The Mole
The Mole
• A mole is defined as the number of particles in
exactly 12g of Carbon-12. This number = 6.02 x
1023 particles.
• 6.02 x 1023 is called Avagadro’s Number after
Amadeo Avagardro who determined the volume
of 1 mol of gas in 1811
• Memorize this number:
– 1 mol = 6.02 x 1023 of something
The Mass of a Mole
• Atomic Mass is the relative mass of single atom of an element in
amu
• Atomic masses on the periodic table can be used to determine
molar masses.
• Molar Mass is the mass of 1 mole of something in grams
• If the atomic mass of carbon is 12.011 amu, then the molar mass of
carbon is 12.011 g/mol
• IN SHORT: The molar mass of any element is exactly equal to its
atomic mass – just change the units
Moles of Compounds
• In a mole of a chemical compound (molecule or
formula unit), there is also one or more moles of
each atom.
• For example:
– In 1 mol of CCl2F2 (Freon), there is:
• 1 mol C
• 2 mol Cl
• 2 mol F
– If you have 2 mol of MgCl2, there is:
• 2 mol Mg
• 4 mol Cl
• (Multiply the coefficient x the subscript!)
The Mole: In Short
• A mole is 6.02 x 1023 of anything, just like a
dozen is 12 of anything.
• Atomic mass units (amu) is a way of
comparing the sizes of different elements.
• atomic mass in amu = molar mass in g
Using the Mole
• You should be able to:
1.
2.
3.
4.
5.
6.
Convert Moles to Particles
Convert Particles to Moles
Convert Moles to Mass
Convert Mass to Moles
Convert Mass to Particles
Convert Particles to Mass
– Determine percent composition
– Determine empirical formula
– Determine molecular formula
• We are going to practice this a lot!!!!!
Preliminary: Moles to Mass of an Element
• Determine the numbers of moles of each element in 1 mole
of CaF2.
– 1 mole of Ca
– 2 mole of F
• Determine the number of grams of each element in 1 mol of
CaF2.
– 1 mol Ca x 40.1 g/mol = 40.1 grams Ca
– 2 mol F x 18.9 g/mol = 37.8 grams F
1 mol Ca 40.1g

 40.1g Ca
1
1 mol
Preliminary: Moles to Mass of Molecule
• Determine the number of grams of each element in 1 mol of
CaF2.
– 1 mol Ca x 40.1 g/mol = 40.1 grams Ca
– 2 mol F x 18.9 g/mol = 37.8 grams F
• What is the molar mass of CaF2?
– 1 mol Ca + 2 mol F = 1 mole CaF2
– 40.1 g Ca + 37.8g F = 77.9 g CaF2
Preliminary: Mass of a Compound
To find the mass of a compound, simply add up
the molar mass of each of its atoms. For
example, calculate the molar mass of CO2.
- molar mass C  12.01 g
- molar mass O  16.00 g
- 1 mol C  2 mol O  1 mole CO2
12.01 g
16.00 g

2(
)  44.01 g/mole CO2
1 mol
1 mol
Reminders!!
• Particles can mean: atoms, molecules,
formula units, etc.
• Covalent compounds = molecules
• Ionic compounds = formula units
• Use the factor-label method!!!
• Learn how to multiply and divide with
exponents on your calculator! (Check with me
if you need help).
The Flow Chart
Part
Pieces
Atoms
Molecules
Ions
Etc.
÷ Avo’s #
x Molar Mass
Mass (g)
Moles
x Avo’s #
÷ Molar Mass
#1 Moles to Particles
Determine the number of formula units in 4.3
moles of NaCl.
1. Multiply the number of moles by the
appropriate conversion factor for Avogadro’s
number. (Multiply by Avogadro’s #).
4.3 mol 6.02 10

