Chemistry Chapter
10
The Mole
10.1 Measuring Matter
• Objectives:
• 1. Explain how a mole is used
indirectly to count the number of
particles of matter
• 2. Relate the mole to a common
everyday unit
• 3. Convert between moles and
number of representative
particles
Counting Particles
• Counting, mass and volume are
the usual methods of
measurement
• Q: What is the mass of 90 average
sized apples if 1 dozen apples has
a mass of 2.0 kg?
• Use conversion factors & cross
cancel: #apples=dozens=mass
• 90 apples x 1dz/12apples x 2.0 kg/dz
• A: 15 kg= mass of 90 average
apples
Avogadro’s number
• The SI unit for particle
measurement is the mole
(abbreviated mol)
• Definition: a mole of a substance is
6.02 x 1023 representative
particles
• 6.02 x 1023 is Avogadro’s number
• Avogadro was an Italian scientist
who lived from 1776-1856 & helped
clarify the distinction between
atoms and molecules
Representative Particles
• Definition: a representative
particle is the type of substance
• The substance is usually atoms,
molecules or ions (which are
called formula units)
• Diatomics exist as molecules, so
their representative particle is the
molecule
• A mole of any substance contains
Avogadro’s number of
representative particles, or 6.02 x
1023 representative particles
Converting between Moles and
Representative Particles
• To convert from particles to
moles: #particles x
1mol/6.02x1023 particles
• Moles to particles: mol x
6.02x1023 particles/1mol
• Q: How many moles of magnesium
is in 1.25 x 1025 atoms of Mg?
• 1.25 x 1025 atoms Mg x
1mol/6.02x1023 atoms
• A: 0.208 mol
10.2 Mass and the Mole
• Objectives:
• 1. Relate the mass of an atom
to the mass of a mole of
atoms
• 2. Convert between number of
moles and the mass of an
element
• 3. Convert between number of
moles and number of atoms of
an element
Atomic Mass & Molar Mass
• The atomic mass of an
element expressed in grams
(instead of amu) is the mass
of one mole of the element
• Definition: Molar mass is the
mass of one mole of an
element or a substance
• Q: How many representative
particles are in one mole?
• Q: How many atoms are in one
mole of magnesium?
Calculating Molar Mass
• To calculate the molar mass
of a compound, find the
number of grams of each
element in one mole of the
compound, then add the
masses together
• What is the molar mass of
H2O2?
• What is the molar mass of
Ca(OH)2?
Mol-Mass & Mass-Mol
Conversions
• Use the molar mass of an
element or compound to
convert between the mass of
a substance and the moles of
a substance
• Mass= # mol x mass(g)/1mol
• Q: What is the mass of 3 mol
of NaCl?
• A: 3 mol x 58.5 g/mol = 176 g
• Moles = mass(g) x mol/mass(g)
• Q: What is the number of
moles in 92.2 g of Fe2O3?
• A: Mol= 92.2 g Fe2O3 x
mol/159.6g Fe2O3 = 0.578 mol
• Q: What is the conversion
factor to convert mass to
moles?
• A: 1mol/mass(g)
• Q: What is the mass of of 9.45
mol Al2O3?
Mass-mol-atom & Atommol-mass
• Using moles & Avogadro’s number
makes conversions easy
• Let’s try: How many atoms of gold
are in a gold alloy coin with a mass
of 31.1 g Au?
• A: convert to mol first, then go from
mol to atoms
• Mass(g) Au X 1 mol Au/grams Au=
mol Au NOW GO TO ATOMS & ANS.
• A: 9.51 x1022 atoms Au
Volume of Gas at STP
• Avogadro’s hypothesis says
that equal volumes of gases at
the same temperature and
pressure contain equal
numbers of particles
• STP means 0°C and 101.3 kPa
or 1 atm
• At STP, 1 mol or 6.02x1023
particles (atoms, molecules or
formula units) of any gas
occupies a volume of 22.4 L
• This gives a new conversion
factor: 1 mole = 22.4 L
• Keep in mind, 1mol=22.4L=
6.02x1023 particles
• You may use this relationship
to create any number of
conversion factors depending
on the problem
• Volume of gas=mol of gas x
22.4L/mol
• Q: What is the volume in liters
of 0.60 mol of SO2 at STP?
• A: 0.60 mol x 22.4 L/mol= 13 L
• Q: What is meant by standard
temperature and pressure?
• Challenge question:
• Q: What is the number of
molecules in 12.28L of H2?
(hint: convert L to mol first,
then go from mol to #
particles)
• A: 3.3 x 1023 molecules of H2
• You can also calculate molar mass
from density
• Recall that the units for density
include g/L among others
• Molar mass=density at STP x molar
volume at STP
• g/mol=g/L x 22.4L/mol
• Q: What is the molar mass of a
gaseous compound with a density of
1.964g/L at STP?
