Solution Manual for Chemistry, 10th Edition

CHAPTER 2 ATOMS, MOLECULES, AND IONS Questions 18. Some elements exist as molecular substances. That is, hydrogen normally exists as H2 molecules, not single hydrogen atoms. The same is true for N2, O2, F2, Cl2, etc. 19. A compound will always contain the same numbers (and types) of atoms. A given amount of hydrogen will react only with a specific amount of oxygen. Any excess oxygen will remain unreacted. 20. The halogens have a high affinity for electrons, and one important way they react is to form anions of the type X−. The alkali metals tend to give up electrons easily and in most of their compounds exist as M+ cations. Note: These two very reactive groups are only one electron away (in the periodic table) from the least reactive family of elements, the noble gases. 21. Law of conservation of mass: Mass is neither created nor destroyed. The total mass before a chemical reaction always equals the total mass after a chemical reaction. Law of definite proportion: A given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g H for every 8 g oxygen. Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers: For CO2 and CO discussed in Section 2.2, the mass ratios of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio. 22. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons and neutrons, which can be broken down into quarks. For our purpose, electrons, neutrons, and protons are the key smaller parts of an atom. b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have 0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions were included, then different ions/atoms of H could have different numbers of electrons. c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2 protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium. d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1 g hydrogen for every 16 g of O present. These are distinctly different compounds, each with its own unique relative number and types of atoms present. 27 28 CHAPTER 2 ATOMS, MOLECULES, AND IONS e. A chemical equation involves a reorganization of the atoms. Bonds are broken between atoms in the reactants, and new bonds are formed in the products. The number and types of atoms between reactants and products do not change. Because atoms are conserved in a chemical reaction, mass is also conserved. 23. J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negatively charged particles that we now call electrons. Thomson also postulated that atoms must contain positive charge in order for the atom to be electrically neutral. Ernest Rutherford and his alpha bombardment of metal foil experiments led him to postulate the nuclear atom−an atom with a tiny dense center of positive charge (the nucleus) with electrons moving about the nucleus at relatively large distances away; the distance is so large that an atom is mostly empty space. 24. The atom is composed of a tiny dense nucleus containing most of the mass of the atom. The nucleus itself is composed of neutrons and protons. Neutrons have a mass slightly larger than that of a proton and have no charge. Protons, on the other hand, have a 1+ relative charge as compared to the 1– charged electrons; the electrons move about the nucleus at relatively large distances. The volume of space that the electrons move about is so large, as compared to the nucleus, that we say an atom is mostly empty space. 25. The number and arrangement of electrons in an atom determine how the atom will react with other atoms, i.e., the electrons determine the chemical properties of an atom. The number of neutrons present determines the isotope identity and the mass number. 26. Density = mass/volume; if the volumes are assumed equal, then the much more massive proton would have a much larger density than the relatively light electron. 27. For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number of neutrons. When the number of protons and neutrons is equal to each other, the mass number (protons + neutrons) will be twice the atomic number (protons). Therefore, for lighter isotopes, the ratio of the mass number to the atomic number is close to 2. For example, consider 28Si, which has 14 protons and (28 – 14 =) 14 neutrons. Here, the mass number to atomic number ratio is 28/14 = 2.0. For heavier isotopes, there are more neutrons than protons in the nucleus. Therefore, the ratio of the mass number to the atomic number increases steadily upward from 2 as the isotopes get heavier and heavier. For example, 238U has 92 protons and (238 – 92 =) 146 neutrons. The ratio of the mass number to the atomic number for 238U is 238/92 = 2.6. 28. Some properties of metals are (1) conduct heat and electricity; (2) malleable (can be hammered into sheets); (3) ductile (can be pulled into wires); (4) lustrous appearance; (5) form cations when they form ionic compounds. Nonmetals generally do not have these properties, and when they form ionic compounds, nonmetals always form anions. CHAPTER 2 ATOMS, MOLECULES, AND IONS 29 29. Carbon is a nonmetal. Silicon and germanium are called metalloids because they exhibit both metallic and nonmetallic properties. Tin and lead are metals. Thus metallic character increases as one goes down a family in the periodic table. The metallic character decreases from left to right across the periodic table. 