Chapter02rw-final

# Chapter02rw-final - 2 2.1(a(b(c(d 2.2(a Atoms Molecules and Ions Visualizing Concepts The path of the charged particle bends because it is repelled

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Unformatted text preview: 2 2.1 (a) (b) (c) (d) 2.2 (a) Atoms, Molecules, and Ions Visualizing Concepts The path of the charged particle bends because it is repelled by the negatively charged plate and attracted to the positively charged plate. Like charges repel and opposite charges attract, so the sign of the electrical charge on the particle is negative. The greater the magnitude of the charges, the greater the electrostatic repulsion or attraction. As the charge on the plates is increased, the bending will increase. As the mass of the particle increases and speed stays the same, linear momentum (mv) of the particle increases and bending decreases. (See A Closer Look: The Mass Spectrometer.) % abundance = # of mass number × particles total number of particles × 100 12 red 293Nv particles 8 blue 295Nv particles 20 total particles % abundance 293 Nv = % abundance 295 Nv = (b) 12 × 100 = 60% 20 8 × 100 = 40% 20 Atomic weight (AW) is the same as average atomic mass. Atomic weight (average atomic mass) = ∑ fractional abundance × mass of isotope AW of Nv = 0.60(293.15) + 0.40(295.15) = 293.95 (Since % abundance was calculated by counting exact numbers of particles, assume % abundance is an exact number. Then, the number of significant figures in the AW is determined by the number of sig figs in the masses of the isotopes.) 18 2 Atoms, Molecules, and Ions 2.3 metals: red and green alkaline earth metal: red 2.4 noble gas: yellow Solutions to Exercises In general, metals occupy the left side of the chart, and nonmetals the right side. nonmetals: blue and yellow Since the number of electrons (negatively charged particles) does not equal the number of protons (positively charged particles), the particle is an ion. The charge on the ion is 2–. Atomic number = number of protons = 16. The element is S, sulfur. Mass number = protons + neutrons = 32 32 2 − 16 S 2.5 In a solid, particles are close together and their relative positions are fixed. In a liquid, particles are close but moving relative to each other. In a gas, particles are far apart and moving. All ionic compounds are solids because of the strong forces among charged particles. Molecular compounds can exist in any state: solid, liquid, or gas. Since the molecules in ii are far apart, ii must be a molecular compound. The particles in i are near each other and exist in a regular, ordered arrangement, so i is likely to be an ionic compound. 2.6 2.7 Formula: IF 5 Name: iodine pentafluoride Since the compound is composed of elements that are all nonmetals, it is molecular. See Figure 2.22. yellow box: 1+ (group 1A); blue box: 2+ (group 2A) black box: 3+ (a metal in Group 3A); orange box: 2– (a nonmetal in group 6A); green box: 1– (a nonmetal in group 7A) 2.8 Cations (red spheres) have positive charges; anions (blue spheres) have negative charges. There are twice as many anions as cations, so the formula has the general form CA 2 . Only Ca(NO 3 ) 2 , calciu m ni tra te, is consistent with the diagram. Atomic Theory and the Discovery of Atomic Structure 2.9 Postulate 4 of the atomic theory is the law of constant composition. It states that the relative number and kinds of atoms in a compound are constant, regardless of the source. Therefore, 1.0 g of pure water should always contain the same relative amounts of hydrogen and oxygen, no matter where or how the sample is obtained. (a) (b) (c) 6.500 g compound – 0.384 g hydrogen = 6.116 g sulfur Conservation of mass According to postulate 3 of the atomic theory, atoms are neither created nor destroyed during a chemical reaction. If 0.384 g of H are recovered from a compound that contains only H and S, the remaining mass must be sulfur. 2.10 19 2 Atoms, Molecules, and Ions 2.11 (a) 17.60 g oxygen 30.82 g nitrogen 35.20 g oxygen 30.82 g nitrogen 70.40 g oxygen 30.82 g nitrogen 88.00 g oxygen 30.82 g nitrogen = = = = 0.5711 g O 1g N 1.142 g O 1g N 2.284 g O 1g N 2.855 g O 1g N ; 1.142/0.5711 = 2.0 Solutions to Exercises ; 0.5711/0.5711 = 1.0 ; 2.284/0.5711 = 4.0 ; 2.855/0.5711 = 5.0 (b) These masses of oxygen per one gram nitrogen are in the ratio of 1:2:4:5 and thus obey the law of multiple proportions. Multiple proportions arise because atoms are the indivisible entities combining, as stated in Dalton’s theory. Since atoms are indivisible, they must combine in ratios of small whole numbers. 1: 2: 3: 3.56 g fluorine 4.75 g iodine 3.43 g fluorine 7.64 g iodine 9.86 g fluorine 9.41 g iodine = 0.749 g fluorine/1 g iodine = 0.449 g fluorine/1 g iodine = 1.05 g fluorine/1 g iodine 2.12 (a) (b) To look for integer relationships among these values, divide each one by the smallest. If the quotients aren’t all integers, multiply by a common factor to obtain all integers. 1: 0.749/0.449 = 1.67; 1.67 × 3 = 5 2: 0.449/0.449 = 1.00; 1.00 × 3 = 3 3: 1.05/0.449 = 2.34; 2.34 × 3 = 7 The ratio of g fluorine to g iodine in the three compounds is 5:3:7. These are in the ratio of small whole numbers and, therefore, obey the law of multiple proportions. This integer ratio indicates that the combining fluorine “units” (atoms) are indivisible entities. 2.13 Evidence that cathode rays were negatively charged particles was (1) that electric and magnetic fields deflected the rays in the same way they would deflect negatively charged particles and (2) that a metal plate exposed to cathode rays acquired a negative charge. Since the unknown particle is deflected in the opposite direction from that of a negatively charged beta (β) particle, it is attracted to the (–) plate and repelled by the (+) plate. The unknown particle is positively charged. The magnitude of the deflection is less than that of the β particle, or electron, so the unknown particle has greater mass than the electron. The unknown is a positively charged particle of greater mass than the electron. 2.14 20 2 Atoms, Molecules, and Ions 2.15 (a) (b) Solutions to Exercises If the positive plate were lower than the negative plate, the oil drops “coated” with negatively charged electrons would be attracted to the positively charged plate and would descend much more quickly. The more times a measurement is repeated, the better the chance of detecting and compensating for experimental errors. That is, if a quantity is measured five times and four measurements agree but one does not, the measurement that disagrees is probably the result of an error. Also, the four measurements that agree can be averaged to compensate for small random fluctuations. Millikan wanted to demonstrate the validity of his result via its reproducibility. The droplets carry different total charges because there may be 1, 2, 3, or more electrons on the droplet. The electronic charge is likely to be the lowest common factor in all the observed charges. Assuming this is so, we calculate the apparent electronic charge from each drop as follows: A: B: C: D: 1.60 × 10 – 19 / 1 = 1.60 × 10 – 19 C 3.15 × 10 – 19 / 2 = 1.58 × 10 – 19 C 4.81 × 10 – 19 / 3 = 1.60 × 10 – 19 C 6.31 × 10 – 19 / 4 = 1.58 × 10 – 19 C 2.16 (a) (b) (c) The reported value is the average of these four values. Since each calculated charge has three significant figures, the average will also have three significant figures. (1.60 × 10 – 19 C + 1.58 × 10 – 19 C + 1.60 × 10 – 19 C + 1.58 × 10 – 19 C) / 4 = 1.59 × 10 – 19 C Modern View of Atomic Structure; Atomic Weights 2.17 (a) 1.9 Å × 1 × 10 −10 m 1Å 1 × 10 −10 m 1Å × 1 nm 1 × 10 −9 m 1 pm 1 × 10 −12 = 0.19 nm 1.9 Å × × m = 1.9 × 10 2 or 190 pm (1 Å = 100 pm) (b) Aligned Kr atoms have diameters touching. d = 2r = 2(1.9 Å) = 3.8 Å 1.0 mm × 1m 1Å 1 Kr atom × × = 2.6 × 10 6 Kr atoms −10 1000 mm 1 × 10 m 3 .8 Å (c) V = 4 / 3 π r 3 . r = 1.9 Å × 1 × 10 −10 m 1Å × 100 cm = 1.9 × 10 −8 cm m V = (4/3)(π)(1.9 × 10 – 8) 3 cm 3 = 2.9 × 10 – 23 cm 3 21 2 Atoms, Molecules, and Ions 2.18 (a) r = d/2; r = 2.8 × 10 −8 cm 2 × × 1Å 1 × 10 −8 cm r= 2.8 × 10 −8 cm 2 1m = 1.4 × 10 −10 m 100 cm Solutions to Exercises = 1.4 Å (b) Aligned Sn atoms have diameters touching. d = 2.8 × 10 8 cm = 2.8 × 10 – 10 m 6.0 μm × 1 × 10 −6 m 1 μm × 1 Sn atom 2.8 × 10 −10 m = 2.1 × 10 4 Sn atoms (c) V = 4/3 π r 3 ; r = 1.4 × 10 – 10 m V = (4/3)[(π(1.4 × 10 – 10) 3 ] m 3 = 1.149 × 10 – 29 = 1.1 × 10 – 29 m 3 2.19 (a) (b) (c) (d) proton, neutron, electron proton = +1, neutron = 0, electron = –1 The neutron is most massive. (The neutron and proton have very similar masses). The electron least massive. The nucleus has most of the mass but occupies very little of the volume of an atom. True The number of electrons in an atom is equal to the number of protons in the atom. True Atomic number is the number of protons in the nucleus of an atom. Mass number is the total number of nuclear particles, protons plus neutrons, in an atom. The mass number can vary without changing the identity of the atom, but the atomic number of every atom of a given element is the same. 32 and 16 X are isotopes of the same element, because they have identical atomic numbers. 