Moles
• Text reference for moles:
• PP 83 to 87 and
• PP 237 to 249
The Mole
• A number of atoms, ions, or molecules that
is large enough to see and handle.
• A mole = number of things
– Just like a dozen = 12 things
– One mole = 6.022 x 1023 things
• Avogadro’s constant = 6.022 x 1023
– Symbol for Avogadro’s constant is NA.
2
The Mole
• How do we know when we have a mole?
– count it out
– weigh it out
• Molar mass - mass in grams numerically equal to the
atomic weight of the element in grams.
• H has an atomic weight of 1.00794 g
– 1.00794 g of H atoms = 6.022 x 1023 H atoms
• Mg has an atomic weight of 24.3050 g
– 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
3
The Mole
• Example 2-1: Calculate the mass of a single
Mg atom in grams to 3 significant figures.
? g Mg 
4
The Mole
• Example 2-1: Calculate the mass of a single
Mg atom in grams to 3 significant figures.
? g Mg  1 Mg atom
5
The Mole
• Example 2-1: Calculate the mass of a single
Mg atom in grams to 3 significant figures.


1 mol Mg atoms
 
? g Mg  1 Mg atom
23
 6.022 10 Mg atoms 
6
The Mole
• Example 2-1: Calculate the mass of a single
Mg atom, in grams, to 3 significant figures.

1 mol Mg atoms
? g Mg  1 Mg atom 
23
6.022

10
Mg atoms

 24.30gMg

 1 mol Mg atoms

 


  4.04  10  23 g Mg

7
The Mole
• Example 2-2: Calculate the number of atoms
in one-millionth of a gram of Mg to 3
significant figures.
? Mg atoms 
8
The Mole
• Example 2-2: Calculate the number of atoms
in one-millionth of a gram of Mg to 3
significant figures.
 1 mol Mg 

? Mg atoms  1.00  10 g Mg
 24.30 g Mg 
6
9
The Mole
• Example 2-2: Calculate the number of atoms
in one-millionth of a gram of Mg to 3
significant figures.
? Mg atoms  1.00  10
6
 6.022  1023 Mg atoms

1 mol Mg atoms

 1 mol Mg 

g Mg 
 24.30 g Mg 



10
The Mole
• Example 2-2: Calculate the number of atoms
in one-millionth of a gram of Mg to 3
significant figures.
 1 mol Mg 

? Mg atoms  1.00  10 g Mg
 24.30 g Mg 
6
 6.022  1023 Mg atoms 

  2.48  1016 Mg atoms
 1 mol Mg atoms

11
The Mole
• Example 2-3. How many atoms are contained
in 1.67 moles of Mg?
? Mg atoms 
12
The Mole
• Example 2-3. How many atoms are contained
in 1.67 moles of Mg?
? Mg atoms  1.67 mol Mg
13
The Mole
• Example 2-3. How many atoms are contained
in 1.67 moles of Mg?
 6.022 10 23 Mg atoms 

? Mg atoms  1.67 mol Mg 
1 mol Mg


14
The Mole
• Example 2-3. How many atoms are contained
in 1.67 moles of Mg?
 6.022 10 23 Mg atoms 

? Mg atoms  1.67 mol Mg 
1 mol Mg


 1.00 10 24 Mg atoms
15
The Mole
• Example 2-3. How many atoms are contained
in 1.67 moles of Mg?
 6.022  1023 Mg atoms 

? Mg atoms  1.67 mol Mg
1 mol Mg


 1.00  1024 Mg atoms
16
The Mole
• Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
You do it!
17
The Mole
• Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
? mol Mg  73.4 g Mg
18
The Mole
• Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
 1 mol Mg atoms 

? mol Mg  73.4 g Mg 
 24.30 g Mg 
19
The Mole
• Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
 1 mol Mg atoms 

? mol Mg  73.4 g Mg 
 24.30 g Mg 
 3.02 mol Mg
IT IS IMPERATIVE THAT YOU KNOW
HOW TO DO THESE PROBLEMS
20
Formula Weights, Molecular Weights,
and Moles
• How do we calculate the molar mass of a
compound?
– add atomic weights of each atom
• The molar mass of propane, C3H8, is:
• 3C X 12.01g = 36.03g
• 8H X 1.01g = 8.08g
•
44.11 g/mole
21
Formula Weights, Molecular Weights,
and Moles
• The molar mass of calcium nitrate, Ca(NO3)2 ,
is:
You do it!
22
Formula Weights, Molecular Weights,
and Moles
• One Mole of
– Cl2 or
Contains
70.90g 6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
– C3H8
You do it!
23
Formula Weights, Molecular Weights,
and Moles
• One Mole of
Contains
– Cl2 or
70.90g 6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
– C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules
3 (6.022 x 1023 ) C atoms
8 (6.022 x 1023 ) H atoms
24
Formula Weights, Molecular Weights,
and Moles
• Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane.
? C3H8 molecules  74.6 g C3H8 
25
Formula Weights, Molecular Weights,
and Moles
• Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane.
? C3 H 8 molecules  74.6 g C3H 8 
 1 mole C3H 8 


