CHEMICAL QUANTITIES:
THE MOLE
CHEM I/IH: CHAPTER 10

MEASURING MASS
 A mole is a quantity of things, just as…
1 dozen = 12 things
1 gross = 144 things
1 mole = 6.02 x 10 23 things
 “Things” usually measured in moles are atoms, molecules, ions, and formula units
2

 You can measure
mass, or volume, or you can count pieces


We measure mass in grams
We measure volume in liters
 We count pieces in
MOLES
3

A MOLE…



is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon-
12
1 mole = 6.02 x 10 23 of the representative particles
Treat it like a very large dozen
6.02 x 10 23 is called: Avogadro’s number
4

 Similar Words for an amount:
 Pair : 1 pair of shoelaces = 2 shoelaces
 Dozen : 1 dozen oranges = 12 oranges
 Gross : 1 gross of pencils= 144 pencils
 Ream : 1 ream of paper= 500 sheets of paper
5

What are Representative
Particles (“RP”)?
 The smallest pieces of a substance:
1.
For a molecular compound: it is the molecule.
2.
3.
For an ionic compound: it is the formula unit (made of ions)
For an element: it is the atom
 Remember the 7 diatomic elements? (made of molecules)
6



Practice Counting Particles
How many oxygen atoms in the following?
1.
2.
CaCO3 3 atoms of oxygen
Al2(SO4)3 12 (4 x 3) atoms of oxygen
How many ions in the following?
1.
CaCl2
 3 total ions (1 Ca2+ ion and 2 Cl1- ions)
2.
NaOH
 2 total ions (1 Na1+ ion and 1 OH1- ion)
3.
Al2(SO4)3
 5 total ions (2 Al3+ + 3 SO4 ions)
7

CONVERSION FACTOR
 RPs x ____1 mole___ = MOLES
6.02 x 10 23 RPs
8

EXAMPLES: ATOMS

MOLES
 How many moles of B are in 3.15 x 10 23 atoms of B?
 Conversion: 1 mole B = 6.02 x 10 23 atoms B
(b/c the atom is the RP of boron)
3.15 x 10 23 atoms of B
1 mole B
6.02 x 10 23 atoms B
=
0.523 mole
9

EXAMPLES: MOLES

ATOMS
 How many atoms of Al are in 1.5 mol of Al?
 Conversion: 1 mole = 6.02 x 10 23 atoms
1.5 mol of Al 6.02 x 10 23 atoms Al
=
1 mole Al
9.03 x 10 23 atoms of Al
10


CAUTION: Identify RPs
Carefully!
See next slide!
11

EXAMPLES: MOLECULES

MOLES
How many atoms of H are there in 3 moles of
H
2
O? (HINT: Are atoms the RP for water?)
Conversions:
1 mole = 6.02 x 10 23 molecules
(b/c molecules are the RP for H
2
O)
H
2
O molecule = 2 atoms of Hydrogen
3 moles of H
2
O
6.02 x 10 23 molec H
2
O
2 atoms H
1 mole H
2
O
1 H
2
O molecule
= 3.612 x 10 24 atoms H
12

MOLAR MASS
Def: The mass of a mole of representative particles of a substance.
Each element & compound has a molar mass.
13

MOLAR MASS OF AN ELEMENT
Determined simply by looking at the periodic table
Molar mass ( g ) = Atomic Mass ( amu )
20
Ca
40.08
* Thus,
1 mol Ca = 40 g
1 atom of Ca weighs 40.08 amu
1 mole of Ca atoms weighs 40.08 grams
14

MOLAR MASS FOR COMPOUNDS


To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound
Then add the masses within the compound
Example: H
2
O
H= 1.01
2 ( 1.01
) + 1 ( 15.999
)= 18.02 g/mol
O= 15.999
15

SOME PRACTICE PROBLEMS
 How many atoms of O are in 3.7 mol of O?
 2.2 X 10 24 atoms of oxygen

 How many atoms of Ca are there in 2.5 moles of CaCl
2
?
 1.5 x 10 24 atoms Ca

How many atoms of P are in 2.3 mol of P?
 1.4 x 10 24 atoms of phosphorus
How many atoms of O are there in 1.7 moles of
SO
4
?
 4.1 x 10 24 atoms of oxygen
16



 The molar mass of any substance (in grams) equals 1 mole
This applies to ALL substance: elements, molecular compounds, ionic compounds
Use molar mass to convert between mass and moles
 Ex: Mass, in grams, of 6 mol of MgCl
2
?
mass of MgCl
2
= 6 mol MgCl
2
92.21 g MgCl
2
1 mol MgCl
2
= 571.26 g MgCl
2
17

VOLUME AND THE MOLE
 Volume varies with changes in temperature & pressure


Gases are predictable, under the same physical conditions
Avogadro’s hypothesis helps explain:
 equal volume of gases, at the same temp and pressure contains equal number of particles
Ex: helium balloon
18




MOLAR VOLUME
Gases vary at different temperatures, makes it hard to measure
To make it easier to measure gases, we use something called “STP”


“Standard Temperature and Pressure”
Temperature = 0 ° C
 Pressure = 1 atm (atmosphere)
19

Molar Volume
 At STP:1 mole, 6.02 x 10 23 atoms, of any gas has a volume of 22.4 L
1 mole gas = 22.4 L gas
 Use it to convert between # of moles and vol of a gas @ STP
 Ex: what is the volume of 1.25 mol of sulfur gas?
1.25 mol S 22.4 L S = 28.0 L
1 mol S
20




MOLAR MASS FROM DENSITY
Different gases have different densities
Density of a gas measured in g/L @ a specific temperature
Can use the following formula to solve : grams = grams X 22.4 L

 mole L 1 mole
Ex: Density of gaseous compound containing oxygen and carbon is 1.964 g/ L, what is the molar mass?
grams = 1.964 g X 22.4 L then you solve mole 1 L 1 mole = 44.o g/mol
21

Atoms, molecules
, etc.
22

Molarity


Def: the concentration of a solution. How many moles/liter
Can be used to calculate # of moles of a solute
 Ex: Household laundry bleach is a dilute aqueous solution of sodium hypochlorite
(NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO?
23

Calculating Percent Composition of a Compound
 Like all percent problems: a part ÷ the whole
1.
Find the mass of each of the components (the elements)
2.
Next, divide by the total mass of the compound
3.
Then X 100 % = percent
Formula:
% Composition = Mass of element X 100%
Mass of compound
24

Method #1: % Comp When Actual Masses are Given
A compound is formed when 9.03 g of Mg combines completely with 3.48 g of N.
What is the percent composition of the compound?
1.
First add the 2 mass of the 2 compounds to reach the total mass 9.03 g Mg + 3.48 g N =
12.51 g Mg
3
N
2
2. Find the % of each compound
% Mg= 9.03 g Mg X 100 = 72.2 %
12.51 g Mg
3
N
2
% N= 3.48 g N X 100 = 27.8 %
12.51 g Mg
3
N
2
25



Method 2: % Comp When Only The

Formula is Known
Can find the percent composition of a compound using just the molar mass of the compound and the element
% mass=mass of the element 1 mol cmpd X100% molar mass of the compound
Example:
Find the percent of C in CO
2
12.01 g C X 100 = 27.30% C
44.01 g CO
2
26




Using % Composition
Can use % composition as a conversion factor just like the mole
After finding the % comp. of each element in a cmpd. can assume the total compound = 100g
Example: C= 27.3% 27.3 g C

O= 72.7 % 72.7 g O
In 100 g sample of compound there is 27.3 g of C & 72.7 g of O
How much C would be contained in 73 g of CO
2
?
73 g CO
2
27.3 g C = 19.93 g C
100 g CO
2
27

 Empirical formulas are the lowest
WHOLE number ratios of elements contained in a compound
28

REMEMBER…
 Molecular formulas tells the actual number of of each kind of atom present in a molecule of the compound
 Ex:
H
2
O
2
Molecular
Formula
CO
2
Molecular
Formula
HO
Empirical
Formula
CO
2
Empirical
Formula
For CO
2 they are the same
29

 Formulas for ionic compounds are
ALWAYS empirical (the lowest whole number ratio = can not be reduced)
 Examples:
NaCl MgCl2 Al2(SO4)3 K2CO3
Simplest whole number ratio for NaCl
30

 A formula is not just the ratio of atoms, it is also the ratio of moles
 In 1 mole of CO
2 of oxygen there is 1 mole of carbon and 2 moles
 In one molecule of CO
2 there is 1 atom of C and 2 atoms of O
 Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio)

Molecular: H
2
O C
6
H
12
O
6
C
12
H
22
O
11
(Correct formula)
 Empirical:
(
Lowest whole
H
2
O CH
2
O number ratio)
C
12
H
22
O
11
31

 STOP HERE!!!
32

CALCULATING EMPIRICAL
 We can get a ratio from the percent composition
1.
Assume you have a 100 g sample the percentage become grams (75.1% = 75.1 grams)
2.
Convert grams to moles
3.
Find lowest whole number ratio by dividing each number of moles by the smallest value
33

Example calculations


Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N
Assume 100 g sample, so
38.67 g C x 1 mol C = 3.22 mole C
12.0 g C
16.22 g H x 1 mol H =
1.0 g H
16.22 mole H
45.11 g N x 1 mol N = 3.22 mole N
14.0 g N
*Now divide each value by the smallest value
34

…Example 1
 The ratio is 3.22 mol C = 1 mol C
3.22 mol N 1 mol N
 The ratio is 16.22 mol H = 5 mol H
3.22 mol N 1 mol N
C
1
H
5
N
1 which is = CH
5
N
35

MORE PRACTICE


A compound is 43.64 % P and 56.36 % O
What is the empirical formula?
PO3
Caffeine is 49.48% C, 5.15% H, 28.87% N and
16.49% O
What is its empirical formula?
C
4
H
5
N
2
O
36

EMPIRICAL TO MOLECULAR
 Since the empirical formula is the lowest ratio, the actual molecule would weigh more
 Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula
37


EXAMPLE
Caffeine has a molar mass of 194 g, what is its molecular formula?
1.
Find the mass of the empirical formula, C
4
H
5
N
2
O
2.
Divide the molar mass by the empirical mass:
194.0 g/mol = 2
97.1 g/mol
3.
Now multiply the entire empirical formula by 2
2(C
4
H
5
N
2
O) = final molecular formula C
8
H
10
N
4
O
2
38