31 16 X 2.20 (a) (b) (c) (d) 2.21 (a) (b) 2.22 (a) (b) These are isotopes of the element sulfur, S, atomic number = 16. 2.23 p = protons, n = neutrons, e = electrons (a) (c) (e) 40 70 Ar has 18 p, 22 n, 18 e Ga has 31 p, 39 n, 31 e W has 74 p, 110 n, 74 e (b) (d) 65 Zn has 30 p, 35 n, 30 e Br has 35 p, 45 n, 35 e (f) 2 43 80 1 84 32 60 Am has 95 p, 148 n, 95 e 2.24 (a) (c) (e) P has 15 p, 17 n (b) (d) (f) 51 Cr has 24 p, 27 n Tc has 43 p, 56 n TI has 81 p, 120 n Co has 27 p, 33 n I has 53 p, 78 n 99 1 31 2 01 22 2 Atoms, Molecules, and Ions 2.25 S ym b ol Protons Neutrons Electrons Mass no. 2.26 S ym b ol Protons Neutrons Electrons Mass No. 2.27 2.28 (a) 196 78 Pt 65 52 Solutions to Exercises 112 Cr 55 Mn 25 30 25 55 Cd 48 64 48 222 Rn 207 Pb 24 28 24 52 86 136 86 222 82 125 82 207 112 Zn 101 Ru 87 Sr 108 Ag 235 U 30 35 30 65 (b) 84 36 Kr 44 57 44 101 (c) 75 33 As 38 49 38 87 (c) 24 12 Mg 47 61 47 108 92 143 92 235 Since the two nuclides are atoms of the same element, by definition they have the same number of protons, 54. They differ in mass number (and mass) because they have different numbers of neutrons. 1 29Xe has 75 neutrons and 1 30Xe has 76 neutrons. (a) (b) 12 6C 2.29 Atomic weights are really average atomic masses, the sum of the mass of each naturally occurring isotope of an element times its fractional abundance. Each B atom will have the mass of one of the naturally occurring isotopes, while the “atomic weight” is an average value. The naturally occurring isotopes of B, their atomic masses, and relative abundances are: 1 0 B, 10.012937, 19.9 %; 1 1 B, 11 .009305, 80.1%. 2.30 (a) (b) 12 amu The atomic weight of carbon reported on the front-inside cover of the text is the abundance-weighted average of the atomic masses of the two naturally occurring isotopes of carbon, 1 2C, and 1 3C. The mass of a 1 2C atom is exactly 12 amu, but the atomic weight of 12.011 takes into account the presence of some 1 3C atoms in every natural sample of the element. 2.31 Atomic weight (average atomic mass) = Σ fractional abundance × mass of isotope Atomic weight = 0.6917(62.9296) + 0.3083(64.9278) = 63.5456 = 63.55 amu 2.32 Atomic weight (average atomic mass) = Σ fractional abundance × mass of isotope Atomic weight = 0.7215(84.9118) + 0.2785(86.9092) = 85.4681 = 85.47 (The result has 2 decimal places and 4 sig figs because each term in the sum has 4 sig figs and 2 decimal places.) 23 2 Atoms, Molecules, and Ions 2.33 (a) Solutions to Exercises Compare Figures 2.4 and 2.13, referring to Solution 2.14. In Thomson’s cathode ray experiments and in mass spectrometry a stream of charged particles is passed through a magnetic field. The charged particles are deflected by the magnetic field according to their mass and charge. For a constant magnetic field strength and speed of the particles, the lighter particles experience a greater deflection. The x-axis label (independent variable) is atomic weight and the y-axis label (dependent variable) is signal intensity. Uncharged particles are not deflected in a magnetic field. The effect of the magnetic field on moving, charged particles is the basis of their separation by mass. The purpose of the magnet in the mass spectrometer is to change the path of the moving ions. The magnitude of the deflection is inversely related to mass, which is the basis of the discrimination by mass. The atomic weight of Cl, 35.5, is an average atomic mass. It is the average of the masses of two naturally occurring isotopes, weighted by their abundances. The single peak at mass 31 in the mass spectrum of phosphorus indicates that the sample contains a single isotope of P, and the mass of this isotope is 31 amu. Average atomic mass = 0.7899(23.98504) + 0.1000(24.98584) + 0.1101(25.98259) = 24.31 amu (b) (c) 2.34 (a) (b) (c) 2.35 (a) (b) The relative intensities of the peaks in the mass spectrum are the same as the relative abundances of the isotopes. The abundances and peak heights are in the ratio 2 4Mg: 2 5Mg: 2 6Mg as 7.8 : 1.0 : 1.1. 2.36 (a) (b) Three peaks: 1 H – 1 H, 1 H – 2 H, 2 H – 2 H 1 1 H – 1 H = 2(1.00783) = 2.01566 amu H – 2 H = 1.00783 + 2.01410 = 3.02193 amu 24 2 Atoms, Molecules, and Ions 2 Solutions to Exercises H – 2 H = 2(2.01410) = 4.02820 amu The mass ratios are 1 : 1.49923 : 1.99845 or 1 : 1.5 : 2. (c) H – 1 H is largest, because there is the greatest chance that two atoms of the more abundant isotope will combine. 2 1 H – 2 H is the smallest, because there is the least chance that two atoms of the less abundant isotope will combine. The Periodic Table; Molecules and Ions 2.37 (a) (e) 2.38 (a) (d) (g) 2.39 (a) (c) (e) 2.40 2.41 Cr (metal) Mg (metal) calcium (metal) thorium (metal) krypton (nonmetal) K, alkali metals (metal) Mg, alkaline earth metals (metal) S, chalcogens (nonmetal) (b) I, halogens (nonmetal) (d) Ar, noble gases (nonmetal) (b) He (nonmetal) (f) Br (nonmetal) (c) P (nonmetal) (g) As (metalloid) (c) gallium (metal) (f) selenium (nonmetal) (d) Zn (metal) (b) titanium (metal) (e) platinum (metal) C, carbon, nonmetal; Si, silicon, metalloid; Ge, germanium, metalloid; Sn, tin, metal; Pb, lead, metal An empirical formula shows the simplest ratio of the different atoms in a molecule. A molecular formula shows the exact number and kinds of atoms in a molecule. A structural formula shows how these atoms are arranged. Compounds with the same empirical but different molecular formulas differ by the integer number of empirical formula units in the respective molecules. Thus, they can have very different molecular structure, size, and mass, resulting in very different physical properties. (a) (e) AlBr 3 C 3 H 2 Cl (b) C 4 H 5 (f) BNH 2 (c) C 2 H 4 O (d) P 2 O 5 2.42 2.43 2.44 A molecular formula contains all atoms in a molecule. An empirical formula shows the simplest ratio of atoms in a molecule or elements in a compound. (a) (b) (c) (d) molecular formula: C 6 H 6 ; empirical formula: CH molecular formula: SiCl 4 ; empirical formula: SiCl 4 (1:4 is the simplest ratio) molecular: B 2 H 6 ; empirical: BH 3 molecular: C 6 H 1 2O 6 ; empirical: CH 2 O 6 4 (b) 6 (b) 8 (c) 12 (c) 9 2.45 2.46 (a) (a) 25 2 Atoms, Molecules, and Ions 2.47 Solutions to Exercises 2.48 2.49 S ym b ol Protons Neutrons Electrons Net Charge 2.50 S ym b ol Protons Neutrons Electrons Net Charge 2.51 2.52 2.53 (a) (a) (a) (c) Mg 2 + Ga3 + 31 59 Co 3 + 27 32 24 3+ 80 Se 2 – 34 46 36 2– 192 Os 2 + 200 Hg 2 + 76 116 74 2+ 80 120 78 2+ P3– 80Br− 115 In3+ 197 Au 3 + 79 15 16 18 3– (b) Al 3 + (b) Sr 2 + (c) K + 35 45 36 1– 49 66 46 3+ (d) S 2 – 118 76 3+ (e) F – (e) Se 2 – (c) As 3 − (d) Br – GaF 3 , gallium(III) fluoride AlI 3 , aluminum iodide (b) LiH, lithium hydride (d) K 2 S, potassium sulfide 26 2 Atoms, Molecules, and Ions 2.54 2.55 2.56 2.57 Ion Cl− OH− CO32− PO43− *Equivalent to NH3(aq). 2.58 Ion O2− NO3 − Solutions to Exercises ( e) Mg 3 (PO 4 ) 2 (e) (NH 4 ) 2 CO 3 (a) (a) (a) AgI (b) Ag 2 S (c) AgF (c) Al(CH 3 COO) 3 (d) (NH 4 ) 2 SO 4 (c) Hg 2 CO 3 (d) Ca 3 (AsO 4 ) 2 CaBr 2 (b) K 2 CO 3 Cu Br 2 (b) Fe 2 O 3 K+ KCl KOH K2CO3 K3PO4 NH4+ NH4Cl NH4OH* (NH4)2CO3 (NH4)3PO4 Mg2+ MgCl2 Mg(OH)2 MgCO3 Mg3(PO4)2 Fe3+ FeCl3 Fe(OH)3 Fe2(CO3)3 FePO4 Na + N a 2O NaNO3 Na2SO4 Na3AsO4 Ca2+ CaO Ca(NO3)2 CaSO4 Ca3(AsO4)2 Fe2+ FeO Fe(NO3)2 FeSO4 Fe3(AsO4)2 Al3+ Al2O3 Al(NO3)3 Al2(SO4)3 AlAsO4 SO42− AsO42− 2.59 Molecular (all elements are nonmetals): (a) B2H6 (b) CH 3 OH (f) NOCl (g) NF 3 Ionic (formed by a cation and an anion, usually contains a metal cation): (c) 2.60 LiNO 3 (d) Sc 2 O 3 (e) CsBr (h) Ag 2 SO 4 Molecular (all elements are nonmetals): (a) PF 5 (c) SCl (h) N 2 O 4 Ionic (formed from ions, usually contains a metal cation): (b) NaI (d) Ca(NO 3 ) 2 (e) FeCl 3 (f) LaP (g) CoCO 3 Naming Inorganic Compounds; Organic Molecules 2.61 2.62 (a) (a) (d) 2.63 (a) (c) (e) ClO2− selenate (b) Cl − (b) selenide (c) ClO3− (d) ClO4− (e) ClO − (c) hydrogen selenide (biselenide) hydrogen selenite (biselenite) calcium, 2+; oxide, 2− potassium, 1+; perchorate, 1− chromium, 3+; hydroxide, 1− (b) sodium, 1+; sulfate, 2− (d) iron, 2+; nitrate, 1− 27 2 Atoms, Molecules, and Ions 2.64 (a) (c) (e) 2.65 (a) (c) (e) (g) (i) 2.66 (a) (c) (e) (f) (g) (i) 2.67 (a) (e) 2.68 (a) (e) 2.69 (a) (d) 2.70 (a) (d) 2.71 (a) (d) 2.72 (a) (d) 2.73 (a) ( d) 2.74 (a) (d) 2.75 (a) copper, 2+; sulfide, 2− aluminum, 3+; chlorate, 1− lead, 2+; carbonate, 2− magnesium oxide lithium phosphate Solutions to Exercises (b) silver, 1+; sulfate, 2− (d) cobalt, 2+; hydroxide, 1− (b) aluminum chloride (d) barium perchlorate copper(II) nitrate (cupric nitrate) (f) iron(II) hydroxide (ferrous hydroxide) calcium acetate potassium chromate potassium oxide strontium cyanide (h) chromium(III) carbonate (chromic carbonate) (j) ammonium sulfate (b) sodium chlorite (d) cobalt(II) hydroxide (cobaltous hydroxide) iron(III) carbonate (ferric carbonate) chromium(III) nitrate (chromic nitrate) ammonium sulfite potassium permanganate Al(OH) 3 HgBr 2 