CHEMICAL QUANTITIES:
THE MOLE
MEASURING MASS
 A mole is a quantity of things, just as…
1 dozen = 12 things
1 gross
= 144 things
1 mole
= 6.02 x 1023 things
 “Things” usually measured in moles are
atoms, molecules, ions, and formula
units
 You can measure
mass, or volume, or
you can count pieces
 We measure mass in
grams
 We measure volume
in liters
 We count pieces in
MOLES
A MOLE…
 is an amount, defined as the number of
carbon atoms in exactly 12 grams of carbon12
 1 mole = 6.02 x 1023 of the representative
particles
 Treat it like a very large dozen
6.02 x 1023 is called: Avogadro’s number
 Similar Words for an amount:
 Pair: 1 pair of shoelaces = 2 shoelaces
 Dozen: 1 dozen oranges = 12 oranges
 Gross: 1 gross of pencils= 144 pencils
 Ream: 1 ream of paper= 500 sheets of
paper
What are Representative
Particles?
 The smallest pieces of a substance:
1. For a molecular compound: it is the
molecule.
2. For an ionic compound: it is the formula
unit (made of ions)
3. For an element: it is the atom
 Remember the 7 diatomic elements? (made
of molecules)
 How many oxygen atoms in the following?
1. CaCO3
2. Al2(SO4)3

3 atoms of oxygen
12 (4 x 3) atoms of oxygen
How many ions in the following?

CaCl2
 3 total ions (1 Ca2+ ion and 2 Cl1- ions)
 NaOH
 2 total ions (1 Na1+ ion and 1 OH1- ion)
 Al2(SO4)3
 5 total ions (2 Al3+ + 3 SO4 ions)
CONVERSION FACTOR
 MOLES = representative particles x
____________1 mole_____________
6.02 x 1023 representative particles
EXAMPLES: ATOMS  MOLES
 How many atoms of Al are in 1.5 mol of Al?
 Conversion: 1 mole = 6.02 x 1023 atoms
1.5 mol of Al 6.02 x 1023 atoms
1 mole
23 atoms
9.03
x
10
=
of Al
EXAMPLES: MOLECULES  MOLES
How many atoms of H are there in 3 moles of
H2O?
Conversions:
1 mole = 6.02 x 1023 molecules
H2O molecule = 2 atoms of Hydrogen
3 moles of H2O 6.02 x 1023 molec
1 mole
2 atoms H
1 H2O molecule
= 3.612 x 1024 atoms H
MOLAR MASS
Determined simply by looking at the periodic chart
Molar mass = Atomic Mass
20
Ca
40.08
* Thus,
1 mol Ca = 40 g
Atomic Mass same as Molar
Mass
MOLAR MASS
 To calculate the molar mass of a compound,
find the number of grams in each element in
one mole of the compound
 Then add the masses within the compound
Example: H2O
H= 1.01
2 (1.01) + 1 (15.999)= 18.02
O= 15.999
SOME PRACTICE PROBLEMS
 How many atoms of O are in 3.7 mol of O?
 2.2 X 1024 atoms of oxygen
 How many atoms of P are in 2.3 mol of P?
 1.4 x 1024 atoms of phosphorus
 How many atoms of Ca are there in 2.5 moles
of CaCl2?
 1.5 x 1024 atoms Ca
 How many atoms of O are there in 1.7 moles of
SO4?
 4.1 x 1024 atoms of oxygen
Remember!!!!
 The molar mass of any substance (in grams)
equals 1 mole
 This applies to ALL substance: elements,
molecular compounds, ionic compounds
 Use molar mass to convert between mass and
moles
 Ex: Mass, in grams, of 6 mol of MgCl2 ?
mass of MgCl2 = 6 mol MgCl2
= 571.26 g MgCl2
92.21 g MgCl2
1 mol MgCl2
VOLUME AND THE MOLE
 Volume of 1 mol of solid and liquids are not
the same
 But gases are more predictable, under the
same physical conditions
 Avogadro’s hypothesis helps explain:
equal volume of gases, at the same temp and
pressure contains equal number of particles
 Volume varies with changes in temperature
 Ex: helium balloon
 Gases vary at different temperatures, makes
it hard to measure
 Because of variation use STP
 Standard Temperature and Pressure
 Temperature = 0° C
 Pressure = 1 atm (atmosphere) or 101.3 kPal
Standard Temperature
Pressure
 At STP:
1 mole, 6.02 x 1023 atoms, of any gas has a
volume of 22.4 L
 Called Molar Volume
 Used to convert between # of moles and vol of a
gas @ STP
 Ex: what is the vol of 1.25 mol of sulfur
Vol of S= 1.25 mol S 22.4 L = 28 L
1 mol
MOLAR MASS FROM DENSITY
 Different gases have different densities
 Density of a gas measured in g/L @ a specific
temperature
 Can use the following formula to solve :
grams = grams X 22.4 L
mole
L
1 mole
 Ex: Density of gaseous compound containing oxygen
and carbon is 1.964 g/ L, what is the molar mass?
