Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition

CHAPTER 2 ATOMS AND THE ATOMIC THEORY PRACTICE EXAMPLES 1A (E) The total mass must be the same before and after reaction. mass before reaction = 0.382 g magnesium + 2.652 g nitrogen = 3.034 g mass after reaction = magnesium nitride mass + 2.505 g nitrogen = 3.034 g magnesium nitride mass = 3.034 g  2.505 g = 0.529 g magnesium nitride 1B (E) Again, the total mass is the same before and after the reaction. mass before reaction = 7.12 g magnesium +1.80 g bromine = 8.92 g mass after reaction = 2.07 g magnesium bromide + magnesium mass = 8.92 g magnesium mass = 8.92 g  2.07 g = 6.85 g magnesium 2A (M) In Example 2-2 we are told that 0.500 g MgO contains 0.301 g of Mg. With this information, we can determine the mass of magnesium needed to form 2.000 g of magnesium oxide. mass of Mg = 2.000 g MgO  0.301 g Mg = 1.20 g Mg 0.500 g MgO The remainder of the 2.00 g of magnesium oxide is the mass of oxygen. mass of oxygen = 2.00 g magnesium oxide  1.20 g magnesium = 0.80 g oxygen 2B (M) In Example 2-2, we see that a 0.500 g sample of MgO has 0.301 g Mg, hence, it must have 0.199 g O2. From this we see that if we have equal masses of Mg and O2, the oxygen is in excess. First we find out how many grams of oxygen reacts with 10.00 g of Mg. 0.199 g O 2  6.61 g O 2 (used up) 0.301 g Mg Hence, 10.00 g  6.61 g  3.39 g O 2 unreacted. Mg is the limiting reactant. mass oxygen  10.00 g Mg  MgO(s) mass  mass Mg + Mass O 2  10.00 g  6.61 g  16.61 g MgO. There are only two substances present, 16.61 g of MgO (product) and 3.39 g of unreacted O2 3A (E) Silver has 47 protons. If the isotope in question has 62 neutrons, then it has a mass number of 109. This can be represented as 109 47 Ag . 3B (E) Tin has 50 electrons and 50 protons when neutral, while a neutral cadmium atom has 48 electrons. This means that we are dealing with Sn2+. We do not know how many neutrons tin has. so there can be more than one answer. For instance, 116 2+ 117 2+ 118 2+ 119 2+ 120 2+ are all possible answers. 50 Sn , 50 Sn , 50 Sn , 50 Sn , and 50 Sn 4A (E) The ratio of the masses of 202Hg and 12C is: 202 12 Hg 201.97062 u   16.830885 C 12 u 27 M02_PETR5044_11_CSM_C02.pdf 1 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 4B (E) Atomic mass is 12 u × 13.16034 = 157.9241 u. The isotope is 158 64 Gd . Using an atomic 16 mass of 15.9949 u for 16O, the mass of 158 O is 64 Gd relative to relative mass to oxygen-16 = 157.9241 u  9.87340 15.9949 u 5A (E) The atomic mass of boron is 10.81, which is closer to 11.0093054 than to 10.0129370. Thus, boron-11 is the isotope that is present in greater abundance. 5B (E) The average atomic mass of indium is 114.82, and one isotope is known to be 113In. Since the weighted-average atomic mass is almost 115, the second isotope must be larger than both In-113 and In-114. Clearly, then, the second isotope must be In-115 (115In). Since the average atomic mass of indium is closest to the mass of the second isotope, In-115, then 115 In is the more abundant isotope. 6A (M) Weighted-average atomic mass of Si = (27.9769265325 u × 0.9223)  25.80 u (28.976494700 u × 0.04685)  1.358 u (29.973377017 u × 0.03092)  0.9268 u 28.0848 u We should report the weighted-average atomic mass of Si as 28.08 u. 6B (M) We let x be the fractional abundance of lithium-6. 6.941 u =  x  6.01512 u  + 1  x   7.01600 u  = x  6.01512 u + 7.01600 u  x  7.01600 u 6.941 u  7.01600 u = x  6.01512 u  x  7.01600 u =  x 1.00088 u x 7A 6.941 u  7.01600 u  0.075 Percent isotopic abundances : 7.5% lithium-6, 92.5% lithium-7 1.00088 u (M) We assume that atoms lose or gain relatively few electrons to become ions. Thus, elements that will form cations will be on the left-hand side of the periodic table, while elements that will form anions will be on the right-hand side. The number of electrons “lost” when a cation forms is usually equal to the last digit of the periodic group number; the number of electrons added when an anion forms is typically eight minus the last digit of the group number. Li is in group 1(1A); it should form a cation by losing one electron: Li + . S is in group 6(6A); it should form an anion by adding two electrons: S2 . Ra is in group 2(2A); it should form a cation by losing two electrons: Ra 2+ . F and I are both group 17(7A); they should form anions by gaining an electron: F  and I  . A1 is in group 13(3A); it should form a cation by losing three electrons: Al 3+ . 