Na 3 PO 4 Co(HCO 3 ) 2 bromic acid HClO HBr carbonic acid sulfur hexafluoride N2O4 dinitrogen monoxide dinitrogen pentoxide ZnCO 3 , ZnO, CO 2 PH 3 NaHCO 3 Mg(OH) 2 (b) K 2 SO 4 (f) Fe 2 (CO 3 ) 3 (b) Zn(NO 3 ) 2 (f) Cr(CH3COO−)3 (h) sodium dihydrogen phosphate (j) silver dichromate (c) Cu 2 O (g) NaBrO (c) Ba(BrO 3 ) 2 (g) K 2 Cr 2 O 7 (c) phosphoric acid (f) H 2 SO 3 (c) HNO 2 (f) acetic acid (c) xenon trioxide (f) P 4 S 6 (c) nitrogen dioxide (d) Fe(ClO 4 ) 2 (d) Zn(NO 3 ) 2 (b) hydrobromic acid (e) HIO 3 (b) H 2 S (e) chloric acid (b) iodine pentafluoride (e) HCN (b) nitrogen monoxide (e) dinitrogen tetroxide (b) HF, SiO 2 , SiF 4 , H 2 O (e) HClO 4 , Cd, Cd(ClO 4 ) 2 (b) Ca(ClO) 2 (e) SnF (c) SO 2 , H 2 O, H 2 SO 3 (f) VBr 3 (c) HCN (f) CdS, H 2 SO 4 , H 2 S A hydrocarbon is a compound composed of the elements hydrogen and carbon only. 28 2 Atoms, Molecules, and Ions (b) Solutions to Exercises 2.76 (a) (b) -ane Hexane has 6 carbons in its chain. 2.77 (a) Functional groups are groups of specific atoms that are constant from one molecule to the next. For example, the alcohol functional group is an –OH. Whenever a molecule is called an alcohol, it contains the –OH group. 2.78 (a) (b) They both have two carbon atoms in their molecular backbone, or chain. In 1-propanol one of the H atoms on an outer (terminal) C atom has been replaced by an —OH group. Additional Exercises 2.79 (a) Based on data accumulated in the late eighteenth century on how substances react with one another, Dalton postulated the atomic theory. Dalton’s theory is based on the indivisible atom as the smallest unit of an element that can combine with other elements. By determining the effects of electric and magnetic fields on cathode rays, Thomson measured the mass-to-charge ratio of the electron. He also proposed the “plum pudding” model of the atom in which most of the space in an atom is occupied by a diffuse positive charge in which the tiny negatively charged electrons are imbedded. (b) 29 2 Atoms, Molecules, and Ions (c) (d) Solutions to Exercises By observing the rate of fall of oil drops in and out of an electric field, Millikan measured the charge of an electron. After observing the scattering of alpha particles at large angles when the particles struck gold foil, Rutherford postulated the nuclear atom. In Rutherford’s atom, most of the mass of the atom is concentrated in a small dense region called the nucleus and the tiny negatively charged electrons are moving through empty space around the nucleus. Most of the volume of an atom is empty space in which electrons move. Most alpha particles passed through this space. The path of the massive alpha particle would not be significantly altered by interaction with a “puny” electron. Most of the mass of an atom is contained in a very small, dense area called the nucleus. The few alpha particles that hit the massive, positively charged gold nuclei were strongly repelled and essentially deflected back in the direction they came from. The Be nuclei have a much smaller volume and positive charge than the Au nuclei; the charge repulsion between the alpha particles and the Be nuclei will be less, and there will be fewer direct hits because the Be nuclei have an even smaller volume than the Au nuclei. Fewer alpha particles will be scattered in general and fewer will be strongly back scattered. Droplet D would fall most slowly. It carries the most negative charge, so it would be most strongly attracted to the upper (+) plate and most strongly repelled by the lower (–) plate. These electrostatic forces would provide the greatest opposition to gravity. Calculate the lowest common factor. A: 3.84 × 10 – 8 / 2.88 × 10 – 8 = 1.33; 1.33 × 3 = 4 B: 4.80 × 10 – 8 / 2.88 × 10 – 8 = 1.67; 1.67 × 3 = 5 C: 2.88 × 10 – 8 / 2.88 × 10 – 8 = 1.00; 1.00 × 3 = 3 D: 8.64 × 10 – 8 / 2.88 × 10 – 8 = 3.00; 3.00 × 3 = 9 The total charge on the drops is in the ratio of 4:5:3:9. Divide the total charge on each drop by the appropriate integer and average the four values to get the charge of an electron in warmombs. A: 3.84 × 10 – 8 / 4 = 9.60 × 10 – 9 wa B: 4.80 × 10 – 8 / 5 = 9.60 × 10 – 9 wa C: 2.88 × 10 – 8 / 3 = 9.60 × 10 – 9 wa D: 8.64 × 10 – 8 / 9 = 9.60 × 10 – 9 wa The charge on an electron is 9.60 × 10 – 9 wa 2.80 (a) (b) (c) 2.81 (a) (b) (c) The number of electrons on each drop are the integers calculated in part (b). A has 4 e – , B has 5 e – , C has 3 e – and D has 9 e – . 