Stoichiometry is a big long word which denotes the general process of figuring out how much stuff you need to make something. For example, if you are going to build 100 cars, how many wheels do you need? This is a (nonchemical) stoichiometric problem. We could solve it using Unit Factor Analysis by introducing the wheel/car ratio:

(4 wheels/car)

Then

wheels = 100 cars (4 wheels/car) = 400 wheels

If we want to know how many lug nuts we need, this becomes

nuts = 100 cars (4 wheels/car)(5 nuts/wheel) = 2000 nuts

Chemical stoichiometry problems are not more complicated than this provided that you become familiar with three new hotdogs:

Name | Unit |
---|---|

Mole Ratio | ( X moles A / Y moles B ) |

Formula Weight | ( Z grams A / mole A ) |

Normal Gas Volume | ( 24.4 L gas / mole gas ) |

What is a mole? Well, when we first introduced balanced chemical reactions, we talked in terms of parts and I promised that we would get more specific when the need arose. The need just arose. Conside the balanced chemical reaction for the fermentation of glucose:

C_{6}H_{12}O_{6}(s) --->
2 CO_{2}(g) + 2 C_{2}H_{5}OH(l)

The "2's" in front of CO_{2} and C_{2}H_{5}OH are
called the *stoichiometric coefficients*. This may seem like a long
and cumbersome word, but it is certainly shorter than saying,
"the numbers in front of each substance in a balanced chemical equation!"
The *mole ratio* is simply the ratio of two stoichiometric
coefficients from the same balanced equation. You may have to think about
that a bit before you realize how simple it is.

What is the stoichiometric coefficient of C_{6}H_{12}O_{6}
in this equation? If a number is not explicitely given, it is implicitely
understood to be "1." Now, the mole ratios for this equation are:

(2 moles CO_{2} / 1 mole C_{6}H_{12}O_{6})

(2 moles C_{2}H_{5}OH / 1 mole C_{6}H_{12}O_{6})

(2 moles C_{2}H_{5}OH / 2 moles CO_{2})

Which ratio you need to use depends on the question you are trying to answer.

Let's work some simple problems.

- How many moles of glucose are needed to produce 25 moles of ethanol?

moles glucose = 25 moles ethanol (1 mole glucose / 2 moles ethanol) = 12.5 moles glucose

So whenever you want to make 25 moles of ethanol, you will need 12.5 moles of glucose. But how do you measure out 12.5 moles of glucose? That's where the second hotdog comes in.

The conversion factor from moles to grams is called the formula weight. You get that from looking up the atomic weight for each atom in the formula and adding them all up. If you look on the Periodic Table you will find the following atomic weights:

- Carbon: 12.011 grams/mole
- Hydrogen: 1.008 grams/mole
- Oxygen: 16.000 grams/mole

6*12 + 12*1 + 6*16 = (180 grams glucose / mole glucose)

For carbon dioxide and ethanol, we have:

1*12 + 2*16 = (44 grams CO

2*12 + 6*1 + 1*16 = (46 grams ethanol/ mole ethanol)

We now have everything we need to answer the very important question:

- How many grams of glucose are needed to produce 1000 grams of ethanol?

grams glucose = 1000 grams ethanol (1 mole ethanol/46 grams ethanol)

(1 mole glucose/2 moles ethanol)(180 grams glucose/1 mole glucose)

= 1000*(1*1*180)/(46*2*1)

= 1956 grams glucose

This is a typical stoichiometry problem. Can you answer the question:

- What volume of carbon dioxide is produce when 100 grams of glucose is fermented to ethanol?

To answer this question we need the third hotdog: (24.4 L of gas/ mole of gas). It turns out that under normal conditions (25 Centigrade and normal atmospheric pressure) one mole of gas occupies approximately 24.4 L no matter what the identity of the gas is. The volume depends on temperature and pressure. You may run across the value 22.4 L/mole for gases at 0 Centigrade and standard atmospheric pressure. And there is a relationship called the Ideal Gas Equation which allows you to calculate the volume for a wide range of conditions. But for this course you only need to remember that under the same conditions all gases occupy approximately the same volume and at room temperature and normal pressure this value is 24.4 L. How big is this volume? About a dozen 2L soft drink bottles.

Now to answer the question:

L carbon dioxide = 100 grams glucose (1 mole glucose/180 grams glucose)

(2 moles carbon dioxide/ 1 mole glucose)

(24.4 L carbon dioxide / 1 mole carbon dioxide)

= 100*(1*2*24.4)/(180*1*1) = 27 L carbon dioxide

No wonder you have to leave the cap loose when you brew mead!

Here are some important tips for working stoichiometry problems:

- Resist the temptation to use short cuts, UAYF!
- Always use "moles A," "moles B," never "moles" alone
- Only cancel "moles A" with "moles A" not "moles B"
- Only use 24.4 L/mole for
*gases*

Just as a factory owner is concerned to have the right number of wheels and nuts available for the number of cars he is making, a chemist must be sure to mix reactants in the proper amounts for his purposes. If he uses more of one reactant than is needed, that reactant will be left over and wasted at the end. Sometimes this simply means that he has wasted some of his chemicals. But other times it can cause a different reaction from the one intended to take place. This will be particularly important when we talk about gunpowder and acids.

For millenia alchmists, tradesmen, and later chemists struggled with the general stoichimetric problem: "How much A do I need to react with a given amount of B." Up until the dawn of the Nineteenth Century, there was no satisfactory answer. Relative amounts were determined by trial and lots of error. The struggle to form a satisfactory theoretical foundation is described in detail in From Caveman to Chemist and is worthy of your time and attention. For our practical use, however, we can simply say that a mole is the unit of chemical amount. When you need to know how much of a chemical to use, the mole is the unit of choice. Since we don't have a direct way of measuring this elusive quantity, we need some conversion factors. The formula weight converts moles to grams, and the normal gas volume converts moles of gas to liters of gas.

When you are ready, I will give you a single stoichiometry problem to work by Unit Factor Analysis.
I will expect that you have memorized the table of unit factors from the
Unit Factor Analysis page. In addition,
I will give you a balanced chemical equation.
I may ask for the number of grams needed or produced or I may ask for the volume
of gas needed or produced. You might need to convert to pounds or cubic feet
or whatever units are discussed in the Unit Factor Analysis page.
You will work
this problem without notes, but you may use a calculator. If you do not get the correct
answer, you fail. You may, however, keep taking the test (one per day) until you pass.
*Of course* the problems will be different from day to day.

- Take a practice quiz online.