CHAPTER 10
THE MOLE
DIMENSIONAL ANALYSIS
Also known as factor label method
Problem solving method that focuses on the
UNITS that are used to describe matter
Unit labels are treated as FACTORS that
can be divided out
Uses conversion factors
– Example: 12 inches = 1 foot
Problem:
– 5 ft - _____________ inches
5 ft
12 in
5 X 12
1 ft
1
60 in
10 mm = 1 cm
100 cm = 1 m
14 m = ___________ mm
14 m
100 cm
1m
10 mm
1 cm
14,000
14 x 100 x 10
1
mm
1000 L= 1 kL
1000 mL = 1 L
4 kL = ____________ mL
4kL
1000 L
1 kL
1000 mL
1L
4,000,000 mL
6.02 x 10 23 bananas = 1 mole
4 moles bananas = ___________ bananas
4 mole bananas
6.02 x 1023 bananas
1 mole banana
4 x (6.02 x 1023)
1
2 x 1024 bananas
AVOGADRO’S CONSTANT
The number of particles in a mole
Equal to 6.02 x 1023
Mole = a unit for the chemical quantity of a
substance
– A term that represents a certain # of particles
– Need a large number because atoms and
molecules are so small
– 1 mole = 6.02 x 1023 of anything (particles,
atoms, molecules etc.)
MASS AND THE MOLE
Atomic mass = mass of one atom of a
substance
Measure in atomic mass unit (amu)
1 hydrogen atom = 1.0079 amu
MOLAR MASS
Mass in grams of a mole of any pure
substance
Use the atomic mass (off the periodic table);
change unit to grams/mole
1 atom hydrogen = 1.0079 amu
1 mole hydrogen = 1.0079 grams
M0LAR MASS OF COMPOUNDS
AND MOLECULES
Sum of the masses of all of the atoms of a
substance
Example: H2O
H
2 x 1.0 = 2.0
O
1 x 16.0 = 16.0
18.0 g/mole
masses
KMnO4
–K
– Mn
–O
1 x 39.1 = 39.1
1 x 54.9 = 54.9
4 x 16.0 = 64.0
158.0 g/mole
DON’T FORGET SIG DIGS!!
Molar Mass of Hydrates (added H2O)
– CoCl2·6H2O
• Co 1 x 58.9 = 58.9
• Cl
2 x 35.5 = 71.0
• H2O 6 x 18.0 = 108.0
237.9 g/mole
PERCENT COMPOSITION
A way to determine the formula of newly
invented compounds (ones that you don’t
know the oxidation numbers of)
%composition is a calculation of the % by
mass of each element in a compound (part
from whole)
Use the formula of the compound and the
molar mass of each part
K2CrO4
–K
2 x 39.1
=
78.2
– Cr
1 x 52.0
=
52.0
–O
4 x 16.0
=
64.0
molar mass =
194.2 g /mole
% K = 78.2 / 194.2 x 100 = 40.3 %
%Cr = 52.0 / 194.2 x 100 = 26.8%
%O = 64.0 / 194.2 x 100= 32.9%
CONVERSION OF THE MOLE
Remember:
– 1 mole = 6.02 x 1023 particles, atoms etc
– Molar mass = _________ g/mole
•(
g= 1 mole)
Know:
Molar mass
6.02 x 1023
Grams g/mole moles Particles etc./mole particles,etc
To go from: grams to moles use molar mass
To go from: moles to molecules use Avogadro’s #
To go from: grams to molecules use both
GRAMS TO MOLES
Calculate the moles of 1.5 g of sodium
– Use molar mass sodium = 23.0 g/mole
1.5 g Na
1 mole Na
23.0 g Na
.065 moles Na
MOLES TO GRAMS
Calculate the mass in grams of 50.0 moles
of KCl
– Use molar mass KCl  74.6 g/mole
(K= 39.1; Cl = 35.5)
50.0 mole KCl
74.6 g KCl
1 mole KCl
3730 g KCl
MOLES TO PARTICLES(ATOMS,
MOLECULES, FORMULA UNITS)
How many formula units are in 4.9 moles of
MgSO4?
– Use: 6.02 x 1023 formula units/mole
4.9 moles MgSO4 6.02 x 1023 formula units
1 mole MgSO4
2.9 x 1024 formula units of
MgSO4
PARTICLES TO MOLES
How many moles are in 2.3 x 1025
molecules of CO2?
– Use: 6.02 x 1023 molecules/mole
2.3 x 1025 molecules CO2
1 mole CO2
6.02 x 1023 molecules CO2
= 38 moles CO2
GRAMS TO PARTICLES
How many atoms are in 14.0 g of barium?
– Use: molar mass Ba = 137.3 g/mole
– Use: 6.02 x 1023 atoms/mole
14.0 g Ba
1 mole Ba
137.3 g Ba
6.02 x 1023 atoms Ba
1 mole Ba
= 6.14 x 1022 atoms Ba
PARTICLES TO GRAMS
How many grams are in 4.5 x 1024
molecules of H2O?
– Use: 6.02 x 1023 molecules/mole
– Use: molar mass H2O  18.0 g/mole
4.5 x 1024 molc. H2O
18.0 g H2O
1 mole H2O
6.02 x 1023 molecules
H2O
1 mole H2O
= 134.55 or 130 g H2O
How many grams are in 3.29 x 1024
molecules of chlorine gas?
