The Mole
PART I: MOLE CONVERSIONS
PART II: % COMPOSITION, EMPIRICAL
FORMULAS, & MOLECULAR FORMULAS
Part I:
MOLAR CONVERSIONS
Counting Particles
 Different units are used to count different types of
objects:
DOZEN EGGS
PAIR OF SOCKS
REAM OF PAPER
Counting Particles
 Each counting unit, dozen, pair, ream, etc., is
appropriate for the kinds of objects they’re used to
count……
…..it is easier to buy and sell paper by the ream –
500 sheets – than by the individual sheet
Counting Particles
 Chemists also need a convenient method for accurately
counting the number of atoms, molecules, or formula
units in a sample of substance
 But atoms are SO small, it is impossible to count them
directly – THUS, chemists created a counting units
called the “MOLE”
THE MOLE
 Chemists use the MOLE to count atoms, molecules,
ions and formula units
 THE MOLE abbreviated “mol” is the SI base unit used
to measure the amount of a substance.
 The MOLE is defined as the number of carbon atoms
in exactly 12.00 g of pure carbon-12.
THE MOLE
 Through experimentation,
it has been established
that a MOLE of ANY SUBSTANCE contains
6.0221367 x 1023 representative particles
A representative particles is any kind of particle,
such as an atom, a molecule, a formula unit, an
electron or an ion.
THE MOLE
The representative particles in a mole of water is a
water molecule
The representative particle in a mole of copper is the
copper atom
The representative particle in a mole of
sodium chloride is the NaCl formula unit
THE MOLE
6.0221367
x 1023
This number is known as Avogardro’s number =
602,213,670,000,000,000,000,000
In this class, we abbreviate Avogardro’s number
6.02 x 1023
THE MOLE
Avogardro’s number would NOT be a convenient way to
count something like marbles….
A mole of marbles would cover the surface of Earth to
a depth of more than SIX KILOMETERS…..
If you spent a billion dollars every second
you couldn’t spend a mole of dollars in
your lifetime!
CONVERTING BETWEEN MOLES AND
PARTICLES
Dimensional Analysis/Factor Label
To find the number of particles in 3.5 moles of
glucose….(the representative particle here is the
molecule)
1. Use the relationship: 1 mol = 6.02 x 1023
2. Develop conversion factors:
1 mol
6.02 x 1023
or
6.02 x 1023
1 mol
What we call “particles”
If you have…
We call the particles…
Element
Atoms
Ionic Compound
(metal+non-metal)
Covalent Compound
(non-metal + non-metal)
Formula units
molecules
CONVERTING BETWEEN MOLES AND
# PARTICLES
QUESTION: 3.5 moles of glucose = ? # molecules
3. Set up dimensional analysis and choose the
conversion factor that will solve for the unknown
1 mol
or
6.02 x 1023 molecules
3.5 mol glucose
6.02 x 1023 molecules
1 mol
6.02 x 1023 molecules =
1 mol glucose
21.07 x 1023 molecules = 2.1 x 1024 molecules
How many molecules are in 7.5
moles of carbon dioxide?
A sample of lead contains 1.5 x
1024 atoms. How many moles
of lead atoms are there?
A sample of zinc chloride
contains 2.150 x 1018 formula
units. How many moles of zinc
chloride are there?
MASS AND THE MOLE
A mole of anything always contains the same number of
particles, however, a mole of different substances
have different masses.
A dozen apples would not have the same mass as a
dozen grapes…..
If you put one mole of carbon atoms and one mole of
copper atoms on separate scales, they, too, would
have different masses
USING MOLAR MASS
Suppose you were asked to count the number of candies
in a large bag of skittles……
You could take a FEW minutes and count them one by
one…..
USING MOLAR MASS
OR you could “count” by massing the candy….
If one piece of candy has a mass of 1 g, and the whole
bag of skittles less the mass of the bag is 691 g,
how many skittles are in the bag?
USING MOLAR MASS
We can do the same thing with atoms. They are TOO small
to count, but if we know the mass of one particle, and the
mass of the whole sample, we can “count” the number of
particles present in the sample.
Instead of skittles,
we have atoms, molecules
ions, or formula units
MOLAR MASS
The mass of one mole of any pure substance is called
its MOLAR MASS
The molar mass of any element is numerically equal to
its atomic mass and has the units g/mol
(A mole is defined as the number of carbon-12 atoms
in exactly 12 g of pure carbon-12, thus 12 g of
carbon have 6.02 x 1023 atoms present)
MOLAR MASS
The samples below represent one mole of several
elements:
1 mol Hg = 200.59 g Molar Mass of Hg = 200.59 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of Fe = 55.85 g/mol
Molar mass of C = 12.01 g/mol
CONVERTING MOLES TO MASS
Recall from the previous page……
1 mol Hg = 200.59 g Molar Mass of Hg = 200.59 g/mol
Suppose you had 2.5 mol of Hg, what is its mass?
2.5 mol Hg
93.8 g Hg
200.59 g = 500 g
1 mol Hg
1 mol Hg = 0.468 mol
200.59 g
CONVERTING MASS TO MOLES
Recall from the previous page……
1 mol Hg = 200.59 g Molar Mass of Hg =
200.59 g/mol
Suppose you had 341 g of Hg, how many mols of
Hg do you have?
341 g Hg
1 mol Hg = 341 mol = 1.70 mol
200.59 g
200.59
THE MOLAR MASS OFCOMPOUNDS
Recall that a chemical formula indicates the numbers
and types of atoms contained in one unit of the
compound.
