CHAPTER TWO ATOMS, MOLECULES, AND IONS. For Review

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1 CHAPTER TWO ATOMS, MOLECULES, AND IONS For Review 1. a. Atoms have mass and are neither destroyed nor created by chemical reactions. Therefore, mass is neither created nor destroyed by chemical reactions. Mass is conserved. b. The composition of a substance depends on the number and kinds of atoms that form it. c. Compounds of the same elements differ only in the numbers of atoms of the elements forming them, i.e., NO, N 2 O, NO Deflection of cathode rays by magnetic and electric fields led to the conclusion that they were negatively charged. The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electric field.. J. J. Thomson discovered electrons. Henri Becquerel discovered radioactivity. Lord Rutherford proposed the nuclear model of the atom. Dalton's original model proposed that atoms were indivisible particles (that is, atoms had no internal structure). Thomson and Becquerel discovered subatomic particles, and Rutherford's model attempted to describe the internal structure of the atom composed of these subatomic particles. In addition, the existence of isotopes, atoms of the same element but with different mass, had to be included in the model. 4. If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout), then alpha particles should have traveled through the thin foil with very minor deflections in their path. This was not the case as a few of the alpha particles were deflected at very large angles. Rutherford reasoned that the large deflections of these alpha particles could be caused only by a center of concentrated positive charge that contains most of the atom s mass (the nuclear model of the atom). 5. The proton and neutron have similar mass with the mass of the neutron slightly larger than that of the proton. Each of these particles has a mass approximately 1800 times greater than that of an electron. The combination of the protons and the neutrons in the nucleus makes up the bulk of the mass of an atom, but the electrons make the greatest contribution to the chemical properties of the atom. 6. The atomic number of an element is equal to the number of protons in the nucleus of an atom of that element. The mass number is the sum of the number of protons plus neutrons in the nucleus. The atomic mass is the actual mass of a particular isotope (including electrons). As we will see in Chapter Three, the average mass of an atom is taken from a measurement made on a large number of atoms. The average atomic mass value is listed in the periodic table. 7. A family is a set of elements in the same vertical column. A family is also called a group. A period is a set of elements in the same horizontal row. 22

2 CHAPTER 2 ATOMS, MOLECULES, AND IONS 2 8. AlCl, aluminum chloride; CrCl, chromium(iii) chloride; ICl, iodine trichloride; AlCl and CrCl are ionic compounds following the rules for naming ionic compounds. The major difference is that CrCl contains a transition metal (Cr) which generally exhibits two or more stable charges when in ionic compounds. We need to indicate which charged ion we have in the compound. This is generally true whenever the metal in the ionic compound is a transition metal. ICl is made from only nonmetals and is a covalent compound. Predicting formulas for covalent compounds is extremely difficult. Because of this, we need to indicate the number of each nonmetal in the binary covalent compound. The exception is when there is only one of the first species present in the formula; when this is the case, mono is not used (it is assumed). 9. When in ionic compounds, the metals in groups 1A, 2A, and aluminum form +1, +2, and + charged ions, respectively. The nonmetals in the groups 5A, 6A, and 7A form, 2, and 1 charged ions, respectively, when in ionic compounds. The correct formulas are A 2 S where A is an alkali metal, B N 2 where B is an alkaline earth metal, and AlC where C is a halogen. 10. The polyatomic ions and acids in this problem are not named in the text. However, they are all related to other ions and acids named in the text which contain a same group element. Since HClO 4 is perchloric acid, HBrO 4 is perbromic acid. Since ClO is the chlorate ion, KIO is potassium iodate. Since ClO 2 is the chlorite ion, NaBrO 2 is sodium bromite. And finally, since HClO is hypochlorous acid, HIO is hypoiodous acid. Questions 14. Some elements exist as molecular substances. That is, hydrogen normally exists as H 2 molecules, not single hydrogen atoms. The same is true for N 2, O 2, F 2, Cl 2, etc. 15. 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. 16. 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. 17. Law of conservation of mass: mass is neither created nor destroyed. The mass before a chemical reaction always equals the mass after a chemical reaction. Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass. 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 one gram of the first element can always be reduced to small whole numbers: For CO 2 and CO discussed in section 2.2, the mass ratios of oxygen that react with 1 g of carbon in each compound are in a 2:1 ratio. 18. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons

3 24 CHAPTER 2 ATOMS, MOLECULES, AND IONS 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, H, and 4 He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium. d. Water (H 2 O) is always 1 g hydrogen for every 8 g of O present, while H 2 O 2 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. 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 does not change. Because atoms are conserved in a chemical reaction, mass is conserved. 19. J. J. Thomson s study of cathode-ray tubes led him to postulate the existence of negatively charged particles which we now call electrons. 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. 20. 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 dis-tances. 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. 21. The number and arrangement of electrons in an atom determines how the atom will react with other atoms. The electrons determine the chemical properties of an atom. The number of neutrons present determines the isotope identity. 22. a. A molecule has no overall charge (an equal number of electrons and protons are present). Ions, on the other and, 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 molelcule 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. H 2 is two hydrogen atoms held

4 CHAPTER 2 ATOMS, MOLECULES, AND IONS 25 together by a covalent bond. H 2 is a molecule, but it is not a compound; H 2 is a diatomic element. d. An anion is a negatively charged ion, e.g., Cl, O 2, and SO 4 2 are all anions. A cation is a posively charged ion, e.g., Na +, Fe +, and NH 4 + are all cations. 2. Statements a and b are true. Counting over in the periodic table, element 118 will be the next noble gas (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 AX 2 (alkaline earth metals form +2 ions and halogens 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 AX 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. Exercises 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 Cu 2 O, hence why we must give the charge of most transition metal compounds. Dicopper monoxide is the name if this were a covalent compound. 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 Li 2 O was a covalent compound (a compound composed of only nonmetals). Development of the Atomic Theory 25. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro s hypothesis implies that volume ratios are equal to molecule ratios at constant temperature and pressure. H 2 (g) + Cl 2 (g) 2 HCl(g). From the balanced equation (2 molecules of HCl are produced per molecule of H 2 or Cl 2 reacted), the volume of HCl produced will be twice the volume of H 2 (or Cl 2 ) reacted. 26. From Avogadro s hypothesis, volume ratios are equal to molecule ratios at constant temperature and pressure. Therefore, we can write a balanced equation using the volume data, Cl 2 + F 2 2 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF for a balanced equation. 27. Hydrazine: g H/g N; Ammonia: g H/g N

5 26 CHAPTER 2 ATOMS, MOLECULES, AND IONS Hydrogen azide: 2.40 Let's try all of the ratios: 2 10 g H/g N = 6.00; = 9.00; = 1.50 = All the masses of hydrogen in these three compounds can be expressed as simple whole number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6:9: 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. We can show this supports the law of multiple proportions by comparing 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 ( = mass O) Compound 2: 42.9 g C and 57.1 g O ( = mass O) Reduce this ratio so that each is compared to 1.0 g oxygen: Compound 1: 27.2 g C = 0.74 g C/g O 72.8 g O Compound 2: 42.9 g C = g C/g O 57.1g O ; This supports the law of multiple proportions. 29. To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with g of oxygen by 0.126, i.e., = To get Na, Mg and O on the same scale, we do the same division. Na: = 22.8; Mg: = 11.9; O: = H O Na Mg Relative Value Accepted Value The atomic masses of O and Mg are incorrect; the atomic masses of H and Na are close to the values in the periodic table. Something must be wrong about the assumed formulas of the compounds. It turns out the correct formulas are H 2 O, Na 2 O, and MgO. The smaller discrepancies result from the error in the atomic mass of H.

