an amount of a reactant given in grams (a mass) to determine the mass of a chemical product which would be produced by reacting all of that chemical. The question asks about grams of carbon dioxide produced by burning a certain number of grams of butane. Remember, the key step is the one in the middle where the number of moles of butane is related to the number of moles of carbon dioxide using the ratio between these two chemicals in the balanced chemical equation. First, the number of grams of the reactant is converted to moles using the molar mass of that chemical. Then the molar ratio from the balanced equation is used to calculate the number of moles of the product. Finally, the moles of the product are converted to the mass using the molar mass of the product chemical.

##### Example 4

How many grams of carbon dioxide result from the burning of 392 grams of butane?

 A ``` 1 mol 39.2 g C4H10 × --------- = 0.674 mol C4H10 58.122 g ``` B ``` 8 mol CO2 0.674 mol C4H10 × ---------- = 2.70 mol CO2 2 mol C4H10 ``` C ``` 44.009 g 2.70 mol CO2 × --------- = 119 g CO2 1 mol ```
This could all be done in one calculation as shown below:
 ``` 1 mol 39.2 g C4H10 × --------- 58.122 g ``` ``` 8 mol CO2 × ---------- 2 mol C4H10 ``` ``` 44.009 g × --------- = 119 g CO2 1 mol ```

In step A the mass of butane in grams is changed to the number of moles of butane using the molar mass of butane. In step B the moles of butane is changed to moles of carbon dioxide using the molar ratio from the balanced chemical equation. In step C the moles of carbon dioxide is changed to the mass of carbon dioxide in grams. All of these calculation steps can be combined into one calculation, as shown above. This simplifies matters and can mean fewer entries into a calculator, which can help to avoid errors due to putting the wrong number into your calculator. Just remember to ‘multiply by the top and divide by the bottom’ as your learned doing dimensional analysis.

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### Exercises

The following exercises were written in the order given to help you to develop the skills necessary to master the topic of stoichiometry. You will need a calculator and a periodic table to complete them. You may also use your ions reference sheet.

#### Remember Dimensional Analysis

Dimensional analysis is an important skill to have when you are learning about stoichiometry. These problems are designed to help you to remember and practice this skill. Show all work using conversion factors and cancelling units as appropriate.

1. How many miles could you cover in eight hours of driving at an average speed of 55 miles per hour?
2. If 0.3048 m equals 1 foot and there are 5,280 feet in one mile, then how many meters is equal to 2 miles?
1. Say that grapes cost \$2.99 per pound at the grocery store. If one pound equals 0.4536 kilograms, then how much would it cost to buy 3.0 kg of grapes?
2. Hydrogen is a gas with a very low density. One liter of the gas weighs only 0.082 g. How many liters would you get if you had just 10.0 g of hydrogen gas?

#### Basic Stoichiometry

5C + 2SO2 CS2 + 4CO
Solid carbon reacts with sulfur dioxide to make carbon disulfide and carbon monoxide.
1. How many moles of carbon disulfide form when 5.0 mol of carbon react?
2. How many moles of carbon are needed to react with 4.0 mol of sulfur dioxide?
3. How many moles of carbon monoxide form at the same time that 0.50 mol of carbon disulfide form?
1. How many moles of sulfur dioxide are required to make 5.0 mol of carbon disulfide?
2. Calculate the number of moles of carbon disulfide and the number of moles of carbon monoxide that form when 30 mol of carbon react with enough sulfur dioxide.

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#### More Basic Stoichiometry

Use molar ratios found in this chemical equation to answer the following questions. Show your work for all calculations.

3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)
Aqueous calcium chloride reacts with aqueous sodium phosphate to precipitate solid calcium phosphate and to produce aqueous sodium chloride.
1. If you have 3 moles of CaCl2 then how many moles of Na3PO4 will you need if you react all of the CaCl2? Also, how many moles of the solid product, Ca3(PO4)2, will be formed?
2. Say that you find that 0.25 mol of Ca3(PO4)2 formed in an experiment. How many moles of CaCl2 were required to get this result? Also, how many moles of Na3PO4 were required?
1. Say that you need to generate 3.5 mol Ca3(PO4)2 using this chemical reaction. How many moles of each reactant do you need?
2. How many moles of both products would be made if you reacted 0.125 mol CaCl2 with enough of the other reactant to react all of the CaCl2?

#### Stoichiometry with Masses

2KClO3(l) 2KCl(s) + 3O2(g)
Potassium chlorate decomposes to form potassium chloride and oxygen gas.
N2(g) + 3H2(g) 2NH3(g)
Nitrogen and hydrogen gases react to form ammonia.
1. How many grams of oxygen can be obtained by the decomposition of 15 grams of potassium chlorate?

2. How many grams of potassium chloride are produced in the same reaction?
1. How many grams of ammonia would be produced by the reaction of 5.00 g of nitrogen with excess hydrogen?
2. How many grams of nitrogen must be used to produce 240 grams of ammonia? How many grams of hydrogen were used in the reaction? (There are two ways to calculate the answer to the second question: show both).

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#### More Stoichiometry with Masses

Tin(II) fluoride and hydrogen gas can be produced by the reaction of metallic tin with hydrogen fluoride gas.
Balanced Equation:
Sn(s) + 2HFSnF2(s) + H2(g)
Limestone (calcium carbonate) can be decomposed into carbon dioxide and calcium oxide by heating to high temperature.
Balanced Equation:
CaCO3(s) → CO2(g) + CaO(s)
1. How many grams of tin(II) fluoride can be produced by the reaction of 128 grams of hydrogen fluoride with an excess amount of tin?
2. How many grams of hydrogen would be produced by the same reaction?
3. Is the mass of tin(II) fluoride after the reaction more than, less than, or equal to the mass of tin before the reaction? Why?
4. What mass of tin(II) fluoride is produced by reacting 115 g of tin with excess hydrogen fluoride?
5. What mass in grams of tin is required to make 10.0 g of hydrogen gas?
1. As calcium carbonate is heated, does the mass of the solid material in the container increase, decrease or stay the same? Why?
2. How many grams of calcium oxide are produced by the complete decomposition of 1,000 grams of limestone?
3. How many grams of carbon dioxide are produced at the same time?
4. What is the sum of the amount of carbon dioxide and calcium oxide produced by the complete decomposition of 1,000 g of limestone? What should it be? Why?
5. How many grams of calcium carbonate are required to produce 1,000 g of calcium oxide?
Do these problems together as a class: Stoichiometry Start-up
Here is the homework for this activity.
A related activity about Limiting Reagent and Percent Yield and the homework to go with it.