28 M02_PETR5044_11_CSM_C02.pdf 2 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 7B 8A (M) Main-group elements are in the “A” families, while transition elements are in the “B” families. Metals, nonmetals, metalloids, and noble gases are color coded in the periodic table inside the front cover of the textbook. Na is a main-group metal in group 1(1A). Re is a transition metal in group 7(7B). S is a main-group nonmetal in group 16(6A). I is a main-group nonmetal in group 17(7A). Kr is a nonmetal in group 18(8A). Mg is a main-group metal in group 2(2A). U is an inner transition metal, an actinide. Si is a main-group metalloid in group 14(4A). B is a metalloid in group 13(3A). A1 is a main-group metal in group 13(3A). As is a main-group metalloid in group 15(5A). H is a main-group nonmetal in group 1(1A). (H is believed to be a metal at extremely high pressures.) (E) This is similar to Practice Examples 2-8A and 2-8B. Cu mass = 2.35 1024 Cu atoms  8B (M) Of all lead atoms, 24.1% are lead-206, or 241 206 Pb atoms in every 1000 lead atoms. First we need to convert a 22.6 gram sample of lead into moles of lead (below) and then, by using Avogadro’s constant, and the percent isotopic abundance, we can determine the number of 206 Pb atoms. 1 mol Pb nPb = 22.6 g Pb×  0.109 mol Pb 207.2 g Pb 206 9A Pb atoms = 0.109 mol Pb× 206 6.022×1023 Pb atoms 241 Pb atoms × = 1.58×10 22 206 Pb atoms 1mol Pb 1000 Pb atoms (M) Both the density and the molar mass of Pb serve as conversion factors. atoms of Pb = 0.105cm3 Pb  9B 1 mol Cu 63.546 g Cu  = 248 g Cu 23 6.022 10 atoms 1 mol Cu 11.34 g 1 mol Pb 1 cm  3 207.2 g  6.022  10 23 Pb atoms 1mol Pb = 3.46  1021 Pb atoms (M) First we find the number of rhenium atoms in 0.100 mg of the element. 1g 1 mol Re 6.022  1023 Re atoms 0.100 mg     3.23  1017 Re atoms 1000 mg 186.21 g Re 1 mol Re % abundance 187 2.02  1017 atoms 187 Re Re =  100% = 62.5% 3.23  1017 Re atoms 29 M02_PETR5044_11_CSM_C02.pdf 3 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory INTEGRATIVE EXAMPLE A. (M) Stepwise approach: First, determine the total number of Cu atoms in the crystal by determining the volume of the crystal and calculating the mass of Cu from density. Then we can determine the amount of 63Cu by noting its relative abundance 1 cm3 3  1.5625  10 17 cm3 Volume of crystal =  25 nm   7 3 (1  10 nm) Mass of Cu in crystal = d  V = 8.92 g/cm3 × 1.5625 × 10–17 = 1.3938 × 10–16 g # of Cu atoms = 1 mol Cu 6.022  1023 Cu atoms 1.3938  10 g Cu    1.3208  106 Cu atoms 63.546 g Cu 1 mol Cu 63 Therefore, the number of Cu atoms, assuming 69.17% abundance, is 9.14 × 105 atoms. 16 Conversion pathway approach: 8.92 g Cu 1 cm3 (25 nm)3 1 mol Cu 6.022  1023 Cu atoms     1 cm3 (1  107 nm)3 crystal 63.546 g Cu 1 mol Cu 69.17 atoms of 63Cu   9.14  105 atoms of 63Cu 100 Cu atoms B. (M) Stepwise approach: Calculate the mass of Fe in a serving of cereal, determine mass of 58Fe in that amount of cereal, and determine how many servings of cereal are needed to reach 58 g of 58Fe. Amount of Fe in a serving of cereal = 18 mg × 0.45 = 8.1 mg Fe per serving First calculate the amount of Fe 1 mol Fe 0.0081 g Fe   1.45 104 mol Fe 55.845 g Fe Then calculate 58Fe amount: 0.282 mol 58 Fe –4 = 4.090×10–7 mol 58Fe 1.45 × 10 mol Fe × 100 mol Fe 30 M02_PETR5044_11_CSM_C02.pdf 4 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory Converting mol of 58F to # of servings: 4.090  10 7 mol 58 Fe 57.9333 g 58 Fe   2.37  10 5 g 58 Fe per serving 58 1 serving 1 mol Fe Total # of servings = 58 g total / 2.37×10–5 per serving = 2.4477×106 servings Conversion Pathway Approach: The number of servings of dry cereal to ingest 58 g of 58Fe = 1 mol 58 Fe 100 mol Fe 58.845 g Fe 1 cereal serving 58.0 g 58 Fe     58 58 57.9333 g Fe 0.282 mol Fe 1 mol Fe 0.018 g Fe  0.45  2.4477  106 servings 