Dr. S. M. Condren Atoms, Molecules & Ions Chapter 2

Dr. S. M. Condren Quantum Corral

Dr. S. M. Condren Scanning Tunneling Microscope

Dr. S. M. Condren Scanning Tunneling Microscope

Dr. S. M. Condren Scanning Tunneling Microscope

Dr. S. M. Condren

Dr. S. M. Condren Developed in collaboration with the Institute for Chemical Education and the Magnetic Microscopy Center University of Minnesota

Dr. S. M. Condren Pull Probe Strip Probe Sample Pull Probe Strip

Dr. S. M. Condren (a) (b) NorthSouth (c) Which best represents the poles?

Dr. S. M. Condren Atoms & Molecules Atoms can exist alone or enter into chemical combination the smallest indivisible particle of an element Molecules a combination of atoms that has its own characteristic set of properties

Dr. S. M. Condren Law of Constant Composition A chemical compound always contains the same elements in the same proportions by mass.

Dr. S. M. Condren Law of Multiple Proportions the same elements can be combined to form different compounds by combining the elements in different proportions

Dr. S. M. Condren Dalton’s Atomic Theory Postulates proposed in 1803 know at least 2 for first exam

Dr. S. M. Condren Dalton’s Atomic Theory Postulate 1 An element is composed of tiny particles called atoms. All atoms of a given element show the same chemical properties.

Dr. S. M. Condren Dalton’s Atomic Theory Postulate 2 Atoms of different elements have different properties.

Dr. S. M. Condren Dalton’s Atomic Theory Postulate 3 Compounds are formed when atoms of two or more elements combine. In a given compound, the relative number of atoms of each kind are definite and constant.

Dr. S. M. Condren Dalton’s Atomic Theory Postulate 4 In an ordinary chemical reaction, no atom of any element disappears or is changed into an atom of another element. Chemical reactions involve changing the way in which the atoms are joined together.

Dr. S. M. Condren Radioactivity

Dr. S. M. Condren Radioactivity Alpha – helium-4 nucleus Beta – high energy electron Gamma – energy resulting from transitions from one nuclear energy level to another

Dr. S. M. Condren Alpha Radiation composed of 2 protons and 2 neutrons thus, helium-4 nucleus +2 charge mass of 4 amu creates element with atomic number 2 lower Ra 226  Rn He 4 (  )

Dr. S. M. Condren Beta Radiation composed of a high energy electron which was ejected from the nucleus “neutron” converted to “proton” very little mass -1 charge creates element with atomic number 1 higher U 239  Np  -1

Dr. S. M. Condren Gamma Radiation nucleus has energy levels energy released from nucleus as the nucleus changes from higher to lower energy levels no mass no charge Ni 60*  Ni 60 + 

Dr. S. M. Condren Cathode Ray Tube

Dr. S. M. Condren Thompson’s Charge/Mass Ratio

Dr. S. M. Condren Millikin’s Oil Drop

Dr. S. M. Condren Rutherford’s Gold Foil

Dr. S. M. Condren Rutherford’s Model of the Atom

Dr. S. M. Condren Rutherford’s Model of the Atom atom is composed mainly of vacant space all the positive charge and most of the mass is in a small area called the nucleus electrons are in the electron cloud surrounding the nucleus

Dr. S. M. Condren Structure of the Atom Composed of: protons neutrons electrons

Dr. S. M. Condren Structure of the Atom Composed of: protons neutrons electrons protons –found in nucleus –relative charge of +1 –relative mass of amu

Dr. S. M. Condren Structure of the Atom Composed of: protons neutrons electrons neutrons –found in nucleus –neutral charge –relative mass of amu

Dr. S. M. Condren Structure of the Atom Composed of: protons neutrons electrons –found in electron cloud –relative charge of -1 –relative mass of amu

Dr. S. M. Condren Size of Nucleus If the nucleus were 1” in diameter, the atom would be 1.5 miles in diameter.

