Ch. 2 - CHEM 1331 Fall Semester 2010 Chemistry for Science...

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Unformatted text preview: CHEM 1331 Fall Semester 2010 Chemistry for Science and Engineering Majors Instructor: Prof. Geanangel Office: 123B-Fleming Sections 21308 MWF 10-11 AM 160F ! 21316 MWF 1-2 PM 160F Office Hours: MWF 11:00 AM - 12:00 PM Read your syllabus! Keep it for reference. Please use Blackboard mail to contact me. www.uh.edu/blackboard (Take PA by Fri. 5 PM!) If you have an incomplete in CHEM 1331, send your name and PS number to me today. Syllabus text assignments define our course. Please read them BEFORE you come to class. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 1 Avogadro’s Number (NA) Atomic Mass Unit (u) Electron charge (e) Faraday’s constant (F) Universal gas const. (R) Plank’s constant (h) Rydberg's constant (R) Speed of light (c) UH Department of Chemistry = = = = = = = = 6.02214 1.66054 1.60218 9.64853 8.20578 6.62607 1.09678 2.99792 x x x x x x x x 10 23 mol-1 10 -27 kg 10 -19 C 10 4 C/mol 10 -2 L.atm/(mol.K) = 8.31451 J/(mol.K) 10 -34 J.s 10 7 m -1 10 8 m/s 2 ©CHEM 1331 Professor Geanangel Chemistry, the Central Science PHYS ENGR CHEM HLTH BIOL GEOL UH Department of Chemistry ©CHEM 1331 Professor Geanangel 3 CHEM 1331 Spring Semester 2010 Chapter 2: The Components of Matter 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 Elements, Compounds & Mixtures: An Atomic Overview The Observations That Led to an Atomic View of Matter Dalton’s Atomic Theory The Observations That Led to the Nuclear Atom Model The Atomic Theory Today Elements: A First Look at the Periodic Table Compounds: Introduction to Bonding Compounds: Formulas, Names, and Masses Mixtures: Classification and Separation UH Department of Chemistry ©CHEM 1331 Professor Geanangel 4 Observations that led to an atomic view of matter Conservation of Mass (nothing is lost, unless...) The total mass of substances does not change during chemical reactions. The number of substances may change, but the total amount of matter remains constant. reactant 1 + reactant 2 +... total mass reactants calcium oxide + carbon dioxide CaO 56.08g + + CO2 44.00g = product(s) total mass products calcium carbonate ! ! CaCO3 100.08g UH Department of Chemistry ©CHEM 1331 Professor Geanangel 5 Example: When 0.2250 g Mg was heated with 0.5331 g N2 gas, all the Mg was consumed forming 0.3114 g Mg3N2. What mass of N2 is left over? Mg(s) Start 0.2250g End 0g + N2(g) 0.5331g Xg => Mg3N2(s) (unbalanced) 0g 0.3114 g 0.3114 g - 0.2250 g (Mg) = 0.0864 g N2 used UH Department of Chemistry ©CHEM 1331 Professor Geanangel 6 Definite Composition What’s in it and how much? A chemical compound is always composed of the same elements in the same fractions by mass. Experimental analysis of the elemental mass composition of 20.0 g calcium carbonate: Mass Analysis Mass Fraction (grams/20.0 g) 8.0 g calcium 2.4 g carbon 9.6 g oxygen 20.0 g total (parts/1.00 part) (parts/100 parts) 0.40 calcium 40% calcium 0.12 carbon 12% carbon 0.48 oxygen 48% oxygen 1.00 part by mass 100% by mass Percent by Mass UH Department of Chemistry ©CHEM 1331 Professor Geanangel 7 Mass of an Element X in a Compound mass fraction X in XYZ = mass of X in XYZ mass of XYZ mass of X in sample = mass fraction x mass of sample Galena the mineral consists of Pb and S. A 1.27 g quantity of Galena contains 1.10 g of Pb. How many grams of S are present in a 1 kg sample of Galena? mass fraction of Pb in G x mass of G sample = mass of Pb mass of G sample - mass of Pb = mass S UH Department of Chemistry ©CHEM 1331 Professor Geanangel 8 Problem: A 1.27 g sample of Galena contains 1.10 g of Pb. What is the mass percent of S in Galena? UH Department of Chemistry ©CHEM 1331 Professor Geanangel 9 If elements A and B react to form two compounds, masses of B that combine with a fixed mass of A are found to be in a ratio of small whole numbers. Consider two compounds of elements carbon (C) and oxygen (O); call them carbon oxides I and II. Carbon oxide I: 57.1 mass %O and 42.9 mass %C Carbon oxide II: 72.7 mass %O and 27.3 mass %C Use %C and %O to find masses of C and O The Law of Multiple Proportions UH Department of Chemistry ©CHEM 1331 Professor Geanangel 10 From %, find masses of C and O in 100 g of I and II Carbon Oxide I Carbon Oxide II g O/100g compound 57.1 72.7 g C/100g compound 42.9 27.3 g oxygen/g carbon 57.1/42.9 72.7/27.3 = 1.33 = 2.66 Dividing the g O/g C ratio in II by that in I gives a ratio of small whole numbers: UH Department of Chemistry ©CHEM 1331 Professor Geanangel 11 Dalton’s Atomic Theory (1808) Four postulates explain the emprical “laws” 1. Matter consists of atoms, tiny, indivisible particles of an element that cannot be created or destroyed. 2. Atoms of one element cannot be converted into atoms of another element. 3. Atoms of an element are identical in mass and other properties and are different from atoms of other elements. 