1
1 mol
23
 2.59 10 formula units
24
Practice: Moles to Particles
1. Determine the number of atoms in 2.50 mol
Zn.
2. Determine the number of formula units (i.e.,
particles of an ionic compound) in 3.25 mol
AgNO3.
3. Determine the number of molecules in 11.5
mol H2O.
Practice: Moles to Particles (x)
1. Determine the number of atoms in 2.50 mol
Zn. 1.51 x 10
2. Determine the number of formula units (i.e.,
particles of an ionic compound) in 3.25 mol
AgNO3. 1.96 x 10
3. Determine the number of molecules in 11.5
mol H2O. 6.92 x 10
NOTE: DON’T FORGET SIGNIFICANT DIGITS!!!
24
24
24
#2 Particles to Moles
• How many moles are there in 5.64x1025 atoms
of Fe?
• Multiply the number of atoms by the
appropriate conversion factor for Avogadro’s
number. (Divide by Avogadro’s #).
5.64 10 atoms
1 mole

 93.6 moles
23
1
6.02 10 atoms
25
Practice: Particles to Moles
How many moles contain each of the following?
1. 5.75x1024 atoms of Al
2. 3.75x1024 molecules of CO2
3. 3.58x1023 formula units of ZnCl2
4. 2.50x1020 atoms of Fe
5. 3.4x1017 atoms of Na
Practice: Particles to Moles
How many moles contain each of the following?
1. 5.75x1024 atoms of Al 9.55 mol
2. 3.75x1024 molecules of Co2 6.23 mol
0.594 mol
3. 3.58x1023 formula units of ZnCl2
4.15x10 mol
20
4. 2.50x10 atoms of Fe
5.64x10 mol
17
5. 3.4x10 atoms of Na
-4
-7
Practice
• WS #1 Moles to Particles
#3 Moles to Mass
If you have 0.0450 mol of Cr, how many grams
do you have?
1. Determine the molar mass of Cr from the periodic
table: 52.00 g/mol (same number as the atomic
mass)
2. Multiply the number of moles by the conversion
factor. (Multiply by the molar mass).
0.0450 mol Cr 52.0 g C

r  2.34 g Cr
1
1 mol
Practice: Moles to Mass (x)
Convert from moles to mass:
1. 3.57 mol Al
2. 42.6 mol Si
3. 3.45 mol Co
4. 2.45 mol Zn
5. 8.2 mol Pb
Practice
Convert from moles to mass:
1. 3.57 mol Al 96.3 g
2. 42.6 mol Si 1.2 x 10 g
3. 3.45 mol Co 203 g
4. 2.45 mol Zn 1.60 x 10 g
5. 8.2 mol Pb 1.7 x 10 g
3
2
3
#4 Mass to Moles
How many moles of Ca are in 525g of Ca?
1. Determine the molar mass of Ca from the
periodic table: 40.08 g/mol
2. Multiply the given mass (525g) by the
conversion factor. (Divide by the molar mass).
525 g Ca 1 mol

