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Group Activity: Stoichiometry


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 + 13O2 Arrowsngl 8CO2 + 10H2O

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

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.

chocolate-chip-cookie (39K)

In principle none of this is any different from making a batch of cookies from scratch. All the ingredients are present in a particular proportion: if you add too much sugar or too little flour or too much vanilla or too few chocolate chips then the cookies don’t come out right. If you double the recipe (2 mol of butane instead of one in the chemical equation ‘recipe’) then you double all the other ingredients and you double the amount of your product. You could even multiply the recipe by a number like 3.19 and, as long as there were no eggs to divide, you could make all the ingredients come out in proportion.

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

Using Masses

It should be clear why it is that chemists prefer not to try to measure out individual molecules in the lab. It’s just not possible. But why are moles any better? Moles are better because they can be measured 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. If you are reading this then you have probably already learned about molar mass and how to calculate it. Even so, here is an example of how to calculate molar mass:

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. It is far easier to measure a chemical by mass than by counting atoms and molecules. Moles enable a chemist to correctly understand the proportions of reactants and products in chemical reactions. These proportions are based on numbers of atoms and molecules and not on mass. Yet by measuring masses of chemicals and converting these masses mathematically to numbers of moles the chemist can use a chemical equation to determine how much of each chemical is required for, or produced by, a reaction.

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Example 4

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

              1 mol       
 392 g C4H10 × ---------  = 6.74 mol C4H10
              58.122 g
                8 mol CO2
6.74 mol C4H10 × ---------- = 27.0 mol CO2
                2 mol C4H10
               44.009 g       
27.0 mol CO2 × ---------  = 1.19 × 103 g CO2
                 1 mol
This could all be done in one calculation as shown below:
             1 mol
392 g C4H10 × ---------
             58.122 g
    8 mol CO2
 × ----------
    2 mol C4H10
    44.009 g       
 × ---------  = 1.19 × 103 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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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

  1. How many miles could you cover in eight hours of driving at an average speed of 55 miles per hour?

  2. Mike raised 75 goats. To maximize his return he entered into a series of possibly dubious trades. He turned in all the goats for cows at an exchange rate of 3 goats for 2 cows. Next, he exchanged all the cows for hogs at a rate of 5 cows for 7 hogs weighing 215 lbs. each. He sold all the hogs at a market price of $74.50 per 85.00 lbs. How much money did he earn from the 75 goats?
  1. Tigers may hunt in territories of up to 30 square miles. How many tigers could live in an area of 40,000 hectares? Assume that tiger territories do not overlap. (1 hectare = 10,000 m2)
  2. The dehumidifier tank in your basement fills up once every 24 hours. If the tank can hold 2.5 gallons, determine how many molecules of water the dehumidifier collects from the air per second. (Hints: The density of water is 1.0 g/ml; the molar mass of water is easy to calculate; 3.78 L = 1 gal).

Basic Stoichiometry

5C + 2SO2 Arrowsngl CS2 + 4CO
Solid carbon reacts with sulfur dioxide to make carbon disulfide and carbon monoxide.
  1. How many moles of carbon disulfide form when 2.7 mol of carbon react?
  2. How many moles of carbon are needed to react with 5.44 mol of sulfur dioxide?
  3. How many moles of carbon monoxide form at the same time that 0.246 mol of carbon disulfide form?
  1. How many moles of sulfur dioxide are required to make 118 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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Basic Stoichiometry with Reactions in Words

Write a balanced chemical equation for each problem. Then use stoichiometry to answer the question. (Hint: you are being given an opportunity to practice your nomenclature skills).
  1. How many moles of hydrochloric acid (HCl) can be produced from 6.0 mol of chlorine gas reacting with hydrogen gas?
  2. How many moles of magnesium hydroxide are produced when 5 moles of magnesium oxide react with water?
  1. How many moles of oxygen gas will be required to react with iron to make 5.1 moles of iron(III) oxide? (Pure metals have a monatomic formula).
  2. When a butane lighter burns in oxygen the products are carbon dioxide and water. How many moles of butane (C4H 10) were burned if 3.7 × 10-3 mol of carbon dioxide were collected?

Stoichiometry with Masses

2KClO3 Arrowsngl 2KCl + 3O2
Potassium chlorate decomposes to form potassium chloride and oxygen gas.
N2 + 3H2 Arrowsngl 2NH3
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:

Limestone (calcium carbonate) can be decomposed into carbon dioxide and calcium oxide by heating to high temperature.
Balanced Equation:

  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?
  1. As calcium carbonate is heated, does the mass of the solid increase, decrease or stay the same? Why?
  2. How many grams of calcium oxide are produced by the complete decomposition of 1 kilogram 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 kg of limestone? What should it be? Why?

Stoichiometry with Density

  1. Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide:
    C6H12O6 Arrowsngl 2C2H5OH + 2CO2
    Starting with 500.4 g of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process? (Density of ethanol = 0.798 g/mL.)
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.
See also some additional problems for Stoichiometry.
Credit for some problem set-ups is due to Nathan Guerin.
Last updated: May 17, 2016 Home