30. Yes, 1.0 g H would react with 37.0 g 37Cl, and 1.0 g H would react with 35.0 g 35Cl. No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37Cl and 1 g H/35 g Cl for 35Cl. As long as we had pure 37Cl or pure 35Cl, the ratios will always hold. If we have a mixture (such as the natural abundance of chlorine), the ratio will also be constant as long as the composition of the mixture of the two isotopes does not change. 31. In the paste, sodium chloride will dissolve to form separate Na+ and Cl− ions. With the ions present and able to move about, electrical impulses will be conducted. 32. a. A molecule has no overall charge (an equal number of electrons and protons are present). Ions, on the other hand, have extra electrons added or removed to form anions (negatively charged ions) or cations (positively charged ions). b. The sharing of electrons between atoms is a covalent bond. An ionic bond is the force of attraction between two oppositely charged ions. c. A molecule is a collection of atoms held together by covalent bonds. A compound is composed of two or more different elements having constant composition. Covalent and/or ionic bonds can hold the atoms together in a compound. Another difference is that molecules do not necessarily have to be compounds. H2 is two hydrogen atoms held together by a covalent bond. H2 is a molecule, but it is not a compound; H2 is a diatomic element. d. An anion is a negatively charged ion; e.g., Cl−, O2−, and SO42− are all anions. A cation is a positively charged ion, e.g., Na+, Fe3+, and NH4+ are all cations. 33. a. This represents ionic bonding. Ionic bonding is the electrostatic attraction between anions and cations. b. This represents covalent bonding where electrons are shared between two atoms. This could be the space-filling model for H2O or SF2 or NO2, etc. 34. Natural niacin and commercially produced niacin have the exact same formula of C6H5NO2. Therefore, both sources produce niacin having an identical nutritional value. There may be other compounds present in natural niacin that would increase the nutritional value, but the nutritional value due to just niacin is identical to the commercially produced niacin. 35. Statements a and b are true. Element 118, Og, is a noble gas and will presumably be a nonmetal. For statement c, hydrogen has mostly nonmetallic properties. For statement d, a family of elements is also known as a group of elements. For statement e, two items are incorrect. When a metal reacts with a nonmetal, an ionic compound is produced, and the formula of the compound would be AX2 (alkaline earth metals form 2+ ions and halo-gens form 1– ions in ionic compounds). The correct statement would be: When an alkaline earth metal, A, reacts with a halogen, X, the formula of the ionic compound formed should be AX2. 30 CHAPTER 2 ATOMS, MOLECULES, AND IONS 36. a. Dinitrogen monoxide is correct. N and O are both nonmetals, resulting in a covalent compound. We need to use the covalent rules of nomenclature. The other two names are for ionic compounds. b. Copper(I) oxide is correct. With a metal in a compound, we have an ionic compound. Because copper, like most transition metals, forms at least a couple of different stable charged ions in compounds, we must indicate the charge on copper in the name. Copper oxide could be CuO or Cu2O, hence why we must give the charge of most transition metal compounds. Dicopper monoxide is the name if this were a covalent compound, which it is not. c. Lithium oxide is correct. Lithium forms 1+ charged ions in stable ionic compounds. Because lithium is assumed to form 1+ ions in compounds, we do not need to indicate the charge of the metal ion in the compound. Dilithium monoxide would be the name if Li2O were a covalent compound (a compound composed of only nonmetals). Exercises Development of the Atomic Theory 37. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro’s hypothesis (law) implies that volume ratios are proportional to molecule ratios at constant temperature and pressure. H2(g) + Cl2(g) → 2 HCl(g). From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted. 38. Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at constant temperature and pressure. Here, 1 volume of N2 reacts with 3 volumes of H2 to produce 2 volumes of the gaseous product or in terms of molecule ratios: 1 N2 + 3 H2 → 2 product In order for the equation to be balanced, the product must be NH3. 39. From the law of definite proportions, a given compound always contains exactly the same proportion of elements by mass. The first sample of chloroform has a total mass of 12.0 g C + 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures). The mass percent of carbon in this sample of chloroform is: 12.0 g C × 100 = 10.05% C by mass 119.41 g total From the law of definite proportions, the second sample of chloroform must also contain 10.05% C by mass. Let x = mass of chloroform in the second sample: 30.0 g C × 100 = 10.05, x = 299 g chloroform x CHAPTER 2 40. ATOMS, MOLECULES, AND IONS 31 A compound will always have a constant composition by mass. From the initial data given, the mass ratio of H : S : O in sulfuric acid (H2SO4) is: 2.02 32.07 64.00 = 1 : 15.9 : 31.7 : : 2.02 2.02 2.02 If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O in the second sample of H2SO4. 