2.44477  106 servings  1 year  6706 years 365 servings EXERCISES Law of Conservation of Mass 1. (E) The observations cited do not necessarily violate the law of conservation of mass. The oxide formed when iron rusts is a solid and remains with the solid iron, increasing the mass of the solid by an amount equal to the mass of the oxygen that has combined. The oxide formed when a match burns is a gas and will not remain with the solid product (the ash); the mass of the ash thus is less than that of the match. We would have to collect all reactants and all products and weigh them to determine if the law of conservation of mass is obeyed or violated. 2. (E) The magnesium that is burned in air combines with some of the oxygen in the air and this oxygen (which, of course, was not weighed when the magnesium metal was weighed) adds its mass to the mass of the magnesium, making the magnesium oxide product weigh more than did the original magnesium. When this same reaction is carried out in a flashbulb, the oxygen (in fact, some excess oxygen) that will combine with the magnesium is already present in the bulb before the reaction. Consequently, the product contains no unweighed oxygen. 3. (E) By the law of conservation of mass, all of the magnesium initially present and all of the oxygen that reacted are present in the product. Thus, the mass of oxygen that has reacted is obtained by difference. mass of oxygen = 0.674 g MgO  0.406 g Mg = 0.268 g oxygen 4. (E) Reaction: 2 K(s) + Cl2(g)  2 KCl(s) Mass of Cl2 reacted = 8.178 g – 6.867 g = 1.311 g Cl2(g) mKCl  1.311 g Cl2  1 mol Cl2 2 mol KCl 74.55 g KCl    2.757 g KCl 70.90 g Cl2 1 mol Cl2 1 mol KCl 31 M02_PETR5044_11_CSM_C02.pdf 5 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 5. (M) We need to compare the mass before reaction (initial) with that after reaction (final) to answer this question. initial mass = 10.500 g calcium hydroxide +11.125 g ammonium chloride = 21.625 g final mass = 14.336 g solid residue + (69.605 – 62.316) g of gases = 21.625 g These data support the law of conservation of mass. Note that the gain in the mass of water is equal to the mass of gas absorbed by the water. 6. (M) We compute the mass of the reactants and compare that with the mass of the products to answer this question. reactant mass = mass of calcium carbonate + mass of hydrochloric acid solution 1.148g = 10.00 g calcium carbonate +100.0 mL soln  1 mL soln = 10.00 g calcium carbonate +114.8g solution = 124.8 g reactants product mass = mass of solution + mass of carbon dioxide 1.9769 g = 120.40 g soln + 2.22 L gas  1 L gas = 120.40 g soln + 4.39 g carbon dioxide  The same mass within experimental error,  = 124.79 g products    thus, the law of conservation of mass obeyed.  Law of Constant Composition 7. (E) (a) (b) (c) 8. (0.755  0.455) g  0.397 0.755 g 0.300 g Ratio of O: Mg in MgO by mass   0.659 0.455 g 0.455 g Mg Percent magnesium by mass  ×100%  60.3% 0.755 g MgO Ratio of O: MgO by mass  (M) (a) We can determine that carbon dioxide has a fixed composition by finding the % C in each sample. (In the calculations below, the abbreviation “cmpd” is short for compound.) 3.62 g C 5.91 g C %C   100%  27.3% C %C   100%  27.3% C 13.26 g cmpd 21.66 g cmpd 7.07 g C %C   100%  27.3% C 25.91 g cmpd Since all three samples have the same percent of carbon, these data do establish that carbon dioxide has a fixed composition. 32 M02_PETR5044_11_CSM_C02.pdf 6 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory (b) 9. Carbon dioxide contains only carbon and oxygen. As determined in (a), carbon dioxide is 27.3% C by mass. The percent of oxygen in carbon dioxide is obtained by difference. %O = 100.0 % – (27.3 %C) = 72.7 %O (M) In the first experiment, 2.18 g of sodium produces 5.54 g of sodium chloride. In the second experiment, 2.10 g of chlorine produces 3.46 g of sodium chloride. The amount of sodium contained in this second sample of sodium chloride is given by mass of sodium = 3.46 g sodium chloride  2.10 g chlorine = 1.36 g sodium. We now have sufficient information to determine the % Na in each of the samples of sodium chloride. 