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