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```					The Mole
Diatomic elements
• – elements that do not exist on their own,
they must be in a covalent bond with
themselves to be more stable.
•   N2
•   O2
•   F2
•   Cl2
•   Br2
•   I2
•   H2
Dimensional Analysis Review
• How many seconds are in 5.0 hours?
• 5.0 hr
• 5.0 hr x 60 min x 60 sec = 18000 sec
1 hr    1 min
Dimensional Analysis Review
• Calculate the number of inches in 26 yards
• 26 yards
• 26 yards x 3 ft x 12 inches = 940 inches
1 yd 1 ft
The Mole
• Chemists need a convenient method for
counting accurately the number of atoms,
molecules, or formula units in a sample of
a substance.

• The mole, commonly abbreviated mol, is
the SI base unit used to measure the
amount of a substance.
• A mole is used to represent the amount of
a substance or compound.
• H2 means 1 mole of H2
• NaCl means 1 mole of NaCl.
• Hydrates – compounds that contain water.
Written as NaCl • 2H2O
The Mole
• A mole of anything contains 6.02 x 1023
representative particles.
• A representative particle is any kind of
particle such as atoms, molecules, formula
units, electrons, or ions.
• 6.02 x 1023 is called Avogadro’s number
Molecular or Formula Mass
Calculations

• You have to add up the masses of every
element found in the compound or molecule.
• Ex. (NH4)3PO4 Ammonium Phosphate
• Make a list of how many of each element
you have.
• N=3
• H = 12
• P=1
• O=4
• Multiply how many of each element you
have by their mass. (Look it up on the
periodic table.) Then add up the total. The
units always grams.
•   N = 3 x 14 = 42
•   H = 12 x 1 = 12
•   P = 1 x 31 = 31
•   O = 4 x 16 = 64
•               149 g
example
• NaCl
• Na = 23g
• Cl = 35 g

• Total 58 g
Hydrates-
• are just a molecule or compound that has
a certain number of water molecules
attached to it.
• They are written like NaCl • 5H2O.
• When you are asked to calculate the mass
you have to understand that there are 5
H2O molecules attached to 1 NaCl.
• That means: 10 H & 5 O in addition to the
Na & Cl.
•   H = 10 x 1 = 10
•   O = 5 x 16 = 80
•   Na = 1 x 23 = 23
•   Cl = 1 x 35 = 35
•    Total       148 amu or grams
Conversion Factor # 1

1 mole
Molecular weight (g)

The molecular mass comes
from the periodic table!
Mole – Mass Calculations
• What is the mass of 4.21 moles of iron (III)
oxide?
• Start with your given:
• 4.21 moles Fe2O3
• Draw your line
• 4.21 moles Fe2O3 x _____________
Mole – Mass Calculations
• Place conversion factors
• 4.21 moles Fe2O3 x 159.70 g Fe2O3
1 moles Fe2O3
• Cross out units & work the problem
• 672 g Fe2O3
Another Example
• Calculate the mass of 1.630 moles of Na

• 1.630 moles x 22.9g
•               1 mole

• 37.47 g Na
Mole – Mass Calculations
• How many moles of Ca(OH)2 are in 325
grams?

• 325 g x 1 mole
•         74 g

• 4.39 moles Ca(OH)2
Another Example
• How many moles are in 62.17 g of sodium
chloride?

• 62.17g x 1 mole
•            58 g
• 1.064 mol NaCl
Conversion Factor #2
Representative Particles
•   Anything - Representative particles
•   Elements – Atoms
•   Covalent Compounds – Molecules
•   Ionic Compounds – Formula Units
•   Ions - Ions
Mole – Representative Particle
Calculations
• Calculate the number of atoms in 3.50
moles of copper
• Start with your given
• 3.50 mol Cu
• Draw your line
• 3.50 mol Cu x _________
Mole – Representative Particle
Calculations
• Place the conversion factors
• 3.50 mol Cu x 6.02 x 1023 atoms
1 mol Cu
• Work the problem
• 2.11 x 10 24 atoms Cu
Another Example

• Calculate the number of molecules in 2.6
moles of H2O
• 2.6 mol H2O x 6.02 x 1023 mlcs = 1.6 x 1024
1 mol H2O         mlcs H2O
Another Example
• Calculate the number of formula units in
5.23 moles of NaCl

• 5.23 mole x 6.02 x 1023 f. units
•                   1 mole
• 3.15 x 10 24 formula units NaCl
Mole – Representative Particle
Calculations
• How many moles of MgO are in
• 9.72 x 10 23 molecules of MgO?

• 9.72 x 10 23 mlcs x    1 mole
–                     6.02 x 10 23 mlcs

• 1.61 moles MgO
Another Example
• How many moles are in 4.50 x 1024 atoms
of Zinc?

• 4.5 x 1024 atoms x 1 mole
•                   6.02 x 10 23 atoms

• 7.48 moles Zn
Mass – Particle Conversions

• How many atoms of gold are in 25.0 g of
gold?

• 25.0 g Au x 1 mole Au    x 6.02 x 1023 atoms Au
196.79 g Au        1 mole Au

• 7.65 x 1022 atoms of Au
Mass – Particle Conversions
• How many grams of He are in 5.50 x 1022
atoms of He?

• 5.50 x 10 22 atoms x 1 mole           x      4g
•                    6.02 x 10 23 atoms     1 mole
• 0.366 g He
example
• How many formula units are in 35.6 g of
NaCl?

