Download Chapter 2: Atoms, Molecules, and Ions

Chapter 2: Atoms, Molecules,
and Ions
1. Atomic Structure
2. Atomic Number & Mass
3. Isotopes
4. Atomic Weight
5. Periodic Table
7. Molecular Formulas & Nomenclature
8. Molecular Compounds
9. Atoms, Molecules & Mole
10. Compound Formulas
11. Hydrated Compounds
2.1 Structure of The Atom
Modern View
Particle
Relative Mass
(g)
Relative Charge
(amu)
Proton
1.6726 x 10-24
1.0073
+1
Neutron
1.6749 x 10-24
1.0087
0
Electron
9.1094x10-28
0.0005486
-1
Structure of The Atom
Modern View
Three Fundamental Particles
1. Proton (Positive Charged Particle)
2. Neutron (Neutral Particles)
3. Electron - Negative Charged Essentially
Mass Less Particle, Occupies Majority of
Atomic Space
Nucleus - Region of Atomic Mass Density,
Contains Neutrons and Protons
Protons, Neutrons & Electrons:
Historical Perspective
Electricity
Labeled 2 types of charge as
Positive and Negative
Ben Franklin
(1706-1790)
Like Charges Repel
Opposite Charges Attract
Noted that charge is balanced, so for
every [+] charge there exists a [-] charge
Types of Radioactive Processes
1. Alpha Decay
-Heavy [+] Particle
2. Beta Decay
-Light [-] Particle
3. Gamma Decay
-Charge less
-Release of Energy
Some History on The Discovery of the Atom
Discovery of The Electron:
J.J. Thomson and the Cathode Ray Tube
(1897)
Calculated
e-/m
Some History on The Discovery of the Atom
Discovery of The Electron:
Robert Millikan (1911) Measured the
Charge of the Electron
Charge of electron
= 1.6 x 10-19C.
Some History on The Discovery of the Atom
Devised an Atomic Theory From the Following
Postulates:
1. Atoms Are Neutral and Have Negative
Tiny Electrons.
2. Atoms Must Therefor Have Positive Particles to
Balance the Charge
3. Electrons Can Not Account for the Mass of the Atom
Plumb-Pudding Model, Small Electrons Are in the
Middle of a Large Sphere of Positive Charge
Some History on The Discovery of the Atom
Rutherford’s Gold Foil Experiment and
the Discovery of the Nucleus
Believed in Plum Pudding Model
Wanted to see How Large Atoms are
Atomic Number & Mass Number
The Mass of an Atom Can Be Determined From the
Masses of the Neutrons and Protons in It’s Nucleus
Atomic Mass Unit (AMU)
1amu = 1.66 X 10-24g
This Is a Very Small Number
1 Amu = Mass of 1 Carbon Atom With
6 Neutrons & 6 Protons
Atomic Number
# of Protons in an Atom
Know How to Determine Atomic
Numbers & Masses From the
Periodic Table
Atomic Masses
The Mass of an Atom Can Be Determined From the
Masses of the Neutrons and Protons in It’s Nucleus
-This is a very small number
One 12C atom weights 1.99 X 10-23 grams
Atomic Mass Unit (AMU)
1amu = 1.66 X 10-24g
Atomic Masses
Note: the Atomic Weights of the Elements (the Mass
of 1 Atom in Amu’s) Are Given in the Periodic
Table.
Why Is the Atomic Weight of a Carbon = 12.01 Amu?
Note: Carbon has 3 Isotopes
12C, 13C,
& 14C
12.01 Is the Weight of the
Natural Occurring Mixture of
These Isotope
Atomic Masses
What is the Mass of a 100 Iron Atoms?
1 Iron atom = 55.85 amu
55.85amu
(100 Fe)(
)  5,585amu
Fe
Always Include Units in Your Calculations!
Atomic Masses
Why is This So Important?
Because by knowing the Mass of an
Element (a Measurable Quantity), We can
Determine the Number of Atoms Present!