Dr. S. M. Condren Ions charged single atom charged cluster of atoms

Dr. S. M. Condren Ions cations –positive ions anions –negative ions ionic compounds –combination of cations and anions –zero net charge

Dr. S. M. Condren Atomic number, Z the number of protons in the nucleus the number of electrons in a neutral atom the integer on the periodic table for each element

Dr. S. M. Condren Isotopes atoms of the same element which differ in the number of neutrons in the nucleus designated by mass number

Dr. S. M. Condren Mass Number, A integer representing the approximate mass of an atom equal to the sum of the number of protons and neutrons in the nucleus

Dr. S. M. Condren Masses of Atoms Carbon-12 Scale

Dr. S. M. Condren Isotopes of Hydrogen H-1, 1 H, protium 1 proton and no neutrons in nucleus only isotope of any element containing no neutrons in the nucleus most common isotope of hydrogen

Dr. S. M. Condren Isotopes of Hydrogen H-2 or D, 2 H, deuterium 1 proton and 1 neutron in nucleus

Dr. S. M. Condren Isotopes of Hydrogen H-3 or T, 3 H, tritium 1 proton and 2 neutrons in nucleus

Dr. S. M. Condren Isotopes of Oxygen O-16 8 protons, 8 neutrons, & 8 electrons O-17 8 protons, 9 neutrons, & 8 electrons O-18 8 protons, 10 neutrons, & 8 electrons

Dr. S. M. Condren The radioactive isotope 14 C has how many neutrons? 6, 8, other

Dr. S. M. Condren The identity of an element is determined by the number of which particle? protons, neutrons, electrons

Dr. S. M. Condren Mass Spectrometer

Dr. S. M. Condren Mass Spectra of Neon

Dr. S. M. Condren Measurement of Atomic Masses Mass Spectrometer a simulation is available at OChem/DEMOS/MassSpec.html

Dr. S. M. Condren Atomic Masses and Isotopic Abundances natural atomic masses = sum[(atomic mass of isotope) *(fractional isotopic abundance)]

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35y = fraction Cl-37 x + y = 1y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = Thus: *x *y =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( )

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x = x =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x = x = x = % Cl-35

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x = x = x = % Cl-35 y = 1 - x

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x = x = x = % Cl-35 y = 1 - x =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x = x = x = % Cl-35 y = 1 - x = =

Dr. S. M. Condren Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of and amu, respectively. The natural atomic mass of chlorine is amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = *x *y = *x *(1-x) = ( )x = ( )x = ( ) x = x = x = % Cl-35 y = 1 - x = = % Cl-37

Dr. S. M. Condren Development of Periodic Table Newlands - English Law of Octaves - every 8th element has similar properties

Dr. S. M. Condren Development of Periodic Table Dmitri Mendeleev - Russian Periodic Law - allowed him to predict properties of unknown elements

Dr. S. M. Condren Mendeleev’s Periodic Table the elements are arranged according to increasing atomic weights

Dr. S. M. Condren Missing elements: 44, 68, 72, & 100 amu Mendeleev’s Periodic Table

Dr. S. M. Condren Properties of Ekasilicon

Dr. S. M. Condren Modern Periodic Table Moseley, Henry Gwyn Jeffreys 1887–1915, English physicist. Studied the relations among bright-line spectra of different elements. Derived the ATOMIC NUMBERS from the frequencies of vibration of X-rays emitted by each element. Moseley concluded that the atomic number is equal to the charge on the nucleus. This work explained discrepancies in Mendeleev’s Periodic Law.

Dr. S. M. Condren Modern Periodic Table the elements are arranged according to increasing atomic numbers

Dr. S. M. Condren Periodic Table of the Elements

Dr. S. M. Condren Organization of Periodic Table period - horizontal row group - vertical column

Dr. S. M. Condren Family Names Group IAalkali metals Group IIAalkaline earth metals Group VIIAhalogens Group VIIIAnoble gases transition metals inner transition metals lanthanum seriesrare earths actinium seriestrans-uranium series

Dr. S. M. Condren Types of Elements metals nonmetals metalloids - semimetals

Dr. S. M. Condren Elements, Compounds, and Formulas Elements can exist as single atoms or molecules Compounds combination of two or more elements molecular formulas for molecular compounds empirical formulas for ionic compounds

Dr. S. M. Condren Organic Compounds Organic Chemistry branch of chemistry in which carbon compounds and their reactions are studied. the chemistry of carbon-hydrogen compounds

Dr. S. M. Condren Inorganic Compounds Inorganic Chemistry field of chemistry in which are studied the chemical reactions and properties of all the chemical elements and their compounds, with the exception of the hydrocarbons (compounds composed of carbon and hydrogen) and their derivatives.