4. Compounds result from the chemical combination of a specific ratio of atoms of different elements. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 12 Dalton’s postulates help explain mass laws Mass conservation Atoms cannot be created or destroyed (post.1) or converted into other types of atoms (post. 2) Each type of atom has a fixed mass (post. 3). In a chemical reaction atoms are just rearranged, so this cannot result in a mass change. Definite composition A compound is a combination of a specific ratio of different atoms (post. 4), each of which has a particular mass (post. 3). UH Department of Chemistry ©CHEM 1331 Professor Geanangel 13 Dalton’s Theory and Multiple Proportions Different numbers of B atoms combine with each A atom in the two oxides giving a small, whole-number ratio. A simple arrangement consistent with the mass data has 1 atom of O combined with 1 atom of C in I. Two atoms of O combine with one atom of C in II. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 14 Workshop What set of these statements contains all that correctly describe the change shown below and no others? Assume A = orange atoms and B = blue atoms i) A mixture of A2 and B2 molecules reacts to form the compound AB ii) The Law of Conservation of Mass is obeyed during the change iii) Since 3 A2 and 3 B2 combine to form 6 AB, the change illustrates the Law of Multiple Proportions UH Department of Chemistry ©CHEM 1331 Professor Geanangel 15 Atomic Masses (relatively speaking) Dalton first assigned an atomic mass of 1 to hydrogen, the lightest known substance. He found relative mass of atoms of another element in a compound from ratio of its mass to that of H. 20 g HF → 1g H + 19g F Lavoisier showed by experiment that water contains 8 g of oxygen for each 1 g of hydrogen. Dalton assumed that there are the same number of O atoms as H atoms in water molecules (HO). Thus, he mistakenly assigned a relative atomic mass of 8 to O. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 16 UH Department of Chemistry ©CHEM 1331 Professor Geanangel 17 Fact: 2 L hydrogen gas combine exactly with 1 L of oxygen gas giving 2 L of water vapor. What is the formula of water? If HO: 1 L hydrogen gas + 1 L oxygen gas = 1 L water vapor H + O = HO If H2O: 2 L hydrogen gas + 1 L oxygen gas = 1 L water vapor 2H + O = H2O Neither match 2 L hydrogen+ 1 L oxygen = 2 L water vapor. The result did not seem to make sense. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 18 Avogadro (New Ideas) In 1811, Amadeo Avogadro (1776-1856) made a proposal to explain these conflicting (?) results: 1. Equal volumes of a gas contain equal numbers of gas particles (under the same conditions). 2. Hydrogen gas particles and oxygen gas particles are actually molecules composed of two atoms. 2 L H2(g) contain twice the number of particles as 1 L O2 gas; the same number as 2 L H2O vapor. 2 H2 (g) + O2 (g) => 2 H2O (g) 2L + 1L => 2L Over time, the atomic weights of atoms were found. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 19 Observations that Led to Nuclear Model of the Atom William Crookes built a glass tube fitted with metal electrodes and pumped most of the air from it. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 20 With power on, Crookes saw a bright spot due to a “ray” striking the phosphor on the end of the tube. Cathode “rays” moved from the negative electrode (cathode) to the positive electrode in a straight line. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 21 Cathode rays were deflected by magnetic fields and electric fields. mass !1 2 All cathode rays behaved the same, no matter what metal was used for the cathode. Particles in the beam were attracted to a positive plate, so they must be negatively charged. charge = !5.7 x10 kg / C Identified by J. J. Thomson, (1897) as electrons. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 22 Milikan Oil Drop Exp’t (what’s the electron charge?) 1 Fine mist of oil is sprayed into apparatus 2 Oil droplets fall through hole in positively charged plate (+) X-ray source (–) 3 X-rays knock electrons from surrounding air, which stick to droplet 4 Electrically charged plates influence droplet’s motion 5 Observer times droplet’s motion and controls electric field 9 8 7 6 5 4 0 1 2 3 UH Department of Chemistry ©CHEM 1331 Professor Geanangel Charges on all droplets were some whole-number multiple of a minimum charge. Oil drops pick up different numbers of electrons; the minimum charge must be that of one electron. Millikan’s value was within 1% of the modern value of the electron’s charge, -1.602 x 10 -19 C (coulomb) From Thomson’s mass/charge ratio value and the charge, Millikan determined the electron’s mass: mass of e _ = mass x charge charge Electron mass ! 