How Do We Do This On the
Macroscopic (Everyday) Scale?
The Mole
Mole = SI Standard
Unit for Quantity
1 Mole = 6.022 x 1023
(Amadeo Avogardo)
Avogadro’s Number = 6.022 x 1023
The Atomic Weight of an Element Is the Number of
Grams One Mole of That Element Would Weight
1 mole of Carbon weights 12.011 grams!
The Mole
How Many atoms are in 48.044
grams of Carbon?
1mol  C
48.044 g  C (
)  4.0000mols  C
12.011g  C
OR
6.022 x1023 atoms  C
4mols  C (
)  2.409 x1024 atoms  C
mol  C
The Mole
Group Exercises:
How many mols are in 48 g Tin?
How many atoms is this?
How much does 4.395 mols gold
weight?
What is the mass of a billion
radon atoms?
Atomic Masses
How Can Chemists Measure Atomic Masses?
Mass Spectroscopy
2.3 Isotopes
The Identity of an Element Is Determined by the
Number of Protons It Possesses
Isotopes - Elements With the Same Number of
Protons but Different Numbers of Neutrons.
Different Atoms of Same Element Can Have
Different Masses.
•Isotopic Notation:
A= Mass Number (# of P and N)
A
zX
Z=Atomic Number (# of P)
X= Elemental Symbol
Isotopes
#Protons = Z, the Atomic Number, and Can
Also Be Determined by X, the Atomic
Symbol
(All Carbon Isotopes Have Z=6 and X=C)
#Neutrons = A - Z
A= Mass Number (# of P and N)
A
zX
Z=Atomic Number (# of P)
X= Elemental Symbol
Isotopes
Determine the Number of Protons , Neutrons and
Electrons for the 3 Isotopes of Oxygen
16
8O:
P = 8, N = (16-8) = 8, e = 8
17
8O:
P = 8, N = (17-8) = 9,
18
8O:
P = 8, N = (18-8) =10, e = 8
e=8
Note: For Neutral Atoms, the # of
Protons Equals the # of Electrons
Isotopes of Monatomic Ions
Note: For ions there is one less electron than
proton for each [+] charge and one more
electron than protons for each [–] charge.
Determine the Number of Protons ,
Neutrons and Electrons for the 2 Ions:
16 O-2 :
8
P = 8, N = (16-8) = 8,
55 Mn+7
25
P = 25, N = (17-8) = 30, e = 18
e = 10
Radioactive Isotopes
Many Isotopes Are Unstable and Undergo
Radioactive Decay.
If the Number of Protons Changes During
a Radiative Process, the Identity of the
Element Changes.
We Will Cover These in Chapter 23 of the
Text
2.4 Atomic Weight
Given in the Periodic Table.
Why Is the Atomic Weight of a Carbon = 12.01 Amu?
Note: Carbon has 3 Isotopes
12C, 13C,
& 14C
12.01 Is the Weight of the
Natural Occurring Mixture of
These Isotope
Isotope Abundance
Natural Samples of Pure Elements Contain a
Mixture of that Elements Isotopes
% Isotopic
Abundance
=
# of Atoms of Given Isotope(100)
Total Atoms of All Isotopes
Class Problem
Magnesium has 3 Isotopes with the
following masses and % Abundances,
23.985 amu (78.99%),
24.986 amu (10.00% )
25.983 amu (11.01%)
Determine the average atomic weight of
Magnesium
.7899(23.985)  .1000(24.986)  .1101(25.983)
 24.305amu
Class Problem
Lithium has has only 2 isotopes, 6Li
(with mass of 6.0151) and 7Li (with
mass of 7.0160)
What is the % Abundance for
Each Isotope?
Let
X = fraction of 6Li
Y = fraction of 7Li
X(6.0151) +Y(7.0160) = 6.941
2 Unknowns Requires 2 Equations
X+Y=1
Y=1-X
X(6.0151) +(1-X)7.0160 = 6.941
X
6.941  7.0160
 .07493; Y  1  X  1  .07493  .9251 6Li
6.0151  7.0160
7Li
Note, 7.941-7.0160 = -0.075
= 7.5%
= 92.5%
2.5 Periodic Table
Dmitri Mendeleev, noted
various elements had
similar properties, with a