30 2 Atoms, Molecules, and Ions (d) 9.60 × 10 −9 wa 1e− 3 3 3 Solutions to Exercises = 6.00 × 10 7 wa/C × 1e− 1.60 × 10 −16 C 2.82 (a) (b) He has 2 protons, 1 neutron, and 2 electrons. H has 1 proton, 2 neutrons, and 1 electron. He: 2(1.6726231 × 10 – 24 g) + 1.6749286 × 10 – 24 g + 2(9.1093897 × 10 – 28 g) = 5.021996 × 10 – 24 g 3 H: 1.6726231 × 10 – 24 g + 2(1.6749286 × 10 – 24 g) + 9.1093897 × 10 – 28 g = 5.023391 × 10 – 24 g Tritium, 3 H, is more massive. (c) The masses of the two particles differ by 0.0014 × 10 – 24 g. Each particle loses 1 electron to form the +1 ion, so the difference in the masses of the ions is still 1.4 × 10 – 27. A mass spectrometer would need precision to 1 × 10 – 27 g to differentiate 3 He + and 3 H. 2.83 (a) (b) (c) 2 protons and 2 neutrons the nuclear strong force The charge of an α particle is twice the magnitude of the charge of an electron, with the opposite sign. That is, 2 (+1.6022 × 10 – 19) C = +3.2044 × 10 – 19 C. 3.2044 × 10 −19 C 4.8224 × 10 4 g/C = 6.6448 × 10 − 24 g 1 amu 1.66054 × 10 − 24 g (d) 6.6448 × 10 − 24 g × = 4.0016 amu (e) The sum of the particle masses in an α particle is 2(1.0073) amu and 2(1.0087) amu = 4.0320 amu. The actual particle mass, 4.0016 amu, is less than the sum of the masses of the components. The difference is the nuclear binding energy, the energy released when protons and neutrons combine to form a nucleus. Mass and energy are interchangeable according to the Einstein relationship E = mc 2 . Calculate the mass of a single gold atom, then divide the mass of the cube by the mass of the gold atom. 2.84 (a) 1g 197.0 amu × = 3.2713 × 10 − 22 = 3.271 × 10 − 22 g/gold atom gold atom 6.022 × 10 23 amu 19.3 g cube × 1 gold atom 3.271 × 10 − 22 g = 5.90 × 10 22 Au atoms in the cube 31 2 Atoms, Molecules, and Ions (b) Solutions to Exercises The shape of atoms is spherical; spheres cannot be arranged into a cube so that there is no empty space. The question is, how much empty space is there? We can calculate the two limiting cases, no empty space and maximum empty space. The true diameter will be somewhere in this range. No empty space: volume cube/number of atoms = volume of one atom V = 4/3π r 3 ; r = (3π V/4) 1 /3; d = 2r vol. of cube = (1.0 × 1.0 × 1.0) = 1.0 cm 3 5.90 × 10 22 Au atoms = 1.695 × 10 − 23 = 1.7 × 10 − 23 cm 3 r = [π (1.695 × 10 – 23 cm 3 )/4] 1 /3 = 3.4 × 10 – 8 cm; d = 2r = 6.8 × 10 – 8 cm Maximum empty space: assume atoms are arranged in rows in all three directions so they are touching across their diameters. That is, each atom occupies the volume of a cube, with the atomic diameter as the length of the side of the cube. The number of atoms along one edge of the gold cube is then (5.90 × 10 2 2) 1 /3 = 3.893 × 10 7 = 3.89 × 10 7 atoms/1.0 cm. The diameter of a single atom is 1.0 cm/3.89 × 10 7 atoms = 2.569 × 10 – 8 = 2.6 × 10 – 8 cm. The diameter of a gold atom is between 2.6 × 10 – 8 cm and 6.8 × 10 – 8 cm (2.6 – 6.8 Å). (c) Some atomic arrangement must be assumed, since none is specified. The solid state is characterized by an orderly arrangement of particles, so it isn’t surprising that atomic arrangement is required to calculate the density of a solid. A more detailed discussion of solid-state structure and density appears in Chapter 11. In arrangement A the number of atoms in 1 cm 2 is just the square of the number that fit linearly in 1 cm. 1.0 cm × 1 atom 4.95 Å × 1 × 10 10 Å 1m × 1m = 2.02 × 10 7 = 2.0 × 10 7 atoms/cm 100 cm 2.85 (a) 1.0 cm 2 = (2.02 × 10 7 ) 2 = 4.081 × 10 1 4 = 4.1 × 10 1 4 atoms/cm 2 (b) In arrangement B, the atoms in the horizontal rows are touching along their diameters, as in arrangement A. The number of Rb atoms in a 1.0 cm row is then 2.0 × 10 7 Rb atoms. Relative to arrangement A, the vertical rows are offset by 1/2 of an atom. Atoms in a “column” are no longer touching along their vertical diameter. We must calculate the vertical distance occupied by a row of atoms, which is now less than the diameter of one Rb atom. Consider the triangle shown below. This is an isosceles triangle (equal side lengths, equal interior angles) with a side-length of 2d and an angle of 60°. Drop a bisector to the uppermost angle so that it bisects the opposite side. 32 2 Atoms, Molecules, and Ions Solutions to Exercises The result is a right triangle with two known side lengths. The length of the unknown side (the angle bisector) is 2h, two times the vertical distance occupied by a row of atoms. Solve for h, the “height” of one row of atoms. (2h) 2 + d 2 = (2d) 2 ; 4h 2 = 4d 2 – d 2 = 3d 2 ; h 2 = 3d 2 /4 h = (3d 2 /4) 1 /2 = (3(4.95 Å) 