Ch. 6: Chemical Composition
Dr. Namphol Sinkaset
Chem 152: Introduction to
General Chemistry
I. Chapter Outline
I. Introduction
II. Counting by Weighing
III. Chemical Formulas as Conversion
Factors
IV. Mass Percent Composition
V. Finding Formulas from Mass Data
I. Introduction
• The FDA RDA for Na =
2.4 g.
• Na usually consumed
as NaCl.
• But 2.4 g NaCl ≠ 2.4 g
Na.
• How do we figure out a
problem like this?
II. Selling Nails
• Nails can be sold two
ways: as individual
units or by the pound.
• By the pound much
more convenient if
want to buy hundreds
of nails.
• Analogy extends to
atoms/molecules.
II. Sample Problem
• If a dozen large nails weighs 0.275 lb.,
how many nails are contained in 5.5 lb?
II. Counting Atoms
• Finding the number of atoms in a
sample of matter is similar to selling
nails by the pound.
• We used a dozen as a unit of count, but
12 is way too small for atom.
• In chemistry, the mole is used instead –
it’s the chemist’s “dozen.”
II. The Mole
• The mole is the SI unit for amount and
is defined as the amount of material
containing 6.0221421 x 1023 particles.
• This # is known as Avogadro’s number.
• This # is defined by the number of
atoms in exactly 12 g of carbon-12.
• One mole of atoms, ions, or molecules
generally makes up a macroscopically
meaningful size.
II. How Much is a Mole?
II. Avogadro’s Number
• Every time you hear “mole,” you should
think 6.022 x 1023.
• We can use Avogadro’s number as a
conversion factor to calculate # of
atoms.
6.022 x 1023 atoms
1 mole atoms
OR
1 mole atoms
6.022 x 1023 atoms
II. Sample Problem
• How many atoms of gold are in a pure
gold ring containing 8.83 x 10-2 moles of
Au?
II. Grams and Moles
• In the nail example, we had a
relationship between a dozen nails and
a mass.
• We need the same relationship for the
mole.
• The mass of 1 mole of atoms of an
element is its molar mass. The value of
an element’s molar mass in g/mole is
numerically equal to the element’s
atomic mass in amu (from periodic
table).
II. Mole of Atoms & Compounds
• 1 Fe atom weighs 55.85 amu, so 1 mole
of Fe atoms weighs 55.85 g.
• 1 O atom weighs 16.00 amu, so 1 mole
of O atoms weighs 16.00 g.
• Same applies for compounds.
 1 molecule H2O weighs 18.02 amu, so 1
mole H2O weighs 18.02 g.
II. Sample Problem
• Calculate the molar mass of calcium
sulfate.
II. The Mass of a Mole
Depends on Unit Size
II. Mole Equivalencies
• Since mass and moles are related, we
can set up conversion factors.
• e.g. 1 mole Fe weighs 55.85 g.
55.85 g Fe
1 mole Fe
OR
1 mole Fe
55.85 g Fe
II. Sample Problem
• Graphite, a crystalline form of carbon, is
used in pencils. How many moles of
carbon are in 0.315 g of graphite?
II. Sample Problem
• How many grams of sulfur are in 2.78
moles of sulfur?
II. Sample Problem
• How many moles of NO2 are in 1.18 g
of NO2?
II. Sample Problem
• How many molecules of H2O are in a
sample of water with a mass of 3.64 g?
II. Sample Problem
• If a sample of molecular bromine
weighs 2100 g, how many molecules of
molecular bromine are in the sample?
III. Inherent Conversion Factors
• Any object that can
be broken down into
parts or pieces has
an inherent
conversion factor.
• For example, a
clover.
III. 3 Leaves : 1 Clover
• Since there are 3 leaves in one clover, we
can write a conversion factor between them.
III. Other Examples
III. Scaling Up
• The ratios hold as long as we keep the
unit the same for both the parts and the
whole.
• 3 leaves : 1 clover  3 dozen leaves : 1
dozen clovers.
• In chemistry, we scale up to the mole.
• 2 H atoms : 1 H2O molecule  2 moles
H atoms : 1 mole H2O molecules.
III. Breaking Down CCl4
III. Sample Problem
• List all the possible atom : formula unit
mole relationships in barium nitrate.
III. Mole Relationships
• A chemical formula gives the
relationships between moles of
substances, NOT grams.
• Always convert to moles then use a
mole relationship to get to the
substance of interest.
III. Sample Problem
• Determine the number of moles of
oxygen in 2.45 moles of nickel(II)
phosphate.
III. Sample Problem
• Calculate the mass of carbon in 25.0 g
of C6H12O6.
III. Sample Problem
• How many grams of sodium chloride
contains 2.00 grams of sodium?
IV. Mass Percent
• The mass percent expresses an element’s
percentage of the total mass of the
compound.
IV. Sample Problem
• Calculate the mass percent of nitrogen
in ammonium nitrate.
IV. Mass Percent as a
Conversion Factor
• The key to using mass percents is
realizing that they can be expressed as
conversion factors.
• NaCl is 39% Na can be expressed as
39 g Na : 100 g NaCl.
39 g Na
100 g NaCl
100 g NaCl
39 g Na
IV. Sample Problem
• If CCl2F2 is 58.64% Cl by mass, how
many grams of CCl2F2 contains 22 g of
Cl?
V. Finding Formulas from
Mass Data
• Given elemental mass % data of a
compound, it’s possible to find the
formula of the compound.
• Elemental analysis is a common test
performed on newly synthesized
compounds.
V. Empirical vs. Molecular
• Often experiments will yield the relative
masses of each element in a
compound.
• With just the relative masses, the best
we can do is find the empirical formula.
• Additional data is needed to go from an
empirical formula to a molecular
formula.
V. The Key to Finding
Empirical Formulas
• In any formula, subscripts represent the
number of atoms of a given element in
that compound.
• However, these ratios scale up!
• Therefore, the subscripts can also be
interpreted as mole ratios between
atoms of a given element in the
compound.
V. Finding Empirical Formulas
• To find an empirical formula, you must
calculate the number of moles of each
atom present in a certain sample of the
compound.
• These moles become the temporary
subscripts in the formula.
• You then use math to convert to whole
numbers.
V. Sample Problem
• Calculate the empirical formula of ethyl
butyrate (pineapple oil) if its mass
composition is 62.04% C, 10.41% H,
and 27.55% O.
V. Sample Problem
• A certain compound is 50.66% C,
4.25% H, and 45.09% S by mass.
Determine the empirical formula.
V. Determining Molecular
Formulas
• To find a molecular formula from an
empirical formula, you must know the
molecular molar mass.
• The molecular molar mass is always a
multiple of the empirical molar mass.
• To find the multiple, we divide the
molecular molar mass by the empirical
molar mass.
V. Sample Problem
• Butane has an empirical formula of
C2H5. If its molar mass is 58.12 g/mole,
determine the molecular formula of
butane.
V. Sample Problem
• The carcinogen benzo[a]pyrene (MW =
252.30 g/mole) is found to be 95.21% C
and 4.79% H by mass. What is its
molecular formula?