Atomic Mass | Isotopes | Atomic Weight |

The Mole | Molecular Weight | Avogadro's Constant |

Converting Grams Into Moles |

The units in which the mass of an atom are expressed are **atomic
mass units**. At one time, the lightest atom was assigned a
mass of 1 amu and the mass of any other atom was expressed in
terms of this standard. Today atomic mass units are defined in
terms of the ^{12}C isotope, which is assigned a mass of
exactly 12.000... amu.

**Isotopes** are atoms of the same element with different
numbers of neutrons, such as the ^{20}Ne and ^{22}Ne
isotopes of neon or the ^{35}Cl and ^{37}Cl
isotopes of chlorine.

The** atomic weight** of an element is the weighted average
of the atomic masses of the different isotopes of an element.
Naturally occurring carbon, for example, is a mixture of two
isotopes, ^{12}C (98.89%) and ^{13}C (1.11 %).
Individual carbon atoms therefore have a mass of either 12.000 or
13.03354 amu. But the average mass of the different isotopes of
carbon is 12.011 amu.

The **molecular weight** of a compound is the sum of the
atomic weights of the atoms in the molecules that form these
compounds.

Example: The molecular weight of the sugar molecule found in
cane sugar is the sum of the atomic weights of the 12 carbon
atoms, 22 hydrogen atoms, and 11 oxygen atoms in a C_{12}H_{22}O_{11}
molecule.

12 C atoms = 12(12.011) amu = | 144.132 amu |

22 H atoms = 22(1.0079) amu = | 22.174 amu |

11 O atoms = 11(15.9994) amu = | 175.993 amu |

342.299 amu |

C_{12}H_{22}O_{11} has a molecular
weight of 342.299 amu. A mole of C_{12}H_{22}O_{11}
would have a mass of 342.299 grams.This quantity is known as the **molar
mass**, a term that is often used in place of the terms *atomic
weight*** **or

The term *mole* literally means a small mass. It is
used as the bridge between chemistry on the atomic and
macroscopic scale. If the mass of a single ^{12}C atom is
12.000 amu, then one mole of these atoms would have a mass of
12.000 grams. By definition, a**a mole of any substance
contains the same number of elementary particles as there are
atoms in exactly 12 grams of the **^{12}**C
isotope of carbon. **

Example: A single ^{12}C atom has a mass of 12 amu,
and a mole of these atoms would have a mass of 12 grams.

A mole of any atoms has a mass in grams
equal to the atomic weight of the element. The term mole can be
applied to any particle: atoms, a mole of atoms, a mole of ions,
a mole of electrons, or a mole of molecules. Each time we use the
term, we refer to a number of particles equal to the number of
atoms in exactly 12 grams of the ^{12}C isotope of
carbon.

Practice Problem 1:Predict the mass of a mole of magnesium atoms. |

Practice Problem 2:Calculate
the molecular weights of carbon dioxide (CO |

**Avogadro's number **(or Avogadro's constant) is the
number of elementary particles in a mole of any substance For
most calculations, four significant figures for Avogadro's
constant are enough: 6.022 x 10^{23}.

A mole of any substance contains
Avogadro's number of elementary particles. It doesn't matter
whether we talk about a mole of atoms, a mole of molecules, a
mole of electrons, or a mole of ions. By definition, a mole
always contains 6.022 x 10^{23} elementary particles.

Practice Problem 3:Describe the difference between the mass
of a mole of oxygen atoms and a mole of O |

Practice Problem 4:Calculate
the mass in grams of a single |

The mole is a powerful tool, which enables chemists armed with nothing more than a table of atomic weights and a balance to determine the number of atoms, ions or molecules in a sample.

Example: Let's calculate the number of C_{12}H_{22}O_{11}
molecules in a pound of cane sugar.

We need two pieces of information to do this calculation. We need to know the number of grams in a pound and the molecular weight or molar mass of this sugar. A pound of sugar has a mass of 453.6 grams and one mole of this sugar has a mass of 342.3 grams.

A pair of unit factors can be constructed from the molecular weight or molar mass.

By paying attention to the units during the calculation, it is easy to choose the correct unit factor to convert grams of sugar into moles of sugar.

We then use Avogadro's number to determine the number of C_{12}H_{22}O_{11}
molecules in our sample.

Practice Problem 5:What is the formula of magnesium chloride if 2.55 grams of magnesium combines with 7.45 grams of chlorine to form 10.0 grams of this compound? |