41. Compound 1: 21.8 g C and 58.2 g O (80.0 – 21.8 = mass O) Compound 2: 34.3 g C and 45.7 g O (80.0 – 34.3 = mass O) The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: 21.8 g C = 0.375 g C/g O 58.2 g O Compound 2: 34.3 g C = 0.751 g C/g O 45.7 g O 0.751 2 = ; this 0.375 1 supports the law of multiple proportions because this carbon ratio is a small whole number. The ratio of the masses of carbon that combine with 1 g of oxygen is 42. Let’s determine the ratios of the masses of fluorine that combine with 1 g of S: 1.188 = 1.000; 1.188 2.375 = 1.999; 1.188 3.563 = 2.999 1.188 Because the masses of the fluorine that combine with 1.000 g of sulfur in the three compounds are all in whole number multiples, this illustrates the law of multiple proportions. 43. For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with 1 g of carbon. From the formulas (two oxygen atoms per carbon atom in CO2 versus one oxygen atom per carbon atom in CO), CO2 will have twice the mass of oxygen that combines per gram of carbon as compared to CO. For CO2 and C3O2, it is easiest to concentrate on the mass of carbon that combines with 1 g of oxygen. From the formulas (three carbon atoms per two oxygen atoms in C3O2 versus one carbon atom per two oxygen atoms in CO2), C3O2 will have three times the mass of carbon that combines per gram of oxygen as compared to CO2. As expected, the mass ratios are whole numbers as predicted by the law of multiple proportions. 44. Compound I: 14.0 g R 4.67 g R ; = 3.00 g Q 1.00 g Q compound II: 7.00 g R 1.56 g R = 4.50 g Q 1.00 g Q The ratio of the masses of R that combine with 1.00 g Q is: 4.67 = 2.99 ≈ 3 1.56 As expected from the law of multiple proportions, this ratio is a small whole number. Because compound I contains three times the mass of R per gram of Q as compared with compound II (RQ), the formula of compound I should be R3Q. 32 CHAPTER 2 ATOMS, MOLECULES, AND IONS 45. Mass is conserved in a chemical reaction because atoms are conserved. Chemical reactions involve the reorganization of atoms, so formulas change in a chemical reaction, but the number and types of atoms do not change. Because the atoms do not change in a chemical reaction, mass must not change. In this equation we have two oxygen atoms and four hydrogen atoms both before and after the reaction occurs. 46. Mass is conserved in a chemical reaction. Mass: ethanol + oxygen → water + carbon dioxide 46.0 g 96.0 g 54.0 g ? Mass of reactants = 46.0 + 96.0 = 142.0 g = mass of products 142.0 g = 54.0 g + mass of CO2, mass of CO2 = 142.0 – 54.0 = 88.0 g 47. To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with 1.00 0.126 g of oxygen by 0.126; that is, = 1.00. To get Na, Mg, and O on the same scale, we do 0.126 the same division. Na: 2.875 1.500 1.00 = 22.8; Mg: = 11.9; O: = 7.94 0.126 0.126 0.126 H O Na Mg Relative value 1.00 7.94 22.8 11.9 Accepted value 1.008 16.00 22.99 24.31 For your information, the atomic masses of O and Mg are incorrect. The atomic masses of H and Na are close to the values given in the periodic table. Something must be wrong about the assumed formulas of the compounds. It turns out the correct formulas are H2O, Na2O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H. 48. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or: 4.784 g In A , A = atomic mass of In = 76.54 = 1.000 g O 16.00 If the formula is In2O3, then two times the atomic mass of In will combine with three times the atomic mass of O, or: 2A 4.784 g In , A = atomic mass of In = 114.8 = (3)16.00 1.000 g O The latter number is the atomic mass of In used in the modern periodic table. The Nature of the Atom 49. From section 2.5, the nucleus has “a diameter of about 10−13 cm” and the electrons “move about the nucleus at an average distance of about 10−8 cm from it.” We will use these CHAPTER 2 ATOMS, MOLECULES, AND IONS 33 statements to help determine the densities. Density of hydrogen nucleus (contains one proton only): 4 4 Vnucleus = π r 3 = (3.14) (5 × 10 −14 cm) 3 = 5 × 10 − 40 cm 3 3 3 1.67 × 10 −24 g d = density = 5 × 10 − 40 cm = 3 × 1015 g/cm 3 3 Density of H atom (contains one proton and one electron): Vatom = d= 50. 4 (3.14) (1 × 10 −8 cm) 3 = 4 × 10 − 24 cm 3 3 1.67 × 10 −24 g + 9 × 10 −28 g 4 × 10 − 24 cm 3 = 0.4 g/cm 3 Because electrons move about the nucleus at an average distance of about 1 × 10 −8 cm, the diameter of an atom will be about 2 × 10 −8 cm. Let’s set up a ratio: diameter of nucleus 1 mm 1 × 10 −13 cm = = ; solving: diameter of atom diameter of model 2 × 10 −8 cm diameter of model = 2 × 105 mm = 200 m 1 electron charge 51. 5.93 × 10 −18 C × 52. First, divide all charges by the smallest quantity, 6.40 × 10 −13 . 1.602 × 10 −19 C 2.56 × 10 −12 6.40 × 10 −13 = 4.00; = 37 negative (electron) charges on the oil drop 7.68 3.84 = 12.0; = 6.00 0.640 0.640 Because all charges are whole-number multiples of 6.40 × 10 −13 zirkombs, the charge on one electron could be 6.40 × 10 −13 zirkombs. However, 6.40 × 10 −13 zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 × 10 −13 zirkombs or an integer fraction of 6.40 × 10 −13 zirkombs. 53. Sn−tin; Pt−platinum; Hg−mercury; Mg−magnesium; K−potassium; Ag−silver 54. fluorine−F; chlorine−Cl; bromine−Br; sulfur−S; oxygen−O; phosphorus−P 55. a. 6; the group 2A elements are Be, Mg, Ca, Sr, Ba, and Ra. b. 6; the group 6A elements are O, S, Se, Te, Po, and Lv. c. 4; the nickel family elements are Ni, Pd, Pt, amd Ds. d. 7; the noble gas group 8A elements are He, Ne, Ar, Kr, Xe, Rn, and Uuo. 56. a. 6; the halogen group 7A elements are F, Cl, Br, I, At, and Uus. 34 CHAPTER 2 ATOMS, MOLECULES, AND IONS b. 6; the alkali group 1A elements are Li, Na, K, Rb, Cs, and Fr. H is not considered an alkali metal. c. 14; elements Ce through Lu are the 14 lanthanide series elements. d. 40; the block of elements from Sc to Zn to Cn to Ac and back to Sc are all transition metal elements. 57. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, and Am. Nonmetals: Si, B, At, Rn, and Br. b. Si, Ge, B, and At. The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At. Aluminum has mostly properties of metals, so it is generally not classified as a metalloid. 58. a. The noble gases are He, Ne, Ar, Kr, Xe, and Rn (helium, neon, argon, krypton, xenon, and radon). Radon has only radioactive isotopes. In the periodic table, the whole number enclosed in parentheses is the mass number of the longest-lived isotope of the element. b. Promethium (Pm) has only radioactive isotopes. 59. 60. a. transition metals b. alkaline earth metals d. noble gases e. halogens c. alkali metals Use the periodic table to identify the elements. a. Cl; halogen b. Be; alkaline earth metal c. Eu; lanthanide metal d. Hf; transition metal e. He; noble gas f. U; actinide metal g. Cs; alkali metal 61. 62. a. Element 8 is oxygen. A = mass number = 9 + 8 = 17; b. Chlorine is element 17. 17 Cl d. Z = 26; A = 26 + 31 = 57; 26 Fe f. Lithium is element 3. 60 c. Cobalt is element 27. 27 Co 57 e. Iodine is element 53. 131 53 I 7 3 Li 58 a. Cobalt is element 27. A = mass number = 27 + 31 = 58; 27 Co b. 63. 37 17 8O 10 5B c. 23 12 Mg d. 132 53 I e. 47 20 Ca f. 65 29 Cu Z is the atomic number and is equal to the number of protons in the nucleus. A is the mass number and is equal to the number of protons plus neutrons in the nucleus. X is the symbol of the element. See the front cover of the text which has a listing of the symbols for the various elements and corresponding atomic number or see the periodic table on the cover to CHAPTER 2 ATOMS, MOLECULES, AND IONS 35 determine the identity of the various atoms. Because all of the atoms have equal numbers of protons and electrons, each atom is neutral in charge. a. 23 11 Na b. 199 F c. 168 O 64. The atomic number for carbon is 6. 14C has 6 protons, 14 − 6 = 8 neutrons, and 6 electrons in the neutral atom. 12C has 6 protons, 12 – 6 = 6 neutrons, and 6 electrons in the neutral atom. The only difference between an atom of 14C and an atom of 12C is that 14C has two additional neutrons. 65. a. 79 35 Br: 35 protons, 79 – 35 = 44 neutrons. Because the charge of the atom is neutral, the number of protons = the number of electrons = 35. 66. 67. b. 81 35 Br: 35 protons, 46 neutrons, 35 electrons c. 239 94 Pu: 94 protons, 145 neutrons, 94 electrons d. 133 55 Cs: 55 protons, 78 neutrons, 55 electrons e. 3 1 H: 1 proton, 2 neutrons, 1 electron f. 56 26 Fe: 26 protons, 30 neutrons, 26 electrons a. 235 92 U: 92 p, 143 n, 92 e b. 27 13 Al: 13 p, 14 n, 13 e c. 57 26 Fe: 26 p, 31 n, 26 e d. 208 82 Pb: 82 p, 126 n, 82 e e. 86 37 Rb: 37 p, 49 n, 37 e f. 41 20 Ca: 20 p, 21 n, 20 e a. Ba is element 56. Ba2+ has 56 protons, so Ba2+ must have 54 electrons in order to have a net charge of 2+. b. Zn is element 30. Zn2+ has 30 protons and 28 electrons. c. N is element 7. N3− has 7 protons and 10 electrons. d. Rb is element 37. Rb+ has 37 protons and 36 electrons. e. Co is element 27. Co3+ has 27 protons and 24 electrons. f. Te is element 52. Te2− has 52 protons and 54 electrons. g. Br is element 35. Br− has 35 protons and 36 electrons. 68. a. 24 Mg: 12 protons, 12 neutrons, 12 electrons 12 b. 24 Mg2+: 12 p, 12 n, 10 e 12 c. 59 Co2+: 27 p, 32 n, 25 e 27 d. 59 Co3+: 27 e. 59 Co: 27 p, 32 n, 27 e 27 27 p, 32 n, 24 e 36 69. CHAPTER 2 ATOMS, MOLECULES, AND IONS f. 79 Se: 34 p, 45 n, 34 e 34 g. 79 2− Se : 34 p, 45 n, 36 e 34 h. 63 Ni: 28 p, 35 n, 28 e 28 i. 59 2+ Ni : 28 p, 31 n, 26 e 28 Atomic number = 63 (Eu); net charge = +63 − 60 = 3+; mass number = 63 + 88 = 151; symbol: 151 3+ 63 Eu Atomic number = 50 (Sn); mass number = 50 + 68 = 118; net charge = +50 − 48 = 2+; symbol: 70. 118 2+ 50 Sn Atomic number = 16 (S); net charge = +16 − 18 = 2−; mass number = 16 + 18 = 34; 34 symbol: 16 S2− Atomic number = 16 (S); net charge = +16 − 18 = 2−; mass number = 16 + 16 = 32; 32 symbol: 16 S2− 71. Number of protons in nucleus Number of neutrons in nucleus Number of electrons Net charge 238 92 U 92 146 92 0 40 2+ 20 Ca 20 20 18 2+ 51 3+ 23 V 23 28 20 3+ 89 39 Y 39 50 39 0 79 − 35 Br 35 44 36 1− 31 3 − 15 P 15 16 18 3− Symbol CHAPTER 2 ATOMS, MOLECULES, AND IONS 37 72. Number of protons in nucleus Number of neutrons in nucleus Number of electrons Net charge 26 27 24 2+ 59 3+ 26 Fe 26 33 23 3+ 210 − 85 At 85 125 86 1– 27 3+ 13 Al 13 14 10 3+ 128 2− 52 Te 52 76 54 2– Symbol 53 2 + 26 Fe 73. 74. In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A, and 3A metals form stable 1+, 2+, and 3+ charged cations, respectively. Group 5A, 6A, and 7A nonmetals form 3−, 2−, and 1− charged anions, respectively. a. Lose 2 e − to form Ra2+. b. Lose 3 e − to form In3+. c. Gain 3 e − to form P 3− . d. Gain 2 e − to form Te 2− . e. Gain 1 e − to form Br−. f. Lose 1 e − to form Rb+. See Exercise 73 for a discussion of charges various elements form when in ionic compounds. a. Element 13 is Al. Al forms 3+ charged ions in ionic compounds. Al3+ b. Se2− c. Ba2+ d. N3− e. Fr+ f. Nomenclature 75. 