– Use: 6.02 x 1023 molecules/mole
– Use: molar mass of chlorine 71.0 g/mole
3.29 x 1024 molecule Cl2 1 mole Cl2
71.0 g Cl2
6.02 x 1023 moleculeCl2 1 mole Cl2
= 388 g Cl2
EMPIRICAL FORMULAS
Definition: the simplest whole # ratio for a
formula
You can calculate this if you know the %
composition of each element of a newly
made compound
STEPS FOR WRITING EMPIRICAL
FORMULAS
1. Given the grams or % composition
(cross off the % and change unit to grams)
2. Convert to moles for each element (use
molar mass)
3.Find ratio of moles compared to other
elements (divide by smallest # of moles)
4.From ratios, write the empirical formula
Example:
What is the empirical formula for a
compound that contains .900 g of Ca and
1.60 g of Cl?
– 1. Determine moles from molar mass for each
element
.900g Ca
1.60 g Cl
1 mole Ca
40.1 g Ca =
1 mole Cl
35.5 g Cl
.0224 mole Ca
=
.0451 mole Cl
– 2. Find simplest ratio (take smallest # of moles
found in step 1 and divide each by that)
• Ca = .0224
.0224
• Cl = .0451
.0224
=
1
=
2
– 3. Write empirical formula using ratios as
subscripts
CaCl2
FRACTIONS: if the ratios divide out to be
fractions, do the following:
– 1.5:1 ratio difference is .5  multiply both by 2
– If the difference is .33 or .67 multiply both by 3
– If the difference is .25 or .75 multiply both by 4
What is the empirical formula of a
compound that is 66.0 % Ca and 34.0 % P?
66.0g Ca
1 mole Ca
1.65 mole
40.1 g Ca
34.0 g P
1 mole P
31.0 g P
1.10 mole
Ca
P
1.65/1.10 = 1.5
1.10/1.10 = 1
Since there is a fraction, multiply both by 2
Ca 1.5 x 2 = 3
P
1 x 2 = 2
The formula is then Ca3P2 a 3:2 ratio
MOLECULAR FORMULA
Shows the actual # of atoms of each
element
– Example: Empirical = HO
Molecular = H2O2
To solve, you need one more piece of
information  the molar mass of the
molecular formula
Use that to find the whole # multiplier
The empirical formula of a compound is
found to be CH2O. The molar mass of the
molecular formula is 120.1 g/mole. What is
the molecular formula of the compound.
Example:
Calculate the empirical formula and
molecular formula for a compound that is
56.4% P and 43.7% O. The molar mass of
the molecular formula is 220 g/mole.
56.4 g P
1 mole
= 1.82 moles
31.0 g P
43.7 g O
1 mole
= 2.73 moles
16.0 g 0
Since 1.82 is the smallest amount of moles,
we divide both by that
–P
–O
1.82 / 1.82 = 1
2.73 / 1.82 = 1.5
Since we have a fraction, multiply both by 2
–P
1x2=2
–O
1.5 x 2 = 3
– The empirical formula is then P2O3
To find the molecular formula:
– Get molar mass of Empirical:
• P2O3 62 + 48.0 = 110g/mole
– Divide that into molar mass of molecular which
is given in the problem
• 220/110 = 2 which equal the whole # multiplier
Empirical = P2O3
Molecular = P4O6
HYDRATES
Crystals that contain water
Forms when crystals form in water solution
(water molecules stick to the crystal in a
specific ratio)
Ex: CuSO4 · 5H2O
HO
HO
2
2
Copper (II) sulfate
pentahydrate
H2O
CuSO4
H2O
H2O
Hydrate problems
Write the formula for the hydrate with
89.2% BaBr2 and 10.8% H2O.
89.2 g BaBr2
1 mole BaBr2
297.0 g BaBr2
10.8 g H2O
.300 .300
.300
1 mole H2O
18.0 g H2O
BaBr2 · 2 H20
.600
.600
.300
Barium bromide dihydrate
1
2
ANHYDROUS
Without water
Determine the hydrate formula for:
.391 g of Li2SiF6 and .0903 g H20
Determine the hydrate formula for:
76.9% CaSO3 and 23.1% H20
Ch. 10 Test
15 multiple choice: definitions, conversions
9 calculations
– Molar mass
– Percent composition
– Grams<-> moles, moles<-> particles, grams<->
particles
– Empirical and molecular formulas
– Hydrate formulas
Periodic table and conversion “cheat”
formula will be provided
I am a Mole
- Videos, more practice and tutorials on
webpage – Chapter Practice section
Popcorn
lab
(video slow mo) (how its made)
Lab prep
Data (2 brands- get
another groups info)
Calculations (show
work for both groups)
Questions 1-6 (use
articles
Conclusion (full- what
learned, error,
application)
1.
2.
3.
4.
5.
Calculations (show work-both
brands)
Mass of unpopped kernels=
Data #2 - Data #1
Mass of popped kernels =
Data #4 - Data #1
Difference in volume = Data
#5 - Data#3
Mass of water in popcorn=
Calc. #1 – Calc #2
% composition of water=
Calc #4 / Calc#1 x 100