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Chapter 2 Atoms, Molecules, and Ions 國防醫學院 生化學科 王明芳老師 2011-9-13 1 Conservation of Mass and the Law of Definite Proportions Law of Conservation of Mass: Mass is neither created nor destroyed in chemical reactions. Aqueous solutions of mercury(II) nitrate and potassium iodide will react to form a precipitate of mercury(II) iodide and aqueous potassium iodide. 3.25 g + 3.32 g = 6.57 g Hg(NO3)2(aq) + 2KI(aq) HgI2(s) + 2KNO3(aq) 4.55 g + 2.02 g = 6.57 g Chapter 2/2 Conservation of Mass and the Law of Definite Proportions 3.25 g + 3.32 g = 6.57 g Hg(NO3)2(aq) + 2KI(aq) HgI2(s) + 2KNO3(aq) 4.55 g + 2.02 g = 6.57 g Chapter 2/3 Conservation of Mass and the Law of Definite Proportions Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass. By mass, water is: 88.8 % oxygen 11.2 % hydrogen Chapter 2/4 The Law of Multiple Proportions and Dalton’s Atomic Theory Law of Multiple Proportions: Elements can combine in different ways to form different substances, whose mass ratios are small whole-number multiples of each other. nitrogen monoxide: 7 grams nitrogen per 8 grams oxygen nitrogen dioxide: 7 grams nitrogen per 16 grams oxygen Chapter 2/5 The Law of Multiple Proportions and Dalton’s Atomic Theory Law of Multiple Proportions: Elements can combine in different ways to form different substances, whose mass ratios are small whole-number multiples of each other. Insert Figure 2.2 p37 Chapter 2/6 The Law of Multiple Proportions and Dalton’s Atomic Theory Taking all three laws together --- the law of mass conservation, the law of definite proportions, and the law of multiple proportions --- ultimately led Dalton to propose a new theory of matter. • Elements are made up of tiny particles called atoms. • Each element is characterized by the mass of its atoms. Atoms of the same element have the same mass, but atoms of different elements have Chapter 2/7 different masses. The Law of Multiple Proportions and Dalton’s Atomic Theory • The chemical combination of elements to make different chemical compounds occurs when atoms join in small whole-number ratios. • Chemical reactions only rearrange how atoms are combined in chemical compounds; the atoms themselves don’t change. Not everything that Dalton proposed was correct. He though, for instance, that water had the formula HO rather than H2O Chapter 2/8 Example 2.1 Using the Law of Multiple Proportions Methane and propane are both constituents of natural gas. A sample of methane contains 5.70 g of carbon atoms and 1.90 g of hydrogen atoms combined in a certain way, whereas a sample of propane contains 4.47 g of carbon atoms and 0.993 g of hydrogen atoms combined in a different way. Show that the two compounds obey the law of multiple proportions. Solution Find the C : H mass ratio in each compound, and then compare the ratios to see whether they are small whole-number multiples of each other. Atomic Structure: Electrons Cathode-Ray Tubes: J. J. Thomson (1856-1940) proposed that cathode rays must consist of tiny negatively charged particles. We now call them electrons. Chapter 2/10 Atomic Structure: Electrons Deflection of electron depends on three factors: 1. Strength of electric or magnetic field 2. Size of negative charge on electron 3. Mass of the electron Thomson calculated the electron’s charge to mass ratio as 1.758820 x 108 Coulombs per gram (C/g). e/m = 1.758820 x 108 C/g Chapter 2/11 Atomic Structure: Electrons Chapter 2/12 Atomic Structure: Electrons Oil Drop Experiment (Millikan, 1868–1953): ● Applied a voltage to oppose the downward fall of charged drops and suspend them. ● Voltage on plates place 1.602176 x 10-19 C of charge on each oil drop. ● Millikan calculated the electron’s mass as 9.109382 x 10-28 grams. e/m = 1.758820 x 108 C/g m = e/(1.758820 x 108 C/g) = (1.602176 x 10-19 C )/(1.758820 x 108 C/g) = 9.109382 x 10-28 g Chapter 2/13 Atomic Structure: Protons and Neutrons Rutherford’s scattering experiment Chapter 2/14 Atomic Structure: Protons and Neutrons Atomic