 44.11 g C3 H 8 
26
Formula Weights, Molecular Weights,
and Moles
• Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane.
? C3H8 molecules  74.6 g C3H8 
 1 mole C3H8  6.022 10 23 C3H8 molecules


1 mole C3H8
 44.11 g C3H8 

 

27
Formula Weights, Molecular Weights,
and Moles
• Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane.
? C3 H 8 molecules  74.6 g C3 H 8 
 1 mole C3H 8  6.022 10 23 C3H 8 molecules


1 mole C3 H 8
 44.11 g C3 H 8 
24
1.02  10 molecules

 

28
Formula Weights, Molecular Weights,
and Moles
• Example 2-6. What is the mass of 10.0 billion
propane molecules?
You do it!
29
Formula Weights, Molecular Weights,
and Moles
• Example 2-6. What is the mass of 10.0
billion propane molecules?
1 mole C3H8


? g C3H8 molecules  1.00  10 molecules 

23
 6.022 10 molecules 
10
 44.11 g C3H8

 1 mole C3H8

  7.32 10 13 g of C3H8

30
Formula Weights, Molecular Weights,
and Moles
• Example 2-7. How many (a) moles, (b) molecules,
and (c) oxygen atoms are contained in 60.0 g of
ozone, O3? The layer of ozone in the stratosphere is
very beneficial to life on earth.
You do it!
31
Formula Weights, Molecular Weights,
and Moles
• Example 2-7. How many (a) moles, (b) molecules, and (c)
oxygen atoms are contained in 60.0 g of ozone, O3? The layer
of ozone in the stratosphere is very beneficial to life on earth.
(a)
 1 mole 
  1.25 moles
? moles O3  60.0 g O3 
 48.0 g O3 
32
Formula Weights, Molecular Weights,
and Moles
• Example 2-7. How many (a) moles, (b) molecules, and (c)
oxygen atoms are contained in 60.0 g of ozone, O3? The layer
of ozone in the stratosphere is very beneficial to life on earth.
(b)
 6.022 10 23 molecules
? molecules O 3  1.25 moles 
1 mole

 7.53 10 23 molecules O 3



33
Formula Weights, Molecular Weights,
and Moles
• Example 2-7. How many (a) moles, (b) molecules, and (c)
oxygen atoms are contained in 60.0 g of ozone, O3? The layer
of ozone in the stratosphere is very beneficial to life on earth.
(c)
 3 O atoms
? O atoms  7.53 10 molecules O3 
 1 O3 molecule
 2.26 10 24 atoms O
23



34
Formula Weights, Molecular Weights,
and Moles
• Example 2-8. Calculate the number of O atoms
in 26.5 g of Li2CO3.
You do it!
35
Formula Weights, Molecular Weights,
and Moles
• Example 2-8. Calculate the number of O atoms
in 26.5 g of Li2CO3.
1 mol Li 2 CO3
? O atoms  26.5 g Li 2 CO3 

73.8 g Li 2 CO3
6.022 10 23 form. units Li 2 CO3
3 O atoms


1 mol Li 2 CO3
1 formula unit Li 2 CO3
6.49 10 23 O atoms
36
Percent Composition and Formulas of
Compounds
• % composition = mass of an individual
element in a compound divided by the total
mass of the compound x 100%
• Determine the percent composition of C in
C3H8.
mass C
%C
100%
mass C3 H 8
3 12.01 g

100%
44.11 g
 81.68%
37
Percent Composition and Formulas of
Compounds
• What is the percent composition of H in C3H8?
You do it!
38
Percent Composition and Formulas of
Compounds
• What is the percent composition of H in C3H8?
mass H
%H
 100%
mass C3 H 8
8 H

