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1
The Mole
2
The mass of a single atom is too small to
measure on a balance.
mass of hydrogen atom = 1.673 x 10-24 g
4
This is an
infinitesimal
1.673 x 10-24 g
mass
5
• Chemists require a unit for counting
which can express large numbers of
atoms using simple numbers.
• Chemists have chosen a unit for
counting atoms.
• That unit is the
6
1 mole = 6.02 x
23
10 objects
7
6.02 x
23
10
is a very
LARGE
number
8
6.02 x
23
10
is
Avogadro’s Number
number
9
Amadeo (Amedeo) Avogadro
10
If 10,000 people started to count
Avogadro’s number and counted at the
rate of 100 numbers per minute each
minute of the day, it would take over 1
trillion years to count the total number.
11
1 mole of any element contains
6.02 x 1023
particles of that substance.
12
The atomic weight in grams
of any element23
contains 1 mole of atoms.
13
This is the same number of particles
6.02 x 1023
as there are in exactly 12 grams of
12
6
C
14
Examples
15
Species
Quantity
Number of H
atoms
H
1 mole
6.02 x
23
10
16
Species
Quantity
Number of
H2 molecules
H2
1 mole
6.02 x
23
10
17
Species
Quantity
Number of
Na atoms
Na
1 mole
6.02 x
23
10
18
Species
Quantity
Number of
Fe atoms
Fe
1 mole
6.02 x
23
10
19
Species C6H6
Quantity 1 mole
Number of
23
6.02 x 10
C6H6 molecules
20
1 mol of atoms = 6.02 x 1023 atoms
1 mol of molecules = 6.02 x 1023 molecules
1 mol of ions = 6.02 x 1023 ions
21
• The mole weight of an element is its
atomic weight in grams.
• It contains 6.02 x 1023 atoms
(Avogadro’s number) of the element.
22
H
Atomic
Number of
Mole weight
mass
atoms
1.008 amu
1.008 g
6.02 x 1023
Mg
24.31 amu
24.31 g
6.02 x 1023
Na
22.99 amu
22.99 g
6.02 x 1023
Element
23
Problems
24
Convert To
Moles!
25
How many moles of iron does 25.0 g of iron
represent?
Atomic weight iron = 55.85
Conversion sequence: grams Fe → moles Fe
Set up the calculation using a conversion factor
between moles and grams.
 1 mol Fe 
(grams Fe) 

 55.85 g Fe 
 1 mol Fe 
(25.0 g Fe) 
 0.448 mol Fe

 55.85 g Fe 
26
How many iron atoms are contained in 40.0 grams of
iron?
Atomic weight iron = 55.85
Conversion sequence: grams Fe → atoms Fe
Set up the calculation using a conversion factor
between atoms and grams.
 6.02 x 1023 atoms Fe 
(grams Fe) 

55.85
g
Fe


 6.02 x 1023 atoms Fe 
23
(40.0 g Fe) 

4.31
x
10
atoms Fe

55.85 g Fe


27
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
Mole weight Na = 22.99 g
Conversion sequence: atoms Na → grams Na
Set up the calculation using a conversion factor
between grams and atoms.
22.99 g Na


(atoms Na) 

23
 6.02 x 10 atoms Na 
22.99 g Na


(3.01 x 10 atoms Na) 
 11.5 g Na
23
 6.02 x 10 atoms Na 
28
23
What is the mass of 0.365 moles of tin?
Atomic weight tin = 118.7
Conversion sequence: moles Sn → grams Sn
Set up the calculation using a conversion factor
between grams and atoms.
mole weight Sn 

(moles Sn) 

 1 mole Sn 
118.7 g Sn 

(0.365 moles Sn) 
  43.3 g Sn
 1 mole Sn 
29
How many oxygen atoms are present in 2.00 mol of
oxygen molecules?
Two conversion factors are needed:
 6.02 x 10 molecules O2 