10-31 kg! UH Department of Chemistry ©CHEM 1331 Professor Geanangel 24 Atoms are electrically neutral. So, what positive charges balance the negative electrons? If electrons have such tiny masses, what accounts for the remainder of an atom’s mass? positive charge • Thomson proposed a spherical atom model composed of diffuse, positively charged matter, in which electrons were embedded like “plums in pudding.” electron UH Department of Chemistry ©CHEM 1331 Professor Geanangel 25 Rutherford Scattering Experiment A Hypothesis: Expected result based on “plum pudding” model Incoming ! particles B Experiment 1 Radioactive sample emits beam of ! particles 2 Beam of ! particles strikes gold foil Almost no deflection Lead block Cross section of gold foil composed of “plum pudding” atoms Gold foil 5 Major deflections of ! particles are seen very rarely 4 Minor deflections of ! particles are seen occasionally 3 Flashes of light produced when ! particles strike zinc-sulfide screen show that most ! particles are transmitted with little or no deflection. UH Department of Chemistry ©CHEM 1331 Professor Geanangel Data showed that most α particles weren’t deflected but that 1 in 20,000 was deflected by more than 90°. Could only happen if nearly all mass and positive charge reside in a tiny region within the atom. 27 UH Department of Chemistry ©CHEM 1331 Professor Geanangel Rutherford’s results suggested: -volume of atoms are mostly occupied by electrons; within that lies a tiny region, the atomic nucleus. - atomic nucleus contains all the positive charge and essentially all the mass of the atom. - positive particles (protons) occupy the nucleus. Rutherford’s model explained the charged nature of matter but didn’t account for all the atom’s mass. Later, Chadwick discovered the neutron: an uncharged particle in nucleus; about same mass as proton, accounted for the “missing mass”. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 28 Approximately 10–10 m Approximately 10–15 m Nucleus Electrons, e– (negative charge) Proton, p+ (positive charge) Neutron, n0 (no charge) A Atom B Nucleus UH Department of Chemistry ©CHEM 1331 Professor Geanangel ++ " " UH Department of Chemistry ©CHEM 1331 Professor Geanangel " 30 Isotopes (Dalton got this wrong) Atoms with the same number of protons (p+) but different numbers of neutrons (no) are called isotopes. Mass number (p+ + n0) Atomic Number (p+) A Z 6e– An atom of carbon-12 6p+ 6n0 8e– 12 6 C An atom of oxygen-16 8p+ 8n0 16 8 O 92e– 92p+ 143n0 235 92 X Atomic symbol An atom of uranium-235 U isotopes 92p+ 146n0 92e– 238 92 An atom of uranium -238 U UH Department of Chemistry ©CHEM 1331 Professor Geanangel How to determine the number of neutrons A Z X Mass number, A, is sum of protons and neutrons Number of neutrons = A - Z Chlorine-35 atoms have A = 35, Z = 17, and so they have 35 -17 = 18 no Problem: An atom of boron-11 has Z = 5. How many p+, e– and no does it contain? UH Department of Chemistry ©CHEM 1331 Professor Geanangel 32 Ne + e– ! Ne+ + 2e– Ar Kr UH Department of Chemistry ©CHEM 1331 Professor Geanangel 33 Mass Spectrum of Ne Mass spectra show the abundance (%) of each isotope. How to calculate the average mass of neon from its isotope masses weighted according to their abundances. Weighted average mass = "(isotope mass x fract. abund.) UH Department of Chemistry ©CHEM 1331 Professor Geanangel 34 Problem: Silver (Z =47) has only two isotopes, 107Ag and 109Ag. Given the mass spectrometric data below, calculate the atomic mass of silver: ISOTOPE MASS (amu) ABUNDANCE (%) 107Ag 106.90509 51.84 109Ag 108.90476 48.16 Find atomic mass contribution of each isotope: mass contrib = isotopic mass x fractional abund. For 107Ag: = 106.90509 amu x 0.5184 = 55.42 amu For 109Ag: = 108.90476 amu x 0.4816 = 52.45 amu 107.9 Ag 47 UH Department of Chemistry ©CHEM 1331 Professor Geanangel 35 The atomic mass standard is the carbon-12 atom; its mass is defined as exactly 12 atomic mass units. So one atomic mass unit (amu or u) is defined as 1/12th the mass of a carbon-12 atom. On this scale, H atoms have a mass of 1.008 amu. Atomic Mass Scale Today One amu equals 1.661 x 10 -24 g (PT/const. sheet) What is the mass of one atom of H? UH Department of Chemistry ©CHEM 1331 Professor Geanangel 36 Learn names and symbols of the first 36 elements MAIN–GROUP ELEMENTS 1A (1) 1 H 1 1.008 Metals (main-group) Metals (transition) Metals (inner transition) Metalloids Nonmetals MAIN–GROUP ELEMENTS 8A (18) 2 He 3A 4A 5A 6A 7A (13) (14) (15) (16) (17) 4.003 5 6 7 8 9 10 B C N O F Ne 10.81 12.01 14.01 16.00 19.00 20.18 2A (2) 3 4 Be 2 Li 6.941 9.012 3 11 12 Na Mg 22.99 24.31 TRANSITION ELEMENTS 13 14 15 16 17 18 Al Si P S Cl Ar 3B 4B 5B 6B 7B 8B 1B 2B (8) (9) (10) (11) (12) 26.98 28.09 30.97 32.07 35.45 39.95 (3) (4) (5) (6) (7) 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 5 Rb 85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 6 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) 87 88 89 104 105 106 107 108 109 110 111 112 Fr Ra Ac Rf Db Sg Bh Hs Mt (223) (226) (227) (261) (262) (266) (262) (265) (266) (269) (272) (277) 114 (285) PERIOD 7 Learn which elements are metals, nonmetals and