Chapter Four, Section One | |

**OBJECTIVES**

- Learn how to determine the formula weight of compounds.
- Understand the meaning of the word "mole" and learn how to apply its meaning.
- Learn how to balance chemical equations.

**I. The Mole**

The mole is nothing more than the basic counting unit in chemistry just as the dozen is the basic counting unit in eating donuts. The number is

N_{0} = 6.022045 x 10^{23} "particles"/mole

Where "particles" can be atoms, molecules, ions, or whatever. This number is called Avogadro's number to honor Avogadro (a Swede) for his many contributions to science. This number is so huge that it is difficult to imagine. One mole of marbles would
cover the entire Earth to a depth of 50 miles (where, by definition, space begins)! But because atoms and molecules are so small, this number is a suitable quantity with which to work. A mole of something contains the same number of "particles" as
there are atoms in 12.00000 g of ^{12}C. The number is not a whole number because we are looking at averages of measurements of such a very large number of "particles". Usually 6.022 x 10^{23} is of sufficient accuracy for us so
**memorize** this number.

So using this standard, N_{o} atoms of C-12 will weigh 12.0 g which is one mole of C-12. Similarly one mole of F atoms weighs 19.0 g containing the same number of atoms as one mole of C-12, but of course it weighs more because each atom of F
weighs more than each atom of C-12. The mass of one mole of atoms is called the molar mass and is equal to the atomic mass (or weight) in amu with the units of g/mol. So 6.022045 x 10^{23} amu is equal to 1 gram (an exact number).

So the molar mass of O is 16.0 g/mol, that of F is 19.0 g/mol, that of Pb is 207.2 g/mol, etc.

Let's work some examples to clarify this concept:

*How many grams are in 2.50 moles of sulfur?*

2.50 mol S * (32.06g/(mol S)) = 80.2 g (the mol S cancel in denominator and numerator).

*How many moles are in 1.00 pound of Si?*

1.00 lb Si * (454 g/lb) * (1 mol Si/(28.1 gm Si)) = 16.2 mol Si

*How many atoms are in 1.00 pound of Si?*

16.2 mol Si * (6.022 x 10^{23} atom Si/(mol Si)) = 9.76 x 10^{24} atoms

*How much does 1 atom of Si weigh?*

28.1 gm Si/(mol Si) * (1 mol Si/(6.022 x 10^{23} atoms Si)) = 4.67 x 10^{-23} gm

**II. Formula and Molecular Weight**

A molecule is a collection of atoms in a fixed ratio. For example, a molecule of water contains 1 oxygen atom and 2 hydrogen atoms. Now what about 14 molecules of water? Well 14 molecules of water would contain 14 oxygen atoms and 28 hydrogen atoms. Similarly 1 million molecules of water would contain 1 million oxygen atoms and 2 million hydrogen atoms. See where we are going with this? One mole of water will contain one mole of oxygen atoms and 2 moles of hydrogen atoms. So what would be the mass of one mole of water?

1 mole of oxygen atoms = 16.00 gms

2 moles of hydrogen atoms = 2.016 gms

So one mole of water = 16.00g + 2.016 g = 18.02 g.

This is called the molar mass of the molecule (or the molecular weight). This is also the formula weight. Strictly speaking the molecular weight only refers to molecules, whereas the formula weight can apply to either ions or molecules. Then we can get the following table (be sure that you understand how):

compound | MW(amu) | Molar Mass(g/mol) | Mass of 1 molecule(g) |
---|---|---|---|

NH_{3} | 17.04 | 17.04 | 2.830 x 10^{-23} |

H_{2}O | 18.02 | 18.02 | 2.992 x 10^{-23} |

CH_{2}Cl_{2} | 84.93 | 84.93 | 1.410 x 10^{-23} |

Let's do some examples:

What is the molar mass of caffeine (C_{8}H_{10}N_{4}O_{2})?
caffeine

8 moles C * (12.011 g/(mol C)) = 96.088 g

10 moles H * (1.0079 g/(mol H)) = 10.079 g

4 moles N * (14.0067 g/(mol N)) = 56.0268 g

2 mol O * (15.9994 g/(mol O)) = 31.9988 g

And the total is 194.193 g which is the molar mass of caffeine.

A can of artificially sweetened drink may contain 70 mg of aspartame(Nutra Sweet). How many moles is this? The formula is C_{14}H_{18}N_{2}O_{5}

First let's calculate the formula weight so that we can then calculate how many moles .070 g of the substance is.

C: 14 moles * (12.011 g/mol) = 168.154 g

H: 18 moles * (1.0079 g/mol) = 18.1422 g

N: 2 moles * (14.0067 g/mol) = 28.0134 g

O: 5 moles * (15.9994 g/mol) = 79.997 g

This gives the molar mass to be 294.307 g/mol. So now we can calculate the number of moles:

.070 g/(294.307g/mol) = 2.4 x 10^{-4} mol.

How many molecules is this?

2.4 x 10^{-4} mole * (6.022 x 10^{23} molecules/mol) = 1.4 x 10^{20}molecules.

**III. Chemical Equations**

We have already seen that a chemical change is a change in which the atoms in the molecules are rearranged in some way. For example if methane is burned the result is the formation of carbon dioxide and water and the methane molecule is no more. Chemical equations afford us a way of keeping track of what we started with and with what we end. We can write such an equation in general as

R_{1} + R_{2} P_{1} + P_{2}

where the R's are the reactants and the P's are the products. This equation will tell us nothing about the time required to go from reactants to products, how we get from reactants to products, or how complete the reaction is (may still have a significant amount of reactants left after a reasonable time). So the chemical reaction just gives us the "big picture". Many details are lacking and some of the details will be discussed when we study Chemical Kinetics. The reaction may also give off (or absorb) heat and we can indicate that as:

R_{1} + R_{2} P_{1} + P_{2} + heat

or

R_{1} + R_{2} P_{1} + P_{2} - heat

It is extremely important to understand that the total mass of all reactants entering into a chemical reaction must equal the total mass of all the products formed in the reaction. We can't create mass by a chemical reaction! We make sure that this
mass balance occurs by *balancing* the chemical equation which means that we must have the same number of each type of atom on each side of the equation -- thus ensuring mass balance.

So let's look at some examples.

H_{2}(g) + O_{2}(g) H_{2}O(l)

I have also introduced another terminology with the (g) and (l -- this is a small "L") which tell us the *physical state* of the substance (solid, liquid, or gas). Does this equation satisfy our mass balancing requirement? We have two hydrogen
atoms on the left and two on the right, but we have two oxygen atoms on the left (remember that oxygen and hydrogen exist as a diatomic) and only one oxygen atom on the right. So we need to *balance* the equation. This one is easy for we can put a
two in front of the water on the right and a two in front of the hydrogen on the left and the equation is balanced. The numbers that we put in front of the atoms or compounds are called the *stoichiometric coefficients*. **You can't change the
subscripts for if you do you are changing what the substance is and you don't have that kind of power!**

2H_{2}(g) + O_{2}(g) 2H_{2}O(l)

We can check the masses and we see on the left (2*(2.02g h_{2}) + 32.00 g O_{2}) = 36.04 g reactants and on the right (2*(18.02 g H_{2}O)) = 36.04 g products. So the equation is mass balanced.

So let's balance some more equations and perhaps learn some other concepts.

Al(s) + Br_{2}(g) Al_{2}Br_{6}(s)

This is not balanced so think about how to balance it before looking at how I do it.

This is balanced as follows:

2Al(s) + 3Br_{2} Al_{2}Br_{6}

Check it out yourself.

Here is one a little harder:

C_{4}H_{10}(g) + O_{2}(g) CO_{2}(g) + H_{2}O(g)

This one is obviously not balanced so play with it a little to see what you can do toward balancing it.

The easiest way to balance it is to not think that you have to use integers all the time. I first balance the atoms which only appear once on each side (C and H in this case). Attacking C first is done by putting a 4 in front of CO_{2}. Then I
balance the hydrogens by putting a 5 in front of H_{2}O so that now I have balanced the C's and H's but now I have 13 oxygen atoms on the right and 2 on the left.

C_{4}H_{10}(g) + O_{2}(g) 4CO_{2}(g) + 5H_{2}O(g)

Now I put 13/2 in front of the O_{2} on the left and the equation is balanced!

C_{4}H_{10}(g) + 13/2O_{2}(g) 4CO_{2}(g) + 5H_{2}O(g)

Even though this is a perfectly legitimate way to end this problem, we just don't like to see fractions sitting in the equation. So we multiply both sides by 2 and then we have the neat result:

2C_{4}H_{10}(g) + 13O_{2}(g) 8CO_{2}(g) + 10H_{2}O(g)

We did that so well, let's try another one:

B_{4}H_{10}(s) + O_{2}(g) B_{2}O_{3}(s) + H_{2}O(g)

I will start here with the elements which only appear once on each side (B and H). We can balance the B by putting a two before the B_{2}O_{3}(s). Now I see that if I put a 5 in front of the H_{2}O the H's will be balanced which
leaves me with my ol' nemesis oxygen. I see that I have 11 oxygens on the right and 2 on the left. Want to guess what I am going to do? I will put 11/2 in front of the O_{2} to balance the equation:

B_{4}H_{10}(s) + 11/2O_{2}(g) 2B_{2}O_{3}(s) + 5H_{2}O(g)

And I will leave this in a neat form by multiplying both sides by 2 to give:

2B_{4}H_{10}(s) + 11O_{2}(g) 4B_{2}O_{3}(s) + 10H_{2}O(g)

Now you need to **practice, practice, practice**. You have homework problems to work and the site at the top of the page has a lot of equations on which to practice.

- Now you are ready to practice some of these problems.
- I also recommend practicing balancing equations.
- Here is an excellent Tutoring Site for this topic. It is for high school chemistry and AP chemistry, but don't be insulted for it gives excellent tutoring help and practice problems for the novice.

After you have studied this material and practiced some problems, take quiz one. If you score at least 80 on the test then you are ready to continue to the next section.

*Web Author: Dr. Leon L. Combs*
*Copyright ©2001 by Dr. Leon L. Combs - ALL RIGHTS RESERVED*