periodic relationship to the
increasing atomic number.
This resulted in the
Modern Periodic Table
Periodic Table
Periodic Table
Be Able to Identify the Following Groups or
Families From the Periodic Table
1. Alkali Metals (Group 1)
2. Alkaline Earth Metals (Group 2)
3. Halogens (Group 7)
4. Noble Gases (Group 8)
5. Transition Metals
6. Lanthanides
7. Actinides
Metals, Nonmetals & Metalloids
Periodic Table
Physical Properties of Metals
1. Conduction of Heat and Electricity
2. Malleability (can be hammered into
thin sheets)
3. Ductility (can be pulled into wires)
4. Lustrous (shiny) appearance
Nonmetals
-Poor Conductors
-Nonlusterous
-Nonmalleable
-Solids are brittle
-Some are brightly
colored
Natural States of The Elements
Liquids - Hg, Br2
Gasses - Noble Gasses and lighter
diatomics, H2 , N2 , O2 , F2 , Cl2
Solids - All other elements
Natural States of The Elements
Elements Which Often Appear in Pure Form
A) Noble Metals (Ag, Au, & Pt)
B) Noble Gases (He, Ne, Ar, Kr Xe & Rn)
“Noble” Implies “Unreactive”
All elements with Atomic # > 83
are Radioactive
7 Diatomics
Know the Diatomic’s and Their Natural States
Hydrogen (H2)
gas
(colorless)
Nitrogen (N2)
gas
(colorless)
Oxygen
(O2)
gas
(pale blue)
Fluorine
(F2)
gas
(pale yellow)
Chlorine (Cl2)
gas
(pale green)
Bromine (Br2)
Liquid (red/brown)
Iodine
Solid
(I2)
(purple)
Group 1A: Alkali Metals
H, Li, Na, K, Rb, Cs & Fr
Very Reactive – Do Not Exist as
Pure Elements in Nature
Form Oxides of Formula
A2O
Group 2A: Alkali Earth Metals
Be, Mg, Ca, Sr, Ba & Ra
Reactive – Form Alkaline
Solutions (Basic)
Form Oxides of Formula
AO
Group 3A
B, Al, Ga, In, Tl
Boron comes from Borax
Form Oxides of Formula
A2O3
Group 3A
B, Al, Ga, In, Tl
Boron comes from Borax
Form Oxides of Formula
A2O3
Group 4A
C, Si, Ge, Sn, Pb
C-non metal
Si, Ge – metalloids
Sn, Pb - metals
Group 4A
C, Si, Ge, Sn, Pb
Carbon forms Allotropes
(Different Forms of the Same Element)
Diamond
Graphite
Buckminsterfullerene
Group 5A
N, P As, Sb, Bi
N, P – Non metals (essential to life, N is
most
abundant element in atmosphere
As, Sb – metalloids
Bi – metal, heaviest non radioactive element
Group 6A
O, S, Se, Te, Po
Chalcogens
Chalk Formers
O, S, Se – nonmetals
Te – metalloid
Po – Radioactive metal
Group 7A
F, Cl,Br, I, At
Halogens
Salt Formers
F,Cl, Br, I – nonmetal diatomics
At – often classified as
radioactive metalloid
Group 8A
He, Ne, Ar, Kr, Xe
Noble Gases
“Inert Gases” – do not tend to
form compounds
Part II: Compounds
4 Types of Chemical Bonds
1. Ionic Bonds - Ionic compounds between metals and
nonmetals consisting of [+] cations and [-] anions.
2. Covalent Bonds - Classical “molecules” where
valence electrons are shared between two
nonmetals
3. Polar Covalent Bonds - Covalent bonds with ionic
character in that the electrons are not equally shared.
4. Metallic Bonds - Pure metals and alloys where
delocalized free electrons hold together the positive
nuclei.
2.6 Molecules & Compounds
-Pure Substance Composed of More Than 1 Atom
-Formulas Describe the Elemental Composition
-Common Names Like Baking Soda or Sugar
Tell Us Nothing About Their Composition
-Their Formulas Provide Information on Their
Chemical Composition
Chemical Formulas
Sugar = C12H22O11
Baking Soda = NaHCO3