2 /4) 1 /2 = 4.2868 = 4.29 Å The number of rows of atoms in 1 cm is then 1.0 cm × 1 row 4.2868 Å × 1 × 10 10 Å 1m × 1m = 2.333 × 10 7 = 2.3 × 10 7 100 cm The number of atoms in a 1.0 cm 2 square area is then 2.020 × 10 7 atoms 1 row × 2.333 × 10 7 rows = 4.713 × 10 14 = 4.7 × 10 14 Note that we have ignored the loss of “1/2” atom at the end of each horizontal row. Out of 2.0 × 10 7 atoms per row, one atom is not significant. (c) The ratio of atoms in arrangement B to arrangement A is then 4.713 × 10 1 4 atoms/4.081 × 10 1 4 = 1.555 = 1.2:1. Clearly, arrangement B results in less empty space per unit area or volume. If extended to three dimensions, arrangement B would lead to a greater density for Rb metal. diameter of nucleus = 1 × 10 – 4 Å; diameter of atom = 1 Å V = 4/3 π r 3 ; r = d/2; r n = 0.5 × 10 – 4 Å; r a = 0.5 Å volume of nucleus = 4/3 π (0.5 10 – 4) 3 Å 3 volume of atom = 4/3 π (0.5) 3 Å 3 volume of nucleus 4/3 π (0.5 × 10 −4 ) 3 Å 3 = = 1 × 10 −12 volume of atom 4/3 π (0.5) 3 Å 3 diameter of atom = 5 Å, r a = 2.5 Å volume fraction of nucleus = volume fraction of nucleus = 4/3 π (0.5 × 10 −4 ) 3 Å 3 4/3 π (2.5) 3 Å 3 = 8 × 10 −15 2.86 (a) Depending on the radius of the atom, the volume fraction of the nucleus is between 1 × 10 – 12 and 8 × 10 – 15, that is, between 1 part in 10 1 2 and 8 parts in 10 1 5. 33 2 Atoms, Molecules, and Ions (b) mass of proton = 1.0073 amu Solutions to Exercises 1.0073 amu × 1.66054 × 10 – 24 g/amu = 1.6727 × 10 – 24 g diameter = 1.0 × 10 −15 m, radius = 0.50 × 10 −15 m × 100 cm = 5.0 × 10 −14 cm 1m Assuming a proton is a sphere, V = 4/3 π r 3 . density = g cm 3 = 1.6727 × 10 −24 g 4/3 π (5.0 × 10 −14 ) 3 cm 3 = 3.2 × 10 15 g/cm 3 2.87 The integer on the lower left of a nuclide is the atomic number; it is the number of protons in any atom of the element and gives the element’s identity. The number of neutrons is the mass number (upper left) minus atomic number. (a) (b) (c) (d) As, 33 protons, 41 neutrons I, 53 protons, 74 neutrons Eu, 63 protons, 89 neutrons Bi, 83 protons, 126 neutrons 16 17 18 8 O, 8 O, 8 O 2.88 (a) (b) All isotopes are atoms of the same element, oxygen, with the same atomic number (Z = 8), 8 protons in the nucleus and 8 electrons. Elements with similar electron arrangements have similar chemical properties (Section 2.5). Since the 3 isotopes all have 8 electrons, we expect their electron arrangements to be the same and their chemical properties to be very similar, perhaps identical. Each has a different number of neutrons (8, 9, or 10), a different mass number (A = 16, 17, or 18) and thus a different atomic mass. 2.89 F = k Q 1 Q 2 /d 2 ; k = 9.0 × 10 9 N m 2 /C 2 ; d = 0.53 × 10 – 10 m; Q (electron) = –1.6 × 10 – 19 C; Q (proton) = –Q (electron) = 1.6 × 10 – 19 C 9.0 × 10 9 N m 2 F= C2 × − 1.6 × 10 −19 C × 1.6 × 10 −19 C (0.53 × 10 −10 ) 2 m 2 = 8.202 × 10 − 8 = 8.2 × 10 − 8 N 2.90 Atomic weight (average atomic mass) = Σ fractional abundance × mass of isotope Atomic weight = 0.014(203.97302) + 0.241(205.97444) + 0.221(206.97587) + 0.524(207.97663) = 207.22 = 207 amu (The result has 0 decimal places and 3 sig figs because the fourth term in the sum has 3 sig figs and 0 decimal places.) 2.91 (a) The 68.926 amu isotope has a mass number of 69, with 31 protons, 38 neutrons 69 and the symbol 31 Ga. The 70.925 amu isotope has a mass number of 71, 31 71 protons, 40 neutrons, and symbol 31Ga. (All Ga atoms have 31 protons.) 34 2 Atoms, Molecules, and Ions (b) 2.92 (a) (b) There are 24 known isotopes of Ni, from The five most abundant isotopes are 58 60 62 61 64 51 Solutions to Exercises The average mass of a Ga atom is 69.72 amu. Let x = abundance of the lighter isotope, 1–x = abundance of the heavier isotope. Then x(68.926) + (1–x)(70.925) = 69.72; x = 0.6028 = 0.603, 6 9Ga = 60.3%, 7 1Ga = 39.7%. Ni to 74 Ni. Ni, 57.935346 amu, 68.077% Ni, 59.930788 amu, 26.223% Ni, 61.928346 amu, 3.634% Ni, 60.931058 amu, 1.140% Ni, 63.927968 amu, 0.926% Data from Handbook of Chemistry and Physics, 74th Ed. [Data may differ slightly in other editions.] 