6 CHAPTER 2 ATOMS, MOLECULES, AND IONS If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or: A 4.784g In, A = atomic mass of In = g O If the formula is In 2 O, then two times the atomic mass of In will combine with three times the atomic mass of O, or: 2 A () g In, A = atomic mass of In = g O The latter number is the atomic mass of In used in the modern periodic table. The Nature of the Atom 1. Density of hydrogen nucleus (contains one proton only): V nucleus = πr (.14)(5 10 cm) 5 10 cm d = cm g g / cm Density of H-atom (contains one proton and one electron): V atom = (.14)(1 10 cm) 4 10 cm d = g cm 28 g 0.4 g / cm 2. Because electrons move about the nucleus at an average distance of about 1 diameter of an atom will be about 2 diameterof nucleus diameterof atom 1mm diameterof model diameter of model = mm = 200 m 8 10 cm. Let's set up a ratio: cm, Solving: cm 8 10 cm, the electroncharge.9 10 C C = 7 negative (electron) charges on the oil drop 4. First, divide all charges by the smallest quantity,

7 28 CHAPTER 2 ATOMS, MOLECULES, AND IONS = 4.00; = 12.0; = Because all charges are whole number multiples of zirkombs, the charge on one 1 electron could be zirkombs. However, zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of 1 an electron is zirkombs or an integer fraction of zirkombs. 5. sodiumna; radiumra; ironfe; goldau; manganesemn; leadpb 6. fluorinef; chlorinecl; brominebr; sulfurs; oxygeno; phosphorusp 7. Sntin; Ptplatinum; Hgmercury; Mgmagnesium; Kpotassium; Agsilver 8. Asarsenic; Iiodine; Xexenon; Hehelium; Ccarbon; Sisilicon 9. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, Am. Nonmetals: Si, B, At, Rn, Br. b. Si, Ge, B, At. The elements at the boundary between the metals and the nonmetals are: B, Si, Ge, As, Sb, Te, Po, At. Aluminum has mostly properties of metals. 40. 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) and technetium (Tc) 41. a. Six; Be, Mg, Ca, Sr, Ba, Ra b. Five; O, S, Se, Te, Po c. Four; Ni, Pd, Pt, Uun d. Six; He, Ne, Ar, Kr, Xe, Rn 42. a. Five; F, Cl, Br, I, and At b. Six; Li, Na, K, Rb, Cs, Fr (H is not considered an alkali metal.) c. 14; Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, and Lu d. 40; All elements in the block defined by Sc, Zn, Uub, and Ac are transition metals. 4. a. 79 5Br: 5 protons, 79 5 = 44 neutrons. Since the charge of the atom is neutral, the number of protons = the number of electrons = 5.

8 CHAPTER 2 ATOMS, MOLECULES, AND IONS 29 b. c. d. e. f. 81 5Br: 5 protons, 46 neutrons, 5 electrons 29 94Pu: 94 protons, 145 neutrons, 94 electrons 1 55Cs: 55 protons, 78 neutrons, 55 electrons 1 H: 1 proton, 2 neutrons, 1 electron 56 26Fe: 26 protons, 0 neutrons, 26 electrons 44. a U: 92 p, 14 n, 92 e b. 1 6 C: 6 p, 7 n, 6 e c Fe: 26 p, 1 n, 26 e d Pb: 82 p, 126 n, 82 e e Rb: 7 p, 49 n, 7 e f Ca: 20 p, 21 n, 20 e 45. a. Element 8 is oxygen. A = mass number = = 17; 17 8 O b. Chlorine is element Cl c. Cobalt is element Co d. Z = 26; A = = 57; Fe e. Iodine is element 5. 5 I f. Lithium is element. 7 Li 46. a. Cobalt is element 27. A = mass number = = 58; Co b B c Mg d I e F f Cu 47. Atomic number = 6 (Eu); Charge = = +; Mass number = = 151; Symbol: Eu+ Atomic number = 50 (Sn); Mass number = = 118; Net charge = = +2; The symbol is Sn Atomic number = 16 (S); Charge = = -2; Mass number = = 4; Symbol: 4 16 S2 Atomic number = 16 (S); Charge = = -2; Mass number = = 2; Symbol: 2 16 S2