This tells us that:
-one molecule of sugar has 12 carbon atoms,
22 hydrogen atoms and 11 oxygen atoms
-Baking Soda has 3 atoms of Oxygen and 1
atom each of Sodium, Hydrogen and Carbon
Molecular Formulas
Molecular Formulas - only tell of the
elemental composition
-both ethanol and diethyl ether have the same
molecular formula, C2H6O
-Yet they are different
Structural Formula
Ethanol = CH3CH2OH
Dimethyl Ether = CH3OCH3
HH
H
H
H-C-C-O-H
H-C-O-C-H
bonds
H H
HH
Molecular Models
behind
plane
in front of
plane
Ball & Stick
Persective Drawing
METHANE (CH4)
Molecular Models - Water
space-filling models
ICE
2.7 Ions & Nomenclature
Charged Atoms Occur When an Atom Losses
or Gains Valence Electron(s)
Two Types:
Anions - Negative Ions (Gain
Electrons)
Cations - Positive Ions (Lose Electrons)
Note: Charge Is Indicated by a +/- Post
Superscript
Ions
Ionic Charge and the Periodic Table:
Metals form Cations (tend to lose electrons)
Nonmetals for Anions (tend to gain electrons)
Noble Gasses Do Not Tend to Form Ions
Ions
Cations
Anions
+1 Alkali Metals, H, Ag
-1 Halogens, Hydrogen
+2 Alkaline Earths, Zn, Cd
-2 Group VI Nonmetals
+3 Aluminum
-3 Group V Nonmetals
Note - Monatomic Ions Tend to Have Charge Required to
Become Isoelectronic With Nearest Nobel Gas
(Have Same # of Electrons As the Noble Gas)
Polyatomic Ions
-Charged Covalent Compounds
+
H
H-N-H
H
NH4
+
O
O-S-O
O
SO4-2
-2
Ionic Compounds & Nomenclature
Ionic Compounds Form Crystal Lattices
Resulting From Electrostatic Interactions
Which Maximize the +/- Attractions While
Minimizing the +/+ and -/- Repulsions
Formula Unit
-Ionic Compounds Are Identified by
Their Formula Unit
-Formula Unit Represents the Lowest
Whole # Ratio of Ions in the
Crystal Lattice
Nomenclature
1. Ionic Compounds (between metals and
nonmetals)
2. Covalent Compounds (between two
nonmetals)
3. Compound with Polyatomic Ions
4. Acids
5. Compound Formulas from Names
Binary Compounds
Binary Compounds - Contain two types
of atoms, can be ionic or covalent
Examples:
H2O, CO, CO2, CoCl2, FeO, Fe2O3, NaCl
Binary Covalent
Binary Ionic
Binary Ionic Compounds
Form Between a Metal and a Nonmetal,
-Have Monatomic Ions
-Form Crystal Lattice Type Solid Structures
-Oxidation # of a Monatomic Ion Is It’s Charge
Cation - [+] ion, (Metals Form Cations)
Type 1 Metal (forms only 1 cation) Na+,
Type 2 Metal (forms 2 or more cations) Fe+2, Fe+3,...
Anion - [-] ion, (Nonmetals Form Anions)
ex: Cl-, O-2, N-3,...
Binary Ionic Compounds
Type I Metals -have single charge in all ionic
compounds (Invariant Oxidation #)
Know Following Invariant Oxidation #’s:
+1 Alkali Metals, H, Ag
-1 Halogens, Hydrogen
+2 Alkaline Earths, Zn, Cd
-2 Group VI Nonmetals
+3 Aluminum
-3 Group V Nonmetals
Note - “A” Group Monatomic Ions Tend to Have Charge
Required to Become Isoelectronic With Nearest Nobel
Gas (Have Same # of Electrons As the Noble Gas)
Binary Ionic Compounds
Compounds with Type I Metals
Rules:
1. Name Metal First
2. Name Nonmetal 2nd With -ide Suffix
Ex: Table Salt, NaCl = Sodium Chloride
BaCl2 = Barium Chloride
Binary Ionic Compounds
Ionic Formulas from Names
Rules:
1. Ionic Formula Must Be Neutral (Principal of
Charge Balance)
2. Subscripts After Elemental Symbol Indicate #