 13.1 mol Ca
1
40.08 g
Practice: Mass to Moles
Convert from mass to moles:
1. 25.5 g Ag
2. 300.0 g S
3. 125 g Zn
4. 99g O
5. 1.0 kg Fe
Practice: Mass to Moles
Convert from mass to moles:
1. 25.5 g Ag 0.236 mol
2. 300.0 g S 9.355 mol
3. 125 g Zn 1.91 mol
4. 99g O 6.2 mol
5. 1.0 kg Fe 17.9 mol
Practice
• WS #2 Mole to Mass
• WS #3 Mixed One-Step Mole WS
Multi-Step Problems
How many molecules are there
in 24 g of FeF3?
How many grams are there in 7.5
x 1023 molecules of AgNO3?
1.
1.
2.
3.
Calculate the molar mass of
FeF3
Divide the mass you have by the
molar mass. This gives you the
number of moles.
Multiply the number of moles
by Avogadro's number.
2.
3.
Calculate the molar mass of
AgNO3.
Divide the number of molecules
by Avogadro’s number. This gives
you the number of moles.
Multiply the number of moles by
the molar mass.
1. Fe (55.8 g)  3F (3 x 19.0)  55.8 g  57.0 g  112.8 g 1. Ag (107.9 g)  N (14.0 g)  3O (3 x 16.0)  169.9 g/mol
2. 24 g  112.8 g/mole  0.21 mol
2. 7.4 x 10 23 atoms  6.02 x 10 23 atoms/mol  1.2 mol
3. 0.21 x 6.02 x 10 23  1.26 x 10 23 atoms
3.1.2 mol x 169.9 g/mol  209 g
#5 Mass to Particles
Calculate the number of atoms in a 25.0 g nugget
of pure gold (Au).
1. Convert mass to moles (mass ÷ molar mass)
2. Convert moles to atoms (moles x 6.02 x 1023)
25g Au
1.
 0.127 mol Au
196.7 g/mol Au
2. 0.127 mol Au  6.02  10 23 atoms/mol  7.65  10 22 atoms Au
Practice: Mass to Particles
• How many atoms are there in 14 g of C?
#6 Particles to Mass
What is the mass of 5.50 x 1022 He atoms?
1. Convert atoms to moles (atoms ÷ 6.02 x 1023)
2. Convert moles to mass (moles x molar mass)
5.50  10 atoms He
1.
 0.0914 mol He
23
6.02  10 atoms/mol
2. 0.0914 mol He  4.00 g/mol He  0.366 g He
22
Practice: Particles to Mass
• What is the mass of 7.6 x 1024 atoms of Ca?
Practice: Mass to Atoms and Atoms to
Mass
Convert to Atoms
1. 55.2 g Li
2. 0.230 g Pb
3. 11.5 g Hg
4. 45.6 g Si
5. 0.120 kg Ti
Convert to Mass
1. 6.02 x 1024 atoms Bi
2. 1.00 x 1024 atoms Mn
3. 3.40 x 1022 atoms He
4. 1.50 x 1015 atoms N
5. 1.50 x 1015 atoms U
Practice: Mass to Atoms and Atoms to
Mass
Convert to Atoms
1. 55.2 g Li 4.79 x 1024 atoms
2. 0.230 g Pb 6.68 x 1020 atoms
3. 11.5 g Hg 3.45 x 1022atoms
4. 45.6 g Si 9.77 x 1023atoms
5. 0.120 kg Ti 1.51 x 1024 atoms
Convert to Mass
1. 6.02 x 1024 atoms Bi 2.09 x 103 g
2. 1.00 x 1024 atoms Mn 91.3 g
3. 3.40 x 1022 atoms He 0.226 g
4. 1.50 x 1015 atoms N 3.49 x 10-8 g
5. 1.50 x 1015 atoms U 5.93 x 10-7 g
Practice
• WS #4 Two-Step Mole Conversions
• WS #5 Mixed Problems w/ Naming 
• WS #6 Mixed Problems
Percent Composition by Mass
• The percent composition by mass tells you
what percent an element is out of a
compound.
• For example:
– Hydrogen (H) is 11.2% of H2O by mass
mass of element

 100%  percent by mass
mass of compound
2.02 g H

 100%  11.2 % H
18.02 g H 2 O
The Steps: Percent by Mass
1. Find the molar mass of the elements in the
compound.
2. Find the molar mass of the compound.
3. Divided the molar mass of each element by
the molar mass of the compound and
multiply by 100%.
– (Make sure you multiply the mass of the element
by the number of moles present in the
compound)
Example:
Find the percent composition of each of the
elements in NaHCO3
molar mass of NaHCO 3  84.01 g/mol
1 mole Na  22.9 g/mol Na 22.9

