1. Jonathan Seibert PNHS - Chemistry
The Mole
Jonathan Seibert
PNHS - Chemistry

2. How do we count things?

3. What happens when Too Small or Too Many?

4. A Mole How do we Count Things? 1 mole = 6.02 x 1023 1 dozen = 12
1 gross =
144
1 ream =
500
1 mole =
6.02 x 1023
A Mole

5. I didn’t discover it. Its just named after me!
Avogadro’s Number
6.02 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ).
I didn’t discover it. Its just named after me!
Amadeo Avogadro

6. What is the Mole?
1 mole of hockey pucks would equal the mass of the moon!
1 mole of basketballs would fill a bag the size of the earth!
1 mole of pennies would cover the Earth 1/4 mile deep!

7. Molar Mass Mass of 1 mole of an element or compound.
Atomic mass tells the...
atomic mass units per atom (amu)
grams per mole (g/mol)

8. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol

9. Molar Mass Examples
CO2
H2O
NaCl
44.01 g/mol
18.02 g/mol
58.44 g/mol

10. The Mole
Molar Conversions

11. Molar Conversions molar mass 6.02  1023 MOLES (g/mol) (particles/mol)IN
GRAMS
MOLES
ATOMS
(g/mol)
(particles/mol)

12. Calculations with Moles: Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
= mol Li
2.62
6.94 g Li

13. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 1.8 moles of
lithium?
1.8 mol Li
6.02 x 1023 atoms Li
1 mol Li
(1.8)(6.02 x 1023)/1
= atoms Li
1.08 x 1024

14. Conversion Examples How many moles of carbon are in 26 g of carbon?
26 g C
1 mol C
12.01 g C
= 2.2 mol C

15. Conversion Examples How many molecules are in 2.50 moles of C12H22O11?
6.02  1023
molecules
1 mol
2.50 mol
= 1.51  1024
molecules
C12H22O11

16. Molar Conversions molar mass 6.02  1023 MOLES (g/mol) (particles/mol)IN
GRAMS
MOLES
ATOMS
(g/mol)
(particles/mol)

17. Conversion Examples Find the mass of 2.1  1024 atoms of Carbon.
1 mol
6.02  1023
atoms
12.01 g
1 mol
= 42 g Carbon

18. Percentage Composition
the percentage by mass of each element in a compound

19. Percentage Composition
Find the % composition of Cu2S.
g Cu
g Cu2S
%Cu =
 100 =
79.9% Cu
32.06 g S
g Cu2S
%S =
 100 =
20.1% S

20. Percentage Composition
Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.
28 g
36 g
%Fe =
 100 =
78% Fe
8.0 g
36 g
%O =
 100 =
22% O

21. Percentage Composition
How many grams of copper are in a 38.0-gram sample of Cu2S?
Cu2S is % Cu
(38.0 g Cu2S)( ) = 30.3 g Cu

22. Percentage Composition
Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?
36.04 g g
%H2O =
 100 =
24.51%
H2O

23. C2H6 CH3 Empirical Formula
Smallest whole number ratio of atoms in a compound
C2H6
reduce subscripts
CH3

24. Empirical Formula 1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to find subscripts.
4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

25. Empirical Formula
Find the empirical formula for a sample of 25.9% N and 74.1% O.
25.9 g
1 mol
14.01 g
= 1.85 mol N
= 1 N
1.85 mol
74.1 g
1 mol
16.00 g
= 4.63 mol O
= 2.5 O

26. N2O5 N1O2.5 Empirical Formula
Need to make the subscripts whole numbers
 multiply by 2
N2O5

27. CH3 C2H6 Molecular Formula
“True Formula” - the actual number of atoms in a compound
CH3
empirical
formula ?
C2H6
molecular
formula

28. Molecular Formula 1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the empirical mass.
4. Multiply each subscript by the answer from step 3.

29. (CH2)2  C2H4 Molecular Formula empirical mass = 14.03 g/mol
The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is g/mol?
empirical mass = g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4