76. a. sodium bromide b. rubidium oxide c. calcium sulfide d. aluminum iodide e. SrF2 f. g. K3N h. Mg3P2 a. mercury(I) oxide b. iron(III) bromide c. cobalt(II) sulfide d. titanium(IV) chloride e. Sn3N2 f. g. HgO h. CrS3 Al2Se3 CoI3 Br− 38 77. CHAPTER 2 a. cesium fluoride ATOMS, MOLECULES, AND IONS b. lithium nitride c. silver sulfide (Silver only forms stable 1+ ions in compounds, so no Roman numerals are needed.) d. manganese(IV) oxide 78. 79. 80. 81. e. titanium(IV) oxide f. strontium phosphide a. ZnCl2 (Zn only forms stable +2 ions in compounds, so no Roman numerals are needed.) b. SnF4 c. Ca3N2 e. Hg2Se f. a. barium sulfite b. sodium nitrite c. potassium permanganate d. potassium dichromate a. Cr(OH)3 b. Mg(CN)2 c. Pb(CO3)2 d. NH4C2H3O2 a. dinitrogen tetroxide b. iodine trichloride c. sulfur dioxide d. diphosphorus pentasulfide d. Al2S3 AgI (Ag only forms stable +1 ions in compounds.) 82. a. B2O3 83. a. copper(I) iodide b. copper(II) iodide d. sodium carbonate e. sodium hydrogen carbonate or sodium bicarbonate f. tetrasulfur tetranitride g. selenium tetrachloride i. barium chromate j. 84. b. AsF5 c. N2O d. SCl6 c. cobalt(II) iodide h. sodium hypochlorite ammonium nitrate a. acetic acid b. ammonium nitrite c. cobalt(III) sulfide d. iodine monochloride e. lead(II) phosphate f. potassium chlorate g. sulfuric acid h. strontium nitride i. aluminum sulfite j. k. sodium chromate l. hypochlorous acid tin(IV) oxide Note: For the compounds named as acids, we assume these are dissolved in water. 85. In the case of sulfur, SO42− is sulfate, and SO32− is sulfite. By analogy: SeO42−: selenate; SeO32−: selenite; TeO42−: tellurate; TeO32−: tellurite 86. From the anion names of hypochlorite (ClO−), chlorite (ClO2−), chlorate (ClO3−), and perchlorate (ClO4−), the oxyanion names for similar iodine ions would be hypoiodite (IO−), iodite (IO2−), iodate (IO3−), and periodate (IO4−). The corresponding acids would be hypoiodous acid (HIO), iodous acid (HIO2), iodic acid (HIO3), and periodic acid (HIO4). 87. a. SF2 b. SF6 c. NaH2PO4 d. Li3N e. Cr2(CO3)3 f. SnF2 CHAPTER 2 ATOMS, MOLECULES, AND IONS g. NH4C2H3O2 j. 88. 89. h. NH4HSO4 Hg2Cl2; mercury(I) exists as Hg22+ ions. a. CrO3 b. S2Cl2 d. K2HPO4 e. AlN f. 39 k. KClO3 NH3 (Nitrogen trihydride is the systematic name.) Co(NO3)3 l. NaH c. NiF2 g. MnS2 h. Na2Cr2O7 i. a. Na2O b. Na2O2 c. KCN d. Cu(NO3)2 e. SeBr4 f. g. PbS2 h. CuCl (NH4)2SO3 i. j. CI4 HIO2 i. GaAs (We would predict the stable ions to be Ga3+ and As 3− .) j. CdSe (Cadmium only forms 2+ charged ions in compounds.) k. ZnS (Zinc only forms 2+ charged ions in compounds.) l. 90. 91. 92. HNO2 m. P2O5 a. (NH4)2HPO4 b. Hg2S c. SiO2 d. Na2SO3 e. Al(HSO4)3 f. NCl3 g. HBr h. HBrO2 i. HBrO4 j. KHS k. CaI2 l. CsClO4 a. nitric acid, HNO3 b. perchloric acid, HClO4 d. sulfuric acid, H2SO4 e. phosphoric acid, H3PO4 c. acetic acid, HC2H3O2 a. Iron forms 2+ and 3+ charged ions; we need to include a Roman numeral for iron. Iron(III) chloride is correct. b. This is a covalent compound, so use the covalent rules. Nitrogen dioxide is correct. c. This is an ionic compound, so use the ionic rules. Calcium oxide is correct. Calcium only forms stable 2+ ions when in ionic compounds, so no Roman numeral is needed. d. This is an ionic compound, so use the ionic rules. Aluminum sulfide is correct. e. This is an ionic compound, so use the ionic rules. Mg is magnesium. Magnesium acetate is correct. f. Phosphide is P3−, while phosphate is PO43−. Because phosphate has a 3− charge, the charge on iron is 3+. Iron(III) phosphate is correct. g. This is a covalent compound, so use the covalent rules. Diphosphorus pentasulfide is correct. 40 CHAPTER 2 ATOMS, MOLECULES, AND IONS h. Because each sodium is 1+ charged, we have the O22− (peroxide) ion present. Sodium peroxide is correct. Note that sodium oxide would be Na2O. i. HNO3 is nitric acid, not nitrate acid. Nitrate acid does not exist. j. H2S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name). H2SO4 is sulfuric acid. Additional Exercises 93. There should be no difference. The composition of insulin (the number and types of atoms) from both sources will be the same and therefore, it should have the some activity regardless of the source. As a practical note, trace contaminants in the two types of insulin may be different. These trace contaminants may be important towards the activity of insulin in the body. 94. The formula of glucose is C6(H2O)6, or C6H12O6. 95. a. 131 53 I has 53 protons and 131 – 53 = 78 neutrons. b. 201 81 Tl has 81 protons and 201 – 81 = 120 neutrons. 96. Carbon (C); hydrogen (H); oxygen (O); nitrogen (N); phosphorus (P); sulfur (S) For lighter elements, stable isotopes usually have equal numbers of protons and neutrons in the nucleus; these stable isotopes are usually the most abundant isotope for each element. Therefore, a predicted stable isotope for each element is 12C, 2H, 16O, 14N, 30P, and 32S. These are stable isotopes except for 30P, which is radioactive. The most stable (and most abundant) isotope of phosphorus is 31P. There are exceptions. Also, the most abundant isotope for hydrogen is 1H; this has just a proton in the nucleus. 2H (deuterium) is stable (not radioactive), but 1H is also stable as well as most abundant. 97. 53 2+ has 26 protons, 53 – 26 = 27 neutrons, and two fewer electrons than protons (24 26 Fe electrons) in order to have a net charge of 2+. 