In 1811 Amedeo Avogadro proposed the volume of a gas is dependent on the number of atoms of gas at a fixed temperature and pressure. If more gas is added at this fixed temperature and pressure, the volume will increase due to more atoms being added. He proposed this was the same for all gases.

Avogadro’s proposal lay dormant for approximately 50 years when other scientists began to see the value of his work. **Namely the relation between the mass of an element and the number of atoms of the element in that mass**.

The number of atoms in a volume of gas was first determined by Johann Josef Loschmidt in 1865. Other attempts to determine the number followed.

Atomic weights were known and it was decided to determine the number of atoms in the mass of an element equal to it’s atomic weight. For example, how many helium (He) atoms would be in 4.0026 g grams of He? 4.0026 is the atomic weight of He. The experimentally determined number has changed over the years as measurement methods improved. The currently accepted value is 6.022 X 10^{23}. This is a very large number . Written out it would be 602200000000000000000000. The exponent representation (6.022 X 10^{23}) is more convenient to use. (See http://www.mathsisfun.com/exponent.html or http://www.mathgoodies.com/lessons/vol3/exponents.html for help with exponents).

Other examples. There are 6.022 X 10^{23} atoms of C in 12.0107 grams of C. 12.0107 is the atomic weight of C. There are 6.022 X 10^{23} atoms of Al in 26.9812 grams of Al. 26.9816 is the atomic weight of Al.

In 1897 the term **“mole”** was proposed as another way to express this fixed number or atoms. So a mole of carbon ( C ) would have a mass of 12.0107 g of mass and contain 6.022 X 1023 atoms of carbon.

The number 6.022 X 1023 was given the name **Avogadro’s Number** by Jean Perrin in 1909.

See the video What is a Mole

**The mass of an element equal to its atomic weight would be 1 mole of that element and that mass would have 6.022 X 10 ^{23} atoms of that element.**

Recalling the Math Skills section of the Units tab, there is a mass-mole conversion factor for each element. For carbon ( C ) the conversion factor is:

12.0107 g C= 1 mole C

For calcium (Ca) it is 40.078 g Ca = 1 mole Ca

The unit mole is comparable to the unit dozen. The mole represents a number. A dozen represents 12 of something. A mole represents 6.022 X 10^{23} of something.

A dozen eggs would be 12 eggs, likewise a dozen doughnuts would be 12 doughnuts. A mole (Fe) of iron would be 6.022 X 10^{23} atoms of iron, likewise a mole of electrons would be 6.022 X 10^{23} electrons.

Just as one could have 0.5 (1/2) dozen eggs, one could have 0.5 (1/2) mole of copper (Cu). Obviously it is possible to have any fraction or any multiple of a mole of a chemical.

**Calculations involving moles**

**Elements**

A piece of copper (Cu) weighs 15.886 g. How many moles of Cu would be in this mass? How many atoms of Cu would be in this mass?

From the Periodic Table, the atomic weight of Cu is 63.546 g/mol. Recalling the Math Skills section of the Units tab, the conversion factor for copper (Cu) is 63.546 g Cu = 1 mole Cu.

In this problem we want moles to be the final unit, the conversion factor would be.

1 mole/63.546 g

and the calculation would be:

14.886 X 1 mole/63.546 g = 0.250 moles

The number of atoms of Cu would be:

0.250 moles x 6.022 X 10^{23} atoms/mole = 1.506 x 10^{23} atoms

The conversion between grams and moles is the most frequent calculation that will be done in this class. Rarely will the number of atoms be calculated. I**t is important however to establish that a mole represents a number of entities**. One can talk about a mole of protons, a mole of electrons, a mole of atoms, a mole of ions (to be defined later) and a mole of molecules (to be defined later).

**Compounds
**Compounds are combinations of elements. The basic unit of a compound can either be an ionic unit or a molecular unit depending on the type of bonding between the atoms of the compound. The distinction between the two will be presented in the Topics Compound section of this website. The purpose of this section is

**to show how to calculate**the formula mass (ionic bonding) or the molecular mass (covalent bonding) of any compound.

The mass of a formula or molecular unit would be the sum of the masses of the number of atoms that in that formula or molecular unit.

** ****Examples
**

**NaCl**

The formula for sodium chloride is NaCl. A unit of sodium chloride is made of one atom of sodium (Na) and 1 atom of chlorine (Cl). The atomic mass of Na is 22.989 g/mole and the atomic mass of Cl is 35.45 g/mole. Adding these two values together give the formula mass of 58.44 g/mole for NaCl.

** ****Mg _{3}N_{2
}**The formula for magnesium nitride is Mg

_{3}N

_{2}. This compound has three (3) magnesium (Mg) atoms and two (2) atoms of nitrogen (N) atoms in each formula unit. To calculate the formula mass, the atomic mass of Mg (24.305 g/mole) is multiplied by 3 and added to the atomic mass of N (14.007 g/mole) multiplied by 2.

Formula mass Mg_{3}N_{2} = 3 x 24.305 g/mole + 2 x 14.007 g/mole = 100.929 g/mole

**Ca(NO _{3})_{2
}**The formula for calcium nitrate is Ca(NO

_{3})

_{2}

**.**In this example the there are two nitrate units. Each unit of nitrate contains one nitrogen atom and three oxygen atoms. This formula unit contains one Ca atom, two nitrogen atoms and six oxygen atoms.

The following is an easy format to use;

- Count all the atoms in the formula
- Make a table like the following:

Ca 1 x 40.078 g/mole = 40.078 g/mole

N 2 x 14.007 g/mole = 28.014 g/mole

O 6 x 15.999 g/mole = 95.994 g/mole

** Formula mass =164.086 g/mole**

** ****C _{5}H_{10}O_{2
}**The formula of pentanoic acid is C

_{5}H

_{10}O

_{2}. The previous compounds have ionic bonding. Pentanoic acid has covalent bonding hence the use of molecular mass. The calculation to determine the molecular mass is the same as for the previous compounds.

Follow the format shown above to make the following table

C 5 x 12.011 g/mole = 60.055 g/mole

H 10 x 1.008 g/mole = 10.080 g/mole

O 2 x 15.999 g/mole = 31.998 g/mole

** Molecular mass = 102.133 g/mole**

Calculate the formula mass or molecular mass for the following;

FeBr_{3} Mg(OH)_{2} Ca_{3}(PO4)_{2} C_{6}H_{14}O_{2}

The Dynamic Periodic Table is the one used by the author to calculate the formula mass or the molecular mass in the examples. The author suggests you use this table.

As noted in the opening paragraph, Avogadro’s number and moles represent a certain number of atoms or molecules. **The important take away from this discussion is that the student needs to understand that number of entities is crucial in chemistry.** The will be reinforced in the Topics Reactions section of this website.

Addendum 1:

A third term molar mass is also used in the discussion regarding atomic mass, formula mass, or molecular mass. It is the same as the formula mass and molecular mass discussed above and is calculated the same.

Addendum 2:

The answers to the four compounds above are:

FeBr_{3 } 295.567 g/mole

Mg(OH)_{2} 58.320 g/mole

Ca_{3}(PO4)_{2} 310.172 g/mole

C_{6}H_{14}O_{2} 118.174 g/mole