Lecture

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Lecture 7 - (a) The Mole Concept, (b) Formula
of an Unknown

• Chemistry 142 B
• James B. Callis, Instructor
• Autumn Quarter, 2004

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Figure 3.1 Mass spectrometer
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Figure 3.2 (a) Peaks of neon injected
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Figure 3.2 (b) Bar graph of neon injected
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Problem 7-1 Isotopic Composition
The two isotopes of potassium with significant
abundance in nature are 39K (isotopic mass
38.9637 amu, 93.258) and 41K (isotopic mass
40.9618 amu, 6.730). Fluorine has only one
naturally occurring isotope, 19F (isotopic mass
18.9984 amu). Use this information to calculate
the formula mass of potassium fluoride.
Solution
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MOLE

• Definition The amount of substance that
  contains as many elementary particles (atoms,
  molecules, ions, or other ?) as there are atoms
  in exactly 12 grams of carbon 12.
• 1 Mole 6.022145 x 1023 particles (atoms,
  molecules, ions, electrons, or) NA particles
• NA is Avogados Number. (100 million x 100
  million x 100 million)

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The Mole is a Chemical Concept

• It represents a fixed number of chemical entities
  
• A mole of a chemical entity has a fixed, unique
  mass. (Molar Mass)
• Thus, the mole allows the mass balance to count
  chemical entities.

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Counting objects of fixed relative mass
12 red marbles _at_ 7g each 84g 12 yellow marbles
_at_4e each48g
55.85g Fe 6.022 x 1023 atoms Fe 32.07g S
6.022 x 1023 atoms S
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Mole - Mass Relationships of Elements
Element Atomic Mass Molar
Mass Number of Atoms
1 atom of H 1.008 amu 1 mole of H 1.008 g
6.022 x 1023 atoms 1 atom of Fe 55.85 amu
1 mole of Fe 55.85 g 6.022 x 1023atoms 1
atom of S amu 1 mole of S
g atoms 1 atom of O
amu 1 mole of O g
atoms Molecular mass 1
molecule of O2
amu 1 mole of O2
g
molecules 1 molecule of S8
amu
1 mole of S8 g
molecules
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Mole - Mass Relationships of Elements
Element Atomic Mass Molar
Mass Number of Atoms
1 atom of H 1.008 amu 1 mole of H 1.008 g
6.022 x 1023 atoms 1 atom of Fe 55.85 amu
1 mole of Fe 55.85 g 6.022 x 1023 atoms 1
atom of S 32.07 amu 1 mole of S 32.07 g
6.022 x 1023 atoms 1 atom of O 16.00 amu
1 mole of O 16.00 g 6.022 x 1023
atoms Molecular mass 1 molecule of O2 16.00 x
2 32.00 amu
1 mole of O2 32.00 g 6.022 x 1023
molecule 1 molecule of S8 32.07 x 8 256.56
amu 1 mole of
S8 256.56 g 6.022 x 1023 molecules
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Molecular Mass - Molar Mass ( M )
The Molecular mass of a compound expressed in amu
is numerically the same as the mass of one
mole of the compound expressed in grams,
called its molar mass.
For water H2O Molecular mass (2 x
atomic mass of H ) atomic mass of O
2 ( amu)
amu amu
Mass of one molecule of water
amu Molar mass ( 2 x
molar mass of H ) (1 x molar mass of O)
2 ( g )
g g
g H2O 6.022 x 1023 molecules of water 1 mole
H2O
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Molecular Mass - Molar Mass ( M )
The Molecular mass of a compound expressed in amu
is numerically the same as the mass of one
mole of the compound expressed in grams ,
called its molar mass.
For water H2O Molecular mass (2 x
atomic mass of H ) atomic mass of O
2 ( 1.008 amu) 16.00 amu
18.02 amu
Mass of one molecules of water 18.02 amu
Molar mass ( 2 x molar mass of H ) (1 x molar
mass of O) 2 ( 1.008 g
) 16.00 g 18.02 g 18.02 g
H2O 6.022 x 1023 molecules of water 1 mole H2O
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One mole of common substances CaCO3 100.09
g Oxygen, O2 32.00 g Copper 63.55 g Water
18.02 g
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Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem 7-2 Tungsten (W) is the element used as
the filament in light bulbs,
and has the highest melting point of any element
3680oC. How many moles of
tungsten, and atoms of the
element are contained in a 35.0 mg sample of the
metal? Plan Convert mass into moles by dividing
the mass by the atomic mass of the
metal, then calculate the number of atoms by
multiplying by Avogadros
number. Solution Converting from mass of W to
moles Moles of W No. of W atoms
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Calculating the Moles and Number of Formula
Units in a given Mass of Compound
Problem 7-3 Trisodium phosphate is a component
of some detergents. How many
moles and formula units are in a 38.6 g
sample? Plan We need to determine the formula,
and the molecular mass from the
atomic masses of each element multiplied by the
coefficients. Solution The formula is Na3PO4.
Calculating the molar mass MM
Converting mass to moles
Formula units
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Flow Chart of Mass Percentage Calculation
Moles of X in one mole of Compound
Multiply by M (g / mol of X)
Mass (g) of X in one mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100
Mass of X
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Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem 7-4 Sucrose (C12H22O11) is common table
sugar. ( a) What is the mass percent of each
element in sucrose? ( b) How many grams of
carbon are in 24.35 g of sucrose? (a)
Determining the mass percent of each element
mass of C per mole sucrose mass of H /
mol mass of O / mol
total mass per mole
Finding the mass fraction of C in Sucrose C