 grams = 1.964 g X
22.4 L then you solve
mole 1 L
1 mole
= 44.o g/mol
Calculating Percent Composition
of a Compound
 Like all percent problems: a part ÷ the whole
1. Find the mass of each of the components (the
elements)
2. Next, divide by the total mass of the compound
3. Then X 100 % = percent
Formula:
% Composition = Mass of element X 100%
Mass of compound
Example:
A compound is formed when 9.03 g of Mg combines
completely with 3.48 g of N.
What is the percent composition of the compound?
1. First add the 2 mass of the 2 compounds to reach
the total mass 9.03 g Mg + 3.48 g N =
12.51 g Mg3N2
1. Find the % of each compound
% Mg= 9.03 g Mg X 100% = 72.2 %
12.51 g Mg3N2
% N= 3.48 g N X 100%
= 27.8 %
12.51 g Mg3N2
% Composition from Chemical
Formula
 Can find the percent composition of a compound
using just the molar mass of the compound and the
element
 % mass=mass of the element 1 mol cmpd X100%
molar mass of the compound
 Example:
Find the percent of C in CO2
12.01 g C X 100% = 27.3% C
44.01 g CO2
Can find O % by subtracting 27.3% from 100%
Using % Composition
 Can use % composition as a conversion factor just
like the mole
 After finding the % comp. of each element in a
cmpd. can assume the total compound = 100g
 Example: C= 27.3%
27.3 g C
O= 72.7 %
72.7 g O
 In 100 g sample of compound there is 27.3 g of C & 72.7 g
of O
How much C would be contained in 73 g of CO2?
73 g CO2 27.3 g C
= 19.93 g C
100 g CO2
EMPIRICAL FORMULAS
 Empirical formulas are the lowest
WHOLE number ratios of
elements contained in a
compound
REMEMBER…
 Molecular formulas tells the actual number of
of each kind of atom present in a molecule of
the compound
 Ex:
H2O2
HO
Molecular
Formula
Empirical
Formula
CO2
CO2
Molecular
Formula
Empirical
Formula
For CO2 they are the same
 Formulas for ionic compounds are
ALWAYS empirical (the lowest whole
number ratio = can not be reduced)
 Examples:
NaCl MgCl2 Al2(SO4)3 K2CO3
Simplest whole number ratio for NaCl
 A formula is not just the ratio of atoms, it is also the
ratio of moles
 In 1 mole of CO2 there is 1 mole of carbon and 2 moles
of oxygen
 In one molecule of CO2 there is 1 atom of C and 2
atoms of O
 Formulas for molecular compounds MIGHT be
empirical (lowest whole number ratio)
 Molecular:
H2O
C6H12O6
H2O
CH2O
C12H22O11
(Correct formula)
 Empirical:
(Lowest whole
number ratio)
C12H22O11
CALCULATING EMPIRICAL
 We can get a ratio from the percent
composition
1. Assume you have a 100 g sample the
percentage become grams (75.1% = 75.1 grams)
2. Convert grams to moles
3. Find lowest whole number ratio by dividing
each number of moles by the smallest value
Example calculations
 Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and 45.11 %N
 Assume 100 g sample, so
38.67 g C x 1 mol C = 3.22 mole C
12.0 g C
16.22 g H x 1 mol H = 16.22 mole H
1.0 g H
45.11 g N x 1 mol N = 3.22 mole N
14.0 g N
*Now divide each value by the smallest value
…Example 1
 The ratio is 3.22 mol C = 1 mol C
3.22 mol N 1 mol N
 The ratio is 16.22 mol H = 5 mol H
3.22 mol N 1 mol N
C1H5N1 which is = CH5N
MORE PRACTICE
 A compound is 43.64 % P and 56.36 % O
What is the empirical formula?
PO3
 Caffeine is 49.48% C, 5.15% H, 28.87% N and
16.49% O
What is its empirical formula?
C4H5N2O
EMPIRICAL TO MOLECULAR
 Since the empirical formula is the lowest ratio,
the actual molecule would weigh more
 Divide the actual molar mass by the empirical
formula mass – you get a whole number to
increase each coefficient in the empirical
formula
EXAMPLE
 Caffeine has a molar mass of 194 g, what is its
molecular formula?
1. Find the mass of the empirical formula, C4H5N2O
2. Divide the molar mass by the empirical mass:
194.0 g/mol =2
97.1 g/mol
3. Now multiply the entire empirical formula by 2
2(C4H5N2O) =
C8H10N4O2
final molecular formula