metalloids INNER TRANSITION ELEMENTS 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 6 Lanthanides Ce 140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 7 Actinides 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 (231) 238.0 (237) (242) (243) (247) (247) (251) (252) (257) (258) (259) (260) ©CHEM 1331 Professor Geanangel UH Department of Chemistry Compounds: Introduction to Bonding The Formation of Ionic Compounds Ionic compounds made of ions, charged particles formed when atom(s) gain or lose electron(s). Ionic compounds typically form when a metal reacts with a nonmetal. Each metal atom loses 1, 2 or 3 of its electrons and becomes a cation, a positively charged ion. Nonmetal atoms gain electrons lost by the metal atoms becoming anions, negatively charged ions. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 38 siL48593_ch02_040-088 58 Ionic compounds always contain equal numbers of positive and negative charges. Ionic compounds are neutral; i.e., zero net charge. A grain of table salt consists of a large number of Na+ ions and an equal number of Cl– ions.58 How can we predict the number of electrons an atom Chapter The Components of will lose or gain 2when it Matter forms a monoatomic ion? siL48593_ch02_040-088 30:11:07 30:11:07 10:53pm Page 58 10:53pm Page 58 Chapter 2 The C MAIN–GROUP ELEMENTS 1A 1+ (1) 1 1 Often, ions are formed with the same number of electrons as in the nearest noble gas [Group 8A] 1A (1) 1 MAIN–GROUP ELEMENTS Metals (main-group) Metals (transition) Metals (inner transition) Metalloids Nonmetals MAIN–GROUP ELEMENTS 8A (18) 2 2+ 2A (2) 4 Metals (mai Metals (tran Metals (inne Metalloids Nonmetals 1 H 1.008 3 3- 2- 15A (15) 7 6A (16) 8 7A (17) 9 H 3 1.008 2 2A (2) 4 3A (13) 5 4A (14) 6 He 4.003 10 10 2 Li 11 Be 12 6.941 9.012 3 TRANSITION EL 3B (3) 21 4B (4) 22 5B (5) 23 6B (6) 24 7B (7) 25 Li 11 Be 12 B TRANSITION ELEMENTS 3B (3) 21 4B (4) 22 5B (5) 23 6B (6) 24 7B (7) 25 (8) 26 8B (9) 27 (10) 28 1B (11) 29 2B (12) 30 13 C 14 N 15 O 16 F 17 Ne Period 18 Na 19 Mg 20 6.941 9.012 3 Period 10.81 12.01 14.01 16.00 19.00 20.18 22.99 24.31 4 ( 2 Na 19 Mg 20 Al 31 Si 32 P 33 S 34 Cl 35 Ar 36 K 37 Ca 38 Sc 39 Ti 40 V 41 Cr 42 Mn 43 F 22.99 24.31 4 26.98 28.09 30.97 32.07 35.45 39.95 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55 5 4 K 37 Ca 38 Sc 39 Ti 40 V 41 Cr 42 Mn 43 Fe 44 Co 45 Ni 46 Cu 47 Zn 48 Ga 49 Ge 50 As 51 Se 52 Br 53 Kr 54 Rb 55 Sr 56 Y 57 Zr 72 Nb 73 Mo 74 Tc (98) 75 R 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 5 63.55 65.41 69.72 72.61 74.92 78.96 79.90 83.80 85.47 87.62 88.91 91.22 92.91 95.94 6 10 7 Rb 55 Sr 56 Y 57 Zr 72 Nb 73 Mo 74 Tc (98) 75 Ru 76 Rh 77 Pd 78 Ag 79 Cd 80 In 81 Sn 82 Sb 83 Te 84 I 85 Xe 86 Cs 87 Ba 88 La 89 Hf 104 Ta 105 W 106 Re 107 O 85.47 87.62 88.91 91.22 92.91 95.94 6 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 132.9 137.3 138.9 178.5 180.9 183.9 186.2 19 7 1 Cs 87 (223) Ba 88 (226) La 89 (227) Hf 104 (263) Ta 105 W 106 Re 107 Os 108 Ir 109 (268) Pt 110 Au 111 Hg 112 Tl 113 Pb 114 (289) Bi 115 (288) Po (209) 116 (292) At (210) Rn (222) Fr (223) Ra (226) Ac (227) Rf (263) Db (262) Sg (266) Bh (267) H 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 7 Fr Ra Ac Rf UH Department of Chemistry (2 Db (262) Sg (266) Bh (267) Hs (277) Mt ©CHEMDs Professor Geanangel 1331 Rg (281) (272) (285) (284) siL48593_ch02_040-088 30:11:07 10:53pm Page 58 6 65 66 67 68 69 70 71 39 INNER TRANSITION ELEM 58 Lanthanides 59 60 61 62 6 Exercise: What monatomic ions do the elements below form? Chapter 2 The Components of Matter 58 Figure 2.9 The modern periodic table. a) iodine (Z= 53) b) strontium (Z = 38)(main-group) Metals MAIN–GROUP MAIN–GROUP Figure 2.9 The modern periodic table. ELEMENTS ELEMENTS Metals (transition) c) aluminum (Z = 13) transition) Metals (inner 58 59 60 61 62 63 64 INNER TRANSITION ELEMENTS Ce 90 Pr 91 Nd 92 Pm (145) 93 Sm 94 E 140.1 140.9 144.2 150.4 15 9 6 Lanthanides Ce 90 Pr 91 Nd 92 Pm 93 Sm 94 Eu 95 Gd 96 Tb 97 Dy 98 Ho 99 Er Tm 101 Yb Lu 7 Actinides Th Pa U Np (237) Pu (242) A 140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 100 102 103 232.0 (231) 238.0 (2 The table consists o ment boxes arranged by increasing atomic number into groups cal columns) and periods (horizontal rows). Each box contain atomic number, atomic symbol, and atomic mass. (A mass in p The table consists of eleplaced below the main body of the table but actually fit between the ment boxes arranged by increasing atomic number into groups (vertielements indicated. Metals lie below and to the left of theses is the mass number of the most stable isotope of that elem the thick cal columns) and periods (horizontal rows). Each box contains the “staircase” line [top of 3A(13) to bottom of 6A(16) in PeriodThe periods are numbered 1 to 7. The groups (sometimes 6] and infamilies) have a number-letter designation and a new group num atomic number, atomic symbol, and atomic mass. (A mass in parenclude