 100%
C3H 8
8  1.01 g

 100%  18.32%
44.11 g
or
18.32%  100%  81.68%
39
Percent Composition and Formulas of
Compounds
• Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 significant
figures.
You do it!
40
Percent Composition and Formulas of
Compounds
• Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 sig. fig.
2  Fe
2  55.8 g
% Fe 
100% 
100%  27.9% Fe
Fe 2 (SO 4 ) 3
399.9 g
3 S
3  32.1 g
%S 
100% 
100%  24.1% S
Fe 2 (SO 4 ) 3
399.9 g
12  O
12 16.0 g
% O 
100% 
100%  48.0% O
Fe 2 (SO 4 ) 3
399.9 g
Total
 100%
41
Derivation of Formulas from Elemental
Composition
• Example 2-11: A compound contains 24.74% K,
34.76% Mn, and 40.50% O by mass. What is its
empirical formula?
• Make the simplifying assumption that we have 100.0
g of compound.
• In 100.0 g of compound there are:
– 24.74 g of K
– 34.76 g of Mn
– 40.50 g of O
42
Derivation of Formulas from Elemental
Composition
1 mol K
? mol K  24.74 g K 
 0.6327 mol K
39.10 g K
43
Derivation of Formulas from Elemental
Composition
? mol K
1 mol K
 24.74 g K 
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
44
Derivation of Formulas from Elemental
Composition
1 mol K
? mol K  24.74 g K 
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
1mol O
? mol O  40.50 g O 
 2.531 mol O
16.00 g O
obtain smallest whole number ratio
45
Derivation of Formulas from Elemental
Composition
1 mol K
? mol K  24.74 g K 
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
1mol O
? mol O  40.50 g O 
 2.531 mol O
16.00 g O
obtain smallest whole number ratio
0.6327
for K 
1K
0.6327
0.6327
for Mn 
 1 Mn
0.6327
46
Derivation of Formulas from Elemental
Composition
? mol K
 24.74 g K 
1 mol K
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
1mol O
? mol O  40.50 g O 
 2.531 mol O
16.00 g O
obtain smallest whole number ratio
0.6327
for K 
1K
0.6327
0.6327
for Mn 
 1 Mn
0.6327
2.531
for O 
4O
0.6327
thus the chemical formula is KMnO 4
47
Derivation of Formulas from Elemental
Composition
• Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O. What
is the empirical formula for this compound?
You do it!
48
Derivation of Formulas from Elemental
Composition
• Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is the empirical formula for this
compound?
1 mol Co
? mol Co  6.541 g Co 
 0.1110 mol Co
58.93 gCo
1mol O
? mol O  2.368 g O 
 0.1480 mol O
16.00 g O
find smallest whole number ratio
49
Derivation of Formulas from Elemental
Composition
• Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is the empirical formula for this
compound?
0.1110
0.1480
for Co 
 1 Co for O 
 1.333O
0.1110
0.1110
multipy both by 3 to turn fraction to whole number
1 Co  3  3 Co 1.333 O  3  4 O
Thus the compound' s formula is :
Co3O 4
50
Determination of Molecular Formulas
• The formulas determined by the previous
method produce empirical or simplest
formulas. In the case of covalent compounds
the empirical formula may or may not
represent a real molecule. For example:
• CH2 is an empirical formula by no molecule
with this formula exists. C2H4 is a real
molecule with empirical formula CH2 !
Determination of Molecular Formulas
• In order to determine a molecular formula
from the empirical formula you need some
additional information, usually the molar mass
of the molecule. Since a molecular formula is
a multiple of its empirical formula, you can
divide the molar mass of the molecular
formula by the molar mass of the empirical to
find the multiplier. For example:
Determination of Molecular Formulas
• Suppose the empirical formula is CH2 and we
were told that the molar mass of a molecule
of this substance is 70 g/mole.
• The molar mass of CH2 is 14 g/mole
• So 70/14 = 5 so the molecular formula is 5 X
the empirical or C5H10
Determination of Molecular Formulas
• Example 2-13: A compound is found to
contain 85.63% C and 14.37% H by mass. In
another experiment its molar mass is found to
be 56.1 g/mol. What is its molecular formula?
»You do it
Determination of Molecular Formulas
• Example 2-13: A compound is found to contain
85.63% C and 14.37% H by mass. In another
experiment its molar mass is found to be 56.1
g/mol. What is its molecular formula?
– short cut method
1 mol contains 56.1 g
85.63% is C and 14.37% is H
56.1 g  0.8563  48.0 g of C
56.1 g  0.1437  8.10 g of H
55
Determination of Molecular Formulas
convert masses to moles
1 mol C
48.0 g of C 
 4 mol C
12.0 g C
1 mol H
8.10 g of H 
 8 mol H
1.01 g H
Thus the formula is :
C4H8
56