mass of C per mole
mass of 1 mole sucrose To find mass
of C
Mass Fraction of C

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Calculating Mass Percents and Masses of Elements
in a Sample of Compound - II
7-4 (a) continued
Mass of H
x 100 Mass of O
x 100 7-4 (b)
Determining the mass of carbon Mass (g)
of C mass of sucrose x ( mass fraction of C in
sucrose) Mass (g) of C
mol H x M of H mass of
1 mol sucrose
mol O x M of O mass of 1 mol sucrose

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Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a
compound that agrees with the elemental
analysis. The smallest set of whole numbers of
atoms.
Molecular Formula - The formula of the compound
as it exists. It may be a multiple of the
Empirical formula.
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Some Examples of Compounds with the same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 ,
C3H6 , C4H8 OH or HO
H2O2 S

S8 P
P4
Cl
Cl2 CH2O
(carbohydrates)
C6H12O6
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Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
Use no. of moles as subscripts.
Preliminary Formula
Change to integer subscripts smallest,
conv. to whole .
Empirical Formula
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Determining Empirical Formulas from
Masses of Elements - I
Problem 7-5 The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na
Moles of Cr Moles of O



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Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula
Converting to integer subscripts (dividing all by
smallest subscript)
Rounding off to whole numbers
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m 2
m 2
CnHm (n )O2(g) n CO2(g)
H2O(g)
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Determining a Chemical Formula from
Combustion Analysis - I
Problem 7-6 Erythrose (M 120 g/mol) is an
important chemical compound
used often as a starting material in chemical
synthesis, and contains Carbon, Hydrogen, and
Oxygen. Combustion analysis of a 700.0 mg
sample yielded 1.027 g CO2 and 0.4194 g
H2O. From this data calculate the molecular
formula. Plan We find the masses of Hydrogen
and Carbon using the mass fractions of
H in H2O, and C in CO2. The mass of Carbon and
Hydrogen are subtracted from the sample
mass to get the mass of Oxygen. We then
calculate moles, and construct the empirical
formula, and from the given molar mass we can
calculate the molecular formula.
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Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements
Mass fraction of C in CO2
Mass fraction
of H in H2O
Calculating masses of C and
H Mass of Element mass of compound x
mass fraction of element


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Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C Mass (g) of H Calculating
the mass of O Calculating moles of each
element C H O
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Answers to Problems in Lecture 7

• 58.09 amu
• 1.90 x 10 - 4 mol, 1.15 x 1020 atoms
• 0.23545 mol, 1.46 x 1023 formula units
• (a) 42.10 C, 6.479 H, 51.417 O (b) 10.25 g C
• Na2CrO4
• C4H8O4