2.18g Na 1.36 g Na %Na = 100% = 39.4% Na %Na = 100% = 39.3% Na 5.54 g cmpd 3.46 g cmpd Thus, the two samples of sodium chloride have the same composition. Recognize that, based on significant figures, each percent has an uncertainty of 0.1% . 10. (E) If the two samples of water have the same % H, the law of constant composition is demonstrated. Notice that, in the second experiment, the mass of the compound is equal to the sum of the masses of the elements produced from it. 3.06 g H 1.45 g H %H =  100%  11.2% H %H   100%  11.2% H 27.35g H 2 O 1.45 +11.51 g H 2O Thus, the results are consistent with the law of constant composition. 11. (E) The mass of sulfur (0.312 g) needed to produce 0.623 g sulfur dioxide provides the information required for the conversion factor. 0.312 g sulfur sulfur mass = 0.842 g sulfur dioxide  = 0.422 g sulfur 0.623g sulfur dioxide 12. (M) (a) From the first experiment we see that 1.16 g of compound is produced per gram of Hg. These masses enable us to determine the mass of compound produced from 1.50 g Hg. 1.16 g cmpd mass of cmpd = 1.50 g Hg  = 1.74 g cmpd 1.00 g Hg (b) Since the compound weighs 0.24 g more than the mass of mercury (1.50 g) that was used, 0.24 g of sulfur must have reacted. Thus, the unreacted sulfur has a mass of 0.76 g (= 1.00 g initially present 24 g reacted). 33 M02_PETR5044_11_CSM_C02.pdf 7 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory Law of Multiple Proportions 13. (M) By dividing the mass of the oxygen per gram of sulfur in the second sulfur-oxygen compound (compound 2) by the mass of oxygen per gram of sulfur in the first sulfur-oxygen compound (compound 1), we obtain the ratio (shown to the right): 1.497 g of O (cmpd 2) 1.500 1.000 g of S  0.998 g of O 1 (cmpd 1) 1.000 g of S To get the simplest whole number ratio we need to multiply both the numerator and the denominator by 2. This gives the simple whole number ratio 3/2. In other words, for a given mass of sulfur, the mass of oxygen in the second compound (SO3) relative to the mass of oxygen in the first compound (SO2) is in a ratio of 3:2. These results are entirely consistent with the Law of Multiple Proportions because the same two elements, sulfur and oxygen in this case, have reacted together to give two different compounds that have masses of oxygen that are in the ratio of small positive integers for a fixed amount of sulfur. 14. (M) This question is similar to question 13 in that two elements, phosphorus and chlorine in this case, have combined to give two different compounds. This time, however, different masses have been used for both of the elements in the second compound. To see if the Law of Multiple Proportions is being followed, the mass of one of the two elements must be set to the same value in both compounds. This can be achieved by dividing the masses of both phosphorus and chlorine in reaction 2 by 2.500: 2.500 g phosphorus “normalized” mass of phosphorus = = 1.000 g of phosphorus 2.500 14.308 g chlorine = 5.723 g of chlorine “normalized” mass of chlorine = 2.500 Now the mass of phosphorus for both reactions is fixed at 1.000 g. Next, we will divide each amount of chlorine by the fixed mass of phosphorus with which they are combined. This gives 3.433 g of Cl (cmpd 1) 1.000 g P  0.600  6 :10 or 3 : 5 5.723 g of Cl (cmpd 2) 1.000 g P 15. (M) (a) First of all we need to fix the mass of nitrogen in all three compounds to some common value, for example, 1.000 g. This can be accomplished by multiplying the masses of hydrogen and nitrogen in compound A by 2 and the amount of hydrogen and nitrogen in compound C by 4/3 (1.333): Cmpd A “normalized” mass of nitrogen “normalized” mass of hydrogen = 0.500 g N  2 = 1.000 g N = 0.108 g H  2 = 0.216 g H Cmpd C “normalized” mass of nitrogen “normalized” mass of hydrogen = 0.750 g N  1.333 = 1.000 g N = 0.108 g H  1.333 = 0.144 g H 34 M02_PETR5044_11_CSM_C02.pdf 8 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory Next, we divide the mass of hydrogen in each compound by the smallest mass of hydrogen, namely, 0.0720 g. This gives 3.000 for compound A, 1.000 for compound B, and 2.000 for compound C. The