1. The Mole Concept
Goal: To develop the concept of the mole as a useful measurement and to apply this in calculations involving mass and volume.

2. Objectives
1. Define the mole in terms of Avogadro's number, formula mass or molecular mass.
2. Given the number of moles, determine the number of molecules of a covalent compound, formula units of an ionic compound or atoms of an element.
3. Given the number of particles present in a sample, determine the number of moles.
4. When given the formula, calculate the molecular mass of a covalent compound or the formula mass of an ionic compound.
5. Given the number of moles in a substance, determine the mass of the substance.
6. Given the mass of a substance, determine the number of moles of the substance.
7. When given the formula, calculate the percentage composition of each of the elements in the compound.

3. Molecular and Formula Mass
Molecular Mass is determined by the sum of the atomic masses of each element in a compound.
The atomic mass of hydrogen in atomic mass units is 1, and the atomic mass of oxygen is 16.
Therefore, the total mass of a water molecule, H2O, is , or 18 amu.
This name is incorrect when applied to an ionic substance.
E.g. Sodium chloride, NaCl, is an ionic substance which does not exist in molecular form.
A better name for the mass of ionic substances is formula mass.
Formula mass is determined by the sum of the atomic masses of all atoms in the formula unit of an ionic compound.

4. EXAMPLE: Formula Mass Find the formula mass of sodium sulfate, Na2SO4.
Solving process:
Add the atomic masses of all the atoms in the Na2SO4 unit.
2 Na atoms, 2 x 23 = 46 amu
1 S atom, 1 x 32 = 32 amu
4 O atoms, 4 x 16 = 64 amu
The total formula mass = 142 amu

5. EXAMPLE: Formula Mass
Find the formula mass of calcium nitrate, Ca(NO3)2
Solving process:
Add the atomic masses of all the atoms in the Ca(NO3)2 formula unit. Remember that the subscript applies to the entire polyatomic ion.
1 Ca atom, 1 x 40 = 40 amu
2N atoms, 2 x 14 = 28 amu
6 O atoms, 6 x 16 = 96 amu
Total formula mass is 164 amu

6. Groups of Atoms
There is one problem in using the molecular and formula masses of substances.
These masses are in atomic mass units, which is only 1.66 x g.
The mass of a single molecule is so small that it is impossible to measure it in the laboratory.
For everyday use in chemistry, a larger unit, such as a gram, is needed.
One helium atom has a mass of 4 amu, and one nitrogen atom has a mass of 14 amu. The ratio of the mass of one helium atom to one nitrogen atom is 4 to 14, or 2 to 7.
No matter what number of atoms we compare, equal numbers of helium and nitrogen atoms will have a mass ratio of 2 to 7.
In other words, the numbers in the atomic mass table give us the relative masses of the atoms of the elements.

7. Avagadro’s Number The standard laboratory unit of mass is the gram.
We would like to choose a number of atoms which would have a mass in grams equivalent to the mass of one atom in atomic mass units.
The same number would fit all elements since equal numbers of different atoms always have the same mass ratio.
Chemists have found that 6.02 x 1023 atoms of an element have a mass in grams equivalent to the mass of one atom in amu.
E.g. 1 hydrogen atom has a mass of amu
6.02 x 1023 atoms of hydrogen have a mass of g.
This number, 6.02 x 1023 is called Avogadro's Number in honor of a 19th century Italian scientist.

8. What is the Mole?
6.02x1023 atoms of an element = 1 mole of that element
E.g. 6.02x1023 atoms of Na = 1 mole of Na
6.02x1023 molecules of a compound = 1 mole of that compound
E.g. 6.02x1023 molecules of NaCl = 1 mole of NaCl

9. Mole Trivia
If there were a mole of rice grains, all the land area in the whole world would be covered with rice to a depth of about 75 meters.
One mole of rice grains is more grains than all the grain that has been grown since the beginning of time. (1)
One mole of rice would occupy a cube about 120 miles on an edge! (1)

10. Mole Trivia
A mole of marshmallows would cover the United States to a depth of 600 miles (3)
In order to put a mole of rain drops in a 30 meter (about 100 feet) diameter tank, the sides of the tank would have to be 280 times the distance from the Earth to the Sun. (4)
A mole of hockey pucks would be equal to the mass of the Moon.

