1. The Mole
Chapter 10.1

2. Why do we use units to measure?
Dozen eggs
Pair of boots
Ream of paper
Gross of pencils

3. What is the mole
y/moles// mole video

4. The Mole
Mole (mol): SI base unit used to measure the amount of a substance.
Contains 6.02 x 1023 representative particles
Representative Particles are: atoms, molecules, formula units, electrons, ions, etc.
Also called Avogadro’s Number

5. Moles to Particles Converting moles to particles:
# of moles x (6.02 x 1023 particles / 1 mole)
Ex:
How many atoms in 3.2 mols of Na?
3.2 mol Na x (6.02 x 1023 atoms Na) =1.93 x 1024 atoms Na
1 mole

6. Moles to particles more examples
Determine the number of atoms in 2.50 mol Zn?
2.50mol Zn X (6.02 x 1023) = 1.51x1024 atoms Zn
1 mol Zn

7. Particles to Moles # of particles x (1 mole / 6.02 x 1023 particles)
Converting Particles to moles:
# of particles x (1 mole / 6.02 x 1023 particles)
Example:
5 atoms of Na is how many moles?
5 atoms Na x (1 mole Na) = x moles Na
6.02x1023 atoms Na

8. Particles to moles more examples
How many moles contain 5.75 x1024 atoms Al?
5.75 x1024 atoms X (1 mole Al) = 9.55 mol Al
6.02 x 1023

9. Ch. 10.2
Mass and the Mole

10. Masses…
Atomic Mass Unit (amu) – unit of mass (much smaller than a gram) used to describe the mass of an atom.
Atomic Mass – mass of an element in amu.
Formula Mass – mass of an ionic compound formula in amu.
Molecular Mass – mass of a covalent compound formula in amu.

11. Calculating Masses
Atomic, Formula and Molecular Masses are all found the same way.
Add up all the masses for each atom in the compound of a single chemical formula.
Round masses on the periodic table to
2 decimal places.
Example: CO2
1 C x amu = amu
2 O x amu = amu
44.01 amu

12. Molar Mass & Molar Mass of Compounds
Molar Mass is the mass in grams of one mole of substance.
One mole of substance is identified by a single element or a single chemical formula for that compound.
Na – one mole of sodium
2Na or Na2 – two moles of sodium
Molar Mass is calculated by adding up all the masses for each atom in a formula
(use the masses on the periodic table with units of grams)
Example Cu: 1Cu x g
Example Cl2: 2Cl x 35.45g
Example H2O: 2H + 1O (2 x 1.01g) + (1 x 16.00g)

13. Moles to Mass Conversions
Mole to Mass Conversions:
# of mol x (grams of substance / 1 mol of substance)
Mass to Mole Conversions:
# of grams x (1 mol of substance / grams of substance)

14. Mole to mass 3.57 mole Al = ? grams Al
3.57 mole Al x grams = 96.3 g Al
mole

15. Mass to mole more examples
How many mols will a mass of 25.0 g Au have?
25.0g Au X (1 mol Au) = mol Au
196.7g Au

16. Mass to atom
To convert from mass to atoms (or any particles) you must convert the mass to moles then the moles to atoms - Mass to mole to atom.
Known Mass in grams x (1 mole ) = Moles
[ molar mass in grams from PT]
Moles x ( 6.02 x 1023 ) = particles
1 mole
Can be combined into one equation

17. Atoms to mass
To convert from atoms to mass you must convert atoms to moles and then moles to mass – atoms to mass to moles
Known atoms x ( 1 mole ) = Moles of element
6.02x1023atoms
Moles of element x [g of element from PT] = grams of elem.
1 mole of element
May be combined into one equation

18. Examples 3.5 mole CO2 = ? grams CO2 (1 mole = molar mass grams)
3.5 mole CO2 x grams = g CO2
mole
9.01 grams H2O = ? molecules H2O (molar mass grams = 6.02x1023 “particles”)
9.01 grams H2O x x1023 molecules = 3.01x1023 molecules H2O
grams
How many atoms of oxygen are in 2.0 moles of water?
2.0 moles H2O  ? Molecules H2O  ? Atoms Oxygen
2.0 moles O x 6.02x1023 molecules O x atom O = 1.2x1024 atoms O
mole moleculeO

19. Ch. 11.3
Moles of compound

20. Molar mass of a compound
Determine the molar mass of CaCl2.
1 x g = 40.08g Ca
2 x g = 70.90g 2Cl
40.08g g = g CaCl2

21. Converting moles of a compound to mass
The process is the same as when finding the mass of an element except you use the mass of the compound.
# of mol x (mass in g / 1 mol) = mass in g
3.5 mole CO2 = ? grams CO2
3.5 mole CO2 x grams = g CO2
1 mole

22. Converting mass of a compound to moles
The process is the same as converting mass of an element to moles
Given mass x (1 mole of compound) = Moles of compound
[mass of compound from PT]
Example:400g of CCl2F2 is how many moles?
1 mol C atoms x 12.01g C
2 mol Cl atoms x 35.45g Cl
2 mol F atoms x 19.00g
400g CCl2F2 x (1 mole CCl2F2) = moles CCl2F2
120.91g CCl2F2

23. Empirical and Molecular Formula
Ch. 11.4
Empirical and Molecular Formula

25. % Composition What is the % composition of Na in NaCl?
22.99g Na x 100 = %
58.44g NaCl

27. How to determine EF Element – given in the problem so list them
ELEMENT - % - g – Mole –Smallest Ratio
Element – given in the problem so list them
% or g – one OR the other will be given in the problem. If %, change to g dropping the % and adding a g .
Moles – convert the grams to moles by using grams given divided by the mass of element from the periodic table.
Smallest Ratio-Divide the number of moles for all the elements by the smallest number of moles – these become your subscripts.

28. Methyl Acetate is 48. 64% C, 8. 16%H and 43. 20%O
Methyl Acetate is 48.64% C, 8.16%H and 43.20%O. What is the empirical formula?
*% to g
48.64% to 48.64g
8.16% to 8.16g
43.20% to 43.20g
*g to mole
48.64 g C x 1 mole C = 4.05 mol C
12.01 g C
8.16 g H x 1 mol H = 8.08 mol H
1.01 g H
43.20 g O x 1 mol O= 2.7 mol O
16.00g O

29. *smallest ratio 4.05/2.7 = /2.7= /2.7 = 1 But we can’t have decimals as subscripts so…

30. What if the smallest ratio numbers are decimals?
If the decimal is less than .2 round down
If the decimal is greater than .75 round up
If the decimal is between .2 and .75, round the decimal closest to .25, .33, .5, .66, or .75 and then multiply all of them by the same number to get a whole number. ( .25 x 4, .33 x 3, .5 x 2 …)

31. So you will multiply these ratios by the same number to get whole numbers 4.05/2.7 = /2.7= /2.7 = 1 So you will get 1.5 x 2 = x 2 = x 2 = 2 These values are now the subscripts C3H6O2 and this is the empirical formula.

32. Molecular Formula (MF)
Formulas for covalent compounds whose subscripts are not in simplest ratios
Determine the molar mass of the formula(either from the formula given in the problem or the calculated formula from the experimental data)
Given mass molecular formula = X
calculated mass empirical formula
X is the multiple by which all subscripts must be increased

33. Using the previous problem, if you started with 148
Using the previous problem, if you started with g of the substance, what would the Molecular (True) formula be?
Given molecular mass = X
Calculated empirical mass
148.18g = MF is C6H12O2
74.09g