This activity is designed to introduce a convenient unit used by chemists and to illustrate uses of the unit.

**Part I: What Is a Mole And Why Are Chemists Interested in It?**

Counting things is a normal part of everyday life. How many days left until vacation? How many eggs do I need for the recipe? If large numbers of things are involved, we use grouping

strategies to make the numbers easier to manage. For example, 4 more weeks until vacation, tells

us that there are twenty-eight days. One dozen eggs is the common way of expressing the quantity 12. Half of a dozen of anything would be 6 units. One gross is 144 items (12 dozen) and a ream of paper contain 500 sheets.

Chemist are faced with a unique problem when dealing with numbers of atoms or molecules. The particles are so small that any amount of them that we are able to physically handle contains a number of particles so large that there is nothing else in our experience that contains so many units. This incredibly large number calls for a special counting group – the MOLE.

**A** **MOLE** **is** **6.022** **x** **1023** **particles**. This is often referred to as Avogadro=s number. Let=s make sure we understand how big this is. One mole of the element carbon has a mass of 12.01 grams. The smallest particle of an element is an atom. So one mole of carbon contains 602,200,000,000,000,000,000,000 atoms of carbon.

Look for the element carbon on the periodic table. Do you notice anything special about the value 12.01? Explain

The mass of one mole of the element magnesium is 24.30 grams. How many atoms does a sample of magnesium with a mass of 24.30 grams contain?

Stated in general terms, the mass of one mole of any element is equal to the

of that element expressed in grams. The mass of a mole of any element can be found by looking on .

The mass of 0.5000 moles of carbon is and contains

atoms of carbon.

Remember when dividing numbers written in scientific notation the number portion is divided

normally and the exponents are subtracted. 6.022×1023 divided by 2 is the same as 6.022 x1023/2 x 100. So the answer is found by dividing 6.022 by 2 = 3.011 and the subtracting exponent 0 from exponent 23. The answer in scientific notation is 3.011×1023 atoms of carbon.

**Calculator tip**: for exponential notation use the EE or EXP key (not 10^)

If you have a bottle containing 8.10 grams of magnesium, how many Mg atoms are present in the bottle? Show your work. What is different about this problem compared to the last one involving carbon?

Remember that some elements, when alone, exist in the form of diatomic molecules: H2, O2, N2, I2, F2, Cl2, Br2, Their smallest piece is a molecule containing two atoms. If one mole of oxygen were required for an experiment you would be using O2 the gas. One mole of O2 would have a mass of

and contain particles (molecules).

The characteristic unit of the compound CO2 is a molecule. Each CO2 molecule has atoms. In order to find the mass of one molecule of CO2, it is necessary to add together the atomic masses of each of the atoms in the compound. Atomic masses are expressed in atomic mass units (a.m.u.)

C = 12.01 a.m.u. per atom x 1 atom = 12.01 a.m.u. O = 16.00 a.m.u. per atom x 2 atoms = 32.00 a.m.u.

44.01 a.m.u.

44.01 a.m.u x 6.022 x 1023 molecules CO2

x 1.66 x 10-24 g

= 44.00g CO2 (close!)

1 molecule CO2 1 mole CO2

1 a.m.u.

1 mole CO2

**The** **molar mass of any substance is the mass of 6.022 x** **1023** **units** **of** **the** **substance**. Find the molar masses of the following compounds. Show your work.

H2SO4 Al (NO3)3

52 Exploring the Chemical World, PGCC, 2003

If you know that a dozen tennis balls has a mass of 1.00 lbs, how would you find the mass of one tennis ball?

Using the same mathematical approach, what is the mass in grams of one magnesium atom? Show all work. What is the mass in grams of one molecule of H2SO4?

**Part II. Using Moles to Find Formulas**

Suppose you had exactly one mole of carbon, 12.01 grams and you chemically combined all of it with oxygen. In other words, you burned it completely.

Write the chemical equation for the reaction.

For every single atom of carbon, how many single atoms of oxygen are needed to form one molecule of the product carbon dioxide?

How many atoms of carbon are in one mole of carbon?

How many atoms of oxygen would be needed to form one mole of CO2?

What is the mass of the oxygen needed to react with one mole of carbon?

What would be the final mass of the CO2?

Let=s suppose that you did not know ahead of time what the formula for the carbon dioxide product was. Maybe when carbon burns it forms CO or CO3 or perhaps C2O3. If you had burned 12.01 grams of carbon, collected all the gas that was formed and found its mass to be 44.01 grams, you would then know that the mass of the oxygen that added to the carbon was

grams.

32.00 grams of oxygen x 1 mole of oxygen atoms = 2 moles of oxygen atoms

16.00 g of oxygen

This information can be used to figure out the formula of the gas.

1 mole of carbon and 2 moles of oxygen combined to form a compound. For every one atom of

carbon there are two atoms of oxygen. The formula could be CO2. **The** **subscripts** **in a formula** **tell** **us** **the** **relative** **number** **of** **moles** **of** **the** **elements** **in** **the** **compound**. Please note that the actual molecular formula could be C2O4 since the relative number of C to O in this compound is also 1:2.

If we know the masses of all of the elements that combined to form a compound and convert those masses to number of moles, and convert the values to whole numbers, we will know the relative number of moles for each element in the compound. These values are the subscripts in the **empirical formula** or the **simplest formula**.

In the following laboratory exercise, you will determine the empirical formula of the compound formed by the combination of the elements magnesium and oxygen. You will find the empirical formula in the way that we Afigured [email protected] the empirical formula of carbon dioxide, that is, by combining magnesium with oxygen and finding the mass of the oxide product. This is easier to do than finding the empirical formula of CO2 because the magnesium oxide is a solid. Because magnesium will react with nitrogen as well as oxygen, we cannot do the procedure in one step.

**PROCEDURE**

1. Before you begin reading the procedure- write down three pieces of data that you know you will have to collect during the activity in order to determine the empirical formula of magnesium oxide.

1.

2.

3.

2. During the experiment, on the data sheet provided, construct a data table, show calculations, and clearly indicate the results and your conclusion. Write balanced equations for the all chemical changes that occur.

3. Clean a crucible and cover. Place the crucible into a triangle on a ring and heat it in the hottest part of the Bunsen burner flame until bottom of crucible is [email protected] hot. Continue heating for another minute (this eliminates any volatile impurities). USING CRUCIBLE TONGS place the crucible and cover on a fireproof fiber board to cool. For our purposes, cool means being able to pick it up comfortably without using the tongs.

4. Find the mass of the heated crucible and cover.

5. Obtain a total of about 10 cm of Mg ribbon and clean it by stroking it with fine steel wool to remove any corrosion (oxidation product). Cut the ribbon into smaller pieces and place all of them into the crucible. Since the Mg tends to fly off as the ribbon is snipped it is recommended that you cut the ribbon at a sharp angle right over the crucible. Put the cover back on the crucible and re-mass.

6. Heat the covered crucible containing the Mg strongly for about **five** **minutes**. During the heating use the tongs to lift the cover off slightly every minute for a few seconds. This admits air into the crucible without allowing any oxide to escape. Next, remove the cover and heat strongly for another minute. Allow the crucible to cool. While heating, write any reactions for magnesium with oxygen and nitrogen on the data sheet.

7. Cool, replace cover, find final mass.

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