• A: 1.964g/L x 22.4L/mol= 44.0g/mol
• The mole road map is on the back of
your periodic table and may be used
to help in these calculations
10.3 Moles of Compounds
• Objectives:
• 1. Recognize the mole relationships
shown by a chemical formula
• 2. Calculate the molar mass of a
compound
• 3. Convert between the number of
moles and mass of a compound
• 4. Apply conversion factors to
determine the number of atoms of
ions in a known mass of a
compound
Mole relationships &
chemical formulas
• This section explains how to
use the mole highway to
convert among moles, mass &
particles (and liters)
• You can convert from mass to
particles, but you must go to
moles first
• The trick is to use mole
conversion factors
• Q: What is the molar mass of
potassium chromate (K2CrO4)?
• A: Find g of each using the
subscripts:
• 2 mol K x 39.10g K/mol K= 78.20 g
• Mol Cr x 52.00g Cr/mol Cr=52.00 g
• 4 mol O x 16.00g O/mol O=64.00 g
• Add them up:
78.2+52+64=194.20g K2CrO4
• Q: Aluminum chloride (AlCl3) has a
mass of 35.6 g.
• a: How many Al ions are present?
• b. How many Cl ions are present?
• c. What is the mass of one
formula unit of AlCl3?
• 1. Find the molar mass of the
compound first
• 1 mol Al x 26.98 g Al/mol
Al=26.98g
• Still finding molar mass:
• 3 mol Cl x 35.45g Cl=106.35 g
• Molar
mass=26.98g+106.35g=133.33g/
mol AlCl3
• 2. Use the conversion factor to
find mol AlCl3 present:
• 35.6 g AlCl3 x mol AlCl3 /133.33g
AlCl3 = 0.267 mol AlCl3
• 3. Use Avogadro’s number:
• 0.267 mol AlCl3 x 6.02x1023
fu/mol AlCl3 = 1.61 x 1023
atoms fu AlCl3
• 4. Use the ratios from the
formula as conversions to get
Al and Cl ions:
• 1.61 x 1023 fu AlCl3 x 1 Al3+/1
AlCl3 fu= 1.61 x 1023 Al3+
• 1.61 x 1023 fu AlCl3 x 3 Cl-/1
AlCl3 fu= 4.83 x 1023 Al3+
• 5. To get mass,
substitute mass AlCl3
= 133.3 g
• 133.3 g AlCl3 /1 mol x
1 mol/ 6.02 x1023 fu =
2.21 x10-22 g AlCl3 /fu
10.4 Empirical & Molecular
Formulas
• Objectives:
• 1. Explain what is meant by
the percent composition of a
compound
• 2. Determine the empirical
and molecular formulas for a
compound from mass percent
and actual mass data
Calculating Percent by Mass
• Definition: percent composition or
percent by mass of each element in
the compound is the relative
amounts of the elements in the
compound
• The percent by mass of an element
in a compound is the number of
grams of the element divided by the
mass in grams of the whole
compound, multiplied by 100%
• %mass element= mass of element x 100%
mass of compound
• Q: Calculate the percent
composition of propane (C3H8).
• Molar mass= 44g/mol
• %C= 36g/44g x 100%= 81.8%
• %H= 8g/44g x 100% = 18.0%
• The elements should add up to
100% if the calculations are
correct
• In this case, with 2 sigfigs the
result is 100%
Empirical Formulas
• Definition: an empirical formula is the
lowest whole number ratio of the
atoms of the elements in a compound
• The empirical formula may or may not
be the same as the molecular formula
• CO2 is both empirical and molecular
since the ratio of C:O is 1:2
• For H2O2, the empirical formula is HO
since the ratio of H:O is 2:2 which
equals 1:1, but the molecular formula
is H2O2
• The empirical formula of a compound
shows the smallest whole-number
ratio of the atoms in the compound
• Q: What is the empirical formula of a
compound that is 25.9% nitrogen & 74.1%
oxygen?
• Assume 100g of compound which gives
25.9g N & 74.1g O
• 25.9g N x molN = 1.85 mol N
14g N
• 74.1g O x mol O = 4.63 mol O
16g O
• Divide each ratio by the smaller number of
moles
• 1.85 mol N =1mol N
1.85
• 4.63 mol O = 2.50 mol O
1.85
• This gives N1O2.5, so multiply ratios by 2
• Final result: N O
Molecular Formulas
• Definition: the molecular formula specifies
the actual number of atoms of each
element in a compound
• The molecular formula of a compound is
either the same as its experimentally
determined empirical formula, or it is a
simple whole number multiple of its
empirical formula
• Q: What is the molecular formula of a
compound whose molar mass is 60g/mol &
empirical formula is CH4N
• The molar mass of the empirical formula is
30g/mol
• 60/30=2, so the molecular formula is C2H8N2