 100%  27.37% Na
84.01 g/mol NaHCO 3
84.01
1 mole H  1.008 g/mol H 1.008

 100%  1.200% H
84.01 g/mol NaHCO 3
84.01
1 mole C  12.01 g/mol C 12.01

 100%  14.30% C
84.01 g/mol NaHCO 3
84.01
3 mol O  16.00 g/mol O 48.00

 100%  57.15% O
84.01 g/mol NaHCO 3
84.01
Practice
1. Determine the percent by mass of each
element in calcium chloride (CaCl2).
2. Calculate the percent by mass of each
element in sodium sulfate (Na2SO4)
3. Calculate the percent composition of
phosphoric acid (H3PO4).
Practice
1. Determine the percent by mass of each
element in calcium chloride (CaCl2).
– Ca = 36.11%, Cl = 63.89%
2. Calculate the percent by mass of each
element in sodium sulfate (Na2SO4)
– Na = 32.37%, S = 22.58%, O = 45.05%
3. Calculate the percent composition of
phosphoric acid (H3PO4).
– H= 3.08%, P = 31.61%, O = 65.31%
Practice: Percent Composition
• WS #7 Percent Composition by Mass
• WS #8 Percent Composition Hard
Molecular Formula vs. Empirical Formula
• A molecular formula indicates the actual number of
elements in a covalent compound.
• We call the mass (in amu) of the molecular formula the
molecular mass.
– Remember: we can convert molecular mass to molar mass by
simply changing the units (amu  g/mol).
• The empirical formula for a molecule is the smallest wholenumber ratio of the elements. It may or may not be the
same as the molecular formula.
• We call the mass (in amu) of the empirical formula the
formula mass.
– Remember: we can convert the formula mass to a molar mass
simply by changing the units (amu  g/mol).
A Few Key Conceptual Points:
• The molecular formula can be the same as or
different from the empirical formula.
• Ionic compounds ONLY have empirical formulas
(lowest whole number ratios) because of their
crystal lattice structure.
• Only covalent compounds can have molecular
formulas that differ from their empirical
formulas.
• For covalent compounds, it is the molecular
formula that occurs in real life.
Example: Empirical vs. Molecular
• The molecular formula for glucose (a product
of photosynthesis) is C6H12O6.
• The empirical formula is CH2O.
• The molecular formula for glucose is 6 times
the empirical formula.
2 New Skills to Learn
1. How to calculate empirical formula from
percent composition.
2. How to calculate molecular formula from
empirical formula and molar mass.
#1 Calculating Empirical Formula from
Percent Composition
1. Assume you have a 100 g sample and convert
your percentages to masses.
2. Divide the masses that you have from step #1 by
their molar masses to determine the mole
ratios.
3. Calculate the simplest ratio by dividing each
mole amount from #2 by the smallest mole
amount that you calculated from #2.
4. Convert to the smallest whole-number ratio by
multiplying all molar amounts by a common
multiple. (No parts of atoms rule).
Example: Calculating Empirical
Formula from Percent Composition
• Example: Imagine that you know the percent
composition of a compound, ethane.
• Problem: Determine the empirical formula
from the percent composition:
• C = 79.9%
• H = 20.1%
The Steps
1. Assume you have 100 g
–
79.9% C = 79.9g C
and
20.1% H = 20.1g H
2. Divide the mass of each element by its molar mass.
–
–
(79.9g ÷ 12.01 g/mol) = 6.65 mol
(20.1g ÷ 1.01g/mol) = 19.9 mol
3. Divide the molar amounts from step 2 by the smallest
molar amount (6.65 mol).
–
–
(6.65 mol C ÷ 6.65 mol) = 1 C
(19.9 mol H ÷ 6.65 mol) = 3 H
4. Determine the ratio of elements and convert to whole
numbers by multiplying by a common multiple if
necessary.
–
Therefore, the empirical formula for ethane is CH3
Practice: Empirical Formula from
Percent Composition
• Determine the empirical formula for the
following compounds based on the percent
composition given.
1. 36.84% nitrogen, 63.16% oxygen N O
2. 35.98% aluminum, 64.02% sulfur Al S
3. 81.82% carbon, 18.18% hydrogen C H
2
3
2 3
3 8
Skill 2: Molecular Formula from
Empirical Formula
• A covalent compounds molecular formula is
some integer multiple of the empirical
formula.
– i.e., Molecular formula = n(empirical formula)
• Where n= natural numbers (1, 2, 3, etc.)
• The molecular formula is found by comparing
the ratio of the molecular mass to the formula
mass.
Example
• Molecular mass of acetylene = 26.04 amu
• Formula mass of acetylene = 13.02 amu
