1. The Mole L

2. How many?
1 Dozen = 12 Roses

3. How Many?
One Pair = Two Eyes

4. How Many?
Two Pairs = 4 shoes

5. How Many?
1 Ream = 500 sheets

6. The Mole
A number like
1 dozen
1 pair
1 ream
1 mole (mol) = 6.02 x 1023 representative particles

7. Converting Mol  particles
Use dimensional analysis

8. How many particles are in:
4.5 mol Sucrose
10 mol HCl
0.5 mol NaOH
mol Pb
1.6 x 10-5 mol HF
2.7 x 1024
6.022 x 1024
3.0 x 1023
6.022 x 1019
9.6 x 1018

9. Converting particles  mol
Use dimensional analysis

10. How many moles are in: 2.7 x 1015 particles of Zn 1 particle of NaCl
9.066 x 1024 particles of Sucrose
5.34 x 1013 particles of Al
4.4 x 10-9 mol of Zn
1.7 x mol of NaCl
mol of Sucrose
8.87 x mol of Al

11. Weights You can obtain the atomic masses from the periodic table
For compounds, add the weights
Molecular weight = covalent compounds
Formula weight = ionic or covalent compounds
What is the formula weights for the following:
NaCl
H2O
Al2(SO4)3
C6H12O6
1 Na 1Cl NaCl
22.99 amu amu = amu
2H O H2O
2(1.01 amu) amu = amu

12. 1 mol of any element will equal the atomic mass in grams
The rule of 12
Mole is bases off of C-12
There is 1 mol of C-12 atoms in 12g of C-12
-Therefore-
1 mol of any element will equal the atomic mass in grams

13. Atomic/Formula Weight in grams
The bridge
The mole is the bridge b/t the microscopic and the macroscopic
Microscopic
Macroscopic
6.022 x 1023
Particles
Atomic/Formula Weight in grams

14. Mol to Mass Atoms are massed in AMU’s
Ca = 40.1 amu
O = amu
H = amu
How many grams would be a mol (1 mol) of each:
Ca = 40.1 g
O = g
H = g

15. Molar Mass Molar Mass – mass of 1 mole of a substance
The atomic weight expressed in g/mol
Molar mass of water (H2O)
g of H2O contains
1 mole of water (H2O)
2 mole of H
1 mole of O H O
Total
1.009 x 2
15.999
18.015amu

16. Mass  Mol Dimensional analysis again
Find the molar mass of the compound, then:

17. How many mol are in: 5.7 g Water 84 g Ammonium Nitrate 13.6 g Gold
0.67 g Silver Nitrate
2.9 g Carbon dioxide
0.317 mol Water
1.06 mol NH4NO3
mol Au
Mol AgNO3
0.066 Mol CO2

18. Mol  Mass Dimensional analysis again
Find the molar mass of the compound, then:

19. How many grams are in: 1 mol of Al(OH)3 0.05 mol H2SO4 4 mol AgNO3
1.5 x 10-9 mol H2CO3
77.98 g Al(OH)3
4.902 g H2SO4
g AgNO3
9.3 x 10-8 g H2CO3

20. Percent Composition The percent by mass of each element in a compound
% = Part/Whole x 100
If a compound contains 55 g of element X and 45g of element Y. What is the percent Composition of element X in compound XY?

21. Percent Composition
If a compound contains 55 g of element X and 45g of element Y. What is the percent Composition of element X in compound XY?
% = Part/Whole x 100
% = 55g/100g x 100
55% = percent composition of X in XY

22. % comp. from Chemical formula
You can use the chemical formula to find percent composition as well.
Assume you have a 1 mol sample
What is the percent composition of H in H2O?
2H O H2O
2(1.01 amu) amu = amu
1 mol H20 = g H2O
Element
Equation
% comp.
H (1.01 g/mol)
(1.01 x 2)/ x 100
11.2% H
O (16.00 g/mol)
(16.00)/ x 100
88.80% O

23. Empirical formula
Smallest whole number mole ratio of elements in a compound
Can be found from two sources of info
Percent comp (just assume a 100g sample)
Actual masses

24. Empirical Formula Find the mols of each element present
Divide all elements involved mol amount by the smallest mol amount.
Must be a whole number! If you get a decimal multiple all by enough to obtain a whole number (ie: X1 Y1.5 Z1  2(X1 Y1.5 Z1) = X2 Y3 Z2

26. Empirical Formula from % comp.
The percent composition of a compound is 40.05% sulfur, and 59.95% oxygen. What is the empirical formula of this compound?
Ratio of S:O is 1.249:3.747  NOT WHOLE #s so…
Divide both by smallest one (S = 1.249)
1.249/1.249 = 1 (S = 1)
3.747/1.249 = 3 (O = 3)
Empirical Formula = S1O3  SO3
S (40.05%  40.05g)
40.05 g S x 1 mol S / g S
1.249 mol S
O (59.95%  59.95g)
59.95 g O x 1 mol O / g O
3.747 mol O

27. Practice
What is the empirical formula for a compound which is 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen.
Multiply by 2 for whole numbers
C1.5H3O1 = 2(C1.5H3O1) = C3H6O2
Element
Mass  Mol
Mol
Mol:Mol
Mol Ratio
C (48.64g)
48.64g C x 1 mol C/ g C
4.050 mol C
4.050 mol C / mol O
1.5
H (8.16g)
8.16g H x 1 mol H/ 1.01 g H
8.10 mol H
8.10 mol H / mol O 3
O (43.20g)
43.20g O x 1 mol O/ g O
2.700 mol O
2.700 mol O / mol O 1

28. Molecular Formula Empirical Formula = simplest ratio
Molecular Formula = actual ratio
To find simply divide molar mass of compound by molar mass of empirical formula, and multiple empirical formula by the product.
(Molar mass of compound/molar mass of empirical) x empirical formula
Compound
Empirical Formula
Molecular Formula
Hydrogen Peroxide HO
H2O2
Sucrose
CH2O
C6H12O6

29. Practice
A compound is 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a molar mass of g/mol. Determine the empirical formula and the molecular formula

30. Practice (empirical)
Element
Mass  Mol
Mol
Mol:Mol
Mol Ratio
C (40.68g)
40.68g C x 1 mol C/ g C
3.387 mol C
3.387 mol C / mol C 1
H (5.08g)
5.08g H x 1 mol H/ 1.01 g H
5.04 mol H
5.04 mol H / mol H
1.49 = 1.5
O (54.24g)
54.24g O x 1 mol O/ g O
3.390 mol O
3.390 mol O / mol C
1.009 = 1
C1H1.5O1  2(C1H1.5O1) = C2H3O2 Empirical Formula = C2H3O2

31. Practice Empirical  Molecular
Formula
Molar Mass
C2H3O2
59.04 g/mol
n(C2H3O2 )
118.1 g/mol
n = Molar/empirical
n = 118.1/59.04 = 2
Molecular formula = n(C2H3O2 ) = 2(C2H3O2 ) = C4H6O4

32. Mol  Volume (of gas at STP)
1 mol of ANY gas has a volume of 22.4L
Again use dimensional analysis

33. What is the volume of (at STP):
1 mol of CN(g)
0.5 mol of H2SO4(g)
10 mol CO2(g)
6 mol O2(g)
5 mol of CCl4(g)

34. Volume (at STP)  Mol
Again 1 mol = 22.4 L
Use dimensional analysis

35. How many mols are in: 6 L of CN(g) 1.0 x 104 L of CO2(g)
0.002 L of C10H10(g)

36. Who’s brain feels like this??