of That Element in Formula
3.Correct Formula Indicates Lowest Whole #
Ratio of Anions to Cations
Magnesium Chloride = MgCl2 not Mg2Cl4
Binary Ionic Compounds
Ionic Formulas from Names
Trick: Set # of Anions = Charge of Cation
and # of Cations = Charge of Anion
Beware That This Is the Lowest Whole # Ratio
Consider Aluminum Oxide,
Charge of Aluminum =+3 and Oxygen = -2,
Al2O3
(2x+3)+(2x-3)=0
Binary Ionic Compounds
1. In first column, chose cations of given charges and
place symbol underneath charge, in first row, place
symbol of appropriate anions underneath each charge.
2. Work in Groups
of 3, each student
-1
-2
-3
picks one row, for
Cl
each cell of their
+1 Sodium Chloride
row, write name in
Na NaCl
upper half, and
formula in lower
+2
half.
3. Check each others +3
work!
Binary Ionic Compounds
Type II Metals -have multiple charges in ionic
compounds (Variable Oxidation #’s)
Know Following Variable Oxidation State Ions
Fe+3 Iron(III) - ferric
Sn+4
Tin(IV) - stannic
Fe+2 Iron(II) - ferrous
Sn+2
Tin(II) - stannous
Cu+2 Copper(II) - cupric
Pb+4 Lead(IV) - plumbic
Cu+1 Copper(I) - cuprous
Pb+2
Co+3 Cobalt(III) - cobaltic
Hg+2 Mercury(II) - mercuric
Co+2 Cobalt(II) - cobaltous
Hg2+2 Mercury(I) - mercurous
Lead(II) - plumbous
Binary Ionic Compounds
Type II Metals
Rules:
1. Name Metal First
2. Indicate Metal’s Charge (Oxidation State)
in Roman Numerals
2. Name Nonmetal 2nd With -ide Suffix
FeCl2 = Iron(II)Chloride
FeCl3 = Iron(III)Chloride
Naming Covalent Compounds
-Compounds Between Nonmetals Do Not Form
Ions but Share Electrons (Forming Covalent Bonds)
- These Are the Classical “Molecules” (in
Contrast to Ionic Crystal Lattices)
- Use Greek Prefixes to Indicate Number of
Each Type of Atom in Molecules
- Write “More Metallic” Nonmetal First
- If More Than 2 Atoms, Write in Order
of Connectivity (SCN vs. SNC)
Naming Covalent Compounds
Greek Prefixes
1 mono-
6 hexa-
2 di-
7 hepta-
3 tri-
8 octa-
4 tetra-
9 nona-
5 penta-
10 deca-
Note: often do not use prefix mono-
Naming Covalent Compounds
Group Problems: Name or write the
following formula’s
CCl4
- Carbon tetrachloride
SO3
- Sulfur trioxide
Dinitrogen trioxide - N2O3
Diphosphorous pentaoxide - P2O5
Polyatomic Ions
Many Ionic Compounds Have Covalently
Bonded Polyatomic Ions
Rules Are Essentially Same, but Use the Name
of the Polyatomic Ion, That Is;
Cation First (Charge If Variable) Anion Last
Polyatomic Ions
NH4+
CH3COONO2NO3OHClO4ClO3ClO2-
MnO4CO3-2
HCO3-
Ammonium
Acetate
Nitrite
Nitrate
Hydroxide
Perchlorate
Chlorate
Chlorite
Hypochlorite
Permanganate
carbonate
bicarbonate
C8H4O4-2
SO4-2
SO3-2
CrO4-2
Cr2O7-2
PO4-3
PO3-3
AsO4-3
CNSCNC2O4-2
O2-2
Phthlate
Sulfate
Sulfite
Chromate
Dichromate
Phosphate
Phosphite
Arsenate ClOCyanide
Thiocyanide
Oxalate
peroxide
Polyatomic Ions
Tip for Remembering Charges - Many Oxyanions
(Polyatomic Ions With Multiple Oxygens) Have the
Same Charge As the Monatomic Nonoxygen Ion
Sulfide = -2 As Does Sulfate & Sulfite
Chloride = -1 As Does Perchlorate, Chlorate, Chlorite
& Hypochlorite
Phosphide = -3 Does Phosphate & Phosphite
Note, Nitrogen Is an Exception:
Nitride = -3, Nitrate & Nitrite = -1
Polyatomic Ions
Note: Oxyanions of Heavier Halogens
(Bromine and Iodine) Form Same Types
of Compounds As Chlorine