98. a. False. Neutrons have no charge; therefore, all particles in a nucleus are not charged. b. False. The atom is best described as having a tiny dense nucleus containing most of the mass of the atom with the electrons moving about the nucleus at relatively large distances away; so much so that an atom is mostly empty space. c. False. The mass of the nucleus makes up most of the mass of the entire atom. d. True. e. False. The number of protons in a neutral atom must equal the number of electrons. CHAPTER 2 99. ATOMS, MOLECULES, AND IONS 41 From the Na2X formula, X has a 2− charge. Because 36 electrons are present, X has 34 protons and 79 − 34 = 45 neutrons, and is selenium. a. True. Nonmetals bond together using covalent bonds and are called covalent compounds. b. False. The isotope has 34 protons. c. False. The isotope has 45 neutrons. d. False. The identity is selenium, Se. 100. a. Fe2+: 26 protons (Fe is element 26.); protons − electrons = net charge, 26 − 2 = 24 electrons; FeO is the formula since the oxide ion has a 2− charge, and the name is iron(II) oxide. b. Fe3+: 26 protons; 23 electrons; Fe2O3; iron(III) oxide c. Ba2+: 56 protons; 54 electrons; BaO; barium oxide d. Cs+: 55 protons; 54 electrons; Cs2O; cesium oxide e. S2−: 16 protons; 18 electrons; Al2S3; aluminum sulfide f. P3−: 15 protons; 18 electrons; AlP; aluminum phosphide g. Br−: 35 protons; 36 electrons; AlBr3; aluminum bromide h. N3−: 7 protons; 10 electrons; AlN; aluminum nitride 101. a. Pb(C2H3O2)2: lead(II) acetate b. CuSO4: copper(II) sulfate c. CaO: calcium oxide d. MgSO4: magnesium sulfate e. Mg(OH)2: magnesium hydroxide f. CaSO4: calcium sulfate g. N2O: dinitrogen monoxide or nitrous oxide (common name) 102. 103. a. This is element 52, tellurium. Te forms stable 2– charged ions in ionic compounds (like other oxygen family members). b. Rubidium. Rb, element 37, forms stable 1+ charged ions. c. Argon. Ar is element 18. d. Astatine. At is element 85. From the XBr2 formula, the charge on element X is 2+. Therefore, the element has 88 protons, which identifies it as radium, Ra. 230 − 88 = 142 neutrons. 42 CHAPTER 2 ATOMS, MOLECULES, AND IONS 104. Because this is a relatively small number of neutrons, the number of protons will be very close to the number of neutrons present. The heavier elements have significantly more neutrons than protons in their nuclei. Because this element forms anions, it is a nonmetal and will be a halogen because halogens form stable 1− charged ions in ionic compounds. From the halogens listed, chlorine, with an average atomic mass of 35.45, fits the data. The two isotopes are 35Cl and 37Cl, and the number of electrons in the 1− ion is 18. Note that because the atomic mass of chlorine listed in the periodic table is closer to 35 than 37, we can assume that 35Cl is the more abundant isotope. This is discussed in Chapter 3. 105. a. Ca2+ and N3−: Ca3N2, calcium nitride b. K+ and O2−: K2O, potassium oxide c. Rb+ and F−: RbF, rubidium fluoride d. Mg2+ and S2−: MgS, magnesium sulfide e. Ba2+ and I−: BaI2, barium iodide f. Al3+ and Se2−: Al2Se3, aluminum selenide g. Cs+ and P3−: Cs3P, cesium phosphide h. In3+ and Br−: InBr3, indium(III) bromide. In also forms In+ ions, but one would predict In3+ ions from its position in the periodic table. 106. These compounds are similar to phosphate (PO43-− ) compounds. Na3AsO4 contains Na+ ions and AsO43− ions. The name would be sodium arsenate. H3AsO4 is analogous to phosphoric acid, H3PO4. H3AsO4 would be arsenic acid. Mg3(SbO4)2 contains Mg2+ ions and SbO43− ions, and the name would be magnesium antimonate. 107. a. Element 15 is phosphorus, P. This atom has 15 protons and 31 − 15 = 16 neutrons. b. Element 53 is iodine, I. 53 protons; 74 neutrons c. Element 19 is potassium, K. 19 protons; 20 neutrons d. Element 70 is ytterbium, Yb. 70 protons; 103 neutrons 108. Mass is conserved in a chemical reaction. Mass: chromium(III) oxide + aluminum → chromium + aluminum oxide 34.0 g 12.1 g 23.3 g ? Mass aluminum oxide produced = (34.0 + 12.1) − 23.3 = 22.8 g 109. The law of multiple proportions does not involve looking at the ratio of the mass of one element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound: Compound 1: 27.2 g C and 72.8 g O (100.0 – 27.2 = mass O) Compound 2: 42.9 g C and 57.1 g O (100.0 – 42.9 = mass O) CHAPTER 2 ATOMS, MOLECULES, AND IONS 43 The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: 27.2 g C = 0.374 g C/g O 72.8 g O Compound 2: 42.9 g C = 0.751 g C/g O 57.1 g O 2 0.751 = ; because the ratio is a small whole number, this supports the law of multiple 0.374 1 proportions. 110. a. Hydrosulfuric acid if dissolved in water or dihydrogen sulfide if a gas. b. sulfur dioxide c. sulfur hexafluoride d. sodium sulfite ChemWork Problems 111. Number of protons in nucleus Number of neutrons in nucleus Symbol 9 10 19 9F 13 14 13 Al 53 74 127 53 I 34 45 79 34 Se 16 16 32 16 S 27 44 CHAPTER 2 ATOMS, MOLECULES, AND IONS 112. Number of protons in nucleus Number of neutrons in nucleus 4 2 He 2 2 20 10 Ne 10 10 48 22 Ti 22 26 190 76 Os 76 114 50 27 Co 27 23 Symbol 113. Number of protons in nucleus Number of neutrons in nucleus Number of electrons 120 50 Sn 50 70 50 25 2+ 12 Mg 12 13 10 56 2+ 26 Fe 26 30 24 79 34 Se 34 45 34 35 17 Cl 17 18 17 63 29 Cu 29 34 29 Symbol 114. a. True b. False; this was J. J. Thomson. c. False; a proton is about 1800 times more massive than an electron. d. The nucleus contains the protons and the neutrons. CHAPTER 2 115. ATOMS, MOLECULES, AND IONS carbon tetrabromide, CBr4; cobalt(II) phosphate, Co3(PO4)2; magnesium chloride, MgCl2; nickel(II) acetate, Ni(C2H3O2)2; 45 calcium nitrate, Ca(NO3)2 116. 