1. The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.02 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.

2. 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

3. Calculations with Moles: Converting moles to grams
How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li
6.9 g Li
= g Li
45.1
1 mol Li

4. 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

5. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium?
3.50 mol
6.02 x 1023 atoms
= atoms
2.07 x 1024
1 mol

6. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.02 x 1023 atoms Li
6.9 g Li
1 mol Li
(18.2)(6.02 x 1023)/6.9
= atoms Li
1.58 x 1024

7. Calculating Formula Mass
Calculate the formula mass of carbon dioxide, CO2.
12.0 g + 2(16.0 g) =
44.0 g
 One mole of CO2 (6.02 x 1023 molecules) has a mass of 44.0 grams

8. Standard Molar Volume 1 mole of a gas occupies 22.4 liters of volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
- Amedeo Avogadro
At STP (Standard Temperature and Pressure):
1 mole of a gas occupies 22.4 liters of volume

9. The Mole Highway Mass (g) Vol (L) Rep Part 6.02x1023 Rep Part
Form Mass
1 mol
22.4 L
1 mol
1 mol x x x x x x
1 mol
Form Mass
1 mol
22.4 L
6.02x1023 Rep Part
1 mol
Mass (g)
Vol (L)
Rep Part

10. Calculations with Moles:
Two-Step Problem
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.02 x 1023 atoms Li
6.9 g Li
1 mol Li
(18.2)(6.02 x 1023)/6.9
= atoms Li
1.58 x 1024

11. Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3.
From previous slide:
24.3 g g + 3(16.0 g) = 84.3 g
100.00

12. Formulas empirical formula = CH
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.
molecular formula = (empirical formula)n [n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH

13. Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3

14. Formulas (continued)
Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11

15. Empirical Formula Determination Rules
Base calculation on 100 grams of compound.
Determine moles of each element in 100 grams of compound.
Divide each value of moles by the smallest of the values.
Multiply each number by an integer to obtain all whole numbers.

16. Empirical Formula Determination Example
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

17. Empirical Formula Determination (part 2)
Divide each value of moles by the smallest of the values.
Carbon:
Hydrogen:
Oxygen:

18. Empirical Formula Determination (part 3)
Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50
Hydrogen: 2.50
Oxygen: 1.00
x 2
x 2
x 2 3 5 2
Empirical formula:
C3H5O2

19. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g

20. Finding the Molecular Formula (continued)
2. Divide the molecular mass by the mass given by the empirical formula.
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
3. Multiply the empirical formula by this
number to get the molecular formula.
(C3H5O2) x 2 =
C6H10O4