ratio of the amounts of hydrogen in the three compounds is 3 (cmpd A) : 1 (cmpd B) : 2 (cmpd C) These results are consistent with the Law of Multiple Proportions because the masses of hydrogen in the three compounds end up in a ratio of small whole numbers when the mass of nitrogen in all three compounds is normalized to a simple value (1.000 g here). (b) 16. The text states that compound B is N2H2. This means that, based on the relative amounts of hydrogen calculated in part (a), compound A might be N2H6 and compound C, N2H4. Actually, compound A is NH3, but we have no way of knowing this from the data. Note that the H:N ratios in NH3 and N2H6 are the same, 3H:1N. (M) (a) As with the previous problem, one of the two elements must have the same mass in all of the compounds. This can be most readily achieved by setting the mass of iodine in all four compounds to 1.000 g. With this approach we only need to manipulate the data for compounds B and C. To normalize the amount of iodine in compound B to 1.000 g, we need to multiply the masses of both iodine and fluorine by 2. To accomplish the analogous normalization of compound C, we must multiply by 4/3 (1.333). Cmpd B: “normalized” mass of iodine = 0. 500 g I  2 = 1.000 g I “normalized” mass of fluorine = 0.2246 g F  2 = 0.4492 g F Cmpd C: “normalized” mass of iodine = 0.750 g I  1.333 = 1.000 g I “normalized” mass of fluorine = 0.5614 g F  1.333 = 0.7485 g F Next we divide the mass of fluorine in each compound by the smallest mass of fluorine, namely, 0.1497 g. This gives 1.000 for compound A, 3.001 for compound B, 5.000 for compound C, and 7.001 for compound D. The ratios of the amounts of fluorine in the four compounds A : B : C : D is 1 : 3 : 5 : 7. These results are consistent with the law of multiple proportions because for a fixed amount of iodine (1.000 g), the masses of fluorine in the four compounds are in the ratio of small whole numbers. (b) 17. As with the preceding problem, we can figure out the empirical formulas for the four iodine-fluorine containing compounds from the ratios of the amounts of fluorine that were determined in 16(a): Cmpd A: IF Cmpd B: IF3 Cmpd C: IF5 Cmpd D: IF7 (M) One oxide of copper has about 20% oxygen by mass. If we assume a 100 gram sample, then ~ 20 grams of the sample is oxygen (~1.25 moles) and 80 grams is copper (~1.26 moles). This would give an empirical formula of CuO (copper(II) oxide). The second oxide has less oxygen by mass, hence the empirical formula must have less oxygen or more copper (Cu:O ratio greater than 1). If we keep whole number ratios of atoms, a plausible formula would be Cu2O (copper(I) oxide), where the mass percent oxygen is 11%. 35 M02_PETR5044_11_CSM_C02.pdf 9 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 18. (M) Assuming the intermediate is “half-way” between CO (oxygen-carbon mass ratio = 16:12 or 1.333) and CO2 (oxygen-carbon mass ratio = 32:12 or 2.6667), then the oxygen- carbon ratio would be 2:1, or O:C = 24:12. This mass ratio gives a mole ratio of O:C = 1.5:1. Empirical formulas are simple whole number ratios of elements; hence, a formula of C3O2 must be the correct empirical formula for this carbon oxide. (Note: C3O2 is called tricarbon dioxide or carbon suboxide). Fundamental Charges and Mass-to-Charge Ratios 19. (M) We can calculate the charge on each drop, express each in terms of 10 19 C, and finally express each in terms of e = 1.6 1019 C.  12.8 1019 C = 8e drops 2 &3 : 1.28 1018  2  0.640 1018 C  6.40 1019 C = 4e drop 4 : 1.28 1018  8  0.160 1018 C  1.60 1019 C = 1e drop 5 : 1.28 1018  4  5.12  1018 C = 32 e drop 1: 1.28 1018  51.2 1019 C We see that these values are consistent with the charge that Millikan found for that of the electron, and he could have inferred the correct charge from these data, since they are all multiples of e . 20. (M) We calculate each drop’s charge, express each in terms of 10 19 C, and then, express each in terms of e = 1.6 1019 C. drop 1: 6.41 1019 C  6.41 1019 C = 4e drop 2 : 6.41 1019  2  3.21 1019 C  3.211019 C = 2 e drop 3 : 6.41 1019  2  1.28 1018 C  12.8 1019 C = 8 e drop 4 : 1.44 1018  14.4  1019 C drop 5 : 1.44  10 18 3  4.8  10 19 = 9e  4.8 10 C 19 C = 3e We see that these values are consistent with the charge that Millikan found for that of the electron. He could have inferred the correct charge from these values, since they are all multiples of e , and have no other common factor. 