• 3.70 x 10 23 formula units
moles
• Al2S3 means you have 1 mole of the
compound Al2S3.

• It is composed of 2 moles of Al atoms and
3 moles of S atoms
Ions in a compound
• How many Cl- ions are there in 35.6 g of
AlCl3?

• 35.6 g AlCl3 x 1 mole AlCl3 x 3 mole Cl3 x 6.02 x 10 23 ions
•                  132 g       1 mole AlCl3   1 mole Cl3

• Remember ions is just a word, atoms particles,
representative units. Anything works
% Composition, Empirical
Formulas, & Molecular Formulas
% composition
• Mass of the element    x 100
• Mass of the compound
% composition-
• gives the relative mass of each atom in a compound.

• Steps
• 1. Calculate formula mass of compound
• 2. Determine % of each atom
• Ex: Calculate the % comp of N in NH4NO3
•                    N x 2 = 28.0
•                    H x 4 = 4.0
•                    O x 3 = 48.0
•                           80.0 g
•                % N = 28.0 g/ 80.0 g x 100 = 35%
• Calculate the % composition of each atom
in
•     a. Fe2O3     b. Ag2O       c. HgO
% Composition
• Calculate the % Composition of iron (III)
oxide
• % Fe = 69.94%
• % O = 30.06%
• Molecular formulas- Used to tell exactly
how many of each atom is present in the
compound.
• Empirical Formulas- The simplest formulas
possible indicating the ratio of atoms found
in the molecule. Think of it as reducing a
fraction to the least common denominator.
Empirical Formula
•   What is the empirical formula for H2O2?
•   HO
•   What is the empirical formula for C6H12O6?
•   CH2O
•   C6H6
•   C2H2
•   C6H12O6
•   C4H10
•   P4O10
•   SO3
•   N2O4
•   NO2
•   Ag2C4H4O6
Empirical Formulas-calculations
• The smallest whole number ratio of
elements in a compound. It is determined
by calculating the:
• Calculate the moles present of each element in the
compound.
• Set up ratios for each element by dividing each mole
amount by the smallest mole amount.
• Use the ratio from step # 2 as the subscripts to write out
the empirical formula.
• Ex: Determine the empirical formula for a
compound with 36.5 % Na, 25.4% S,
38.1% O

• Since each elements % adds up to 100%
treat each % as a gram amount
• Na = 36.5 g Na x 1 mole Na = 1.59 mole Na
•                    23.0 g Na
• S = 25.4 g S x 1 mole S      = .791 mole S
•                   32.1 g S
• O = 38.1 g O x 1 mole O       = 2.23 mole O
•                   16.0 g O
• Na           S            O
• 1.59 = 2.01 .791 = 1      2.23 = 3.01
• .791        .791          .791
•               Ratio = 2:1:3
•                Na2SO3
Empirical Formula Problem
• Calculate the empirical formula of a
compound containing 40.05 % S and
59.95 % O.

40.05 g S x 1 mol S     = 1.249 mol / 1.249 mol = 1
32.07 g S

59.95 g O x 1 mol O   = 3.747 mol / 1.249 mol = 3
16.00 g O

SO3
Ratio problems
• Occasionally one part of your ratio will not
be a whole number ( 1.5 : 1) It has to
be!!!!!
• Since H1.5O can’t exist you have got to fix
it.
• When this occurs you need to multiply by
factor that will make it a whole # (like 2)
• * what you do to one you must do to all, so
every mole amount gets multiplied by 2!!
Empirical Formula Problem

• Calculate the empirical formula for a
compound containing 48.64 g C, 8.16 g
H, and 43.20 g O.
• 48.64g x 1 mole C = 4.04 mole/ 2.7 =1.5 x 2 = 3
•           12 g
• 8.16 g x 1 mole H = 8.16 mole/2.7 = 3.02 x 2 = 6
•            1g
• 43.20 x 1 mole O = 2.7 mole/2.7 = 1x 2 = 2
•            16 g

• C3H6O2
Steps for Calculating Molecular
Formula
1. Calculate the empirical formula
2. Get the molecular mass of the empirical
formula that you just determined
3. Divide the experimentally determined molecular
mass (given) by the molecular mass of the
empirical formula
4. You will get a whole number
5. Multiply everything in the empirical formula by
this number
Molecular Formula Problem
• Calculate the molecular formula of a
compound containing 40.68%C, 5.08%H,
and 54.25%O with an experimentally
determined molecular weight of 118.1
g/mol
Molecular Formula Problem

40.68g C x 1 mol C = 3.387 / 3.387 = (1)2 = 2
12.01 g C
5.08g H x 1 mol H = 5.04 / 3.387 = (1.5)2= 3
1.01 g H
54.25 g Ox 1 mol O = 3.390 / 3.387 = (1)2 = 2
16.00 g O

C 2H 3O 2
Molecular Formula Problem
•   C2H3O2
•   Molecular Mass = 59.04
•   EDMM / EFMM = 118.1 / 59.04 = 2
•   Molecular Formula
•   C4H6O4
Molecular Formula Problem
• Calculate the molecular formula of a
compound containing 57.84 g C, 3.64 g H,
and 38.52 g O with an experimentally
determined molecular mass of 249.21
g/mol
• C12H9O6
Stoichiometry
• Stoichiometry is just a long word for changing
units in chemistry
• Just remember to ALWAYS start with your
given!
• If you can do Dimensional Analysis, you can do
stoichiometry

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