BrO4-
perbromate
IO4-
periodate
BrO3-
bromate
IO3-
Iodate
BrO2-
bromite
IO2-
Iodite
BrO-
hypobromite
IO-
hypoiodite
Acid Nomenclature
Compound which can release H+ in water are
called Acids. (aqueous, aq.,- means dissolved in water)
Nomenclature is related to that of Ionic Compounds
-If salt is binary, it forms binary acids :
-ide => hydro(anions elemental name)-ic acid
If salt has oxyanions it forms oxyacids with endings:
ate => ic acid
ite => ous acid
Sulfur’s Sodium Salts and Acids
Na2S - Sodium Sulfide
H2S(aq) - Hydrosulfuric Acid
Note: H2S(g) - Hydrogen Sulfide (gas)
Na2SO4 - Sodium Sulfate
H2SO4(aq)- Sulfuric Acid
Na2SO3 - Sodium Sulfite
H2SO3(aq)- Sulfurous Acid
Acid Nomenclature
Name or give Formulas for Following Acids:
Acetic Acid (vinegar) -
CH3COOH
Chloric Acid -
HClO3
Hydrochloric Acid -
HCl
H3PO4 -
Phosphoric Acid
HCN -
Hydrocyanic Acid
Acid Salts
-Ions with more than one negative site can
combine with more than one cation.
- Acid salts result when one cation is a
proton
Lets look at all of the compounds that
can form from phosphate
Phosphate = PO4-3
Na3PO4= Sodium Phosphate
Na2HPO4= Sodium Hydrogen Phosphate
NaH2PO4= Sodium Dihydrogen Phosphate
H3PO4 = Phosphoric Acid
Name the compounds that can form from
hydrogen, potassium and phthlate; and
identify them as salts, acids or acid salts
Hydrated Salts
-Salts which incorporate water into their
crystal structure are hydrated salts
-Include the water in the compound formula
-Greek prefixes indicate the # of waters of
hydration
CaCl2.6H2O = Calcium Chloride hexahydrate
What is the name of
CuSO4 .5 H2O ?
Copper(II) sulfate pentahydrate
2.9 Molar Mass
Molar Mass - Is the Mass (in Grams) of
One Mole of a Substance
(often called molecular weight or formula weight)
What is the Molar Mass of Sulfate?
1 mole SO4-2: has 1 mole S and 4 moles O
1 mole S
= 32.07 g
4 mole O = 4(16) = 64.00g
1 mole SO4-2
= 96.07g
Molar Mass
How many moles are in 5.86g of AgCl?
1 mol AgCl = 143.3g, ie.,
MM(AgCl)=143.3g/mol
mol  AgCl
(5.86 g  AgCl )(
)  0.041mol  AgCl
143.3g  AgCl
Note how we use Dimensional Analysis and the
formula weight as a mass-->mole conversion factor
Molar Mass
Calculate the Molar Mass of:
1. Benzene (C6H6)
78g/mol
2. Acetylene (C2H2) :
26g/mol
3. Water
18g/mol
4. Hydrogen Peroxide
34g/mol
Molar Mass
What is the Molar Mass for
Aluminum Sulfate?
Al2(SO4)3 has 2 Al Cations and 3 Sulfate
Anions
2(Al) = 2(26.98) = 53.96g
3(SO4)3 = 3(96.07)
= 288.21g
1 mol Al2(SO4)3
= 342.17g
Note: We used the Molar Mass of Sulfate which we
determined in the Previous Problem
Molar Mass
Calculate Molar Mass for:
1. Sodium Bicarbonate
84.02 g/mol
2. Ammonium Nitrate
80.08 g/mol
3. Potassium Perchlorate
138.54g/mol
4. Silver Chloride
143.32g/mol
Molar Mass
1. What is the mass of 5.80 moles
Sodium Bicarbonate?
420.g
2. What is the mass of 6.82 moles
Ammonium Nitrate
546g
3. How many moles are in 51.50 g of
Potassium perchlorate?
.3717 moles
4. How many moles are in 100.0 g of
Silver Chloride?
.6977 moles
2.10 Percent Composition
What Is the Mass Fraction of an Element in a
Compound?
Mass Fraction
of a given element
Mass of element in compound
=
Mass of compound
What is the Mass Percent of an Element in a
Compound?
Mass Percent = Mass Fraction