main-group metals ( purple-blue), transition elements (blue), and Metalloids 8A 1A parentheses. the theses is the mass number of the most stable isotope of that element.) inner transition elements ( gray-blue). Nonmetals (yellow) lie to the right The A groups are(18) main-group elements; the B g (1) Nonmetals The periods are numbered 1 to 7. The groups (sometimes called of the line. Metalloids ( green) lie along the line. We discuss are the transition elements. Two series of inner transition elemen the place7 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 (231) 238.0 (237) (242) (243) (247) (247) (251) (252) (257) (258) (259) (260) families) have a 1 number-letter designation and a new group number in parentheses. The A groups are the main-group elements; the B groups 1 H 2A are the transition elements. Two series of inner transition elements are 1.008 (2) ment of hydrogen in Chapter 14. As of mid-2007, elements 112–116 had not been named. 3A 4A 5A 2 (13) 5 (14) 6 (15) 7 6A (16) 8 7A (17) 9 He 4.003 10 3 2 4 Li 11 Be 12 6.941 9.012 3 Na 19 Mg 20 22.99 24.31 3B (3) 21 4B (4) 22 4 K 37 Ca 38 Sc 39 Ti 40 39.10 40.08 44.96 47.88 5 Rb 55 Sr 56 Y 57 Zr 72 85.47 87.62 88.91 91.22 6 Cs 87 Ba 88 La 89 Hf 104 B C N At this point in the text, the clearest distinction among 10.81 12.01 is their the elements 14.01 classification as metals, nonmetals, or metalloids. The “staircase” line that runs TRANSITION ELEMENTS 14 13 from the top of Group 3A(13) to the bottom of Group 6A(16) in Period 615 a is S P 5B 6B 7B 8B 1B dividing line for this classification. The metals (three2B shadesAl blue)i appear in of (8) (9) (10) (6) (7) (11) (12) 26.98 elements are the(5) large lower-left portion of the table. About three-quarters of the28.09 30.97 metals, including 25 many main-group elements and all the transition and inner 33 tran29 23 27 24 26 32 28 30 31 sition elements. They are generally shiny solids at room temperatureGe (mercury is V Cr Mn Fe Co Ni Cu Zn Ga As the only 52.00 that conduct heat and electricity well and can69.72 72.61 74.92 50.94 liquid) 54.94 55.85 58.93 58.69 63.55 65.41 be tooled into sheets (malleable) and wires (ductile). The nonmetals (yellow) appear in the small 41 43 45 46 51 48 42 44 47 50 49 upper-right portion of the table. They are generally gases or dull, brittle solids at Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb room temperature (bromine is the only liquid) and conduct heat and electricity 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 poorly. Along the staircase line lie the metalloids (green; also called semimetals), 73 75 77 80 83 74 76 79 82 elements that have properties between78 those of metals and 81 nonmetals. Several Ta W Re Os Ir Pt Au Hg Tl Pb Bi 105 106 107 108 109 110 111 112 (285) 113 (284) 114 (289) 115 (288) O 16 F 17 Ne 18 16.00 19.00 20.18 S 34 Cl 35 Ar 36 32.07 35.45 39.95 Se 52 Br 53 Kr 54 78.96 79.90 83.80 Te 84 I 85 Xe 86 At this po classification a from the top o dividing line f the large lowe metals, includ sition element the only liquid (malleable) an upper-right po room tempera poorly. Along elements that Period 127.6 126.9 131.3 Po (209) 116 (292) At (210) Rn (222) 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 7 Fr (223) Ra (226) Ac (227) Rf (263) Db (262) Sg (266) Bh (267) Hs (277) Mt (268) Ds (281) Rg (272) UH Department of Chemistry ©CHEM 1331 Professor Geanangel 40 INNER TRANSITION ELEMENTS 58 59 60 61 62 63 64 65 66 67 68 69 70 71 e– p+ e– p+ 62 Formation of Covalent Compounds B Attraction begins In covalent compounds, atoms share pairs of electrons. Usually occurs between+ p+ p nonmetals, C-H, Cl-Cl, etc. – Simplest case: electron sharing occursCbetween two hydrogen Covalent bond atoms to form H2 molecule. e– p+ p+ e– e e– UH Department of Chemistry many combinations to ments form: form a The F The simplest case of electr The For C Covalent bond e– TheCovale Form two hydrogen atoms (H; betwee Z Covalent – D Interaction of forces +e + p Covalent A eachp+other,+p in Figure betweenc as 2.14. many e– – p e many T between coh no the p+ e– p+ of the other atom electron Figure 2.14 Formation of a covalent many The com twooth h bond between two H atoms. A, The D Interaction of forces begin to interpenetrate eachhydr e– The sim two distance is too great for the atoms to D Interaction oftwo satoms form a two each o the force covalen hydro each othe the ele D igure 2.14ofFormation of a covalent F Interaction forces affect each other. B, As the distance deeach other, the electr Figure 2.14 Formation of a covalent bond the two nuclei. The begin Hydrogen gas consists of atom begins between two H atoms.covalentto result is creases, the nucleus of each many bond between two H atoms. A, The the electro Figure 2.14 too great for the A, The Formation of a atoms begin to distance is the tw diatomic hydrogen molecules, Hdistanceeach other. Hfor the atoms to to a particula . bond is longer “belongs” to attract the electron of