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Molar Mass,
Empirical & Molecular
Formulas
Molecular Mass & Formula Mass
• A chemical formula is a
convenient way of showing how
many and what type of atoms
make up a particular molecule
of formula unit
Molecular Mass & Formula Mass
• While the mass of an atom can
be found by looking at the
periodic table, the mass of a
molecule or formula unit must
be calculated.
Molecular Mass & Formula Mass
• The molecular mass of a
substance can be found by
finding the sum of the atomic
masses that make up a
particular molecule
• CH2Cl2 = 12 + 2(1) + 2(35.45)
= 84.9 amu
Molecular Mass & Formula Mass
• The formula mass, which is
used for ionic compounds, is
found in the same fashion as
molecular compounds.
• NaCl = 22.99 + 35.45
= 58.44 amu
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
The Avogadro Constant
• Atoms are much too small to
count individually
• The mass of a single molecule is
so small that it is impossible to
measure by ordinary means in
the laboratory.
The Avogadro Constant
• For this reason, scientists had
to come up with a way to
convert atomic mass units to
something that could be used in
the laboratory, grams
The Avogadro Constant
• Chemists have
found that 6.02
x 1023 atoms of
an element has a
mass in gram
equivalent to the
mass of one atom
in atomic mass
units
The Avogadro Constant
• This number, 6.02
x 1023 is known as
the Avogadro
Constant
The Mole
• Just like a dozen donuts is
equivalent to twelve donuts, one
mole of donuts would be 6.02 x
1023 donuts. Unlike the term
dozen, however, moles are used
to quantify very small objects,
like atoms.
The Mole
• A mole is just a
particular number
of atoms, ions,
molecules, or
formula units
• The mole is the SI
base unit
representing the
chemical quantity
of a substance,
and has been given
the symbol NA
Review: The Mole
 The number equal to the number of
carbon atoms in exactly 12 grams of pure
12C.
 1 mole of anything = 6.022  1023 units
of that thing
The Mole
• The mass of
one mole of
molecules,
atoms, ions, or
formula units is
called the
molar mass of
that species
Review: Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole
of the compound.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
Mass-Mole and MoleMass Conversions
• In order to convert the mass of an object
to the number of moles, you divide by the
molar mass of the substance
EXAMPLE:
4.56 g CO2 x 1 mole CO2 = 0.104 mole CO2
44 g CO2
Mass-Mole and Mole-Mass
Conversions
• In order to convert moles to mass, you
multiply by the molar mass of the
substance
EXAMPLE:
0.58 moles NH4NO3 x 80.g NH4NO3 = 46 g NH4NO3
1 mole NH4NO3
Volume-Mole and
Mole-Volume
Conversions
EXAMPLE:
• In order to do any conversions
with moles and volume, the
substance you are dealing with
must be a gas and exist at
STP which is 0°C and 1 atm
• To convert the volume of an
object to the number of
moles, you divide by 22.4 L
54.2 L of N2 x 1 mole N2 = 2.42 moles N2
22.4 L N2
Volume-Mole and
Mole-Volume Conversions
• To convert the number of moles to
volume, you multiply by 22.4 L
EXAMPLE:
0.78 moles of He x 22.4 L He = 17 L He
1 mole He
Molecules/Atoms/Formula Units - Moles
and Vice Versa
• To convert the number of M/A/F to moles,
divide by 6.02 x 1023 M/A/F
• To convert moles to M/A/F multiply by
6.02 x 1023 M/A/F
• Remember . . .
– molecules are for covalent compounds
– formula units are for ionic compounds
Molecules/Atoms/Formula
Units - Moles
and Vice Versa
EXAMPLE:
5.21 X 1024 molecules SO2 x
1 mole SO2 = 8.65 moles SO2
6.02 x 1023 molecules SO2
1.25 moles Cu(NO3)2 x 6.02 x 1023 f.u. = 7.53 x1023 f.u. Cu(NO3)2
1 mole Cu(NO3)2
Converting from Mass-Volume, MassM/A/F, Volume-M/A/F, or Anything in
Between
• Convert the first unit of moles. Then
convert from moles to the unit you’re
looking for.
MOLES ARE THE MIDDLE MEN!
The Mole
Percentage Composition
The percentage composition of a
compound is a statement of the
relative mass each element contributes
to the mass of the compound as a whole
“PART TO WHOLE!”