21. (M) (a) Determine the ratio of the mass of a hydrogen atom to that of an electron. We use the mass of a proton plus that of an electron for the mass of a hydrogen atom. or mass of proton  mass of electron 1.0073u  0.00055 u   1.8  103 mass of electron 0.00055 u mass of electron 1   5.6  104 mass of proton  mass of electron 1.8  103 36 M02_PETR5044_11_CSM_C02.pdf 10 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory (b) The only two mass-to-charge ratios that we can determine from the data in Table 2-1 are those for the proton (a hydrogen ion, H+) and the electron. mass 1.673 1024 g For the proton : = = 1.044 105 g/C 19 charge 1.602  10 C mass 9.109 1028 g = = 5.686  109 g/C 19 charge 1.602  10 C The hydrogen ion is the lightest positive ion available. We see that the mass-to-charge ratio for a positive particle is considerably larger than that for an electron. For the electron : 22. (M) We do not have the precise isotopic masses for the two ions. The values of the mass-tocharge ratios are only approximate. This is because some of the mass is converted to energy (binding energy), that holds all of the positively charged protons in the nucleus together. Consequently, we have used a three-significant-figure mass for a nucleon, rather than the more precisely known proton and neutron masses. (Recall that the term “nucleon” refers to a nuclear particle— either a proton or a neutron.) m 127 nucleons 1 electron 1.67 1024 g 127  I = = 1.32  103 g/C (7.55  102 C/g)   19 1 electron 1.602 10 C 1 nucleon e m 32 nucleons 1 electron 1.67  1024 g 32 2  S = = 1.67 104 g/C (6.00  103C/g)   19 e 2 electrons 1.602  10 C 1 nucleon Atomic Number, Mass Number, and Isotopes 23. 32 59 226 (E) (a) cobalt-60 60 27 Co (b) phosphorus-32 15 P (c) iron-59 26 Fe (d) radium-226 88 Ra 24. (E) The nucleus of 202 80 Hg contains 80 protons and (202 – 80) = 122 neutrons. Thus, the percent of nucleons that are neutrons is given by 122 neutrons % neutrons =  100 = 60.4% neutrons 202 nucleons 25.a Name Symbol Na Si Rb Number of Protons 11 14 37 Number of Electrons 11 14a 37a Number of Neutrons 12 14 48 Mass Number 23 28 85 (E) Sodium Silicon 23 11 28 14 85 37 Potassium 40 19 K 19 19 21 40 Arsenic a 75 33 As 33a 33 42 75 Neon 20 10 Ne 2+ 10 8 10 20 80 35 Br 35 35 45 80 82 82 126 208 Rubidium Bromine Lead b b 208 82 Pb 37 M02_PETR5044_11_CSM_C02.pdf 11 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory a This result assumes that a neutral atom is involved. Insufficient data. Does not characterize a specific nuclide; several possibilities exist. The minimum information needed is the atomic number (or some way to obtain it, such as from the name or the symbol of the element involved), the number of electrons (or some way to obtain it, such as the charge on the species), and the mass number (or the number of neutrons). b 26. 27. (E) (a) Since all of these species are neutral atoms, the numbers of electrons are the atomic numbers, the subscript numbers. The symbols must be arranged in order of increasing 39 58 59 120 112 122 value of these subscripts. 40 18 Ar < 19 K < 27 Co < 29 Cu < 48 Cd < 50 Sn < 52Te (b) The number of neutrons is given by the difference between the mass number and the atomic number, A-Z. This is the difference between superscripted and subscripted values and is provided (in parentheses) after each element in the following list. 39 40 59 58 112 122 120 19 K(20) < 18 Ar(22) < 29 Cu(30) < 27 Co(31) < 50 Sn(62) < 52Te(70) < 48 Cd(72) (c) Here the nuclides are arranged by increasing mass number, given by the superscripts. 39 40 58 59 112 120 122 19 K < 18 Ar < 27 Co < 29 Cu < 50 Sn < 48 Cd 50% more that is at least 2.5 times greater than the atomic number. Only 90 neutrons than protons (number of neutrons = 226 – 90 = 136; number of protons = 90; 90 × 1.5 = 135, less than the total number of neutrons). 