Ex. CF4
1 carbon atom
(12.01 amu)
4 fluorine atoms
(19.00 amu)
What mass would one molecule of CF4 have?
1 C (12.01 amu) + 4 F (19.00 amu x 4) =
1 molecule would have a mass of 88.01 amu
THE MOLAR MASS OF COMPOUNDS
What mass would one molecule of CF4 have?
1 C (12.01 amu) + 4 F (19.00 amu x 4) =
1 molecule would have a mass of 88.01 amu
If one molecule of CF4 has a mass of 88.01 amu, how
many grams would one mole of CF4 have?
1 mole of CF4 has 1 mole of C atoms and 4 moles of F
atoms: 12.01 g + 76.00 g = 88.01 g/mol
DETERMINING THE MOLAR MASS OF A
COMPOUND
1.
2.
3.
4.
Start with the CORRECT formula of the compound
Look up the molar mass of each element present in
the compound
Multiply each molar mass by the number of moles
of that element in the chemical formula (same as
the subscript)
Add up the mass of each element
CONVERTING THE MOLES OF A COMPOUND
TO MASS
Dimensional analysis can be used to convert between
the mols and mass of a compound
EX. Convert 3.25 mol H2SO4 to grams
3.25 mol H2SO4
98.09 g = 319 g
1 mol H2SO4
CONVERTING THE MASS OF A COMPOUND
TO MOLES
Dimensional analysis can be used to convert between
mass and mols of a compound
EX. Convert 84.07 g H2SO4 to mols
84.07 g H2SO4
1 mol H2SO4 = 0.8570 mol
98.09 g
What is the mass of 0.25 moles
of chromium?
What is the molar mass of
lithium nitrate, LiNO3?
What is the molar mass of
calcium acetate, Ca(C2H3O2)2?
How many moles are in 423.1 g
of calcium acetate, Ca(C2H3O2)2?
How many grams are in 12.8
moles of calcium acetate,
Ca(C2H3O2)2?
MOLE RELATIONSHIPS FROM A
CHEMICAL FORMULA
Al2O3: If you had 1.50 mol of Al2O3, how many mols
of Al3+ ions are present?
1.50 mol Al2O3
2 mol Al3+ ion s = 3.0 mol Al3+
1 mol Al2O3
If you had 4.80 mol of Na2SO4,
how many moles of Na1+ are
present? (SO4)2-?
CONVERTING THE MASS OF A COMPOUND TO
NUMBER OF PARTICLES
# particles  mols  grams
To convert from grams to number of particles, one has
to go through “Mole-ville”
Ex. 2.35 g Mg = ? # atoms of Mg
2.35 g Mg
1 mol Mg
24.31 g
6.02 x 1023 atoms Mg =
1 mol
5.82 x 1022 atoms Mg
CONVERTING THE NUMBER OF PARTICLES OF A
COMPOUND TO MASS
# particles  mols  grams
To convert from number of particles to grams,
one has to go through “Mole-ville”
Ex. 8.31 X 1023 molecules of water = ? grams
8.31 x 1023 molecules
1 mol H2O
6.02 x 1023
24.9 g H2O
18.02 g =
1 mol H2O
How many molecules of
ammonia, NH3 are in a 68-g
sample?
What is the mass of the oxygen
atoms in 4 moles of sodium
sulfate?
Mole Conversion Chart
Molar Conversions
molar
mass
6.02  1023
MASS
NUMBER
MOLES
IN
GRAMS
OF
PARTICLES
(g/mol)
(particles/mol)
Part II
% COMPOSITION, EMPIRICAL FORMULAS, &
MOLECULAR FORMULAS
Percentage Composition
 the percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass of compound
 Be sure to include the total mass of the
element (so for like CuCl2, include 2 Cl
masses)
Percentage Composition
 Find the % composition of Cu2S.
%Cu =
%S =
127.10 g Cu
 100 =
159.17 g Cu2S
79.852% Cu
32.07 g S
159.17 g Cu2S
 100 =
20.15% S
Percentage Composition
 Find the percentage composition
of a sample that is 28 g Fe and 8.0
g O.
28 g
 100 = 78% Fe
%Fe =
36 g
%O =
8.0 g
36 g
 100 = 22% O
Percentage Composition
 How many grams of copper are in
a 38.0-gram sample of Cu2S?
Cu2S is 79.852% Cu
(38.0 g Cu2S)(0.79852) = 30.3 g Cu
Percentage Composition
 Find the mass percentage of water
in calcium chloride dihydrate,
CaCl2•2H2O?
%H2O =
36.04 g
 100 = 24.51%
H2O
147.02 g
Empirical Formula
 Smallest whole number ratio of
atoms in a compound
C 2H 6
reduce subscripts
CH3
Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element by using
the molar mass
3. Divide moles by the smallest # of
moles calculated in step 2 to find
subscripts.
4. When necessary, multiply subscripts
by 2, 3, or 4 to get whole #’s.
Empirical Formula
 Find the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
=1N
1.85 mol
14.01 g
74.1 g 1 mol
= 4.63 mol O
= 2.5 O
16.00 g
1.85 mol
Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Molecular Formula
 “True Formula” - the actual
number of atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the
empirical mass.
4. Multiply each subscript by the
answer from step 3.
MF mass
n
EF mass
EF n
Molecular Formula
 The empirical formula for ethylene
is CH2. Find the molecular formula
if the molecular mass is
28.1 g/mol?
empirical mass = 14.03 g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4