9 0 CHAPTER 2 ATOMS, MOLECULES, AND IONS 49. Symbol Number of protons in nucleus Number of neutrons in nucleus Number of electrons Net charge U Ca V Y Br P Symbol Number of protons in nucleus Number of neutrons in nucleus Number of electrons Net charge 5 26 Fe Fe At Al Te a. transition metals b. alkaline earth metals c. alkali metals d. noble gases e. halogens

10 CHAPTER 2 ATOMS, MOLECULES, AND IONS Carbon is a nonmetal. Silicon and germanium are metalloids. 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. 5. In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A and A metals form stable +1, +2 and + charged cations, respectively. Group 5A, 6A and 7A nonmetals form, 2 and -1 charged anions, respectively. a. Lose 2 d. Gain 2 e to form Ra 2+. b. Lose e to form 2 Te. e. Gain 1 e to form In +. c. Gain e to form Br. f. Lose 1 e to form P. e to form Rb See Exercise 5 for a discussion of charges various elements form when in ionic compounds. a. Element 1 is Al. Al forms + charged ions in ionic compounds. Al + b. Se 2 c. Ba 2+ d. N e. Fr + f. Br Nomenclature 55. a. sodium bromide b. rubidium oxide c. calcium sulfide d. aluminum iodide e. SrF 2 f. Al 2 Se g. K N h. Mg P a. mercury(i) oxide b. iron(iii) bromide c. cobalt(ii) sulfide d. titanium(iv) chloride e. Sn N 2 f. CoI g. HgO h. CrS 57. a. cesium fluoride b. lithium nitride c. silver sulfide (Silver only forms stable +1 ions in compounds so no Roman numerals are needed.) d. manganese(iv) oxide e. titanium(iv) oxide f. strontium phosphide 58. a. ZnCl 2 (Zn only forms stable +2 ions in compounds so no Roman numerals are needed.) b. SnF 4 c. Ca N 2 d. Al 2 S e. Hg 2 Se f. AgI (Ag only forms stable +1 ions in compounds.) 59. a. barium sulfite b. sodium nitrite c. potassium permanganate d. potassium dichromate 60. a. Cr(OH) b. Mg(CN) 2

11 2 CHAPTER 2 ATOMS, MOLECULES, AND IONS c. Pb(CO ) 2 d. NH 4 C 2 H O a. dinitrogen tetroxide b. iodine trichloride c. sulfur dioxide d. diphosphorus pentasulfide 62. a. B 2 O b. AsF 5 c. N 2 O d. SCl 6 6. a. copper(i) iodide b. copper(ii) iodide c. cobalt(ii) iodide d. sodium carbonate e. sodium hydrogen carbonate or sodium bicarbonate f. tetrasulfur tetranitride g. sulfur hexafluoride h. sodium hypochlorite i. barium chromate j. ammonium nitrate 64. a. acetic acid b. ammonium nitrite c. cobalt(iii) sulfide d. iodine monochloride e. lead(ii) phosphate f. potassium iodate g. sulfuric acid h. strontium nitride i. aluminum sulfite j. tin(iv) oxide k. sodium chromate l. hypochlorous acid 65. a. SF 2 b. SF 6 c. NaH 2 PO 4 d. Li N e. Cr 2 (CO ) f. SnF 2 g. NH 4 C 2 H O 2 h. NH 4 HSO 4 i. Co(NO ) j. Hg 2 Cl 2 ; Mercury(I) exists as Hg 2+ 2 ions. k. KClO l. NaH 66. a. CrO b. S 2 Cl 2 c. NiF 2 d. K 2 HPO 4 e. AlN f. NH (Nitrogen trihydride is the systematic name.) g. MnS 2 h. Na 2 Cr 2 O 7 i. (NH 4 ) 2 SO j. CI a. Na 2 O b. Na 2 O 2 c. KCN d. Cu(NO ) 2 e. SeBr 4 f. HIO 2 g. PbS 2 h. CuCl i. GaAs (Predict Ga + and j. CdSe (Cadmium only forms +2 charged ions in compounds.) k. ZnS (Zinc only forms +2 charged ions in compounds.) l. HNO 2 m. P 2 O 5 As ions.) 68. a. (NH 4 ) 2 HPO 4 b. Hg 2 S c. SiO 2 d. Na 2 SO e. Al(HSO 4 ) f. NCl g. HBr h. HBrO 2 i. HBrO 4 j. KHS k. CaI 2 l. CsClO 4