11. Mole Trivia
Assuming that each human being has 60 trillion body cells (6.0 x 1013) and the Earth's population is 6 billion (6 x 109), the total number of living human body cells on the Earth at the present time is 3.6 x 1023 or a little over half of a mole.

12. Mole Trivia
If one mole of pennies were divided up among the Earth's population, each person would receive 1 x 1014 pennies. 
Personal spending at the rate of one million dollars a day would use up each persons wealth in about three thousand years. 
Life would not be comfortable because the surface of the Earth would be covered in copper coins to a depth of at least 400 meters.

13. Mole Trivia
If you had a mole of pennies and wanted to buy kite string at the rate of a million dollars per inch, you would get your money's worth.
After stretching your string around the Earth one million times, and to the Moon and back twenty-five times, you would have enough string left over to sell back at a dollar an inch (a decided loss) to gain enough money to buy every man, woman and child in the US a $50,000 automobile and enough gasoline to run it at 55 mph for a year. 
After those purchases, you would still have enough money left over to give every man, woman, and child in the whole world about $5000.

14. Mole Trivia Basis for calculations:
Earth's circumference = 25,000 miles               
Distance to moon = 240,000 miles
Cost of gasoline = $2.50 per gallon
Gasoline mileage = 20 miles per gallon
U.S. population =220,000,000
World population = 6,000,000,000

15. Calculating Molecular Mass
1. Calculate the molecular or formula masses of the following compounds, all in amu (g/mol):
a. C2H6
Amu=30
b. SiCl4
Amu=170
c. MgCO3
Amu=84
d. Ca3(P04)2
Amu=310

16. Calculating Molecular Mass
Calculate the molecular or formula masses of the following compounds, all in amu (g/mol):
K2S
Amu=110
CH2CHCH2OH
Amu=58
Pb3(As04)2
Amu=899
C12H22011
Amu=342

17. Conversions Make the following conversions:
1.00 x 1026 molecules of SnCl2 to moles.
Ans: 1.66 X 102 mol.
0.400 moles of H2O to molecules.
Ans: 2.41 X 1023 molecules.
76.0 grams CaBr2 to moles.
Ans: mol. Or 3.80 X 10-1 mol.
18.0 grams HBr to moles.
Ans: mol. Or 2.22 X 10-1 mol.

18. Conversions Make the following conversions:
9.30 moles SiH4 to molecules.
Ans: 5.60 X 1024 molecules.
Find the mass of one atom of Na.
Ans: 3.82 X 10-23g.
Find the mass of one molecule of H2SO4.
Ans: 1.63 X g.

19. Conversions Make the following conversions: 9.30 moles SiH4 to grams.
Ans: 298 g.
Find the mass of mole of Na2SO4.
Ans: g.
Find the mass of one molecule of H2SO4.
Ans: 1.63 X g.

20. Percent Composition
1 mole of NaCl is comprised of equal parts Na and Cl
But Na and Cl have 23 g/mol and g/mol, respectively!
So they have differing percent composition
The percent composition of Na is
(molar mass Na) x 100
molar mass NaCl
Molar mass NaCl= = 58 g/mol
%Na= (23) x 100
(58)
%Na= 40%
%Cl= (35) x 100
%Cl=60%
So, while NaCl is equal parts NaCl by molar ratios, by mass, it’s a 60/40 split.

21. Percent Composition Find the percentage composition of the following:
CsF
Ans: 87.5%; 12.5%.
Bi203
Ans: 89.7%, 10.3%.
BaH2
Ans: 98.6%, 1.44%.