Nucleus: Ernest Rutherford (1871-1937) bombarded gold foil with alpha particles. Although most of the alpha particles passed through the foil undeflected, approximately 1 in every 20,000 particles were deflected. A fraction of those particles were deflected back at an extreme angle. Rutherford proposed that the atom must consist mainly of empty space with the mass concentrated in a tiny central core—the nucleus. Chapter 2/15 Atomic Structure: Protons and Neutrons a particle: A type of emission, naturally occurring radioactive elements. Chapter 2/16 Atomic Structure: Protons and Neutrons A view of the atom Chapter 2/17 Atomic Structure: Protons and Neutrons Chapter 2/18 Atomic Structure: Protons and Neutrons The mass of the atom is primarily in the nucleus. Chapter 2/19 Atomic Structure: Protons and Neutrons The charge of the proton is opposite in sign but equal to that of the electron. Chapter 2/20 Example 2.2 Calculations Using Atomic Size Ordinary “lead” pencils actually are made of a form of carbon called graphite. If a pencil line is 0.35 mm wide and the diameter of a carbon atom is 1.5 1010 m, how many atoms wide is the line? conversion factors Atomic Numbers Atomic Number (Z): Number of protons in an atom’s nucleus. Equivalent to the number of electrons around an atom’s nucleus Mass Number (A): The sum of the number of protons and the number of neutrons in an atom’s nucleus Isotope: Atoms with identical atomic numbers but different mass numbers Chapter 2/22 Atomic Numbers 99.985% 0.015% Chapter 2/23 Atomic Numbers carbon-12 mass number 12 6C 6 protons 6 electrons 6 neutrons 14 6C 6 protons 6 electrons 8 neutrons atomic number carbon-14 mass number atomic number Chapter 2/24 Example 2.3 Interpreting an Isotope Symbol The isotope of uranium used to generate nuclear power is does an atom of have? . How many protons, neutrons, and electrons Solution An atom of has 92 protons, 92 electrons, and 235 – 92 = 143 neutrons. Example 2.4 Writing an Isotope Symbol Element X is toxic to humans in high concentration but is essential to life in low concentrations. Identify element X, whose atoms contain 24 protons, and write the symbol for the isotope of X that has 28 neutrons. Solution According to the periodic table, the element with atomic number 24 is chromium (Cr). The particular isotope of chromium in this instance has a mass number of 24 + 28 = 52 and is written . Atomic Masses and the Mole The mass of 1 atom of carbon-12 is defined to be 12 amu. Atomic Mass (Atomic Weight): The weighted average of the isotopic masses of the element’s naturally occurring isotopes Atomic mass unit (amu), also called a dalton (Da) 1 amu = 1.660539 X 10-24 g Chapter 2/27 Atomic Masses and the Mole Why is the atomic mass of the element carbon 12.01 amu? carbon-12: 98.89 % natural abundance 12 amu carbon-13: 1.11 % natural abundance 13.0034 amu mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111) = 11.87 amu + 0.144 amu = 12.01 amu Chapter 2/28 Atomic Masses and the Mole Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units. Molar Mass: The mass in grams of one mole of any element. It is numerically equivalent to its atomic mass. These samples of helium, sulfur, copper, and mercury each contain 1 mole. Chapter 2/29 Example 2.5 Calculating an Atomic Mass Chlorine has two naturally occurring isotopes: with a natural abundance of 75.76% and an isotopic mass of 34.969 amu, and , with a natural abundance of 24.24% and an isotopic mass of 36.966 amu. What is the atomic mass of chlorine? Solution Atomic mass = (34.969 amu)(0.7576) + (36.966 amu)(0.2424) = 35.45 amu Example 2.6 Converting from Mass to Numbers of Moles and Atoms How many moles and how many atoms of silicon are in a sample weighing 10.53 g? The atomic mass of silicon is 28.0855 amu. The fact that the atomic mass of silicon is 28.0855 amu means that 1 mol of silicon has a mass of 28.0855 g. Use this molar mass to convert between mass and number of moles, and then use Avogadro’s number to find the number of atoms. Solution . Nuclear Chemistry: The Change of One Element Into Another Nuclear Equation: A reaction that changes an atomic nucleus 14 C 6 carbon-14 nucleus 14 N 7 + 0 e -1 nitrogen-14 nucleus In a nuclear equation, the element symbols represent only the nuclei of atoms Rather than the entire neutral atoms, so the subscript represents only the number of nuclear charges (protons). An emitted electron is written as 0-1e , where the superscript o indicates that the mass of an electron is essentially zero when compared to that of a proton or neutron, and the subscript indicates that the charge is -1. Chapter 2/32 Nuclear Chemistry: The Change of One Element Into Another Comparisons Between Nuclear and Chemical Reactions • A nuclear reaction changes an atom’s nucleus usually producing a different element. A chemical reaction, by contrast, involves only a change in the way that different atoms are combined. • Different isotopes of an elements have essentially the same behavior in chemical reactions but often have completely different behavior in nuclear reactions. • The energy change accompanying a nuclear reaction is far greater than that accompanying a chemical reaction. Chapter 2/33 Radioactivity The effect of an electric field on a, b, and g radiation Chapter 2/34 Radioactivity Alpha (a ) Radiation An alpha particle is a helium-4 nucleus (2 protons and 2 neutrons). Alpha particle, a Chapter 2/35 Radioactivity Beta (b ) Radiation A beta particle is an electron. Beta particle, b Chapter 2/36 Radioactivity Gamma (g ) Radiation A gamma particle is a high-energy photon Gamma radiation almost always accompanies a and b emission as a mechanism for the release of energy, but it is often not shown when writing nuclear equations because it changes neither the mass number nor the atomic number of the product nucleus. Chapter 2/37 Radioactivity Positron Emission A positron has the same mass as an electron but the opposite charge. Positron, b + Chapter 2/38 Radioactivity Electron Capture A process in which the nucleus captures an innershell electron, thereby converting a proton to a neutron. Chapter 2/39 Radioactivity Chapter 2/40 Example 2.7 Balancing Nuclear Equations Write a balanced nuclear equation for each of the following processes: (a) Alpha emission from curium-242: (b) Beta emission from magnesium-28: (c) Positron emission from xenon-118: Solution (a) In a emission, the mass number decreases by 4 and the atomic number decreases by 2, giving plutonium-238: (b) In b emission, the mass number is unchanged and the atomic number increases by 1, giving aluminum-28: (c) In positron emission, the mass number is unchanged and the atomic number decreases by 1, giving iodine-118: Nuclear Stability The band indicates various neutron/proton combinations that give rise to nuclei that are either nonradioactive or that are radioactive but decay slowly enough to exist for a measurable time. The band of nuclear stability Chapter 2/42 Nuclear Stability • Every element in the periodic table has at least one radioactive isotope. • Hydrogen is the only element whose most abundant stable isotope, hydrogen-1, contains more protons (1) than neutrons (0). Chapter 2/43 Nuclear Stability • The ratio of neutrons to protons gradually increases, giving a curved appearance to the band of stability. • All isotopes heavier than bismuth-209 are radioactive, even though they may decay slowly and be stable enough to occur naturally. Chapter 2/44 Nuclear Stability This look at the region from Z = 66 (dysprosium) through Z = 79 (gold) shows the types of radioactive processes that various radioisotopes undergo. A close-up look at the band of nuclear stability Chapter 2/45 Mixtures and Chemical Compounds A mixture is simply a blend of two or more substances added together in some arbitrary proportion without chemically changing the individual substances themselves. (H2 and O2; sugar+ salt) Most substances are chemical compounds, formed when atoms of two or more elements combine a chemical reaction. Chapter 2/46 Molecules