12 CHAPTER 2 ATOMS, MOLECULES, AND IONS 69. a. nitric acid, HNO b. perchloric acid, HClO 4 c. acetic acid, HC 2 H O 2 d. sulfuric acid, H 2 SO 4 e. phosphoric acid, H PO a. Iron forms +2 and + 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. Calcium forms only +2 ions in ionic compounds; no Roman numeral is needed. Calcium oxide is correct. 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. Because phosphate has a charge, the charge on iron is +. Iron(III) phosphate is correct. g. This is a covalent compound so use the covalent rules. Diphosphorus pentasulfide is correct. h. Because each sodium is +1 charged, we have the O 2 2 (peroxide) ion present. Sodium peroxide is correct. Note that sodium oxide would be Na 2 O. i. HNO is nitric acid, not nitrate acid. Nitrate acid does not exist. j. H 2 S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name). H 2 SO 4 is sulfuric acid. Additional Exercises 71. Yes, 1.0 g H would react with 7.0 g 7 Cl and 1.0 g H would react with 5.0 g 5 Cl. No, the mass ratio of H/Cl would always be 1 g H/7 g Cl for 7 Cl and 1 g H/5 g Cl for 5 Cl. As long as we had pure 7 Cl or pure 5 Cl, the above 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. 72. 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.

13 4 CHAPTER 2 ATOMS, MOLECULES, AND IONS 7. From the Na 2 X formula, X has a 2 charge. Since 6 electrons are present, X has 4 p, 79 4 = 45 neutrons, and is selenium. a. True. Nonmetals bond together using covalent bonds and are called covalent compounds. b. False. The isotope has 4 protons. c. False. The isotope has 45 neutrons. d. False. The identity is selenium, Se. 74. a. Fe 2+ : 26 protons (Fe is element 26.); protons electrons = charge, 26 2 = 24 electrons; FeO is the formula since the oxide ion has a 2 charge. b. Fe + : 26 protons; 2 electrons; Fe 2 O c. Ba 2+ : 56 protons; 54 electrons; BaO d. Cs + : 55 protons; 54 electrons; Cs 2 O e. S 2 : 16 protons; 18 electrons; Al 2 S f. P : 15 protons; 18 electrons; AlP g. Br 5 protons; 6 electrons; AlBr h. N : 7 protons; 10 electrons; AlN 75. a. Pb(C 2 H O 2 ) 2 : lead(ii) acetate b. CuSO 4 : copper(ii) sulfate c. CaO: calcium oxide d. MgSO 4 : magnesium sulfate e. Mg(OH) 2 : magnesium hydroxide f. CaSO 4 : calcium sulfate g. N 2 O: dinitrogen monoxide or nitrous oxide 76. a. This is element 52, tellurium. Te forms stable 2 charged ions (like other oxygen family members). b. Rubidium. Rb, element 7, forms stable +1 charged ions. c. Argon. Ar is element 18. d. Astatine. At is element From the XBr 2 formula, the charge on element X is +2. Therefore, the element has 88 protons, which identifies it as radium, Ra = 142 neutrons 78. 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 since they form stable 1 charged ions in ionic compounds. From the halogens listed, chlorine, with an average atomic mass of 5.45, fits the data. The two isotopes are 5 Cl and 7 Cl, and the number of electrons in the 1 ion is 18. Note that since the atomic mass of chlorine listed in the periodic table is closer to 5 than 7, we can assume that 5 Cl is the more abundant isotope. This is discussed in Chapter. 79. a. Ca 2+ and N : Ca N 2, calcium nitride b. K + and O 2 : K 2 O, potassium oxide

