The Mole -just as a _____
pair is always ____,
two a _______
dozen is always _______,
twelve
a ______
gross is always ____,
144 and a _____
ream is always ____,
500 a
_____
mole is always ____________________________,
602,000,000,000,000,000,000,000 or
__________,
6.02 x 1023 and just like it is sometimes easier to buy things
dozen or by the _____,
ream it is sometimes easier to
by the _______
consider numbers of _________
particles of _______
matter by the _____
mole
I. Converting Moles to Particles and Particles to Moles
How do we simplify when we multiply fractions?
__
3 x __
8 = __
24 = __
2
4
9
36
3
3 x __
8 2 = __
2
or 11__
4
93
3
How many eggs in 4½ dozen?
4.5 dozen
x
12 eggs
________
1 dozen
=
54 eggs
conversion factor
The Mole
I. Converting Moles to Particles and Particles to Moles
How many molecules in 4.5 moles of sucrose?
4.5 moles sucrose
x
6.02 x 1023 molecules sucrose
_________________________
1 mole sucrose
2 significant figures
2 significant figures
=
2 significant figures
2.7 x 1024 molecules sucrose
conversion factor
How many moles in 1.75 x 1025 molecules of water?
1.75 x 1025 molecules H2O
3 significant figures
x
1 mole H2O
_________________________
6.02 x 1023 molecules H2O
3 significant figures
=
3 significant figures
29.1 moles H2O
The Mole
II. Molar Mass -the _______
masses of ______
atoms are established ________
relative to
the mass of one __________
Carbon -12 atom, which has ____
six
_________
protons and ____
six __________
neutrons
-____-________
one twelfth of the mass of a __________
Carbon -12 atom,
proton or a ________,
neutron is
or about the mass of a _______
atomic _____
mass _____,
unit or _____
amu
called an _______
-the ______
molar _____
mass is the _____
mass of one _____
mole of an
atomic
element, and is numerically equal to the ________
mass but with units of ______
grams per _____,
mole so the
_____,
_______
atomic _____
mass of a Carbon-12 atom is ___
12 ____,
amu
and the ______
molar _____
mass of Carbon-12 is ___
12 ______
grams
mole
per _____
The Mole
II. Molar Mass
A. Converting Mass to Moles and Moles to Mass
What is the mass of 0.415 moles of Vanadium?
3 significant figures
5 significant figures
0.415 moles V
50.942 grams V
_______________
1 mole V
x
=
21.1 grams V
3 significant figures
How many moles of Tantalum are in 75 grams
of Tantalum ?
75 grams Ta
1 mole Ta
_______________
180.948 grams Ta
x
2 significant figures
6 significant figures
=
0.41 grams V
2 significant figures
The Mole
II. Molar Mass
A. Converting Mass to Atoms and Atoms to Mass
How many atoms are there in 99.838 g Uranium?
3 significant figures
99.838 g U
x
1 mole U
_______________
238.029 g U
x
3 significant figures
6.02 x 1023 atoms U
_______________
1 mole U
5 significant figures
=
2.525 x 1023 atoms U
=
2.52 x 1023 atoms U
6 significant figures
What is the mass, in grams, of 1.1703 x 1024 atoms of Niobium?
5 significant figures
1.1703 x 1024 atoms Nb
x
1 mole Nb
_______________
x
23
6.022 x 10 atoms Nb
5 significant figures
4 significant figures
92.906 g Nb
_______________
1 mole Nb
4 significant figures
=
1.8055 x 102 g Nb
=
1.806 x 102 g Nb
The Mole
II. Molar Mass
A. Converting Mass to Molecules and Molecules to Mass
How many molecules are there in 456 g Silicon dioxide?
456 g SiO2
x
1 mole SiO2
_______________
60.084 g SiO2
x
6.02_______________
x 1023 molecules SiO2
1 mole SiO2
=
4.57 x 1024 molecules SiO2
What is the mass, in grams, of 1.75 x 1026 molecules of cyclohexane?
1.75 x 1026 molecules C6H12 x
1 mole C6H12
_______________
x
6.02 x 1023 molecules C6H12
84.162 g C6H12
_______________
1 mole C6H12
=
2.45 x 104 g C6H12
The Mole
III. Empirical and Molecular Formulas
A. Calculating Percent Composition
What is the percent composition of Sodium, Sulfur, and Oxygen in
Sodium Sulfate?
2 moles Na
x
22.990 g Na
=
1 mole Na
1 mole S
x
32.065 g Na
x
15.999 g O
1 mole O
x
100
=
32.38%
x
100
=
22.58%
x
100
=
45.07%
142.0 g Na2SO4
=
1 mole S
4 moles O
45.98 g Na
32.065 g S
142.0 g Na2SO4
=
63.996 g O
142.0 g Na2SO4
The Mole
III. Empirical and Molecular Formulas
A. Calculating Percent Composition
What is the percent composition of Carbon, Hydrogen, and Oxygen in
fructose, C6H12O6?
6 moles C
x
12.011 g C
=
1 mole C
12 moles H x
1.008 g H
x
15.999 g O
1 mole O
x
100
=
39.99%
x
100
=
6.715%
x
100
=
53.27%
180.2 g C6H12O6
=
1 mole H
6 moles O
72.066 g C
12.10 g H
180.2 g C6H12O6
=
95.994 g O
180.2 g C6H12O6
The Mole
III. Empirical and Molecular Formulas -the empirical formula is the
lowest whole number mole
ratio of elements in a
compound
B. Determining Empirical Formula from Percent Composition
What are the empirical and molecular formulas for Ibuprofen if the molar
mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H,
and 15.5% O?