• The ratio of molecular mass to formula mass:
– (26.04 amu)/(13.02 amu) = 2
• Therefore: the molecular mass is 2x the
formula mass, so the molecular formula has
2x the number of each elements found in the
empirical formula.
The Steps
Note: You will be given the molecular mass!!
Otherwise, you can’t figure this out!!
1. Determine the empirical formula of the
compound.
2. Determine the formula mass of the empirical
formula.
3. Divide the molecular mass by the formula mass.
– (molar mass) ÷ (formula mass) = ????
4. Multiply the subscripts of each element in the
empirical formula by the number determined in
step 3. You’re done.
Example:
• Determine the molecular formula for succinic acid,
which is 40.68% C, 5.08% H, and 54.24% O. It has an
experimentally determined molecular mass of 118.1
amu.
1. Find the empirical formula.
–
C2H3O2
2. Calculate the formula mass of the empirical formula.
–
59.05 amu
3. Divide the molecular mass by the formula mass.
– (118.1 amu) ÷ (59.05 amu) = 2
4. Multiply the empirical formula by the whole number
you found in step 3.
–
2(C2H3O2) = C4H6O4
Practice Problems
1. A compound with a molar mass of 60.01
g/mol is 46.68% N and 53.32% O. What is its
molecular formula? N O
2 2
2. The experimentally determined molar mass
of an unknown compound is 110.0 g/mol,
and it is 65.45% C, 5.45% H, and 29.09% O.
What is its molecular formula?
C6H6O2
Practice
• WS #9 Empirical & Molecular Formula
• WS #10 Empirical & Molecular
Hydrates
• Some salts have the ability to bind water molecules
within their lattice structure. These compounds are
known as hydrated salts or hydrates, written as CoCl2 •
6H2O. The salt with no water in the crystal lattice is
called an anhydrous salt CoCl2.
• Heating a hydrated salt will drive off the water
molecules drying out the crystal and forming the
anhydrous salt.
• Anhydrous salts will absorb water from the air. They are
used to keep products dry by inserting little packets into
shoe boxes, cameras etc.
Formula For a Hydrate
• In the formula for a hydrate, the number of water
molecules associated with each formula unit of
the compound is written following a dot.
• Example:
– Sodium carbonate decahydrate
– Na2CO3 ∙ 10H2O
• The number of water molecules associated with
the formula unit in the compound are indicated
in the prefix.
• The mass of water associated must always be
included in the molar mass
% Composition of Hydrates
% Composition of hydrate tells you how much of the hydrate is salt, and
how much is water.
It’s just like the other percent composition calculations.
1.
Calculate the molar masses for the anhydrous salt, the water and the
total mass.
2.
Divide the molar mass of each part (salt, water) by the total molar
mass and multiply by 100%. Done.
•
% = Part / whole X 100
– % H2O = (mass H2O) ÷ (total mass) x 100
– % salt = (mass salt) ÷ (total mass) x 100
Practice Problem
1. What is the % composition of CaSO4 · 2H2O?
H20 = 21%
CaSO4 = 79%
Analyzing a Hydrate
• To determine the formula for a hydrate you must
determine the moles of water associated with
each mole of hydrate.
• The water is boiled off, and the before and after
masses of the compound are compared. The
difference is the mass of the water in the
compound.
• The differences in grams are converted to molar
masses and the ratio of the formula unit to water
is determined.
Hydrate Example
• You start with 5.00 g of BaCl2 · xH2O. After heating there is only 4.26 g of
BaCl2. What is the formula for the hydrate?
1.
Find the difference between the two masses. This is the mass of
water.
– 5.00g – 4.26g = 0.74g H2O
2.
Divide the mass of each (salt, water) by their molar masses. This
gives you the # of moles.
– BaCl2 : (4.26g)/(208.23g/mol) = 0.0205 mol BaCl2
– H2O: (0.74g)/(18.02g/mol) = 0.041 mol H2O
3.
Divide the moles of water by the moles of salt. This gives you the
mole ratio of the hydrate. You’re done.
– (0.041 mol)/(0.0205 mol) = 2/1 (Note: if you get decimals, round).
– So there are 2 moles of water per mole of barium chloride
– Barium chloride dihydrate = BaCl2 ∙ 2H2O
Practice Problems
1. During a lab, 1.62g of CoCl2 · xH2O is heated.
After heating, 0.88g CoCl2 remains. What is
the formula for the original hydrate?
CoCl2 · 6H2O
Practice
•
•
•
•
WS #11 Hydrate Problems
Mole PRACTICE Quiz
Mole Quiz
Mole Test Review