14 CHAPTER 2 ATOMS, MOLECULES, AND IONS 5 c. Rb + and F : RbF, rubidium fluoride d. Mg 2+ and S 2 : MgS, magnesium sulfide e. Ba 2+ and I : BaI 2, barium iodide f. Al + and Se 2 : Al 2 Se, aluminum selenide g. Cs + and P : Cs P, cesium phosphide h. In + and Br : InBr, indium(iii) bromide. In also forms In + ions, but one would predict In + ions from its position in the periodic table. 80. These compounds are similar to phosphate (PO 4 - ) compounds. Na AsO 4 contains Na + ions and AsO 4 ions. The name would be sodium arsenate. H AsO 4 is analogous to phosphoric acid, H PO 4. H AsO 4 would be arsenic acid. Mg (SbO 4 ) 2 contains Mg 2+ ions and SbO 4 ions, and the name would be magnesium antimonate. 81. 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 is: 2.02 : : = 1 : 15.9 : 1.7 If we have 7.27 g H, then we will have = 116 g S and = 20. g O in the second sample of H 2 SO Mass is conserved in a chemical reaction. chromium(iii) oxide + aluminum chromium + aluminum oxide mass: 4.0 g 12.1 g 2. g? mass aluminum oxide produced = ( ) 2. = 22.8 g Challenge Problems 8. Copper(Cu), silver (Ag) and gold(au) make up the coinage metals. 84. Because the gases are at the same temperature and pressure, the volumes are directly related to the number of molecules present. Let s consider 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 the molecule ratios, which correspond to coefficients in the equation: 2H + O 2HO 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: 2H + O 2 2HO

15 6 CHAPTER 2 ATOMS, MOLECULES, AND IONS This does not require hydrogen to be diatomic. Of course, if we know water has the formula H 2 O, we get: 2H + O 2 2H 2 O. The only way to balance this is to make hydrogen diatomic: 2H 2 + O 2 2H 2 O 85. 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 implies that volume ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule of C x H y produces 8 molecules of CO 2 and 9 molecules of H 2 O. C x H y + O 2 8 CO H 2 O. Because all the carbon in octane ends up as carbon in CO 2, octane contains 8 atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in H 2 O, so one molecule of octane contains 9 2 = 18 atoms of H. Octane formula = C 8 H 18 and the ratio of C:H = 8:18 or 4: From Figure 2.14 of the text, the average diameter of the nucleus is ~10 cm diameter of the volume where the electrons roam about is ~10 cm cm cm = 10 5 ; 1mile 1grape 5280ft 1grape 6,60in 1grape 8 and the average Because the grape needs to be 10 5 times smaller than a mile, the diameter of the grape is 6,60/ = ~0.6 in. This is reasonable. 87. Compound I: 14.0 g R.00g Q 4.67g R ; Compound II: 1.00g Q 7.00g R 4.50g Q 1.56g R 1.00g Q 4.67 The ratio of the masses of R that combine with 1.00 g Q is: = 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 to Compound II (RQ), the formula of Compound I should be R Q. 88. The alchemists were incorrect. The solid residue must have come from the flask. 89. a. Both compounds have C 2 H 6 O as the formula. Because they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties since 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 CO 2 and H 2 O. c. The atom is not an indivisible particle, but is instead composed of other smaller particles, e.g., electrons, neutrons, protons.