the other. C, The between greatB,atoms. A, The de- begin two too two 2affect the to in As the distance the tw distance is nuclei.As the atoms to detoo great for the distance affect covalent bond forms when the two nuclei eachthe nucleusRepulsions between the creases, other. B, of each atom begins the two ato the two affect each other. B, of each atom decreases, the nucleusAs the distancebegins two “b These mutually attract the pair of electronsH2creases, thethe electroneachthe other., C, The the longer behave as independent attoto nitsthe net of of atom begins greater tha u attract nucleus of attraction The longer nu is attract the electron the other. C nuclei covalent bond forms the other. C, The -not separate hydrogen D, The H2 mol- attractbond forms when the two innuclei longer “bel to some optimum distance. atoms. covalent the electron ofwhen the two great nuclei. R covalent bonds nuclei detail mutually attract the pair of two nuclei the ne covalent bond formspair of theelectrons at attract the when ecule is more stable than the separate mutuallyoptimum distance.electrons atmol- nuclei. Rep the net a some attract the pair of electrons H2hydrogen gas D, The at A sample ofmol- the net attr mutually covale some optimum distance. The covalent Otheratoms because the attractive forces ecule is is moredistance. DD,The HH moldiatomics are ecule some optimum stable than the separate , separate covalent A bo more stable than that are the of bo A sa atoms ecule because the attractive forces N2, O(black2arrows),between I2 nucleus and isbecauseatoms the separate chemicallyatom 2, F , Cl2 Br2, each atoms more stable attractive forces the than A sam of atoms because the attractive (black arrows) between each nucleus and nonmeta arate H atoms. and the two electrons are greater than the (black arrows) between each forces Other ofof atoms nucleus atoms th arate H (black arrows) between each nucleus and the electrons are the two electrons greater than the repulsive forces (red arrows) between the two electrons areare greater than the (Narateoxy peraturearrows)nitrogen arate HH are between the 2), ato the two greater than the repulsive forces (red peratu Memorize the “7 Famous Diatomics” forces (between thebetween the perature ar repulsive forces (red arrows) perature electrons and between the nuclei. repulsive (Cl red arrows) between the ), and iodin electrons and 2), bromine (Br2 nuclei. ), (Cl2bro electrons and between the nuclei. (Cl2 brom electrons and between the nuclei. (Cl2), ), ), a ( (P4), and sulfur and selenium sa (P4 P and (P4), ), 4and AtAt roo room At room temperature, covalent At room te ©CHEM 1331 Professor Geanangel 41 SOLUTION inany mem Groups 5 Group p+ p+ SOLUTION ( – – (b) Ca e e Group 8A mentsgform: (a) +16S; (b) any any (Gr + 37Rb; 2 p p B Attraction be ins (b)membe Ca c) GroupGroup gas, in any 8A(1 B Attraction begins (b) (c)2 AlC Ca gas, in thi any Group 2 B Attraction begins elect (c)3 Al3 e– gas, in this 3 electron p + e– p+ (c) Al3 A FOLLO Covalente–compounds form whe p+ e– p+ 3 FOLLOW electrons ments e p+ ments for between– nonmetals. Even thou p+ FOLLOWp e– e– Iodine (53I) is Chapter gai any member of this +group, SAMP it 2 T p PROBL Group 8A(18) member, inSAMPLE this ca (a) Iod e– p+ p+e– (b) Ca2 Calcium (20Ca)PROBLEMW is a m SAMPLE (a)PLAN Iodine A p+ interaction No any Group 2A member, it losesgrou 2 its We p+ PROBLEM W PLAN A No interaction this case, 18Ar. gas, in (a) Iodine to its trons ( group 3 inWe u PLANis Grou trons a a (c) Al Aluminum (13Al) to m A No interaction – its group lie in SOLUTI Groups e– attain the same num e 3+ electrons to tronsany atta to me + e– p+ e– SOLUTION (a) I FOLLOW-UP PROBLEM 2.6 p The Formation of Covale C Covalent bond e– C Covalent bond 2 2 molecules. Figure 2.15 Figure 2.15 Elements that occur a molecu molecules molecules. Figure Figure 2. Diatomic UH Department of Chemistry ©CHEM 1331 Professor Geanangel 42 Tetratomi Octatomic Compounds: Formulas, Names, and Masses Chemical formulas: symbols + numeric subscripts show the type and number of each atom present. Empirical formulas show the ratio of numbers of atoms of each element in the compound. Empirical formula of hydrogen peroxide is HO; note the 1:1 ratio of atoms. Molecular formulas show the number of atoms of each element in a molecule of a compound. Molecular formula of hydrogen peroxide is H2O2 UH Department of Chemistry ©CHEM 1331 Professor Geanangel 43 Structural formulas show the actual number of atoms and the arrangement of the atoms in the molecule. Structural formula of hydrogen peroxide H-O-O-H. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 44 Names of compounds with monatomic ions. Rules for binary ionic compounds, (MX). Ionic compound names give positive ion (cation) followed by negative ion (anion). Metal cation name is the same as that of the metal. Ca Ca 2+ Anion takes the root of the nonmetal name and adds the suffix “-ide.” Br Br – The compound formed from the metal calcium and the nonmetal bromine is “calcium bromide.” Learn all the monatomic ions in Table 2.3 UH Department of Chemistry ©CHEM 1331 Professor Geanangel 45 Exercise (assumes that you learned 1st 36 elements) Name the binary ionic compound that forms from: siL48593_ch02_040-088 a) strontium (Z =38) and N strontium nitride 30:11:07 metal 10:53pm Page 58 nonmetal b) S and Zn zinc sulfide nm m 58 cMAIN–GROUP fluorine ) Al and aluminum fluoride 1A (1) 1 1 m nm Chapter 2 The Components of Matter Metals (main-group) Metals (transition) Metals (inner transition) Metalloids Nonmetals d) oxygen MAIN–GROUP and Li lithium oxide 3A (13) 5 4A (14) 6 5A (15) 7 nm m ELEMENTS ELEMENTS 8A (18) 2 6A (16) 8 7A (17) 9 H 1.008 3 2A (2) 4 He 4.003 10 2 Li 11 Be 12 B TRANSITION ELEMENTS 3B (3) 21 4B (4) 22 5B (5) 23 6B (6) 24 7B (7) 25 (8) 26 8B (9) 27 (10) 28 1B (11) 29 2B (12) 30 13 C 14 N 15 O 16 F 17 Ne 18 6.941 9.012 3 Period 10.81 12.01 14.01 16.00 19.00 20.18 Na 19 Mg 20 Al 31 Si 32 P 33 S 34 Cl 35 Ar 36 22.99 24.31 4 26.98 28.09 30.97 32.07 35.45 39.95 K 37 Ca 38 Sc 39 Ti 40 V 41 Cr 42 Mn 43 Fe 44 Co 45 Ni 46 Cu 47 Zn 48 Ga 49 Ge 50 As 51 Se 52 Br 53 Kr 54 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 5 63.55 65.41 69.72 72.61 74.92 78.96 79.90 83.80 Rb 55 Sr 56 Y 57 Zr 72 Nb 73 Mo 74 Tc (98) 75 Ru 76 Rh 77 Pd 78 Ag 79 Cd 80 In 81 Sn 82 Sb 83 Te 84 I 85 Xe 86 85.47 87.62 88.91 91.22 92.91 95.94 6 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 Cs 87 Ba 88 La 89 Hf 104 Ta 105 W 106 Re 107 Os 108 Ir 109 Pt 110 Au 111 Hg 112 (285) Tl 113 (284) Pb 114 (289) Bi 115 (288) Po (209) 116 (292) At (210) Rn (222) 46 132.9 of Chemistry UH Department 137.3 138.9 7 178.5 180.9 183.9 186.2 CHEM 1331 Professor Geanangel © 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 Fr (223) Ra (226) Ac (227) Rf (263) Db (262) Sg (266) Bh (267) Hs (277) Mt (268) Ds (281) Rg (272) How to Predict Formulas of Ionic Compounds Positive charges of cations in a formula must be balanced by the negative charges of the anions. Calcium bromide is composed of Ca2+ ions + Br – ions. Two Br – ions are needed to balance each Ca2+, so the formula is CaBr2. (empirical formula) Deduce other ionic formulas by the same method. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 47 Problem: Write the empirical formulas for the compounds named in the previous problem. Plan: Write the ions and find the smallest number of each that gives a neutral formula. Solution: a) strontium nitride Sr 2+ and N 3– ; three Sr 2+ (6+) balance two N 3– (6 –) ! Sr3N2 b) zinc iodide Zn 2+ and I– ; one Zn 2+ ion (2+) balances two I– ions (2–) ! ZnI2 c) aluminum fluoride Al 3+ and F– ; one Al 3+ ion (3+) balances three F– ions (3–) ! AlF3 d) lithium oxide Li + and O2– ; two Li+ ions (2+) balance one O2– ion (2–) ! Li2O UH Department of Chemistry ©CHEM 1331 Professor Geanangel 48 Transition elements (B groups), often form more than one ion, with different charges. Naming their compounds: give metal’s ionic charge in Roman numerals after the metal ion’s name. For example, iron forms both Fe 2+ and Fe 3+ ions. The two iron-chlorine compounds are: FeCl2, named iron(II) chloride, and FeCl3, named iron(III) chloride. Learn the transition element ions given in Table 2.4 UH Department of Chemistry ©CHEM 1331 Professor Geanangel 49 Problem: Give systematic names for the formulas or formulas for the names of the following compounds: (a) tin(II) bromide (tin, Z = 50) (b) CrF3 (c) iron(III) oxide (d) MnS UH Department of Chemistry ©CHEM 1331 Professor Geanangel 50 Compounds formed from polyatomic ions Learn the formulas, charges and names of the common polyatomic ions given in Table 2.5 (NO3-) If two or more of a given polyatomic ion are present, put the ion in parentheses with subscript following. Calcium nitrate contains one Ca2+ and two NO3– ions; formula = Ca(NO3)2. Hydrates are ionic compounds having a number of water molecules associated with each formula unit. Epsom salt has the formula MgSO4.7H2O and the name magnesium sulfate heptahydrate. Learn numerical prefixes in Table 2.6. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 51 Workshop: What, if anything, is wrong with each of the following names or formulas? (a) Ca(C2H3O2)2 is called calcium diacetate. (b) Lithium sulfite has the formula (Li)2SO3 . (c) Iron(II) phosphate has the formula Fe2(PO4)3 . (d) Cesium carbonate has the formula Cs2(CO3). UH Department of Chemistry ©CHEM 1331 Professor Geanangel 52 Families of Oxoanions Most polyatomic ions are oxoanions, in which an element is bonded to one or more oxygen atoms. Families of oxoanions exist that differ only in the number of oxygen atoms. Naming convention for these ions: If only two oxoanions are in the family: -the ion with more O atoms takes the nonmetal root and the suffix “-ate.” -the ion with fewer O atoms takes the nonmetal root and the suffix “-ite.” Examples: SO42– = sulfate ion; SO32– = sulfite ion. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 53 With four oxoanions in the family: -the ion with most O atoms has the prefix “per-,” the nonmetal root, and the suffix “-ate.” -the ion with one less O atoms has the suffix “-ate.” -the ion with two less O atoms has the suffix “-ite.” -the ion with three less O atoms has the prefix “hypo-” and the suffix “-ite.” Examples: the four chlorine oxoanions, • ClO4– is perchlorate • ClO2– is chlorite • ClO3– is chlorate • ClO– is hypochlorite UH Department of Chemistry ©CHEM 1331 Professor Geanangel 54 Problem: Give systematic names for the formulas or formulas for the names of the following compounds: a) Fe(ClO3)2 b) Sodium carbonate c) Ba(OH)2.8H2O a) ClO3– is chlorate; since it has a 1- charge, the cation must be Fe 2+. Name is iron(II) chlorate. b) Sodium is Na+ ; carbonate is CO32–. Two Na+ ions balance one CO32– ion. Formula is Na2CO3 . c) Ba2+ is barium; OH– is hydroxide. There are eight water molecules in each formula unit. Name is barium hydroxide octahydrate. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 55 Naming acids Acids are a type of hydrogen-containing compound. To name acids, treat them as “anions” connected to number of H+ ions needed for electrical neutrality. Two main types are binary acids and oxoacids: 1. Gaseous hydrogen chloride (HCl) dissolved in water forms a solution called hydrochloric acid. - name consists of: prefix hydro- + anion nonmetal root + suffix -ic + word acid. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 56 2. Oxoacid names are similar to those of the oxoanions, except for two suffix changes: Anion “-ate” suffix becomes an “-ic” suffix in acid. Anion “-ite” suffix becomes an “-ous” suffix in acid. Oxoanion prefixes “hypo-” and “per-” are retained. BrO4– is perbromate, so HBrO4 is perbromic acid IO2– is iodite, so HIO2 is iodous acid. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 57 Problem: Name the following anions and give the names and formulas of the acid solutions derived from them: (a) Br – (b) IO3– (c) CN– (d) NO2– UH Department of Chemistry ©CHEM 1331 Professor Geanangel 58 Names and Formulas of Binary Covalent Compounds Some simple covalent compounds, have common names: ammonia (NH3) water (H2O) Most are named in a systematic way using 4 rules: 1. The element with the lower group number in the periodic table is named first Important exception: If the compound contains oxygen and a halogen, the halogen is named first 2. If both elements are in the same group, the one lower in the group is named first 3. The second element is named with its root and the suffix “-ide” 4. Use numerical prefixes (Table 2.6) to indicate the number of atoms of each element in the compound UH Department of Chemistry ©CHEM 1331 Professor Geanangel 59 Problem: (a) What is the formula of carbon disulfide? (b) What is the name of AsF5 ? (c) Give the name and formula of the compound formed from two N atoms and five O atoms. UH Department of Chemistry ©CHEM 1331 Professor Geanangel 60 Organic Compounds; Hydrocarbons Hydrocarbons have only H and C atoms Alkanes are one type of hydrocarbon Methane, CH4, is the first alkane Alkane series general formula = CnH2n+2 Ethane, C2H6, (n = 2) is the 2nd member of the alkane series Learn straight-chain alkanes in Table 2.7 Alkanes with branches have a number for branch location. 2-methylbutane UH Department of Chemistry ©CHEM 1331 Professor Geanangel 61 Functional Groups Specific groupings of atoms attached to a carbon chain • Alcohols • Amines • Carboxylic Acids UH Department of Chemistry ©CHEM 1331 Professor Geanangel 62 Exercise: What is the structure of 3-pentanol? Exercise: What is the name of CH3CH2CH2COOH UH Department of Chemistry ©CHEM 1331 Professor Geanangel Problem: Using the PT, calculate the molecular mass of the compound tetraphosphorus trisulfide: Plan: Write the formula, then multiply the number of atoms of each element by its atomic mass and find the sum. Solution: The formula is P4S3 . Molecular mass = (4 x at. mass P) + (3 x at. mass S) = (4 x 30.97 amu) + (3 x 32.07 amu) = 220.1 amu UH Department of Chemistry ©CHEM 1331 Professor Geanangel 64 Section 2.9 (p76-79) is assigned for self study. End of Chapter 2 UH Department of Chemistry ©CHEM 1331 Professor Geanangel 65 Avogadro’s Number (NA) Atomic Mass Unit (u) Electron charge (e) Faraday’s constant (F) Universal gas const. (R) Plank’s constant (h) Rydberg's constant (R) Speed of light (c) UH Department of Chemistry = = = = = = = = 6.02214 1.66054 1.60218 9.64853 8.20578 6.62607 1.09678 2.99792 x x x x x x x x 10 23 mol-1 10 -27 kg 10 -19 C 10 4 C/mol 10 -2 L.atm/(mol.K) = 8.31451 J/(mol.K) 10 -34 J.s 10 7 m -1 10 8 m/s 66 ©CHEM 1331 Professor Geanangel ...
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This note was uploaded on 04/12/2011 for the course CHEM 1331 taught by Professor Bott during the Spring '08 term at University of Houston.

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