Dr. S. M. Condren Molecular and Structural Formulas

Dr. S. M. Condren Bulk Substances mainly ionic compounds –empirical formulas –structural formulas

Dr. S. M. Condren Models of Sodium Chloride NaCl “table salt”

Dr. S. M. Condren How many atoms are in the formula Al 2 (SO 4 ) 3 ? 3, 5, 17

Dr. S. M. Condren Naming Binary Molecular Compounds For compounds composed of two non- metallic elements, the more metallic element is listed first. To designate the multiplicity of an element, Greek prefixes are used: mono => 1; di => 2; tri => 3; tetra => 4; penta => 5; hexa => 6; hepta => 7; octa => 8

Dr. S. M. Condren Common Compounds H2OH2O water NH 3 ammonia N2ON2O nitrous oxide CO carbon monoxide CS 2 carbon disulfide SO 3 sulfur trioxide CCl 4 carbon tetrachloride PCl 5 phosphorus pentachloride SF 6 sulfur hexafluoride

Dr. S. M. Condren Alkanes - C n H 2n+2 methane - CH 4 ethane - C 2 H 6 propane - C 3 H 8 butanes - C 4 H 10 pentanes - C 5 H 12 hexanes - C 6 H 14 heptanes - C 7 H 16 octanes - C 8 H 18 nonanes - C 9 H 20 decanes - C 10 H 22

Dr. S. M. Condren Burning of Propane Gas

Dr. S. M. Condren Butanes

Dr. S. M. Condren Ionic Bonding Characteristics of compounds with ionic bonding: non-volatile, thus high melting points solids do not conduct electricity, but melts (liquid state) do many, but not all, are water soluble

Dr. S. M. Condren Ion Formation

Dr. S. M. Condren Valance Charge on Ions compounds have electrical neutrality metals form positive monatomic ions non-metals form negative monatomic ions

Dr. S. M. Condren Valence of Metal Ions Monatomic Ions Group IA=> +1 Group IIA=> +2 Maximum positive valence equals Group A #

Dr. S. M. Condren Valence of Non-Metal Ions Monatomic Ions Group VIA=> -2 Group VIIA=> -1 Maximum negative valence equals (8 - Group A #)

Dr. S. M. Condren Charges of Some Important Ions

Dr. S. M. Condren Polyatomic Ions more than one atom joined together have negative charge except for NH 4 + and its relatives negative charges range from -1 to -4

Dr. S. M. Condren Polyatomic Ions ammoniumNH 4 + perchlorateClO 4 1- cyanideCN 1- hydroxideOH 1- nitrateNO 3 1- sulfateSO 4 2- carbonateCO 3 2- phosphatePO 4 3-

Dr. S. M. Condren Names of Ionic Compounds 1. Name the metal first. If the metal has more than one oxidation state, the oxidation state is specified by Roman numerals in parentheses. 2. Then name the non-metal, changing the ending of the non-metal to -ide.

Dr. S. M. Condren Nomenclature NaCl sodium chloride Fe 2 O 3 iron(III) oxide N2O4N2O4 dinitrogen tetroxide KI potassium iodide Mg 3 N 2 magnesium nitride SO 3 sulfur trioxide

Dr. S. M. Condren Nomenclature NH 4 NO 3 ammonium nitrate KClO 4 potassium perchlorate CaCO 3 calcium carbonate NaOH sodium hydroxide

Dr. S. M. Condren Nomenclature Drill Available for PCs: –on your disk to use at home or in the dorm – in the Chemistry Resource Center – off the web under Chapter 2, Links

Dr. S. M. Condren How many moles of ions are there per mole of Al 2 (SO 4 ) 3 ? 2, 3, 5

Dr. S. M. Condren Chemical Equation reactants products coefficients reactants -----> products

Dr. S. M. Condren Writing and Balancing Chemical Equations Write a word equation. Convert word equation into formula equation. Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

Dr. S. M. Condren Example Hydrogen gas reacts with oxygen gas to produce water. Step 1. hydrogen + oxygen -----> water Step 2. H 2 + O > H 2 O Step 3. 2 H 2 + O > 2 H 2 O

Dr. S. M. Condren Example Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe 3 O 4 ) and carbon dioxide. iron(III) oxide + carbon monoxide -----> Fe 3 O 4 + carbon dioxide Fe 2 O 3 + CO -----> Fe 3 O 4 + CO 2 3 Fe 2 O 3 + CO -----> 2 Fe 3 O 4 + CO 2