THE MOLE
Chapter 10
10.1 MEASURING MATTER
1. Chemists use the mole to
count atoms, molecules,
and ions.
1. Atoms – individual atoms
2. Molecules – two or more
atoms covalently bonded
3. ions – positively or
negatively charged
particles
4. Formula Units – Ionic
compounds
• A mole is used to compare large
amounts of atoms, particles, or # of
molecules
• 1 mole = 6.02 x 1023
atoms/particles/molecules
AVOGADRO
• Avagadro’s Number = 6.02 x 1023
• Named in Honor of 
In 1811 he determined the volume
of 1 mol of a gas
HOW WAS AVOGODRO’S NUMBER
FIRST CALCULATED?
• The first person to have calculated the
number of molecules in any mass of
substance seems to have been Josef
Loschmidt, (1821-1895), an Austrian high
school teacher, who in 1865, using the new
Kinetic Molecular Theory (KMT) calculated the
number of molecules in one cubic centimeter
of gaseous substance under oridnary
conditions of temperature of pressure, to be
somewhere around 2.6 x 1020 molecules. This
has always been known as the"Loschmidt
Number."
HOW WAS AVOGODRO’S NUMBER
FIRST CALCULATED?
• Use the following conversions to calculate
Avagodro’s number
• 100 ml /1litre
• 22.414 l/1mole
HOW WAS AVOGODRO’S NUMBER
FIRST CALCULATED?
• (2.68 x 10 19 molecules) x (1000 ml) x (22.414 Liters ) =
•
1cm3
1 litre
1 mole
(2.68 x 10 19molecules) x (1000 ml) x (22.414 Liters )
•
1cm3
1 litre
1 mole
=
molecules
1 mole
• The mole is independent of mass and size. It only measures number of
particles
A MOLE COMPARISON
HOW MANY ROSES?
CONVERTING BETWEEN REAL
WORLD UNITS
• To convert use a conversion factor (aka a ratio in math) to convert or
switch from one unit to another unit.
• Ex. How many roses are in 3 dozen?
• The conversion factor is 1 dozen / 12 roses or 12 roses / 1 dozen
• The set up
• What you are given is written first.
• Then use the correct conversion factor
• 3 dozen x 12 roses = 36 roses
1 dozen
Cross out = the units that cancel = one on top and one on bottom
CONVERTING BETWEEN MOLES
AND PARTICLES
• Ex. How many molecules are in 3.50 moles of sucrose?
3.50 mol sucrose x 6.02 x 1023 molecules sucrose =
1 mol sucrose
• =2.11 x 1024 molecules sucrose
• Do questions 1& 2 pg 307, 4&5 pg 309
CONVERTING BETWEEN MOLES
AND PARTICLES
3.50 mol sucrose x 6.02 x 1023 molecules sucrose =
1 mol sucrose
• =2.11 x 1024 molecules sucrose
• Do questions 1& 2 pg 307, 4&5 pg 309
CONVERTING BETWEEN MOLES
AND PARTICLES
• Practice 1. How many particles are in 2.5 moles of Zn?
2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn
1 mole Zn
CONVERTING BETWEEN MOLES
AND PARTICLES
2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn
1 mole Zn
CONVERTING BETWEEN MOLES
AND PARTICLES
2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn
1 mole Zn
MORE PRACTICE
• Silver nitrate (AgNO3) is used to make several different silver
halides used in photographic films. How many formula units
of AgNO3 are there in 3.25 moles of AgNO3.
CONVERTING PARTICLES TO
MOLES
• Start with the number of Representative Particles (aka
atoms, molecules, ion, or formula units)
• Use the correct conversion factor to cancel out unwanted
units.
• How many moles in 2.11 x 1024 molecules of sucrose?
• Ex. 2.11 x 1024 molecules of sucrose x 1 mole sucrose
=
6.02x1023 mol sucrose
= 3.50 mols sucrose
PRACTICE PARTICLES TO MOLES
1. How many moles are in 4.50 x 1024 atoms Zn?
2. How many moles in 2.50 x 1020 atoms Fe?
10.2 MASS AND THE MOLE
1. A mole always contains the same number of particles;
however, moles of different elements/compounds have
different masses
2. Molar mass: the mass in grams of one mole of any pure
substance: units=g/mol
3. Atomic Mass = g/mol
REAL WORLD EXAMPLE
• Was is the mass of 5 dozen jelly beans
• 1 dozen jelly beans = 35 g jelly beans
• 5 dozen jelly beans x 35 g jelly beans = 175 g jelly beans
1 dozen jelly beans
EXAMPLE
• What is the mass of 3.00 moles of Cu?