Percentage Composition
• To determine percent composition of a
particular element in a compound, divide the
mass of the element by the molecular mass
of the compound, and multiply by 100
• If the percent composition were calculated
for all elements of the compound, and
added together, the sum of the percents
should equal 100%
Percentage Composition
FOR EXAMPLE:
Calculate the % composition of carbon in ethanol
(C2H5OH)
1. Calculate the molecular mass of the compound:
2(12) + 5(1) + 1(16) + 1 = 46 amu
2. Calculate the total mass of the element in
question: 2(12) = 24 amu
3. Divide the mass of the element by the mass of
the compound and multiply by 100:
24 amu x 100 = 52 % carbon
46 amu
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
 24.31 
Mg  
  100  28.83%
 84.32 
 12.01 
C 
  100  14.24%
 84.32 
 48.00 
O
  100  56.93%
 84.32 
100.00
Empirical Formulas
The empirical formula of a substance is
the simplest, whole-number ratio
between the atoms of the elements
present in a compound
Empirical Formulas
A step-by-step method for determining empirical
formulas is found below:
1. Convert all quantities to moles (if the quantity
is a percent, remove the percent sign, and
stick in a grams sign!)
2. Divide all numbers by the SMALLEST number
of moles
3. If the quotient is NOT a whole number,
convert the decimal to a fraction, then
multiply by the denominator of the fraction to
make it a whole number
4. Write each element with its corresponding
subscript
Empirical Formulas
Find the empirical formula for a compound
containing 43.9% C, 7.32 % H, 48.78 % O
Step One:
43.9 g C x 1 mole = 3.66 moles
12 g
7.32 g H x 1 mole = 7.32 moles
1 g
48.78 g O x 1 mole = 3.05 moles
16 g
Empirical Formulas
Step 2:
C: 3.66/3.05 =1.2
Step 3:
C: 1.2 = 6/5 x 5 = 6
H: 7.32/3.05 =2.4
H: 2.4 = 12/5 x 5 = 12
O: 3.05/3.05 =1
O: 1 x 5 = 5
Step 4:
C6H12O5
Molecular Formula
• Molecular formulas indicate
the actual number of atoms of
each element making up a
molecule
• To determine a molecular
formula, one must know both
the empirical formula and the
molecular mass of the
substance
Molecular Formula
What is the molecular formula of a substance
that has an empirical formula of AgCO2 and a
formula mass of 304 g?
Step 1: Determine the formula mass of the
empirical formula: 108+12+2(16) = 152 g
Step 2: Divide the formula mass by the mass of
the empirical formula: 304 g/152 g = 2
Step 3: Multiply the subscripts of the empirical
formula by the ratio found in step 2:
Ag2C2O4
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
 molecular formula = (empirical
formula)n [n = integer]
 molecular formula = C6H6 =
(CH)6
 empirical formula = CH
Formulas
(continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas
(continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1. Base calculation on 100 grams of
compound.
2. Determine moles of each element in
100 grams of compound.
3. Divide each value of moles by the
smallest of the values.
4. Multiply each number by an integer to
obtain all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
 49.32 g C 1 mol C 
 4.107 mol C
12.01 g C 
 6.85g H 1 mol H 
 6.78 mol H
1.01 g H 
 43.84 g O 1 mol O 
 2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107 mol C
Carbon:
 1.50
2.74 mol O
Hydrogen:
6.78 mol H
Oxygen:
2.74 mol O
 2.47
2.74 mol O
2.74 mol O
 1.00
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
73
2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
73
2
(C3H5O2) x 2
=
C6H10O4
Hydrates
• Hydrates are
compounds that have
crystallized from a
water solution. In
these compounds, the
water molecules
adhere to the ions in
the compound, and
become part of the
crystal
Hydrates
• The formula for a hydrate indicates the
number of water molecules attached with a
coefficient
CuSO4•5H20 would have 5 water molecules
attached to it
• To name hydrates, name the regular
compound and use a Greek prefix to indicate
the number of waters present
CuSO4•5H20 would be named copper (II)
sulfate pentahydrate
Hydrates
• When calculating the formula mass for a
hydrate, the mass of the water is added to
the mass of the formula unit
CuSO4•5H20 = 63.5 + 32 + 4(16) + 5(18)
= 249.5 amu
• To determine the ratio of compound to
water in a hydrate, the compounds can be
heated to drive off the water. By
comparing the mass of the sample to the
mass of the water, one can determine the
formula of the hydrate
Hydrates
EXAMPLE: A 10.407 g sample of hydrated barium iodide is
heated to drive off the water. The mass of the dry
sample is 9.520 g. What is the formula of the hydrate?
Step 1: Find the mass of the water in the compound:
10.407 g – 9.520 g = 0.887 g water
Step 2: Convert mass of both compounds to moles:
9.520 g BaI2 x 1 mole BaI2 = 0.0243 moles BaI2
391 g BaI2
0.887 g H2O x 1 mole H2O = 0.0493 moles H2O
18 g H2O
Hydrates
Step 3: Divide both results by the smaller number
of moles:
0.0243 moles / 0.0243 moles = 1
0.0493 moles / 0.0243 moles = 2
Step 4: Write the formula for the hydrate:
BaI2•2H20