Today’s focus.
Mole Concept
Avogadro’s Number is 6.02x1023
The mole unit is used to express: 1. A mass quantity 2. A counting quantity
1 water molecule
1 mole of water molecules
Conversion Factors:
Conversion Factors:
6.02x1023 amu = 1 gram (Scale up so that we can weigh in the lab)
(Scale up so that we can weigh in the lab)
6.02x1023 = 1 mole water molecules of water molecules
amu
molecule
g
mole
Information in Chemical Formulae
Sodium carbonate
N 2CO3
Na
What are the parts
p
needed to make ONE formula unit of sodium carbonate?
22.990
12.011
15.999
Na
C
O
11
6
8
Information in Chemical Formulae
N 2CO3
Na
N
Na : C : O
C
O
2
2 : 1 : 3
1
3
Information in Chemical Formulae
N b
Number of ratio of ions: (parts)
f i fi
(
)
N 2CO3
Na
2 : 1
Particles can be: # of atoms, Parts of a formula unit
Parts of a formula unit
# of ions, # of molecules, # of formula units
Conversion Factor
Conversion Conversion
Factor
*
*
*
Conversion Conversion
Factor
*
Particles can be: # of atoms, Parts of a formula unit
Parts of a formula unit
# of ions, # of molecules, # of formula units
*In a problem, look for
In a problem look for
unit given and unit sought.
Q: How many hydrogen atoms are in
y y g
54.15 g of ammonium phosphate?
(NH4)3PO4
Dimensional Analysis:
Dimensional Analysis:
Unit given x conversion = unit sought factor(s)
Do you expect the answer to be
a big number or a small number?
Conversion Factor: Molar Mass = 149.05 Conversion
Factor: Molar Mass = 149 05 g/mole
2
1
4
A: 2.62x1024 H atoms
x
H atoms
( NH ) PO
3
4
( NH ) PO
mole
4
x
3
4
.
3
x
2
0
1
4
3
4
.
2
0
6
5
0
1 9
4
1
54.15 g (NH4)3PO4 x
mole
g ( NH ) PO
= ? H atoms
This is an example of a theoretical calculation of Percent Composition of the elements
in a the chemical formula.
Percent Composition
(NH4)3PO4
Element
Formula Mass = 149.05 g/mole
Mass Contribution
to the formula mass
Percent Composition
p
g
mole
g
mole
x
g
mole
g
mole
l
x
5
0
9
4
1
14.007 x 3 = 42.021 14
007 x 3 = 42 021 g/mole
.
0
0
1
1
2
0
2
4
Nitrogen
.
5
0
9
4
1
Hydrogen
1.008 x 12 = 12.10 g/
.
mole
0
0
1
0
1
2
1
.
= 28.193%
30.9738 x 1 = 30.9738 g/mole
5
0
9
4
1
Phosphorus
g
mole
g
mole
.
0
0
1
8
3
7
9
0
3
.
x
5
0
9
4
1
15.999 x 4 = 63.996 g/mole
g
mole
g
mole
.
= 20.781%
0
0
1
6
9
9
3
6
Oxygen
.
= 8.118%
x
The percent composition of nitrogen in (NH4)3PO4 is 28.193%. etc. = 42.936%
100%
Percent Composition
Difference in mass = mass of oxygen
Experimentally:
ll

Start with KClO3
of a certain mass
+
After heating,
solid remains should
have less mass
Oxygen escaped
Empirical Formula
The simplest formula for a compound that has the SIMPLEST whole
Th
i l f
l f
d h h h SIMPLEST h l
number ratio of the atoms present.
Example: Water
Example:
Water
H2 O
H2O is the chemical
formula of water.
2 : 1
Since this is the SIMPLEST whole number ratio of H:O.
Since
this is the SIMPLEST whole number ratio of H:O
Therefore, H2O is also the empirical formula for water.
Empirical Formula
The simplest formula for a compound that has the SIMPLEST whole
p
p
number ratio of the atoms present.
Example: Hydrogen peroxide
Example: Hydrogen peroxide
H2O2
H2O2 is the chemical
formula of hydrogen peroxide.
2 : 2
The chemical formula is NOT the SIMPLEST whole number ratio of H:O.
Therefore, the empirical formula for hydrogen peroxide is HO.
Empirical Formula
The simplest formula for a compound that has the SIMPLEST whole
p
p
number ratio of the atoms present.
Chemical formula
of the compound
of the compound
H:O
Ratio
Empirical Formula
Multiple
H2O water
2:1
H2O 1x
H2O2
hydrogen peroxide
22
2:2
HO
2
2x
What do we know from this:
1. Sometimes the chemical formula IS the empirical formula,
sometimes it is not.
2. The chemical formula IS a multiple of the empirical formula. When the chemical formula is the empirical formula, the multiple is one.
3. If we know the empirical formula and we know the molar mass of the
compound, then we can determine the chemical formula of the compound!
From Empirical Formula to Chemical Formula
Question:
Q
i
H
Hexane has a molar mass of 86.05 h
l
f 86 05 g/mole. The
Th
empirical formula of hexane is C3H7. What is the chemical formula of hexane?
Chemical formula
of the compound
C:H
Ratio
Empirical Formula
Multiple
Hexane
?
3:7
C3H7
2
A: C6H14
x2
Empirical formula has a mass of: (12.011 x 3) + (1.008x7) = 43.027 g/mole. 
.
.
g
mole
g
mole
7
2
0
3
4
molar mass of compound
Multiple 
empirical formula mass
5
0
6
8
To determine the multiple,
=2
Percent Composition  Empirical Formula  Chemical Formula
Road Map
Question: A sample was sent to the lab and the analysis comes back showing us g
p
that the following mass percent of the elements was found. What is the empirical formula?
15.88% Carbon
2.22 % Hydrogen
63.42% Oxygen
18.50% Nitrogen

1
.
.
.
.
(Nitroglycerine or TNT)


53
 .
3
Answer: C3H5O9N3
7
6
6
1
= 1.320 moles
1
mole
.
g
.
.
x 3 = 3
00
22
33
11
g
l
mole
= 3.964 moles
.
g
Divide
through
by the
smallest ll t
number

40
62
93
31
.
= 2.203 moles
7
0
0
4
1
18.50 g
0
5
8
1
N
g
mole
.
g
.
.
0
22
23
21
.
= 1.322 moles
9
9
9
5
1
63.42 g
63.42 g
2
4
3
6
O
g
.
mole
.
g
0
2
3
1
2.22 g
g
Mole Ratio
2
2
3
1
H
.
8
0
0
1
15.88 g
1
1
0
2
1
C
Moles of Elements
2
2
2
100 gg
8
8
5
1
Elements
x 3 = 9
x 3 = 3
x 3 3
x 3 = 5
Do the Percent Composition, Empirical h
l
Formula, Chemical Formula Worksheet.
Answers will be posted later in the week.
p
Try some practice problems in Maple TA.
Use Dimensional Analysis in ALL your
calculations!!