of a given element
of a given element
X 100
Percent Composition
1. What Is the Sum of the Mass Percents
of All the Elements in a Compound?
100%
2. Does the Percent Composition Change
as the Quantity of Substance Changes?
NO
What is the mass percent of H, C & O
in Acetic Acid (Vinegar)
CH3COOH
Percent Composition
Consider 1 mole CH3COOH
MCarbon = 2(12.01) = 24.02g
MOxygen = 2(16.00) = 32.00g
%C=24.02(100) = 40.00%
60.06
MHydrogen = 4(1.01) = 4.04g %O=32.00(100)=53.28%
60.06
MAcetic Acid =
60.06g
How Do We Get the
Percent Composition
From the Above Data?
%H=4.04(100)=7.0%
60.06
Percent Composition Problems
1. Determine the % Composition of the
Sulfate Ion.
33.3% S, 67.7% O
2. Determine the % Composition of
Aluminum Sulfate 15.8% Al, 28.1% S, 56.1%O
3. Determine the Percent Composition
of Benzene (C6H6)
7.7% H, 92.3% C
4. Determine the Percent Composition
of Acetylene (C2H2) 7.7% H, 92.3% C
Compound Formulas
1. Empirical Formula - Simplest Whole #
Ratio of the Various Elements in a
Compound
2. Molecular Formula - the Actual Ratio of
the Various Elements in a Compound or
Molecule
Note: Benzene (C6H6) & Acetylene (C2H2)
Have Different Molecular Formulas,
but the Same Empirical Formula (CH)!
Empirical Formulas
-Empirical data is based on observation and
experiment, not theory
-Opposite Calculation to Percent Composition
-For Empirical Formulas, You Know the Masses
of Each Element From Experimental Data,
So Determine Simplest Formula
-In % Composition, You Start Knowing the
Formula, and Determine the Mass %
Empirical Formulas
1. Obtain Mass of Each Element (in Grams)
If given % Composition, assume 100 g of substance
2. Calculate # of Moles of Each Element Present
From Masses and Atomic Weights
3. Divide # of Moles by the Moles of Least
Present Element (Forcing It to 1)
4. Multiple the Results of Step 3 by Smallest
Integer Which Will Convert Them All to
Whole Numbers
Empirical Formulas
Determine the Empirical Formula of Aspirin,
Which Was Analyzed and Found to Have
a Mass Percent Composition of 60.0%C,
4.48%H and 35.5%O.
Empirical Formulas
1. Assume 100 grams of sample, giving
MC=60.0g,
MH=4.48g,
MO=35.5g
2. Calculate Moles Present
moles C = 60.0g(1mol/12.0g) = 5
moles H = 4.48g(1mol/1.00g)=4.48
moles O = 35.5g(1mol/16.0g) =2.22
Empirical Formulas
3. Divide By
Smallest Mole
Fraction (2.22)
4.Multiply by smallest
Integer Forcing all
results from step 3 to
be whole numbers
C: 5/2.22 = 2.25
H: 4.48/2.22 = 2.02
O: 2.22 /2.22 = 1.00
C: 2.25(4) = 9
H: 2.02(4) = 8
O: 1.00(4) = 4
Empirical Formula of Aspirin = C9H8O4
Molecular Formulas
In Order to Determine the Molecular
Formula, We Need to Know the
Molar Mass in Addition to the
Empirical Formula
The Molecular Formula will be an Integral
Multiple of the Empirical Formula
For an Ionic Compound, the Formula Weight
Corresponds to the Empirical Formula
Determine the molecular formula of a compound
containing H, O & C. A 50.00 g sample has 23.88g
of both carbon & oxygen and the compound has a
molar mass of 603g/mol.
1. Determine moles in sample
note: mass hydrogen = total mass-mass oxygen –mass carbon
= 50.00g-2(23.88g) = 2.24g
 1mol 
moles C  23.88 gC 
  1.988mol C
 12.01g 
 1mol 
moles O  23.88O 
  1.493mol O
16.00
g


 1.008mol 
moles H  2.24 gH 
  2.26mol H
g


Step 2: Divide by Smallest number
Trick: Convert to Fractions
 1 
H : 2.24 
  1.5
 1.493 
 1 
C : 1.988 
  1.33
 1.493 
 1 
O : 1.493 
 1
 1.493 
 1  3
H : 2.24 

 1.493  2
 1  4
C : 1.988 

1.493

 3
 1 
O : 1.493 
 1
 1.493 
Tips in converting decimals to fractions
1
1
 decimal expresssion  X 
X
decimal expression
Know:
0.50 = 1/2
0.25 = 1/3
0.33 = 1/3
0.20 = 1/5
Step 2: Divide by Smallest number
Trick: Convert to Fractions
 1 
H : 2.24 
  1.5
 1.493 
 1 
C : 1.988 
  1.33
 1.493 
 1 
O : 1.493 
 1
 1.493 
 1  3
H : 2.24 

 1.493  2
 1  4
C : 1.988 

 1.493  3
 1 
O : 1.493 
 1
 1.493 
Step 3: Find lowest whole number ratio by
factoring out the denominators
 1  3
H : 2.24 
  (6)  9
 1.493  2
 1  4
C : 1.988 
  (6)  8
 1.493  3
 1 
O : 1.493 
  1(6)  6
1.493


C8H9O6
Step 4: Calculate Molecular Formula
from Empirical Formula Using Masses
EF = Empirical Formula
MF = Molecular Formula
EM = Empirical Mass
MM = Molecular Mass
MF = n(EF) & MM=n(EM), n=1,2,3…
EM(C8H9O6) = 201g/mol, MM=603g/mol
MM 603 g mol
n

3
g
EM 201 mol
MF = n(EF)=3(C8H9O6) = C24H27O18
Acidum Formicum – Formic Acid
What is the empirical formula
of Formic Acid?
Obtained through the distillation of Formica rufa
Empirical Formula of Formic Acid
from Combustion Data
H2O
CO2
excess
(CxHyOn)
absorber
absorber
oxygen
2.4541g
formic acid
0.9606g
2.3482g
Sample
furnace
Step 1: Determine Mass of each element
mH = .9606g(2gH/18gH2O ) = .1067g H
mC = 2.3482g(12gC/44gCO2 ) = .6404g C
mO = 2.4541g - .1067g - .6404g = 1.7071gO
Step 2: Determine Mole of Each Element
 1mol 
moles H  0.1067 gH 
 0.1067mol H

 1.008 g 
 1mol 
moles C  0.6404 gC 
 0.0533mol C

 12.01g 
 1mol 
moles O  1.7071 
 0.10670mol O

 15.9994 g 
Step 3: Divide by Smallest Number
 1 
H : 0.1067 
2
0.0533


 1 
C : 0.0533 
 1
 0.0533 
 1 
O : 0.1067 
2
0.0533


CH2O2
2.11 Hydrated Salts
Hydrated salts absorb water into
the crystal lattice- this is the
“water of hydrations”
-Stoichiometric Proportions
(water of hydration is of
integer proportions)
Anhydrous Salt – “Without Water”, Form
of Salt without water of hydration, forms
different type of crystal
Hydrated Salts
Sodium Carbonate Forms a Hydrated Salt
Na2SO4.xH2O
What is the Formula if a 0.767g
sample was dried to a constant weight
of 0.284 g?
Hydrated Salts
Mhydrated salt = Manyhdrous salt + M water
M water = Mhydrated salt - Manyhdrous salt
M water = 0.767g – 0.284g = 0.483g
n  mole, M  mass, fw  formula weight
nH 2O
M H 2O  g 
0.483g


 0.0268 mole H 2O
g
g
fwH 2O  mol  18  mol 
n Na2CO 
3
M Na2CO 3  g 
fwNa2CO f
0.284 g

 0.00268 mol Na2CO3
g
g
 mol  106  mol 
Hydrated Salts
Dividing moles water and moles Sodium
carbonate by species with least number of
moles (0.00268 moles Na2CO3):
Na2CO3.10 H2O