16 CHAPTER 2 ATOMS, MOLECULES, AND IONS 7 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). 90. Let X a = formula for the atom/molecule X, Y B = formula for the atom/molecule Y, X c Y d = formula of compound I between X and Y, and X e Y f = 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. X a + 2 Y b 2 X c Y d ; 2 X a + Y b 2 X e Y f From the balanced equations, a = 2c = e and b = d = 2f. Substituting into the balanced equation: X 2c + 2 Y 2f 2 X c Y 2f 2 X 2c + Y 2f 2 X 2c Y f For simplest formulas, assume c = f = 1. So: X Y 2 2 XY 2 and 2 X 2 + Y 2 2 X 2 Y Compound I = XY 2 : if X has relative mass of 1.00, y = 0.04, y = 1.14 Compound II = X 2 Y: if X has relative mass of 1.00, y = 0.664, y = 1.14 The relative mass of Y is 1.14 times that of X. So if X has an atomic mass of 100, then Y will have an atomic mass of 114. Integrated Problems 91. The systematic name of Ta 2 O 5 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 S 2. Therefore, the formula of the sulfur analogue would be Ta 2 S 5. Total number of protons in Ta 2 O 5 : Ta, Z = 7, so 7 protons 2 = 146 protons; O, Z = 8, so 8 protons 5 = 40 protons Total protons = 186 protons Total number of protons in Ta 2 S 5 : Ta, Z = 7, so 7 protons 2 = 146 protons; S, Z = 16, so 16 protons 5 = 80 protons

17 8 CHAPTER 2 ATOMS, MOLECULES, AND IONS Total protons = 226 protons Proton difference between Ta 2 S 5 and Ta 2 O 5 : 226 protons 186 protons = 40 protons 92. 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 fewer electrons than protons. Therefore, the charge on the cation is +. The anion has one-third of the number of protons in 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 Sb + and Cl is SbCl. 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. 9. Number of electrons in the unknown ion: g 1kg 1000g 1electron kg = 28 electrons Number of protons in the unknown ion: g 1kg 1000g 1proton kg = 2 protons Therefore this ion has 2 protons and 28 electrons. This is element number 2, germanium (Ge). The charge is +4 because four electrons have been lost from a neutral germanium atom. The number of electrons in the unknown atom: g 1kg 1000g 1electron kg = 4 electrons In a neutral atom, the number of protons and electrons is the same. Therefore, this is element 4, technetium (Tc). The number of neutrons in the technetium atom: kg 1proton g = 56 neutrons g kg The mass number is the sum of the protons and neutrons. In this atom, the mass number is 4 protons + 56 neutrons = 99. Thus, this atom and its mass number is 99 Tc. Marathon Problem 94. a. For each set of data, divide the larger number by the smaller number to determine relative masses = 2.04, A = 2.04 when B =

18 CHAPTER 2 ATOMS, MOLECULES, AND IONS = 2., C = 2. when B = = 1.17, C = 1.17 when A = To determine whole numbers, multiply the results by. Data set 1: A = 6.1 and B =.0 Data set 2: C = 7.0 and B =.0 Data set : C =.5 and A =.0 or C = 7.0 and A = 6.0 Assuming 6.0 for the relative mass of A, then the relative masses would be A = 6.0, B =.0, and C = 7.0 (if simplest formulas are assumed). b. Gas volumes are proportional to how many molecules are present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is: 6 A 2 + B 4 4 A B B C 4 BC A C 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 A2) 0.602, mass A 2 = 0.40 mass B 4 mass B (mass C) 0.401, mass C = 0.58 mass B 4 mass B (mass C) 0.74, mass A 2 = mass C (mass A ) Assume some relative mass number for any of the masses. We will assume mass B =.0, so mass B 4 = 4(.0) = 12. mass C = 0.58(12) = 7.0, mass C = 7.0/ mass A 2 = 0.570(7.0) = 4.0, mass A = 4.0/2 = 2.0 When we assume a relative mass for B =.0, then A = 2.0 and C = 7.0/. The relative masses having all whole numbers would be A = 6.0, B = 9.0 and C = 7.0. Note: Any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility.