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Chapter 2 Practice Exercises 2.1 The first sample has a ratio of 1.25 g Cd 0.357 g S Therefore, the second sample must have the same ratio of Cd to S: 1.25 g Cd x 0.357 g S 3.50 g S Cross-multiplication gives, (1.25 g Cd)(3.50 g S) = x(0.357 g S) (1.25g Cd)(3.50 g S) x 0.357 g S x = 12.3 g Cd 2.2 Compare the ratios of the mass of the compound before heating and the mass of the iron after heating, if they are the same, the compounds are the same. Sample Ratio A 25.36 g = 1.574 16.11 g B 15.42 g = 1.86 8.28 g C 7.85 g = 1.86 4.22 g D 11.87 g = 1.57 7.54 g Compounds A and D are the same, as are compounds B and C. 2.3 240 94 Pu The bottom number is the atomic number, found on the periodic table (number of protons). The top number is the mass number (sum of the number of protons and the number of neutrons). Since it is a neutral atom, it has 94 electrons. 2.4 35 17 Cl 2.5 We can discard the 17 since the 17 tells the number of protons which is information that the symbol "Cl" also provides. In addition, the number of protons equals the number of electrons in a neutral atom, so the symbol "Cl" also indicates the number of electrons. The 35 is necessary to state which isotope of chlorine is in question and therefore the number of neutrons in the atom. 2.6 2.24845 × 12 u = 26.9814 u 2.7 Copper is 63.546 u 12 u = 5.2955 times as heavy as carbon 2.8 (0.198 × 10.0129 u) + (0.802 × 11.0093 u) = 10.8 u 2.9 (a) (b) (c) (d) contains 17 protons, 17 electrons, and 18 neutrons. 1 Ni, 2 Cl 1 Fe, 1 S, 4 O 3 Ca, 2 P, 8 O 1 Co, 2 N, 12 O, 12 H 15 Chapter 2 2.10 (a) (b) (c) (d) 2 N nitrogen, 1 Fe iron, 1 Mo molybdenum, 6 C carbon, 4 H hydrogen, 1 N nitrogen, 2 N nitrogen, 4 H hydrogen, 3 O oxygen 4 H hydrogen, 11 O oxygen, 1 Cl chlorine, 2 S sulfur, 10 H hydrogen 1 N nitrogen, 8 O oxygen 2 O oxygen 2.11 This is a balanced chemical equation, and the number of each atom that appears on the left is the same as that on the right: 1 Mg, 2 O, 4 H, and 2 Cl. 2.12 Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O 2.13 6 N, 42 H, 2 P, 20 O, 3 Ba, and 12 C are on both the products and the reactants sides of the equation, therefore the reaction is balanced. 2.14 The term "octa' means eight, therefore there are 8 carbon atoms in octane. The formula for an alkane is CnH2n+2, so octane has 8 carbons and ((2 × 8) + 2) = 18 H. The condensed formula is CH3CH2CH2CH2CH2CH2CH2CH3, and the structural format is: H H H H H H H H H C C C C C C C C H H H H H H H H H 2.15 The condensed formula is CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 The structural formula is: H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H 2.16 (a) Propanol: CH3CH2CH2OH H H H H C C C O H H H H (b) Butanol: CH3CH2CH2CH2OH H H H H H C C C C O H H H H H 2.17 (a) (b) (c) (d) Fe: 26 protons and 26 electrons Fe3+: 26 protons and 23 electrons N3–: 7 protons and 10 electrons N: 7 protons and 7 electrons 2.18 (a) (b) (c) (d) O: 8 protons and 8 electrons O2–: 8 protons and 10 electrons Al3+: 13 protons and 10 electrons Al: 13 protons and 13 electrons 2.19 (a) NaF (b) Na2O (c) MgF2 (d) Al4C3 16 Chapter 2 2.20 (a) Ca3N2 (b) AlBr3 (c) Na3P d) CsCl 2.21 (a) (b) CrCl3 and CrCl2, Cr2O3 and CrO CuCl, CuCl2, Cu2O and CuO 2.22 (a) (b) Au2S and Au2S3, Au3N and AuN SnS and SnS2, Sn3N2 and Sn3N4 2.23 (a) KC2H3O2 (b) Sr(NO3)2 2.24 (a) Na2CO3 (b) (NH4)2SO4 2.25 (a) (b) (c) 2.26 (a) AsCl5 (b) SCl6 (c) S2Cl2 2.27 (a) K2O (b) BaBr2 (c) Na3N 2.28 (a) (c) aluminum chloride sodium bromide (b) (d) barium sulfide calcium fluoride 2.29 (a) (c) postassium sulfide nickel(II) chloride (b) (d) magnesium phosphide iron(III) oxide 2.30 (a) Al2S3 2.31 (a) (b) lithium carbonate iron(III) hydroxide 2.32 (a) KClO3 2.33 diiodine pentaoxide 2.34 chromium(III) acetate (c) Fe(C2H3O2)3 phosphorous trichloride sulfur dioxide dichlorine heptaoxide (b) SrF2 (d) Al2S3 (c) TiO2 d) Au2O3 (b) Ni3(PO4)2 Review Questions 2.1 The first law of chemical combination is the law of conservation of mass: no detectable gain or loss of mass occurs in chemical reactions. The other law is the law of definite proportions: in a given chemical compound, the elements are always combined in the same proportions by mass. 2.2 Isotopes of a particular element have nearly identical chemical properties and the average mass of an atom is independent, almost, of the source of the atom. 2.3 Conservation of mass derives from the postulate that atoms are not destroyed in normal chemical reactions. The Law of Definite Proportions derives from the notion that compound substances are always composed of the same types and numbers of atoms of the various elements in the compound. 2.4 This is the Law of Definite Proportions, which guarantees that a single pure substance is always composed of the same ratio of masses of the elements that compose it. 17 Chapter 2 2.5 (a) (b) c) 2.6 To test the law of conservation of mass, a reaction would have to be carried out in which the mass of the reactants and the mass of the products are weighed and shown to be the same. The law of definite proportions could be shown by demonstrating that no matter how a compound is made, the same proportions by mass are used. This could be done by decomposing a compound and showing that the masses of the elements are always in the same ratio. To test the law of multiple proportions, two different compounds made up of the same elements would have to be decomposed. The amount used would have to keep mass of one of the elements constant, and then the masses of other the element from the different samples would have to be in a ratio of small whole numbers. Protons, 11 p + , +1 charge Electron, 00 e , –1 charge Neutron, 01 n , no charge 2.7 Nearly all of the mass is located in the nucleus, because this is the portion of the atom where the proton and the neutron are located. 2.8 A nucleon is a subatomic particle found in the atomic nucleus. We have studied neutrons and protons. 2.9 The atomic number is equal to the number of protons in the nucleus of the atom, and the mass number is the sum of the number of neutrons and the number of protons. The atomic number (symbol Z) is designated by a subscript preceding the chemical symbol and the mass number (symbol A) is a superscript preceding the chemical symbol. 2.10 (a) 2.11 (a) 131 53 I 2.12 For all group IA elements (the alkali metals), the formula is MX, that is one Cl per atom of metal. For all group IIA elements (the alkaline earth metals), the formula is MX 2, that is two Cl atoms per atom of metal. The correspondence in formula and the similarities in chemical behavior allowed Mendeleev to locate theses two series into their separate groups on the periodic table. 2.13 Mendeleev constructed his periodic table by arranging the elements in order of increasing atomic weight, and grouping the elements by their recurring properties. The modern periodic table is arranged in order of increasing number of protons. 2.14 Strontium and calcium are in the same Group of the periodic table, so they are expected to have similar chemical properties. Strontium should therefore form compounds that are similar to those of calcium, including the sorts of compounds found in bone. 2.15 Silver and gold are in the same periodic table group as copper, so they might well be expected to occur together in nature, because of their similar properties and tendencies to form similar compounds. 2.16 Cadmium is in the same periodic table group as zinc, but silver is not. Therefore cadmium would be expected to have properties similar to those of zinc, whereas silver would not. 2.17 The superscript before the symbol indicates the mass number; the superscript after the symbol indicates the charge on the atom; the subscript before the symbol indicates the atomic number; and the subscript after the symbol indicates the number of atoms in the compound. mass number For example: (b) 90 38 Sr (b) atomic number (c) 137 55 Ce (d) 189 F mass number charge atomic number N number of atoms 14 7 N2 Nitrogen has 7 protons and 7 electrons for a neutral atom. The molecule has two atoms in it, and the isotope with 7 neutrons gives it a mass number of 14 18 Chapter 2 2.18 See Figure in the margin of page 50. 2.19 (a) (b) (c) (d) (e) (f) (g) 2.20 Luster, electrical conductivity, thermal conductivity, ductility, and malleability are the characteristic properties of metals. 2.21 Mercury is used in thermometers because it is a liquid, and tungsten is used in light bulbs because is has such a high melting point. 2.22 The noble gases: He, Ne, Ar, Kr, Xe, and Rn 2.23 Mercury and bromine 2.24 They are semiconductors. 2.25 See figure 2.7 2.26 The heavy line separates the metals from the nonmetals, and the metalloids border the line. 2.27 Metals which are used to make jewelry are those that do not corrode, silver, gold, and platinum. Iron would be useless for jewelry because it is susceptible to rusting. Potassium reacts violently with water to form hydrogen and potassium hydroxide. 2.28 Luster, malleability, color, and brittleness are some trends that are mentioned in terms of moving from the metals to the nonmetals across the periodic table, or moving down a group from nonmetals to metals. 2.29 (a) (b) (c) Li I W Xe Sm Pu Mg In general, melting points decrease from left to right across the periodic table and increases from top to bottom. In general, boiling points decrease from left to right across the periodic table and increases from top to bottom. In general, density has a maximum in the middle of the periodic table and falls off to the right and left. Also, the density increases moving down a group. 2.30 This may stand for the name of an element or for the name of one atom of an element. 2.31 The smallest particle that is representative of a particular element is the atom of that element. A molecule is a representative unit that is made up of two or more atoms linked together. 2.32 H2, hydrogen Cl2, chlorine 2.33 A chemical reaction is balanced when there is the same number of each kind of atom on both the reactant and product side of the equation; and the total charges on both the reactant and product sides of the equation are the same. These conditions must be met due to the law of conservation of matter. 2.34 Reactants are the substances to the left of the arrow in a reaction that are present before the reaction begins. Products are the substances to the right of the arrow in a reaction and they are formed during the reaction and are present when the reaction is over. N2, nitrogen, Br2. bromine O2, oxygen I2, iodine F2, fluorine 19 Chapter 2 2.35 (a) (b) (c) (d) Magnesium reacts with oxygen to give (yield) magnesium oxide. The reactants are Mg and O2. The product is MgO. 2Mg(s) + O2(g) 2MgO(s) 2.36 2C8H18(l) + 25O2(g) 18CO2(g) + 18H2O 2.37 The noble gases: He, Ne, Ar, Kr, Xe, and Rn 2.38 S8, P4 2.39 Nonmetals 2.40 (a) CH4 2.41 PH3 2.42 HAt 2.43 SnH4 2.44 (a) CH4, component of natural gas (c) CH3CH2CH3, gas-fired barbecues (b) CH3CH3, component of natural gas (d) CH3CH2CH2CH3, cigarette lighters 2.45 (a) (b) 2.46 C10H22 or CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 2.47 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 or CH3(CH2)21CH3 or C23H48 2.48 All of the elements are nonmetals, and the formula is not in the smallest whole number ratio. 2.49 methane: (b) NH3 CH3OH (c) TeH2 (d) HI CH3CH2OH ethane: propane: 20 Chapter 2 butane: methanol: ethanol: decane (10 carbons): 23 carbon hydrocarbon 2.50 (a) (b) An ionic compound is formed by the transfer of electrons, and it is accompanied by the formation of ions of opposite charge. Molecular compounds arise from the sharing of electrons between atoms, rather than from the complete transfer of electrons as in (a). 21 Chapter 2 2.51 Metals react with nonmetals. 2.52 Nonmetals react with metals, nonmetals, and metalloids. 2.53 Nonmetals are more frequently found in compounds because of the large variety of ways they may combine. A particularly illustrative example is the combination of carbon, a nonmetal, with other elements. So many compounds are possible that there is one entire area of chemistry devoted to the study of carbon compounds, organic chemistry. 2.54 An ion is a charged particle. It can be monatomic or polyatomic, and it can have either a positive or a negative charge. It is derived from an atom or a molecule by gain or loss of electrons. Atoms and molecules are neutral. 2.55 In ionic substances, no molecules exist. Rather we have a continuous array of cations and anions, which are present in a constant ratio. The ratio is given by the formula unit. 2.56 Al2Cl6 is molecular because the smallest whole number ratio of elements is not used in the formula. 2.57 (a) (b) (c) (d) (e) 2.58 A cation is a positively charged ion with one or fewer electrons than its neutral atom. An anion is a negatively charged ion with one or more electrons than its neutral atom. A polyatomic ion is made up of more than one atom; the whole unit is the ion. 2.59 Titanium lost four electrons to form Ti4+; it has 22 protons and 18 electrons. 2.60 Negative 2.61 Nitrogen gained 3 electrons to form N3–; it has 7 protons and 10 electrons. 2.62 Rb forms a +1 cation (Rb+) and Cl forms a –1 anion (Cl–), so the formula should be RbCl. The cation is first in the formula, therefore the formula should be Na 2S. 2.63 The formula should have the smallest whole numbers possible. The formula should be TiO 2. 2.64 (a) (b) (c) (d) (e) (f) 2.65 The incorrect ones are a, d, and e. 2.66 (a) (d) CN– SO32– (b) (e) NH4+ ClO3– (c) f) NO3– SO42– 2.67 (a) (d) OCl– H2PO4– (b) (e) HSO4– MnO4– (c) (f) PO43– C2O42– 2.68 (a) (d) dichromate ion carbonate ion (b) (e) hydroxide ion cyanide ion (c) (f) acetate ion perchlorate ion Na, Na+ These particles have the same number of nueclei. These particles have the same number of protons These particles could have different numbers of neutrons, if they are different isotopes. These particles do not have the same number of electrons; Na + has one less electron. Fe2+, Fe3+ Co2+, Co3+ Hg2+, Hg22+ Cr2+, Cr3+ Sn2+, Sn4+ Cu+, Cu2+ 22 Chapter 2 (a) (b) (c) (d) Ca(s) + Cl2(g) CaCl2(s) 2Mg(s) + O2(g) 2MgO(s) 4Al(s) + 3O2(g) 2Al2O3(s) S(s) + 2Na(s) Na2S(s) 2.70 (a) (b) Fe(OH)3(s) + 3HCl(g) H2O + FeCl3(aq) 2AgNO3(aq) + BaCl2(aq) 2AgCl(s) + Ba(NO3)2(aq) 2.71 (a) (b) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 2Na(s) + 2H2O 2NaOH(aq) + H2(g) 2.72 Binary compounds, such as CCl4 contain two elements only. A diatomic substance is composed of molecules having two atoms, such as HCl or N2. In the latter, the two atoms may or may not be the same. 2.73 When naming the compound, molecular compounds need the prefixes to specify the number of atoms in the molecule. Ionic compounds made with transition metals or post-transition metals need to have the charge of the metal specified. 2.74 For naming ionic compounds of the transition elements it is essential to know the charge on the anion since that will help determine the charge on the transition element. Transition elements can have more than one charge. 2.75 (a) 2.69 (b) S8(s) + 16Na(s) 8Na2S(s) or Greek prefixes are used to specify the number of atoms of each element in a molecular compound (PCl5 is phosphorous pentachloride; specify the number of water molecules in a hydrate (CuSO4·5H2O is copper sulfate pentahydrate); specify the number of atoms in polyatomic ions; and organic compounds use Greek prefixes to specify the number of carbon atoms in the compound (pentane has five carbons). Roman numerals are used in the name of transition metal compounds to specify the charge of the metal. Review Problems 2.76 From the first ratio we see that there is a ratio of 9.33 g of nitrogen to 2.00 g of hydrogen. Multiplying the mass of nitrogen by 3.14 we see that for every 6.28 g hydrogen there will be 29.3 g nitrogen. 9.33 g nitrogen x g nitrogen 2.00 g hydrogen 6.28 g hydrogen x = 29.3 g nitrogen 2.77 From the ratio of phosphorous to chlorine, we see that for every 1.20 g phosphorus, there are 4.12 g chlorine. Dividing the mass of chlorine by 3.43, we find that for every 6.22 g chlorine there will be 1.81 g phosphorus. 1.20 g phosphorous x g phosphorous 4.12 g chlorine 6.22 g chlorine x =1.81 g phosphorous 2.78 5.54 g ammonia. From the ratio, Problem 2.76, above we see that for every 4.56 g nitrogen there needs to be 0.98 g hydrogen. According to the Law of Conservation of Mass 5.54 g ammonia will be produced. 9.33 g nitrogen 4.56 g nitrogen 2.00 g hydrogen x g hydrogen x = 0.977 g hydrogen mass of NH3 = 4.56 g nitrogen + 0.977 g hydrogen = 5.54 g NH3 23 Chapter 2 2.79 55.0 g of the phosphorus chloride. From the ratio above, Problem 2.77, we see that for every 12.5 g phosphorus it is necessary to have 42.9 g chlorine. According to the Law of Conservation of Mass, 55.0 g of the phosphorus chloride compound was formed. 1.20 g phosphorous 12.5 g phosphorous 4.12 g chlorine x g chlorine x = 42.9 g chlorine mass of PCl3 = 12.5 g phosphorous + 42.9 g chlorine = 55.4 g PCl3 2.80 2.286 g of O. The first nitrogen-oxygen compound has an atom ratio of 1 atom of N for 1 atom of O, and a mass ratio of 1.000 g of N for 1.143 g of O. The second compound has an atom ratio of 1 atom of N for 2 atoms of O; therefore, the second compound has twice as many grams of O for each gram of N, or 2.286 g of O. 2.81 This ratio should be 4/2 = 2/1, as required by the formulas of the two compounds. Twice 0.597 g Cl, or 1.19 g Cl. 2.82 Since we know that the formula is CH4, we know that one fourth of the total mass due to the hydrogen atom constitutes the mass that may be compared to the carbon. Hence we have 0.33597 g H 4 = 0.083993 g H and 1.00 g assigned to the amount of C-12 in the compound. Then it is necessary to realize that the ratio 1.00 g C 12 for carbon is equal to the ratio 0.083993 g H X, where X equals the relative atomic mass of hydrogen. 1.000 g C 0.083993 g H 12 u C 1.008 u X 2.83 1 mol O mol O 1.000 g O 0.06250 mol O 15.999 g O 2 mol X mol X 0.6250 mol O 0.04167 mol X 3 mol O 1.125 g g/mol 27.00 g/mol 0.04167 mol The element is aluminum. 2.84 Regardless of the definition, the ratio of the mass of hydrogen to that of carbon would be the same. If C–12 were assigned a mass of 24 (twice its accepted value), then hydrogen would also have a mass twice its current value, or 2.01588 u. 2.85 Taking the mass ratio of 109Ag to 12C and multiplying it by 12 we see that the mass of Ag-109 is 108.90 u. 12 × 9.0754 = 108.90 u 2.86 (0.6917 × 62.9396 u) + (0.3083 × 64.9278 u) = 63.55 u 2.87 (0.7899 × 23.9850 u) + (0.1000 × 24.9858 u) + (0.1101 × 25.9826 u) = 24.31 u 2.88 neutrons 138 protons 88 electrons 88 124 82 82 Carbon-14 8 6 6 23 12 11 11 (a) Radium-226 (b) 206 (c) (d) Pb Na 24 Chapter 2 2.89 electrons 55 protons 55 neutrons 82 92 92 146 Iodine-131 53 53 78 197 79 79 118 (a) Cesium-137 (b) 238 (c) (d) U Au 2.90 1 Cr, 6 C, 9 H, 6 O 2.91 3 Ca, 5 Mg, 8 Si, 24 O, 2 H 2.92 MgSO4 2.93 KNaC4H4O6·4H2O 2.94 (a) (b) (c) (d) (e) 2 K, 2 C, 4 O 2 H, 1 S, 3 O 12 C, 26 H 4 H, 2 C, 2 O 9 H, 2 N, 1 P, 4 O 2.95 (a) (b) (c) (d) (e) 3 Na, 1 P, 4 O 1 Ca, 4 H, 2 P, 8 O 4 C, 10 H 3 Fe, 2 As, 8 O 3 C, 8 H, 3 O 2.96 (a) (b) (c) (d) (e) 1 Ni, 2 Cl, 8 O 1 Cu, 1 C, 3 O 2 K, 2 Cr, 7 O 2 C, 4 H, 2 O 2 N, 9 H, 1 P, 4 O 2.97 (a) (b) (c) (d) (e) 6 C, 12 H, 2 O 1 Mg, 1 S, 14 H, 11 O 1 K, 1 Al, 2 S, 20 O, 24 H 1 Cu, 2 N, 6 O 4 C, 10 H, 1 O 2.98 (a) (b) (c) 6 N, 3 O 4 Na, 4 H, 4 C, 12 O 2 Cu, 2 S, 18 O, 20 H 2.99 (a) (b) (c) 14 C, 28 H, 14 O 4 N, 8 H, 2 C, 2 O 10 K, 10 Cr, 35 O 2.100 (a) 6 (b) 3 (c) 27 2.101 (a) 16 (b) 36 (c) 50 2.102 (a) (d) K+ S2– (b) (e) Br– Al3+ (c) Mg2+ 25 Chapter 2 2.103 (a) (d) Ba2+ Sr2+ (b) (e) O2– Rb+ (c) F– 2.104 (a) (d) NaBr MgBr2 (b) (e) KI BaF2 (c) BaO 2.105 (a) (b) (c) (d) (e) CrCl2 and CrCl3 FeCl2 and FeCl3 MnCl2 and MnCl3 CuCl and CuCl2 ZnCl2 2.106 (a) (d) KNO3 Fe2(CO3)3 (b) (e) Ca(C2H3O2)2 Mg3(PO4)2 (c) NH4Cl 2.107 (a) (d) Zn(OH)2 Rb2SO4 (b) (e) Ag2CrO4 LiHCO3 (c) BaSO3 2.108 (a) (d) PbO and PbO2 FeO and Fe2O3 (b) (e) SnO and SnO2 Cu2O and CuO (c) MnO and Mn2O3 2.109 (a) (d) CdCl2 NiCl2 (b) AgCl (c) ZnCl2 2.110 (a) (c) silicon dioxide tetraphosphorus decaoxide (b) (d) xenon tetrafluoride dichlorine heptaoxide 2.111 (a) (c) chlorine trifluoride dinitrogen pentoxide (b) (d) disulfur dichloride arsenic pentachloride 2.112 (a) (c) (e) calcium sulfide sodium phosphide rubidium sulfide (b) (d) aluminum bromide barium arsenide 2.113 (a) (c) (e) sodium fluoride lithium nitride potassium selenide (b) (d) magnesium carbide aluminum oxide 2.114 (a) (c) iron(II) sulfide tin(IV) oxide (b) (d) copper(II) oxide cobalt(II) chloride hexahydrate 2.115 (a) (c) manganese(III) oxide lead(II) sulfide (b) (d) mercury(I) chloride chromium(III) chloride tetrahydrate 2.116 (a) (c) sodium nitrite magnesium sulfate heptahydrate (b) (d) potassium permanganate potassium thiocyanate 2.117 (a) (c) potassium phosphate iron(III) carbonate (b) (d) ammonium acetate sodium thiosulfate pentahydrate 2.118 (a) (b) (c) (d) (e) (f) ionic molecular ionic molecular ionic molecular chromium(II) chloride disulfur dichloride ammonium acetate sulfur trioxide potassium iodate tetraphosphorous hexaoxide 26 Chapter 2 (g) (h) (i) (j) ionic ionic ionic molecular calcium sulfite silver cyanide zinc(II) bromide hydrogen selenide 2.119 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) ionic ionic ionic ionic molecular ionic ionic molecular molecular molecular vanadium(III) nitrate cobalt(II) acetate gold(III) sulfide gold(I) sulfide germanium tetrabromide potassium chromate iron(II) hydroxide diiodine tetraoxide tetraiodine nonaoxide tetraphosphorus triselenide 2.120 (a) (d) (g) Na2HPO4 S2F10 SbF5 (b) (e) Li2Se Ni(CN)2 (c) (f) Cr(C2H3O2)3 Fe2O3 2.121 (a) (d) (g) Al2Cl6 Cu(HSO4)2 I2O5 (b) (e) As4O10 NH4SCN (c) (f) Mg(OH)2 K2S2O3 2.122 (a) (d) (g) (NH4)2S MoS2 P4S6 (b) (e) Cr2(SO4)3·6H2O SnCl4 (c) (f) SiF4 H2Se 2.123 (a) (d) (g) Hg(C2H3O2)2 Ca3P2 XeF4 (b) (e) Ba(HSO3)2 Mg(H2PO4)2 (c) (f) BCl3 CaC2O4 2.124 diselenium hexasulfide and diselenium tetrasulfide 2.125 diphosphorous pentasulfide Additional Exercises 2.126 (a) Metal (b) (c) (d) 55 25 Mn (e) 2.127 30 neutrons 25 electrons 54.9380 4.578 times heavier than a C-12 atom 12.000 First calculate the number of moles of oxygen that are combined with X: 1 mole O 1.14 g oxygen = 0.0713 moles oxygen 16.00 g O Then calculate the moles of X in the oxygen compound: 1 mole X 0.0713 moles O = 0.0356 moles X 2 moles O Finally, calculate the molar mass of X: 27 Chapter 2 1.00 g X = 28.1 g/mole X 0.0356 moles X Note that X is Si, atomic number 14, and that its oxygen compound is SiO 2. The same number of moles of X are also combined with Y: 4 moles Y 0.0356 moles X = 0.142 moles Y 1 moles X The atomic mass of Y is thus: 5.07 g Y = 35.6 g/mole Y 0.142 moles Y Note that Y is Cl, atomic number 17, and that its compound with X is SiCl 4. 2.128 (0.0580 × 53.9396 u) + (0.9172 × 55.9349 u) + (0.0220 × 56.9354 u) + (0.0028 × 57.9333 u) = 55.85 u 2.129 Let x equal the abundance of 79Br and y equal the abundance of 81Br. We know that x + y = 1 and x(78.9183) + y(80.9163) = 79.904. Substituting y = 1 – x we get x(78.9183) + (1 – x)80.9163 = 79.904. Solving for x we get x = 0.5067 and y = 0.4933. 2.130 The mass ratio of Fe to O in the first compound is 2.325 g Fe to 1.000 g O and the atom ratio is 2 Fe atoms to 3 O atoms. Dividing the masses by the number of atoms gives the mass ratios of Fe to O 2.325 g Fe 1.162 g Fe/atom Fe 2 atoms Fe 1.000 g O 0.3333 g O/ atom O 3 atoms O The second compound has a mass ratio of 2.616 g Fe to 1.000 g O. Dividing the mass of Fe by the mass/atom ratio determined for Fe and dividing the mass of O by the mass/atom ratio determined for O gives the ratio of number of atoms in the compound. 2.616 g Fe 2.251 atom Fe 1.162 g Fe/atom Fe 1.000 g O 3.000 atom O 0.3333 g O/atom O Multiplying the atomic ratios by 4 gives whole numbers to the subscripts in the formula: Fe 9O12. Both 9 and 12 are divisible by 3 to give: Fe3O4. 2.131 (24.31)1.6605389 × 10–24 g = 4.037 × 10–23 g (55.85)1.6605389 × 10–24 g = 9.274 × 10–23 g Comparing answers, we see that both numbers are on the order of 1 × 10–23. We would expect 6 × 1023 atoms of Ca in 40.078 g of Ca. 2.132 Hg2(NO3)2·2H2O 2.133 (a) (b) (c) (d) (e) CaO NO2 Cl2 Cl2 CaO Hg(NO3)2·H2O HBr CuCl2 AsH3 HBr CaO CuCl2 AsH3 HBr NO2 NO2 28 Chapter 2 (f) (g) Cl2 CaO 2.134 (a) (c) (e) (g) (i) auric sulfate plumbic oxide cupric sulfate cobaltous hydroxide stannic sulfide 2.135 (a) (b) (c) (d) (e) (f) (g) copper(II) bromide copper(I) iodide iron(II) sulfate chromium(II) chloride mercury(I) nitrate manganese(II) sulfate lead(II) acetate 2.136 ferric nitrate nonahydrate 2.137 (a) Ca(s) + Br2(l) CaBr2(s) (b) C(s) + 2Cl2(g) CCl4(l) (c) 2Al(s) + 3S(g) Al2S3(s) 2.138 HBr CuCl2 AsH3 NO2 NaNO3 (b) (d) (f) (h) auric nitrate mercurous chloride mercuric chloride stannous chloride CuBr2 CuI FeSO4 CrCl2 Hg2(NO3)2 MnSO4 Pb(C2H3O2)2 Mass of Br 79.904 2 3.99 Mass of Ca 40.078 Mass of Cl 35.453 4 11.8 Mass of C 12.0107 Mass of S 32.065 2 0.792 Mass of Al 26.9815 3 N2O5(g) + 3SO2(g) 3SO3(g) + 2NO(g) The small whole number ratio for oxygen in the nitrogen oxides is 5 to 1. The small whole number ratio for oxygen in the sulfur oxides is 3 to 2. 29