117. Co(NO2)2, cobalt(II) nitrite; AsF5, arsenic pentafluoride; LiCN, lithium cyanide; K2SO3, potassium sulfite; Li3N, lithium nitride; PbCrO4, lead(II) chromate K will lose 1 e− to form K+. Cs will lose 1 e− to form Cs+. Br will gain 1 e− to form Br−. Sulfur will gain 2 e− to form S2 −. Se will gain 2 e− to form Se2 −. 118. a. False; magnesium is Mg. b. True c. Ga is a metal and is expected to lose electrons when forming ions. d. True e. Titanium(IV) oxide is correct for this transition metal ionic compound. Challenge Problems 119. Copper (Cu), silver (Ag), and gold (Au) make up the coinage metals. 120. Because the gases are at the same temperature and pressure, the volumes are directly proportional to the number of molecules present. Let’s assume hydrogen and oxygen to be monatomic gases and that water has the simplest possible formula (HO). We have the equation: H + O → HO But the volume ratios are also equal to the molecule ratios, which correspond to the coefficients in the equation: 2 H + O → 2 HO Because atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correct this, we can make oxygen a diatomic molecule: 2 H + O2 → 2 HO This does not require hydrogen to be diatomic. Of course, if we know water has the formula H2O, we get: 2 H + O2 → 2 H2O The only way to balance this is to make hydrogen diatomic: 2 H2 + O2 → 2 H2O 46 CHAPTER 2 ATOMS, MOLECULES, AND IONS 121. Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equal numbers of molecules. In terms of balanced equations, Avogadro’s hypothesis (law) implies that volume ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule of CxHy produces 8 molecules of CO2 and 9 molecules of H2O. CxHy + n O2 → 8 CO2 + 9 H2O. Because all the carbon in octane ends up as carbon in CO2, octane must contain 8 atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in H2O, so one molecule of octane must contain 9 × 2 = 18 atoms of H. Octane formula = C8H18, and the ratio of C : H = 8 : 18 or 4 : 9. 122. From Section 2.5 of the text, the average diameter of the nucleus is about 10−13 cm, and the electrons move about the nucleus at an average distance of about 10 −8 cm . From this, the diameter of an atom is about 2 × 10 −8 cm . 2 × 10 −8 cm 1 × 10 −13 cm = 2 × 105; 1 mi 5280 ft 63,360 in = = 1 grape 1 grape 1 grape Because the grape needs to be 2 × 105 times smaller than a mile, the diameter of the grape would need to be 63,360/(2 × 105) ≈ 0.3 in. This is a reasonable size for a small grape. 123. The alchemists were incorrect. The solid residue must have come from the flask. 124. The equation for the reaction would be 2 Na(s) + Cl2(g) → 2 NaCl(s). The sodium reactant exists as singular sodium atoms packed together very tightly and in a very organized fashion. This type of packing of atoms represents the solid phase. The chlorine reactant exists as Cl2 molecules. In the picture of chlorine, there is a lot of empty space present. This only occurs in the gaseous phase. When sodium and chlorine react, the ionic compound NaCl forms. NaCl exists as separate Na+ and Cl− ions. Because the ions are packed very closely together and are packed in a very organized fashion, NaCl is depicted in the solid phase. 125. a. Both compounds have C2H6O as the formula. Because they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties because the atoms are bonded together differently. These compounds are called isomers of each other. b. When wood burns, most of the solid material in wood is converted to gases, which escape. The gases produced are most likely CO2 and H2O. c. The atom is not an indivisible particle but is instead composed of other smaller particles, called electrons, neutrons, and protons. d. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Although the compounds are composed of different isotopes, their properties are similar because different isotopes of the same element have similar properties (except, of course, their mass). 126. Let Xa be the formula for the atom/molecule X, Yb be the formula for the atom/molecule Y, XcYd be the formula of compound I between X and Y, and XeYf be the formula of compound II between X and Y. Using the volume data, the following would be the balanced equations for the production of the two compounds. CHAPTER 2 ATOMS, MOLECULES, AND IONS 47 Xa + 2 Yb → 2 XcYd; 2 Xa + Yb → 2 XeYf From the balanced equations, a = 2c = e and b = d = 2f. Substituting into the balanced equations: X2c + 2 Y2f → 2 XcY2f ; 2 X2c + Y2f → 2 X2cYf For simplest formulas, assume that c = f = 1. Thus: X2 + 2 Y2 → 2 XY2 and 2 X2 + Y2 → 2 X2Y 1.00 = 0.3043, y = 1.14. 1.00 + 2 y 2.00 Compound II = X2Y: If X has relative mass of 1.00, = 0.6364, y = 1.14. 2.00 + y Compound I = XY2: If X has relative mass of 1.00, The relative mass of Y is 1.14 times that of X. Thus, if X has an atomic mass of 100, then Y will have an atomic mass of 114. 127. Most of the mass of the atom is due to the protons and the neutrons in the nucleus, and protons and neutrons have about the same mass (1.67 × 10−24 g). The ratio of the mass of the molecule to the mass of a nuclear particle will give a good approximation of the number of nuclear particles (protons and neutrons) present. 