In 1961 (the year that I graduated with a B.S. degree in Chemistry) the modern system of atomic masses was established based upon ^{12}C as the standard with ^{12}C assigned a mass of exactly 12 atomic mass units (amu). The masses of all
other atoms are defined relative to this mass by using a mass spectrometer. Using a mass spectrometer, the ratio of the mass of ^{13}C to that of ^{12} was determined to be 1.0836129 and so the amu of ^{13}C is

1.0836129 * 12 = 13.003355 amu (remember "12" is exact)

Similarly the amu of any element can, and has been, determined as you see the mass units in the periodic table. For the electron, proton, and neutron, the numbers are as follows:

particle | weight in grams | amu |
---|---|---|

electron | 9.1083 x 10^{-28} | 0.0005483 |

proton | 1.6726 x 10^{-24} | 1.007276 |

neutron | 1.6750 x 10^{-24} | 1.008665 |

Ignoring the electron (why?), we can estimate the mass of an atom in amu by letting the proton be 1 amu, the neutron be 1 amu, and then summing over the number of protons and neutrons. For example, consider oxygen which has 8 protons and 8 neutrons.
Then what is the mass number (A)?
Yes, it is 16 (8*1 + 8*1) = 16amu. Therefore we can write oxygen as ^{16}O.

These are the whole numbers that you see in the table. But you know the situation is more complicated than this for the actual masses. The situation is more complicated for two reasons ; the mass defect and the existence of isotopes.

**Consider first the mass defect.** If we sum over all of the masses of the 26 electrons, 26 protons, and 32 neutrons for iron, we find that the sum is 58.48072 amu. The experimental value is 57.933272 amu and the difference is the mass equivalent
of the energy of binding the particles in the nucleus -- the mass defect.

**Now consider the isotope problem.** You have already learned about isotopes so you know that hydrogen has three isotopes and oxygen has three isotopes (O-16, O-17, and O-18). Each isotope of an element occurs with a __percent abundance__ (PA).
To get the atomic mass of an element we need to perform the following arithmetic:

atomic mass = SUM(PA/100)_{i}*(isotopic mass)_{i} where SUM means to sum over all isotopes. For example, ^{35}Cl has an exact mass of 34.96885 and a PA of 75.77 and ^{37}Cl has an exact mass of 36.96590 and a PA of
24.23. So the atomic mass of Cl would be

0.7577*34.96885 + .2423*36.96590 = 35.45 amu

This is the number that you see in the periodic table under the element Cl.

Similarly for oxygen:

0.9889*12 + 0.0111*13.0034 = 12.01 amu

When we work with such atoms in doing calculations, we will use this average atomic mass even though no single Cl atom exists with a mass of 35.45 amu. How can we do this? As you will see later, we are actually working with very large numbers of atoms and the average atomic mass of all of the Cl atoms will be 35.45 amu. Traditionally these numbers are also called the atomic weights.

We can also calculate the **isotopic abundances**. Let's work with Boron for an example. Boron has two isotopes, ^{10}B and ^{11} with exact masses of 10.012939 and 11.009305 respectively. We also know that the atomic weight (or
mass) of boron is 10.811. So what are the isotopic abundances of each?

Now you need to use a little algebra to set up the problem:

X * (10.012939) + Y * (11.009304) = 10.811

and X + Y = 1

So that we have two equations and two unknowns. Of course the X + Y = 1 equation came from knowing that when we have only two fractions of each, the sum of the two fractions must equal one. So just solve these two equations for X and Y and you will have the fractional abundances which any child on the street can change into percent abundances. Solve the second equation for Y = 1 - X and substitute it into the first equation gives

X = (0.198)/(0.996366) = 0.199

Y = 1 - X = 0.801

Therefore the percent abundance of B-10 is 19.9 and the percent abundance of B-11 is 80.1.

**Practice this with homework problems so that you can do it by yourself with no notes and no book.**

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 (I will quit using the subscript 0 because I don't want to type the HTML code a zillion times) 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

Now you are ready to practice some of these problems

Now take a practice quiz to help you understand if you understand the basic concepts. |

You must use your real name when it asks for a name. |

The test will only submit when you have answers all of the questions correctly. |

If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests. |

*Web Author: Dr. Leon L. Combs*
*Copyright ©1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED*