Chapter 2: Atoms, Molecules, and Ions

What is the Mass of a 100 Iron Atoms? 1 Iron atom = 55.85 amu amu. Fe amu. Fe. 585,5). 85.55. )( 100(. = Always Include Units in Your Calculations! ...

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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