7.31 × 10 −23 g = 43.8 ≈ 44 nuclear particles 1.67 × 10 −24 g Thus there are 44 protons and neutrons present. If the number of protons equals the number of neutrons, we have 22 protons in the molecule. One possibility would be the molecule CO2 [6 + 2(8) = 22 protons]. 128. For each experiment, divide the larger number by the smaller. In doing so, we get: experiment 1 X = 1.0 Y = 10.5 experiment 2 Y = 1.4 Z = 1.0 experiment 3 X = 1.0 Y = 3.5 Our assumption about formulas dictates the rest of the solution. For example, if we assume that the formula of the compound in experiment 1 is XY and that of experiment 2 is YZ, we get relative masses of: X = 2.0; Y = 21; Z = 15 (= 21/1.4) and a formula of X3Y for experiment 3 [three times as much X must be present in experiment 3 as compared to experiment 1 (10.5/3.5 = 3)]. However, if we assume the formula for experiment 2 is YZ and that of experiment 3 is XZ, then we get: 48 CHAPTER 2 ATOMS, MOLECULES, AND IONS X = 2.0; Y = 7.0; Z = 5.0 (= 7.0/1.4) and a formula of XY3 for experiment 1. Any answer that is consistent with your initial assumptions is correct. The answer to part d depends on which (if any) of experiments 1 and 3 have a formula of XY in the compound. If the compound in experiment 1 has a formula of XY, then: 21 g XY × 4.2 g Y = 19.2 g Y (and 1.8 g X) (4.2 + 0.4) g XY If the compound in experiment 3 has the XY formula, then: 21 g XY Η 7.0 g Y = 16.3 g Y (and 4.7 g X) (7.0 + 2.0) g XY Note that it could be that neither experiment 1 nor experiment 3 has XY as the formula. Therefore, there is no way of knowing an absolute answer here. Integrated Problems 129. The systematic name of Ta2O5 is tantalum(V) oxide. Tantalum is a transition metal and requires a Roman numeral. Sulfur is in the same group as oxygen, and its most common ion is S2–. Therefore, the formula of the sulfur analogue would be Ta2S5. Total number of protons in Ta2O5: Ta, Z = 73, so 73 protons × 2 = 146 protons; O, Z = 8, so 8 protons × 5 = 40 protons Total protons = 186 protons Total number of protons in Ta2S5: Ta, Z = 73, so 73 protons × 2 = 146 protons; S, Z = 16, so 16 protons × 5 = 80 protons Total protons = 226 protons Proton difference between Ta2S5 and Ta2O5: 226 protons – 186 protons = 40 protons 130. The cation has 51 protons and 48 electrons. The number of protons corresponds to the atomic number. Thus this is element 51, antimony. There are 3 fewer electrons than protons. Therefore, the charge on the cation is 3+. The anion has one-third the number of protons of the cation, which corresponds to 17 protons; this is element 17, chlorine. The number of electrons in this anion of chlorine is 17 + 1 = 18 electrons. The anion must have a charge of 1−. The formula of the compound formed between Sb3+ and Cl– is SbCl3. The name of the compound is antimony(III) chloride. The Roman numeral is used to indicate the charge on Sb because the predicted charge is not obvious from the periodic table. CHAPTER 2 131. ATOMS, MOLECULES, AND IONS 49 Number of electrons in the unknown ion: 2.55 × 10 −26 g × 1 kg 1 electron = 28 electrons × 1000 g 9.11 × 10 −31 kg Number of protons in the unknown ion: 5.34 × 10 −23 g × 1 kg 1 proton = 32 protons × 1000 g 1.67 × 10 − 27 kg Therefore, this ion has 32 protons and 28 electrons. This is element number 32, germanium (Ge). The net charge is 4+ because four electrons have been lost from a neutral germanium atom. The number of electrons in the unknown atom: 3.92 × 10 −26 g × 1 kg 1 electron = 43 electrons × 1000 g 9.11 × 0 −31 kg In a neutral atom, the number of protons and electrons is the same. Therefore, this is element 43, technetium (Tc). The number of neutrons in the technetium atom: 9.35 × 10 −23 g × 1 proton 1 kg = 56 neutrons × 1000 g 1.67 × 10 − 27 kg The mass number is the sum of the protons and neutrons. In this atom, the mass number is 43 protons + 56 neutrons = 99. Thus this atom and its mass number is 99Tc. Marathon Problem 132. a. For each set of data, divide the larger number by the smaller number to determine relative masses. 0.602 = 2.04; A = 2.04 when B = 1.00 0.295 0.401 = 2.33; C = 2.33 when B = 1.00 0.172 0.374 = 1.17; C = 1.17 when A = 1.00 0.320 To have whole numbers, multiply the results by 3. Data set 1: A = 6.1 and B = 3.0 Data set 2: C = 7.0 and B = 3.0 Data set 3: C = 3.5 and A = 3.0 or C = 7.0 and A = 6.0 50 CHAPTER 2 ATOMS, MOLECULES, AND IONS Assuming 6.0 for the relative mass of A, the relative masses would be A = 6.0, B = 3.0, and C = 7.0 (if simplest formulas are assumed). b. Gas volumes are proportional to the number of molecules present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is: 6 A2 + B4 → 4 A3B B4 + 4 C3 → 4 BC3 3 A2 + 2 C3 → 6 AC In any correct set of reactions, the calculated mass data must match the mass data given initially in the problem. Here, the new table of relative masses would be: 6 (mass A 2 ) 0.602 ; mass A2 = 0.340(mass B4) = 0.295 mass B 4 4 (mass C 3 ) 0.401 ; mass C3 = 0.583(mass B4) = mass B 4 0.172 2 (mass C 3 ) 0.374 ; mass A2 = 0.570(mass C3) = 3 (mass A 2 ) 0.320 Assume some relative mass number for any of the masses. We will assume that mass B = 3.0, so mass B4 = 4(3.0) = 12. Mass C3 = 0.583(12) = 7.0, mass C = 7.0/3 Mass A2 = 0.570(7.0) = 4.0, mass A = 4.0/2 = 2.0 When we assume a relative mass for B = 3.0, then A = 2.0 and C = 7.0/3. The relative masses having all whole numbers would be A = 6.0, B = 9.0, and C = 7.0. Note that any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility.