Unit 4 – The Language
of Chemistry: Part Deux
Amadeo Avogadro
II.B.2(f) – Describe Avogadro’s
hypothesis and use it to solve
stoichiometric problems
III.A.2(a) – Explain the meaning
of mole and Avogadro’s number
Avogadro’s Hypothesis
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Equal amounts of gases at the same
temperature contain equal numbers of
molecules
Leads to definition of the “mole”
Mole Def: the number equal to the number of
atoms in 12.01 grams of carbon
We will return to this idea later
Atomic Mass Units (amu)
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Since the mass of one atom is so tiny, it is
more practical to use relative atomic masses
Carbon has been arbitrarily assigned a mass
of exactly 12 atomic mass units
One atomic mass unit is defined as exactly
1/12 the mass of a carbon-12 atom
Avogadro’s Number

Experimentally determined that there are
6.022 x 1023 atoms in exactly one mole
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Remember this number!
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6.022 x 1023
Molar Mass

The mass of one mole of a pure substance

Units: grams/mol

Equal to the mass of 6.022 x 1023 atoms of a pure
substance

Molar mass of an element is numerically equal to
the atomic mass of the element in atomic mass units
Converting Mass, Moles,
& Atoms/Molecules/Ions
III.A.2(b) -Interconvert between
mass, moles, and number of
particles
Gram/Mole/Atom Conversions
Example: What is the mass in
grams of 3.50 mol of copper?
63.55 g Cu
3.50 mol Cu 
 222 g Cu
1 mol Cu
More Examples
What is the mass in grams of 2.25 mol of Fe?
 What is the mass of 0.375 mol of K?
 What is the mass of 16.3 mol of Ni?
OR
 How many moles are in 5.00 g of calcium?
-10
 How many moles are in 3.60 x 10
g Au?
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More Complex Examples
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How many moles of Ag are in 3.01 x 1023
atoms of Ag?
How many atoms of Na are in 36.0 grams of
Na?
How many oxygen atoms are in 180.18 g of
glucose?
Determining Chemical
Formulas
III.A.1(e) - Calculate the percent composition of a
substance, given its formula or masses of each
component element in a sample
III.A.1(f) - Determine the empirical formulas and
molecular formulas of compounds, given
percent composition data or mass composition
data
III.A.2(c) - Distinguish between formula mass,
empirical mass, molecular mass, gram
molecular mass, and gram formula mass
Empirical
Formulas
III.A.1(f) - Determine the
empirical formulas and molecular
formulas of compounds, given
percent composition data or
mass composition data
Calculating an Empirical
Formula
Empirical Formula – consists of the symbols for
the elements combined in a compound, with
subscripts showing the smallest whole-number
ratio of the different atoms in the compound
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Always assume 100.0 g sample
Change percents to grams
Calculate moles of each element
Divide moles by smallest mole amount to
determine ratio
Multiply to get whole numbers (if necessary)
Determine the Empirical
Formula of Aspirin

What is the empirical formula of aspirin? It
has 4.48 % H, 60.00 % C, and 35.52 % O by
mass.
Example #1

An oxide of aluminum is formed by the
reaction of 4.151 g of aluminum with
3.692 g of oxygen. Calculate the
empirical formula for this compound.
Example #2

When a 0.3546 g sample of vanadium
metal is heated in air, it reacts with
oxygen to achieve a final mass of
0.6330 g. Calculate the empirical
formula of this vanadium oxide.
Example #3

A sample of lead arsenate, an
insecticide used against the potato
beetle, contains 1.3813 g of lead,
0.00672 g of hydrogen, 0.4995 g
arsenic, and 0.4267 g of oxygen.
Calculate the empirical formula for lead
arsenate
Molecular
Formulas
III.A.1(f) - Determine the empirical formulas
and molecular formulas of compounds,
given percent composition data or mass
composition data
III.A.2(c) - Distinguish between formula
mass, empirical mass, molecular mass,
gram molecular mass, and gram formula
mass
Calculating a Molecular
Formula
Molecular Formula – the actual formula of a
molecular compound
Example: C2H4 – ethylene
C3H6 – cyclopropane
Both have a 2H:1C ratio
Determining Molecular
Formula
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Must know formula mass to calculate
molecular formula
Divide experimental formula mass by
empirical mass
Multiply subscripts by quotient
Example #4

Find the molecular formula of a compound
with an empirical formula of CH and a
formula mass of 78.110 amu
Answer
C – 12.011
H – 1.008
Total = 13.019
78.110 ÷ 13.019 ≈ 6
6(CH) = C6H6
Example #5

A white powder is analyzed and found
to have an empirical formula of P2O5.
The compound has a molar mass of
283.88g. What is the compounds
molecular formula?
Finding Molecular Formula
from Empirical Formula
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
An unknown compound is found in tree sap.
It has been shown to be composed of 40.0%
carbon, 6.7% hydrogen and 53.3% oxygen by
mass. It was also discovered that 5.00 moles
of the material has a mass of 900 grams.
What is the molecular formula of this
compound?
Combustion
Analysis
Example #6

What is the empirical formula of a
hydrocarbon that produces 2.703 g CO2
and 1.108 g H2O when combusted?
Example #7

What is the empirical formula of a substance
containing carbon, hydrogen, and oxygen if
1.000 g of substance produces 1.467 g CO2
and 0.6003 g H2O upon combustion?

The molar mass of the substance is 120
g/mol. What is the molecular formula?
Example #8

What is the molecular formula of a substance
containing carbon, hydrogen, and oxygen if it
has a molar mass of 234 g/mol and 0.360 g
of substance produces 0.406 g CO2 and
0.250 g H2O upon combustion?