124Sn is close, having 74 neutrons, and 50 protons, so 50 × 1.5 = 75, slightly less than 50%. 39 M02_PETR5044_11_CSM_C02.pdf 13 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 35. (E) If we let n represent the number of neutrons and p represent the number of protons, then p + 4 = n. The mass number is the sum of the number of protons and the number of neutrons: p + n = 44. Substitution of n = p + 4 yields p + p + 4 = 44. From this relation, we see p = 20. Reference to the periodic table indicates that 20 is the atomic number of the element calcium. 36. (M) We will use the same type of strategy and the same notation as we used previously in Equation 35 to come up with the answer. n=p+1 There is one more neutron than the number of protons. n + p = 9 × 3 = 27 The mass number equals nine times the ion’s charge of 3+. Substitute the first relationship into the second, and solve for p. 27  ( p  1)  p  2 p  1 27  1  13 p 2 27 Thus this is the 3+ action of the isotope Al-27  13 Al3+ . 37. (M) The number of protons is the same as the atomic number for iodine, which is 53. There is one more electron than the number of protons because there is a negative charge on the ion. Therefore the number of electrons is 54. The number of neutrons is equal to 70, mass number minus atomic number. 38. (E) For iodine, Z = 53, and so it has 53 protons. Because the overall charge on the ion is 1−, there are 54 electrons in a single ion of iodine-131. The number of neutrons is 131 – 53 = 78. 39. (E) For americium, Z = 95. There are 95 protons, 95 electrons, and 241 − 95 = 146 neutrons in a single atom of americium-241. 40. (E) For cobalt, Z = 27. There are 27 protons, 27 electrons, and 60 − 27 = 33 neutrons in a single atom of cobalt-60. Atomic Mass Units, Atomic Masses 41. (E) There are no chlorine atoms that have a mass of 35.45 u. The masses of individual chlorine atoms are close to integers and this mass is about midway between two integers. It is an average atomic mass, the result of averaging two (or more) isotopic masses, each weighted by its isotopic abundance. 40 M02_PETR5044_11_CSM_C02.pdf 14 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 42. (E) It is exceedingly unlikely that another nuclide would have an exact integral mass. The mass of carbon-12 is defined as precisely 12 u. Each nuclidic mass is close to integral, but none that we have encountered in this chapter are precisely integral. The reason is that each nuclide is composed of protons, neutrons, and electrons, none of which have integral masses, and there is a small quantity of the mass of each nucleon (nuclear particle) lost in the binding energy holding the nuclides together. It would be highly unlikely that all of these contributions would add up to a precisely integral mass. 43. (E) To determine the weighted-average atomic mass, we use the following expression: average atomic mass =   isotopic mass  fractional isotopic abundance Each of the three percents given is converted to a fractional abundance by dividing it by 100. Mg atomic mass =  23.985042 u  0.7899  +  24.985837 u  0.1000  +  25.982593u  0.1101 = 18.95 u + 2.499 u + 2.861u = 24.31 u 44. (E) To determine the average atomic mass, we use the following expression: average atomic mass =   isotopic mass  fractional isotopic abundance Each of the three percents given is converted to a fractional abundance by dividing it by 100. Cr atomic mass =  49.9461 0.0435  +  51.9405  0.8379  +  52.9407  0.0950  +  53.9389  0.0236  = 2.17 u + 43.52 u + 5.03u +1.27 u = 51.99 u If all digits are carried and then the answer is rounded at the end, the answer is 52.00 u. 45. (E) We will use the expression to determine the weighted-average atomic mass. 107.868 u = 106.905092 u  0.5184  +  109 Ag  0.4816  = 55.42 u + 0.4816 109 Ag 107.868 u  55.42 u  0.4816 109 Ag = 52.45 u 46. 