75.7 g C
x
1 mole C
=
12.011 g C
8.80 g H
x
1 mole H
x
1 mole O
15.999 g O
=
6.50 moles C
Empirical Formula
= C13H18O2
=
9.01 moles H
Molar mass of Ibuprofen
0.969
=
1.008 g H
15.5 g O
6.30 moles C
8.73 moles H
Molar mass of Empirical Formula
0.969
=
0.969 moles O
0.969
=
1.00 mole O
Molecular Formula
= C13H18O2
The Mole
III. Empirical and Molecular Formulas -the empirical formula is the
lowest whole number mole
ratio of elements in a
compound
B. Determining Empirical Formula from Percent Composition
What are the empirical and molecular formulas for Glycerol if the molar
mass is 92.11 g/mole and the percent composition of 39.12% C, 8.75% H,
and 52.12% O?
39.12 g C
x
1 mole C
=
12.011 g C
3.257 moles C
=
1.000 moles C
3.257
Empirical Formula
= C3H8O3
Molar mass of Glycerol
8.75 g H
x
1 mole H
=
1.008 g H
8.68 moles H
=
2.66 moles H
Molar mass of Empirical Formula
3.257
92.11 g/mole C?H?O?
52.12 g O
x
1 mole O
15.999 g O
=
3.257 moles O
=
1.000 mole O
92.09 g/mole C3H8O3
3.257
Molecular Formula
= C3H8O3
The Mole
III. Empirical and Molecular Formulas -the empirical formula is the
lowest whole number mole
ratio of elements in a
compound
B. Determining Empirical Formula from Percent Composition
What are the empirical and molecular formulas for Naphthalene if the
molar mass is 128 g/mole and the percent composition of 93.7% C and
6.3% H?
93.7 g C
x
1 mole C
=
12.011 g C
7.80 moles C
=
1.25 moles C
6.25
Empirical Formula
= C5H4
Molar mass of Naphthalene
6.3 g H
x
1 mole H
1.008 g H
=
6.25 moles H
=
1.00 moles H
Molar mass of Empirical Formula
6.25
128 g/mole C?H?
64.09 g/mole C5H4
Molecular Formula
= C10H8
The Mole
III. Empirical and Molecular Formulas -the empirical formula is the
lowest whole number mole
ratio of elements in a
compound
B. Determining Empirical Formula from Percent Composition
What are the empirical and molecular formulas for a Lead chloride
compound if the molar mass of the compound is 349.0 g/mole and the
percent composition is 59.37% Pb?
59.37 g Pb
x
1 mole Pb
=
207.2 g Pb
0.2865 moles Pb =
1.000 moles Pb
0.2865
Empirical Formula
= PbCl4
Molar mass of Lead chloride
40.63 g Cl
x
1 mole Cl
35.453 g Cl
=
1.146 moles Cl
=
4.00 moles Cl
Molar mass of Empirical Formula
0.2865
349.0 g/mole Pb?Cl?
349.0 g/mole PbCl4
Molecular Formula
= PbCl4
The Mole
III. Empirical and Molecular Formulas
A. Calculating Percent Composition -lab
1. Hypothesis -How does the mass of chewed bubblegum compare to the
mass of chewed bubblegum?
2. Prediction
3. Gathering Data -safety
a. Carefully avoid contaminating the unchewed gum on any surface
before chewing
- procedure
a. Carefully unwrap one piece of bubblegum. Using the wrapper as
weighing paper, place the unchewed gum on the top-loading
balance to determine and record its mass to the nearest 0.1 gram.
_____________ g. Save the wrapper.
b. Chew the bubblegum for 5 minutes.
The Mole
III. Empirical and Molecular Formulas
A. Calculating Percent Composition -lab
3. Gathering Data -procedure
c. After chewing for 5 minutes, remove the gum from mouth and,
using the wrapper as weighing paper again, carefully place the
wad of chewed gum on the top-loading balance to determine and
and record its mass to the nearest 0.1 gram. _______________ g
4. Analyzing Data -calculate the percent composition of bubblegum
mass of chewed gum
x
100
=
mass of unchewed gum
5. Drawing Conclusions
percent
composition
x
100
=
_________%
The Mole
IV. Hydrates -hydrates are __________
compounds that have a specific number
of ______
water molecules attached to them
A. Naming Hydrates
Formula
Name
(NH4)2C2O4·H2O
Ammonium oxalate monohydrate
CaCl2·2H2O
Calcium chloride dihydrate
NaC2H3O2·3H2O
Sodium acetate trihydrate
FePO4·4H2O
Iron(III) phosphate tetrahydrate
CuSO4·5H2O
Copper(II) sulfate pentahydrate
CoCl2·6H2O
Cobalt(II) hexahydrate
MgSO4·7H2O
Magnesium sulfate heptahydrate
Ba(OH)2·8H2O
Barium hydroxide octahydrate
Na2CO3·10H2O
Sodium carbonate decahydrate
The Mole
IV. Hydrates
B. Calculating the Formula for a Hydrate
What is empirical formula and for a hydrated compound of Copper(II)
sulfate, if 2.50 grams of blue CuSO4·nH2O is heated in a crucible until
1.59 grams of white anhydrous CuSO4 remains ?
Mass of hydrated Copper(II) sulfate
2.50 g
Mass of anhydrous Copper(II) sulfate - 1.59 g
0.91 g
1.59 g CuSO4
x
1 mole CuSO4
=
159.6 g CuSO4
0.91 g H2O
x
1 mole H2O
18.02 g H2O
0.00996 moles CuSO4
=
1.00 moles CuSO4
0.00996
=
0.050 moles H2O =
0.00996
Empirical Formula
= CuSO4·5H2O
Copper(II) sulfate pentahydrate
5.0 moles H2O