2.93 (a) A Br 2 molecule could consist of two atoms of the same isotope or one atom of each of the two different isotopes. This second possibility is twice as likely as the first. Therefore, the second peak (twice as large as peaks 1 and 3) represents a Br 2 molecule containing different isotopes. The mass numbers of the two isotopes are determined from the masses of the two smaller peaks. Since 157.836 ≈ 158, the first peak represents a 7 9Br— 7 9Br molecule. Peak 3, 161.832 ≈ 162, represents a 81 Br— 8 1Br molecule. Peak 2 then contains one atom of each isotope, 7 9Br— 8 1Br, with an approximate mass of 160 amu. The mass of the lighter isotope is 157.836 amu/2 atoms, or 78.918 amu/atom. For the heavier one, 161.832 amu/2 atoms = 80.916 amu/atom. The relative size of the three peaks in the mass spectrum of Br 2 indicates their relative abundance. The average mass of a Br 2 molecule is 0.2569(157.836) + 0.4999(159.834) + 0.2431(161.832) = 159.79 amu. (Each product has four significant figures and two decimal places, so the answer has two decimal places.) (d) (e) 1 Br2 molecule 159.79 amu × = 79.895 amu avg. Br2 molecule 2 Br atoms (b) (c) Let x = the abundance of 7 9Br, 1 – x = abundance of 8 1Br. From (b), the masses of the two isotopes are 78.918 amu and 80.916 amu, respectively. From (d), the mass of an average Br atom is 79.895 amu. x(78.918) + (1 – x)(80.916) = 79.895, x = 0.5110 79 Br = 51.10%, 81 Br = 48.90% 2.94 (a) Five significant figures. 1 H + is a bare proton with mass 1.0073 amu. 1 H is a hydrogen atom, with 1 proton and 1 electron. The mass of the electron is 5.486 × 10 – 4 or 0.0005486 amu. Thus the mass of the electron is significant in the fourth decimal place or fifth significant figure in the mass of 1 H. 35 2 Atoms, Molecules, and Ions (b) Mass of 1 H = 1.0073 amu 0.0005486 amu 1.0078 amu mass of e − mass of 1 H Solutions to Exercises (proton) (electron) (We have not rounded up to 1.0079 since 49 < 50 in the final sum.) × 100 = 5.486 × 10 −4 amu 1.0078 amu × 100 = 0.05444% Mass % of electron = 2.95 (a) (d) (g) (i) an alkali metal: K a halogen: Br (b) an alkaline earth metal: Ca (e) a metalloid: Ge (c) a noble gas: Ar (f) a nonmetal in 1A: H a metal that forms a 3+ ion: Al an element that resembles Al: Ga 266 106 Sg (h) a nonmetal that forms a 2– ion: O 2.96 (a) (b) has 106 protons, 160 neutrons and 106 electrons Sg is in Group 6B (or 6) and immediately below tungsten, W. We expect the chemical properties of Sg to most closely resemble those of W. chlorine gas, Cl2: ii sulfur trioxide, SO3: iii nickel(II) oxide, 2+ chromium(III) oxide, 3+ IO 3 − 2.97 (a) (d) (b) propane, C3H8: v (c) nitrate ion, NO3− : i (e) methylchloride, CH3Cl: iv (b) manganese(IV) oxide, 4+ (d) molybdenium(VI) oxide, 6+ (d) HIO (e) HIO 4 or (H 5 IO 6 ) 2.98 (a) (c) 2.99 2.100 (a) (a) (c) (b) IO 4 − (c) IO perbromate ion AsO 4 3 − (b) selenite ion (d) HTeO 4 − 2.101 Carbonic acid: H2CO3; the cation is H+ because it is an acid; the anion is carbonate because the acid reacts with lithium hydroxide to form lithium carbonate. Lithium hydroxide: LiOH; lithium carbonate: Li2CO3 (a) (c) (e) sodium chloride sodium hypochlorite ammonium carbonate potassium nitrate hydrochloric acid CaS, Ca(HS) 2 NH 3 , NH 4 + 2.102 (b) sodium bicarbonate (or sodium hydrogen carbonate) (d) sodium hydroxide (f) calcium sulfate (b) sodium carbonate (e) magnesium sulfate (c) calcium oxide (f) magnesium hydroxide (d) FeO, Fe 2 O 3 (h) HClO 3 , HClO 4 2.103 (a) (d) 2.104 (a) (e) (b) HBr, HBrO (f) K 2 SO 3 , KHSO 3 (c) AlN, Al(NO 2 ) 3 (g) Hg 2 Cl 2 , HgCl 2 2.105 (a) In an alkane, all C atoms have 4 single bonds, so each C in the partial structure needs 2 more bonds. All alkanes are hydrocarbons, so 2 H atoms will bind to each C atom in the ring. 36 2 Atoms, Molecules, and Ions Solutions to Exercises (b) The molecular formula of cyclohexane is C 6 H 1 2; the molecular formula of n-hexane is C 6 H 1 4 (see Solution 2.64(d)). Cyclohexane can be thought of as n-hexane in which the two outer (terminal) C atoms are joined to each other. In order to form this C—C bond, each outer C atom must lose 1 H atom. The number of C atoms is unchanged, and each C atom still has 4 single bonds. The resulting molecular formula is C 6 H 1 4–2 = C 6 H 1 2. On the structure in part (a), replace 1 H atom with an OH group. (c) 2.106 Elements are arranged in the periodic table by increasing atomic number and so that elements with similar chemical and physical properties form a vertical column or group. By its position in the periodic chart, we know whether an element is a metal, nonmetal, or metalloid, and the common charge of its ion. Members of a group have the same common ionic charge and combine in similar ways with other elements. 37 ...
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## This note was uploaded on 04/04/2009 for the course CHEM 121 taught by Professor Wyzlouzil during the Spring '07 term at Ohio State.

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