1
mol
O


2
23
 2 atoms O 
 1 molecule O 

2 
Conversion sequence:
moles O2 → molecules O2 → atoms O
 6.02 x 1023 molecules O2   2 atoms O 
(2.00 mol O2 ) 



1 mol O2

  1 molecule O2 
= 2.41 x1024 atoms O
30
Mole Weight of
Compounds
31
The mole weight of a compound can be
determined by adding the mole weights
of all of the atoms in its formula.
32
Calculate the mole weight of C2H6O.
2 C = 2(12.01 g) = 24.02 g
6 H = 6(1.01 g) = 6.06 g
1 O = 1(16.00 g) = 16.00 g
46.08 g
33
Calculate the mole weight of LiClO4.
1 Li = 1(6.94 g) = 6.94 g
1 Cl = 1(35.45 g) = 35.45 g
4 O = 4(16.00 g) = 64.00 g
106.39 g
34
Calculate the mole weight of (NH4)3PO4 .
3 N = 3(14.01 g) = 42.03 g
12 H = 12(1.01 g) = 12.12 g
1 P = 1(30.97 g) = 30.97 g
4 O = 4(16.00 g) = 64.00 g
149.12 g
35
Calculate the mole weight of NaC2H3O2 · 3
H2O
Solids that contain water molecules as
part of their structure are called Hydrates.
1 Na = 1(22.99 g) = 22.99 g
2 C = 2(12.01 g) = 24.02 g
9 H = 9(1.01 g) = 9.09 g
5 O = 5(16.00 g) = 80.00 g
136.10 g
36
Avogadro’s
Number of
Particles
6.02 x 1023
Particles
1 MOLE
Mole Weight
37
Avogadro’s
Number of
Ca atoms
6.02 x 1023
Ca atoms
1 MOLE Ca
40.078 g Ca
38
Avogadro’s
Number of
H2O molecules
6.02 x 1023
H2O
molecules
1 MOLE H2O
18.02 g H2O
39
These relationships are present when
hydrogen combines with chlorine.
H
Cl
HCl
6.02 x 1023 H
atoms
6.02 x 1023 Cl
atoms
6.02 x 1023 HCl
molecules
1 mol H atoms 1 mol Cl atoms
1 mol HCl
molecules
1.008 g H
35.45 g Cl
36.46 g HCl
1 mole weight
H atoms
1 mole weight
Cl atoms
1 mole weight
40
HCl molecules
In dealing with diatomic elements (H2, O2,
N2, F2, Cl2, Br2, and I2), distinguish between
one mole of atoms and one mole of
molecules.
41
Calculate the mole weight of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the mole weight of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g
42
Problems
43
How many moles of benzene, C6H6, are present in
390.0 grams of benzene?
The mole weight of C6H6 is 78.12 g.
Conversion sequence: grams C6H6 → moles C6H6
78.12 grams C6 H 6
Use the conversion factor:
1 mole C6 H 6
 1 mole C6 H 6 
= 5.000 moles C6 H 6
(390.0 g C6 H 6 ) 

 78.12 g C6 H 6 
44
How many grams of (NH4)3PO4 are contained in 2.52
moles of (NH4)3PO4?
The mole weight of (NH4)3PO4 is 149.12 g.
Conversion sequence: moles (NH4)3PO4
→ grams (NH4)3PO4
149.12 grams (NH 4 )3PO4
Use the conversion factor:
1 mole (NH 4 )3PO4
 149.12 g (NH 4 )3 PO4 
(2.52 mol (NH 4 )3PO 4 )) 

 1 mol (NH 4 )3 PO4 
= 376g (NH 4 )3 PO445
56.04 g of N2 contains how many N2 molecules?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors
1 mol N 2
28.02 g N 2
6.02 x 1023 molecules N 2
1 mol N 2
 1 mol N 2   6.02 x 10 molecules N 2 
(56.04 g N 2 ) 



1 mol N 2
 28.02 g N 2  

23
= 1.20 x 1024 molecules46N 2
56.04 g of N2 contains how many N atoms?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
Use the conversion factors
1 mol N 2 6.02 x 1023 molecules N 2
2 atoms N
28.02 g N 2
1 molecule N 2
1 mol N 2
 1 mol N 2   6.02 x 1023 molecules N 2 
(56.04 g N 2 ) 

 
28.02
g
N
1
mol
N

2  

2
 2 atoms N 
 1 molecule N 

2 
= 2.41 x 1024 atoms N
47
Percent Composition
of Compounds
48
Percent composition of a compound is the
mass percent of each element in the compound.
H2O
11.19% H by mass
88.79% O by mass
49
Percent Composition
From Formula
50
If the formula of a compound is known,
a two-step process is needed to calculate
the percent composition.
Step 1 Calculate the mole weight of the
formula.
Step 2 Divide the total mass of each
element in the formula by the
mole weight and multiply by
100.
51
total mass of the element
x 100 = percent of the element
mole weight
52
Calculate the percent composition of hydrosulfuric
acid H2S(aq) .
Step 1 Calculate the mole weight of H2S.
2 H = 2 (1.01 g) = 2.02 g
1 S = 1 (32.07 g) = 32.07 g
34.09 g
53
Calculate the percent composition of hydrosulfuric
acid H2S.
Step 2 Divide the mass of each element
by the mole weight and multiply
by 100.
 2.02 g H 
H: 
 (100) = 5.93%
 34.09 g 
 32.07 g S
S: 
 34.09 g

 (100)  94.07%

S
H
94.07% 5.93%
54
Percent Composition
From Experimental Data
55
Percent composition can be calculated
from experimental data without knowing
the composition of the compound.
Step 1 Calculate the mass of the
compound formed.
Step 2 Divide the mass of each element
by the total mass of the
compound and multiply by 100.
56
A compound containing nitrogen and oxygen is found
to contain 1.52 g of nitrogen and 3.47 g of oxygen.
Determine its percent composition.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g = total mass of product
57
Calculate the percent composition of hydrosulfuric
acid H2S.
Step 2 Divide the mass of each element
by the total mass of the
compound formed.
 1.52 g N 