and Covalent Bonds Covalent Bond: Results when two atoms share several (usually two) electrons. Typically a nonmetal bonded to a nonmetal A covalent bond between atoms is analogous to tug-of-war Chapter 2/47 Molecules and Covalent Bonds Covalent Bond: Results when two atoms share several (usually two) electrons. Typically a nonmetal bonded to a nonmetal Molecule: The unit of matter that results when two or more atoms are joined by covalent bonds. Chapter 2/48 Molecules and Covalent Bonds Insert Figure 2.12 p56 Structural formula: Uses lines between atoms to indicate the covalent bonds 49 Example 2.8 Drawing a Structural Formula Propane, C3H8, has a structure in which the three carbon atoms are bonded in a row, each end carbon is bonded to three hydrogens, and the middle carbon is bonded to two hydrogens. Draw the structural formula, using lines between atoms to represent covalent bonds. Solution Example 2.9 Visual Representations of Mixtures and Compounds Which of the following drawings represents a mixture, which a pure compound, and which an element? Solution Drawing (a) represents a mixture of two diatomic elements, one composed of two red atoms and one composed of two blue atoms. Drawing (b) represents molecules of a pure diatomic element because all atoms are identical. Drawing (c) represents molecules of a pure compound composed of one red and one blue atom. Ions and Ionic Bonds Ionic Bond: A transfer of one or more electrons from one atom to another. A strong electrical attraction between charged particles. Typically a metal bonded to a nonmetal. Ion: A charged particle Cation: A positively charged particle. Metals tend to form cations. Anion: A negatively charged particle. Nonmetals tend to form anions. Chapter 2/52 Ions and Ionic Bonds In the formation of sodium chloride, one electron is transferred from the sodium atom to the chlorine atom. 1 Na + 2 Cl2 Na+ + Cl- Ionic solid Chapter 2/53 Example 2.10 Identifying Ionic and Molecular Compounds Which of the following compounds would you expect to be ionic and which molecular (covalent)? (a) BaF2 (b) SF4 (c) PH3 (d) CH3OH Remember that covalent bonds generally form between nonmetal atoms, while ionic bonds form between metal and nonmetal atoms. Solution Compound (a) is composed of a metal (barium) and a nonmetal (fluorine) and is likely to be ionic. Compounds (b)–(d) are composed entirely of nonmetals and therefore are probably molecular. Naming Chemical Compounds Cation Charges for Typical Main-Group Ions 1+ 2+ 3+ Chapter 2/55 Naming Chemical Compounds Anion Charges for Typical Main-Group Ions 3- 2- 1- A cation bears the same name as the element it is derived Form; An anion name has an –ide ending Chapter 2/56 Naming Chemical Compounds Ionic Compound: A neutral compound in which the total number of positive charges must equal the total number of negative charges. Binary Ionic Compounds Na+ Cl- NaCl magnesium oxide: Mg2+ O2- MgO aluminum sulfide: Al3+ S2- Al2S3 sodium chloride: Chapter 2/57 Naming Chemical Compounds Some transition metals form more than one cation Only ions that exist in aqueous solution are shown. Chapter 2/58 Naming Chemical Compounds Use Roman numerals in parentheses to indicate the charge on metals that form more than one kind of cation. Binary Ionic Compounds iron(III) oxide: Fe3+ O2- Fe2O3 tin(II) chloride: Sn2+ Cl- SnCl2 lead(II) fluoride: Pb2+ F- PbF2 Chapter 2/59 Naming Chemical Compounds Fe2+ Iron(II) ion; Ferrous ion Fe3+ Iron(III) ion; Ferric ion Sn2+ Tin(II) ion; Stannous ion Sn4+ Tin(IV) ion; Stannic ion FeCl2 Iron(II) chloride; Ferrous chloride FeCl3 Iron(III) chloride; Ferric chloride Chapter 2/60 Example 2.11 Naming Binary Ionic Compounds Give systematic names for the following compounds: (a) BaCl2 (b) CrCl3 (c) PbS (d) Fe2O3 Try to figure out the number of positive charges on each cation by counting the number of negative charges on the associated anion(s) Solution (a) Barium chloride No Roman numeral is necessary because barium, a group 2A element, forms only