• Identify the Molar mass of Cu from the periodic table(round
to 4 sig figs)
• Cu = 63.55 g/mol
• Calculation
3.00 moles Cu x 63.55g Cu = 191 g Cu
1 mol Cu
PRACTICE
• Pg. 328 # 15b
• Determine the mass in grams of 42.6 mol Si
• # 16
• 2.45 x 10-2 mol Zn (answer in scientific notation)
MASS TO MOLE CONVERSION
• Ex. Convert 525 g of Ca to mol
• 525 g Ca x 1 mol Ca = 13.1 mol Ca
40.08 g Ca
Do questions 7 & 8 pg 315 9,12-15
PRACTICE
• Pg. 329
• # 18b. How many moles in 300.0 g of S
• #19b. How many moles in 1.00 Kg Fe
10.3 MOLES OF COMPOUNDS
• Identify the molar mass of a compound.
• Add up the molar mass of the atoms.
• Ex. CCl4
• C = 12.01 g/mol
• Cl = 35.45 g/mol x 4 = 141.8 g/mol
• 141.8 g/mol + 12.01 g/mol = 153.81 g/mol = 153.8 g/mol CCl4
• Ex. How many moles are in 500.0 g of CCl4
• 500.0 g CCl4 x 1 mol CCl4
153.8 g CCl4
= 3.251 mol CCl4
PRACTICE
1. What is the Molar Mass of H2SO4?
2. How many grams are in 3.25 moles of H2SO4?
MASS TO MOLE
• Ex. How many moles are in 35.0g of HCl?
• 35.0 g HCl x 1 mol HCl
36.46 g HCl
= 0.960 mol HCl
PRACTICE
• How many moles are in 145 g of AgNO3?
MASS TO PARTICLES
• Mass in g  moles  particles
• Two conversions
• Ex. How many particles are in 35.0 g of AlCl3
• 1st Identify the molar mass of AlCl3.
Al = 26.98 g/mol
Cl = 35.45 g/mol x 3 = 106.35 g/mol
AlCl3 = 26.98 g/mol + 106.35 g/mol = 133.3 g/mol
TO CONVERT FROM MASS TO
PARTICLES
• The given value goes in the upper left spot
35.6 g of AlCl3 x 1 mol AlCl3 = 0.267 mol AlCl3
133.3 g AlCl3
• 0.267 mol AlCl3 x 6.02 x 1023 formula units = 1.61 x 1023 formula units AlCl3
1 mol AlCl3
• OR 35.6 g of AlCl3 x 1 mol AlCl3
133.3 g AlCl3
x 6.02 x 1023 formula units = 1.61 x 1023
1 mol AlCl3
Formula Units
AlCl3
HOW MANY IONS?
• 1.61 x 1023 formula units AlCl3
• Identify the number of ions that make up the compound and multiply by
that number.
• 1.61 x 1023 formula units AlCl3 x 3 Cl- Ions
= 4.83 x1023 Cl- ions
1 AlCl3 formula unit
• 1.61 x 1023 formula units AlCl3 x 1 Al3+ Ions
1 AlCl3 formula unit
= 1.61 x1023 Al3+ ions
MASS OF 1 FORMULA
UNIT/MOLECULE
• 133.33 g/mol AlCl3 (molar mass) divide by Avogadro’s number to get the
mass of 1 Formula unit
• 133.33 g/1 mol AlCl3 x 1 mol/6.02 x 1023 formula units
• = 2.21 x 10-22 g AlCl3/formula unit
10.4 EMPIRICAL AND MOLECULAR
FORMULAS
• Percent composition
1. Synthetic chemists makes the new compounds
2. Analytical chemists identifies what its composition is
• Percent composition from experimental data
1. Identify the percent by mass of a compound
2. A 100 g sample of XY contains 55 g of X and 45 g of Y.
CALCULATING PERCENT BY MASS
• Percent by mass (element) = (mass of element/mass of compound) x 100
• Ex (55g X/ 100g XY) x 100 = 55% X
• Ex (45g Y/ 100g XY) x 100 = 45% Y
• Percent by mass from the chemical formula
• Percent by mass = mass of element in 1 mol of compound
molar mass of compound
x 100
EXAMPLE
• Compound NaHCO3
• Na = 22.99 g/mol
• H = 1.008 g/mol
• C = 12.01 g/mol
• O = 16.00 g/mol x 3 = 48.00 g/mol
• NaHCO3 = 84.01 g/mol
• Find the percent of Na in NaHCO3.
• Percent Na = (22.99 g/mol /84.01 g/mol) x 100 = 27.37%
PRACTICE
• Find the % composition for H, C, and O in NaHCO3.
