1. Moles
Videodisk Unit 4

2. Moles
1 mole of a substance contains x particles (or atoms, formula units, molecules)
So a “mole” is a number, very similar to a dozen.
A dozen = 12
A mole (mol) = x 1023 particles
It can be found in your reference packet on page one
Avogadro’s number = x 1023
Moles are similar to a pair(2) of dozen(12).
See attached page for examples of a mole.

3. Meaning of Avogadro’s Number (The mole)
The mole (abbreviated mol) is the base unit for measuring the amount of a substance.
The definition of a mole comes from how many particles (atoms, in this case) there are in exactly 12 grams of Carbon-12.
Through many years of experimentation, it has been confirmed that a mole of any substance has 6.022*1023 representative particles

5. Molar Mass or Gram Formula Mass (GFM) or Formula Weight
Molar Mass = the mass of 1 mole of a substance in grams (also known as Gram Formula Weight or Formula Weight)
mass given on periodic table is in amu
Because amu and grams are relative we can say the masses on the periodic table are in grams
To find molar mass add all the atomic masses of the elements in the compound.

6. Practice – Find the molar mass of:
MgCl2
Ca(OH)2
C6H12O6
If the molar mass is the final answer to a question, then the significant figures for adding measurements rule should be invoked. If it’s a step along the way, then keep the extra decimal place.
(35.45) = g/mol
(16) + 2(1.008) = g/mol
6(12.01) + 12(1.008) + 6(16) = g/mol

7. Moles are the central unit in chemistry.
They allow us to change units between:
Mass (g)
Number of particles (atoms or molecules or formula units or ions)
Volume (L) - gas at STP only
Moles are a name for a group.
pair = 2
dozen = 12
mole = 6.02 x 1023

8. Moles and Mass 160 g S The conversion factor is
1 mol = Molar Mass (or GFM or FW)
Convert 5.0 moles of Sulfur to grams
160 g S

9. Sample Problem 606 g Zn(NO3)2
To carry out a chemical reaction you need 3.20 moles of zinc nitrate Zn(NO3)2. What is the mass of 3.20 moles of Zn(NO3)2?
(14.01) + 6(16) = g/mol Zn(NO3)2
3.20 mol Zn(NO3)2 x g/mol
606 g Zn(NO3)2

10. Moles and Particles The conversion factor is
1 mol = x 1023 particles (molecules, atoms, etc)
Convert 2.0 moles H2O to molecules
1.2 x 1024 mc H2O

11. Convert 4.3 mol of O2 gas at STP to liters
Moles and Liters of Gas
The conversion factor is 1 mol = 22.4 L
It only applies to Gas at STP (standard temperature and pressure) which is 0 degrees C and 1 atm
Convert 4.3 mol of O2 gas at STP to liters
96 L O2

12. Sample Problem 1.45 mol CaCO3
A piece of marble contains 8.74 x 1023 formula units of calcium carbonate. How many moles of CaCO3 is that?
8.74 x form.un.CaCO3 x 1 mol/6.022 x form.un.
1.45 mol CaCO3

13. Hall Analogy M O L E S Mass Molecules Liters (of Gas At STP)
1 mole = 22.4 L M O L E S
Mass
Molecules
1 mole = x 1023 particles
1 mole = GFM
You must enter the hall to get to the other rooms.

14. Traffic Circle Analogy
To get to volume, mass, or particles, you have to pass through the mole!

15. Sample Problem
You need 250 grams of table sugar, or sucrose (C12H22O11), to bake a cake. How many sucrose molecules will be in the cake?
4.4 x 1023 molecules C12H22O11

16. Sample Problem
If you burned 4.00 x 1023 molecules of natural gas, or methane (CH4) during a laboratory experiment, what mass of methane did you burn?
10.7 g CH4

17. Finding Molarity from Grams of Solute
What is the molarity of mL of a sodium chloride solution that has grams of solute dissolved in it?
Convert grams to moles
Convert mL of solution to L
Use M = mol/L (found in reference packet) to solve for molarity. This was part of Unit 2, but now the skill of converting to moles first can be included in problems.
100.0 mL x 1 L/1000 mL = L
30.56 g NaCl x 1 mol/58.44 g = mol NaCl
mol NaCl/ L =
5.229 mol/L = M NaCl

18. More Solution Examples
How many grams of potassium iodide are in mL of a 0.50 M solution?
A 0.75 M solution has grams of sodium nitrate dissolved in it. What is the volume of the solution?
250.0 mL x 1 L/1000 mL = L
NaNO3 = (16) = 85 g/mol NaNO3
0.50 mol/L KI x L = mol KI
45.62 g NaNO3 x 1 mol/85 g = mol NaNO3
KI = = 166 g/mol
mol KI x 166 g/mol = 21 g KI
mol NaNO3/0.75 mol/L = 0.72 L NaNO3