 (100) = 30.5%
 4.99 g 
 3.47 g O

 4.99 g

 (100) = 69.5%

O
69.5%
N
30.5%
58
Empirical Formula versus
Molecular Formula
59
• The empirical formula or simplest
formula gives the smallest wholenumber ratio of the atoms present in a
compound.
• The empirical formula gives the
relative number of atoms of each
element present in the compound.
60
• The molecular formula is the true
formula of a compound.
• The molecular formula represents the
total number of atoms of each element
present in one molecule of a compound.
61
Examples
62
Molecular Formula
C2H4
Empirical Formula
CH2
Smallest Whole
Number Ratio
C:H 1:2
63
Molecular Formula
C6H6
Empirical Formula
CH
Smallest Whole
Number Ratio
C:H 1:1
64
Molecular Formula
H2O2
Empirical Formula
HO
Smallest Whole
Number Ratio
H:O 1:1
65
66
Two compounds can have identical
empirical formulas and different molecular
formulas.
67
68
Calculating
Empirical Formulas
69
Step 1 Assume a definite starting
quantity (usually 100.0 g) of the
compound, if not given, and
express the mass of each
element in grams.
Step 2 Convert the grams of each
element into moles of each
element using each element’s
mole weight.
70
Step 3 Divide the moles of atoms of
each element by the moles of
atoms of the element that had
the smallest value
– If the numbers obtained are whole
numbers, use them as subscripts
and write the empirical formula.
– If the numbers obtained are not
whole numbers, go on to step 4.
71
Step 4 Multiply the values obtained in
step 3 by the smallest numbers
that will convert them to whole
numbers
Use these whole numbers as the
subscripts in the empirical
formula.
FeO1.5
Fe1 x 2O1.5 x 2
Fe2O3
72
• The results of calculations may differ
from a whole number.
– If they differ ±0.1 round off to the next
nearest whole number.
2.9  3
– Deviations greater than 0.1 unit from a
whole number usually mean that the
calculated ratios have to be multiplied by
a whole number.
73
Problems
75
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
K = 56.58 g
C = 8.68 g
O = 34.73 g
76
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 2 Convert the grams of each element to moles.
 1 mol K atoms 
 1.447 mol K atoms
K:  56.58 g K  

 39.10 g K 
 1 mol C atoms 
C atoms
atoms
C: 8.68 g C  
 0.723 mol C

 12.01 g C  C has the smallest number
of moles
 1 mol O atoms 
 2.171 mol O atoms
O:  34.73 g O  

77
 16.00 g O 
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 3 Divide each number of moles by the smallest
value.
1.447 mol
0.723 mol
K=
= 2.00
C:
= 1.00
0.723 mol
0.723 mol
0.723 mol C atoms
2.171 mol
O=
= 3.00
C has the smallest number
0.723 mol
of moles
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3
78
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
N = 25.94 g
O = 74.06 g
79
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 2 Convert the grams of each element to moles.
 1 mol N atoms 
 1.852 mol N atoms
N:  25.94 g N  

 14.01 g N 
 1 mol O atoms 
O:  74.06 g O  
 4.629 mol C atoms

 16.00 g O 
80
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 3 Divide each number of moles by the smallest
value.
1.852 mol
N=
= 1.000
1.852 mol
4.629 mol
O:
= 2.500
1.852 mol
This is not a ratio of whole numbers.
81
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 4 Multiply each of the values by 2.
N: (1.000)2 = 2.000
O: (2.500)2 = 5.000
Empirical formula N2O5
82
Calculating the Molecular Formula
from the Empirical Formula
83
• The molecular formula can be calculated
from the empirical formula if the mole
weight is known.
• The molecular formula will be equal to the
empirical formula or some multiple n of it.
• To determine the molecular formula evaluate
n.
• n is the number of units of the empirical
formula contained in the molecular formula.
mole weight
n=
=
empirical weight
number of empirical
formula units
84
What is the molecular formula of a compound which
has an empirical formula of CH2 and a mole weight of
126.2 g?
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = 2.02g
14.03g
126.2 g
n
 9 (empirical formula units)
14.03 g
The molecular formula is (CH2)9 = C9H18
85
A compound made of 30.4% N and 69.6% O has a
mole weight of 138 grams/mole. Find the molecular
formula.
Old Way:
30.4 g 
1mole
 2.17 mole of N
14.0 g
1mole
69.6g 
 4.35mole of O
16.0 g
138
 3units
46
N2.17 O4.35  NO2  46g/mole
2.17
2.17
3(NO2 )  N3O6
86
A compound made of 30.4% N and 69.6% O has a
mole weight of 138 grams/mole. Find the molecular
formula.
New Way:
138gcomp'd
30.4 gN
1mole N


 3mole N
1mole comp'd 100 gcomp'd 14.0 g
138gcomp'd
69.6gO
1mole O


 6mole O
1mole comp'd 100 gcomp'd 16.0 g
N3O6
87
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