Ba2+. (b) Chromium(III) chloride The Roman numeral III is necessary to specify the +3 charge on chromium (a transition metal). (c) Lead(II) sulfide The sulfide anion (S2) has a double negative charge, so the lead cation must be doubly positive. (d) Iron(III) oxide The three oxide anions (O2) have a total negative charge of –6, so the two iron cations must have a total charge of +6. Thus, each is Fe(III). Example 2.12 Converting Names into Formulas Write formulas for the following compounds: (a) Magnesium fluoride (b) Tin(IV) oxide (c) Iron(III) sulfide Solution For transition metal compounds, the charge on the cation is indicated by the Roman numeral in the name. Knowing the number of positive charges, you can then figure out the number of necessary negative charges for the associated anions. (a) MgF2 Magnesium (group 2A) forms only a 2+ cation, so there must be two fluoride ions (F) to balance the charge. (b) SnO2 Tin(IV) has a +4 charge, so there must be two oxide ions (O2) to balance the charge. (c) Fe2S3 Iron(III) has a +3 charge and sulfide ion a 2 charge (S2), so there must be two irons and three sulfurs. Naming Chemical Compounds Binary Molecular Compounds Because nonmetals often combine with one another in different proportions to form different compounds, numerical prefixes are usually included in the names of binary molecular compounds. Chapter 2/63 Naming Chemical Compounds N2F4 The prefix is added to the front of each to indicate the number of each atom. dinitrogen tetrafluoride Chapter 2/64 Naming Chemical Compounds Binary Molecular Compounds Whenever the prefix ends in ―a‖ or ―o‖ and the element name begins with a vowel, drop the ―a‖ or ―o‖ in the prefix. N2O4 dinitrogen tetroxide Whenever the prefix for the first element is ―mono,‖ drop it. CO2 carbon dioxide CO carbon monoxide Chapter 2/65 Example 2.13 Naming Binary Molecular Compounds Give systematic names for the following compounds: (a) PCl3 (b) N2O3 (c) P4O7 (d) BrF3 Solution Look at a periodic table to see which element in each compound is more cationlike (farther to the left or lower) and which is more anionlike (farther to the right or higher). Then name the compound using the appropriate numerical prefix. (a) Phosphorus trichloride (c) Tetraphosphorus heptoxide (b) Dinitrogen trioxide (d) Bromine trifluoride Naming Chemical Compounds 1. 2. Cation is identified, then the anion Oxoanions fewer O, -ite more O, -ate less O, hypomore O, per- “H” hydrogen bi- 3. Presence/absence Chapter 2/67 Naming Chemical Compounds Polyatomic Ionic Compounds Na+ OH- NaOH Mg2+ CO32- MgCO3 sodium carbonate: Na+ CO32- Na2CO3 iron(II) hydroxide: Fe2+ OH- Fe(OH)2 sodium hydroxide: magnesium carbonate: Chapter 2/68 Example 2.14 Naming Compounds with Polyatomic Ions Give systematic names for the following compounds: (a) LiNO3 (b) KHSO4 (c) CuCO3 (d) Fe(ClO4)3 Solution Unfortunately, there is no alternative: The names and charges of the common polyatomic ions must be memorized. (a) Lithium nitrate Lithium (group 1A) forms only the Li+ ion and does not need a Roman numeral. (b) Potassium hydrogen sulfate (potassium bisulfate) Potassium (group 1A) forms only the K+ ion. (c) Copper(II) carbonate The carbonate ion has a 2 charge, so copper must be +2. A Roman numeral is needed because copper, a transition metal, can form more than one ion. (d) Iron(III) perchlorate There are three perchlorate ions, each with a 1 charge, so the iron must have a +3 charge. Example 2.15 Writing Formulas of Compounds with Polyatomic Ions Write formulas for the following compounds: (a) Potassium hypochlorite (b) Silver(I) chromate (c) Iron(III) carbonate Solution (a) KClO Potassium forms only the K+ ion, so only one ClO is needed. (b) Ag2CrO4 The polyatomic chromate ion has a 2 charge, so two Ag+ ions are needed. (c) Fe2(CO3)3 Iron(III) has a +3 charge and the polyatomic carbonate ion has a 2 charge, so there must be two iron ions and three carbonate ions. The polyatomic carbonate ion is set off in parentheses to indicate that there are three of them.