EMPIRICAL FORMULA
• The opposite of % composition.
• When you know the % you can determine the empirical
formula
• Empirical Formula: the formula with the smallest wholenumber mole ratio of the elements
• May or may not be the same as the molecular formula
(aka the actual chemical formula)
SOLVING FOR EMPIRICAL
FORMULA
• How to solve for Empirical formula
1. Turn % into mass (g) by assuming you have 100% and therefore
100.0g
2. Convert all masses in g to mole using the molar mass of the
compound.
3. Divide each mole value by the smallest mole value (this gives
the molar ratio)
4. If decimals are present multiply the all mole values by the
smallest number to equal all whole numbers
5 Write the empirical formula with symbols and numbers of atoms
present
EXAMPLE
• Pg. 345 Example Problem 10.11
• Methyl acetate is a solvent commonly used in
some paints, inks, and adhesives. Determine the
empirical formula for methyl acetate, which has
the following chemical analysis: 48.64% C, 8.16%
H, and 43.20% Oxygen
PRACTICE EMPIRICAL FORMULA
• # 58 pg. 346
• The circle graph at the right gives the percent composition
for a blue solid. What is the empirical formula for this solid?
• 63.16% O
• 36.84% N
WARM UP
• Identify the % composition of H and O is H2O.
MOLECULAR FORMULA
• Empirical Formula = smallest whole number ratio. (many compounds have
the same empirical formula
• Molecular formula = actual formula
1. Begin with the steps to identify the empirical formula!
2. Identify the molar mass of the empirical formula by adding up all the molar
masses
3. Compare that to the molar mass of the compound given. Divide the given
molar mass/empirical formula molar mass
4. Distribute the value through your empirical formula to create the molecular
formula
EXAMPLE OF MOLECULAR
FORMULA
• Pg. 348 10.12
• Succinic acid is a substance produced by lichens.
Chemical analysis indicates it is composed of 40.68%
C, 5.08% H, and 54.24 % O and has a molar mass of
118.1g/mol. Determine the empirical then the
molecular formula.
PRACTICE MOLECULAR FORMULA
• Pg. 350 # 62
• A compound was found to contain 49.98 g of C and 10.47 g
of H. The molar mass is 58.12 g/mol. What is the empirical
formula and the molecular formula?
WARM UP
• Pg. 361 #164
• Vitamin D3: Your body’s ability to absorb calcium is aided by vitamin
D3. Chemical analysis yields the data that D3 is made of 84.31% C,
11.53% H, and 4.16% O. Find the Empirical and Molecular Formulas
• Molar mass of D3 = 384 g/mol
• Pg. 361 # 186
• A 1.628 g sample of a hydrate of magnesium iodide is heated until its
mass is reduced to 1.072 g and all water has been removed. What is
the formula of the hydrate?
CHAPTER 10.5 HYDRATES
What is a Hydrate?
a compound that has a specific
number of water molecules bound to its
atoms
NAMING HYDRATES
• The number of water molecules is written following a dot for example
• Na2CO3 * 10H2O = sodium carbonate decahydrate
•
•
•
•
Na = Sodium
CO3 = Carbonate
10 = Deca
H2O = Hydrate
• The first compound is named based on Ionic or covalent naming.
ANALYZING A HYDRATE
• When a hydrate is heated it loses the water molecules and
becomes the anhydrous form.
• How do you determine the formula of a hydrate? (how
many waters it has?)
1. Find the number of moles of water associated with 1 mole of the
hydrate.
2. Dehydrate your sample
3. Calculate the water lost and convert to the value to moles
4. Calculate the mass of the anhydrate and convert to moles
5. Divide the moles of water by the moles of anhydrate
EXAMPLE
• Pg. 352
• How many water molecules are in the hydrate BaCl2 * xH2O?
• A 5.00g sample of BaCl2 * xH2O is dehydrated. The resulting mass is 4.26 g of
anhydrous BaCl2.
• Subtract: 5.00g – 4.26g = 0.74 g H2O lost.
• Convert to moles:
• 4.26g x 1mol BaCl2/208.2g BaCl2 = 0.0205 mol BaCl2
• 0.740g x 1 mol H2O/ 18.02 g H2O =
0.0411 mol H2O
Divide the water by barium chloride
.0411mol H2O/0.0205 mol BaCl2 =2
So 2 moles of water to every 1 mole barium chloride = BaCl2 * 2H20
PRACTICE
• A mass of 2.50 g of blue, hydrated copper (II)
sulfate is placed in a crucible and heated.
After heating, 1.59g of white anhydrous
copper sulfate remains. What is the formula
for the hydrate? (How many waters?) Name
the hydrate.