109 Ag  52.45 u  108.9 u 0.4816 (M) The percent abundance of the two isotopes must add to 100.00%, since there are only two naturally occurring isotopes of gallium. Thus, we can determine the percent abundance of the second isotope by using difference: second isotope = 100.00% − 60.11% = 39.89% From the periodic table, we see that the atomic mass of gallium is 69.723 u. We use this value in the expression to determine the weighted-average atomic mass, along with the isotopic mass of 69Ga and the fractional abundances of the two isotopes (the percent abundances divided by 100). 69.723 u  (0.6011  68.925581 u)  (0.3989  other isotope)  41.43 u  (0.3989  other isotope) 69.723  41.43 u other isotope   70.92 u  mass of 71 Ga, the other isotope 0.3989 41 M02_PETR5044_11_CSM_C02.pdf 15 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 47. (M) Since the three percent abundances total 100%, the percent abundance of 40 K is found by difference. % 40 K = 100.0000%  93.2581%  6.7302% = 0.0117% Then the expression for the weighted-average atomic mass is used, with the percent abundances converted to fractional abundances by dividing by 100. Note that the average atomic mass of potassium is 39.0983 u.  39.0983u =  0.932581  38.963707 u  +  0.000117  39.963999 u  + 0.067302  mass of 41K  = 36.3368 u + 0.00468 u + 0.067302  mass of 41K mass of 41K = 48.   39.0983u   36.3368 u + 0.00468 u  = 40.962 u 0.067302 (M) We use the expression for determining the weighted-average atomic mass, where x represents the fractional abundance of 10 B and 1  x  the fractional abundance of 11 B 10.81u = 10.012937 u  x  + 11.009305  1  x  = 10.012937 x +11.009305  11.009305 x 10.81  11.009305 = 0.20 = 10.012937 x  11.009305 x = 0.996368 x 0.20 = 0.20 x= 0.996368  20% 10 B and 100.0  20 = 80% 11B Mass Spectrometry 49. (M) (a) (b) As before, we multiply each isotopic mass by its fractional abundance, after which, we sum these products to obtain the (average) atomic mass for the element.  0.205  70  +  0.274  72  +  0.078  73 +  0.365  74  +  0.078  76  = 14 + 20. + 5.7 + 27 + 5.9 = 72.6 = average atomic mass of germanium The result is only approximately correct because the isotopic masses are given to only two significant figures. Thus, only a two-significant-figure result can be quoted. 42 M02_PETR5044_11_CSM_C02.pdf 16 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 50. (M) (a) Six unique HCl molecules are possible (called isotopomers): 1 35 H Cl, 2 H 35Cl, 3 H 35Cl, 1H 37 Cl, 2 H 37 Cl, and 3 H 37 Cl The mass numbers of the six different possible types of molecules are obtained by summing the mass numbers of the two atoms in each molecule: H 35 Cl has A  36 1 H 37 Cl has A  38 H 35 Cl has A  37 2 H 37 Cl has A  39 1 (b) 2 H 35 Cl has A  38 3 H 37 Cl has A  40 3 The most abundant molecule contains the isotope for each element that is most abundant. It is 1 H 35Cl . The second most abundant molecule is 1 H 37 Cl . The relative abundance of each type of molecule is determined by multiplying together the fractional abundances of the two isotopes present. Relative abundances of the molecules are as follows. 1 H 35 Cl : 75.76% 2 H 35 Cl : 0.011% 1 H 35 Cl :  0.0008% 3 3 H 37 Cl : 24.23% 2 H 37 Cl : 0.0036% H 37 Cl :  0.0002% The Periodic Table 51. 52. (E) (a) Ge is in group 14 and in the fourth period. (b) Other elements in group 16(6A) are similar to S: O, Se, and Te. Most of the elements in the periodic table are unlike S, but particularly metals such as Na, K, and Rb. (c) The alkali metal (group 1), in the fifth period is Rb. (d) The halogen (group 17) in the sixth period is At. (E) (a) Au is in group 11 and in the sixth period. (b) Ar, Z = 18, is a noble gas. Xe is a noble gas with atomic number (54) greater than 50. (c) If an element forms a stable anion with charge 2–, it is in group 16. (d) If an element forms a stable cation with charge 3+, it is in group 13. 43 M02_PETR5044_11_CSM_C02.pdf 17 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 53. (E) If the seventh period of the periodic table is 32 members long, it will be the same length as the sixth period. Elements in the same family (vertical group), will have atomic numbers 32 units higher. The noble gas following radon will have atomic number = 86 + 32 = 118. The alkali metal following francium will have atomic number = 87 + 32 = 119. 54. (M) There are several interchanges: Ar/K, Co/Ni, Te/I, Th/Pa, U/Np, Pu/Am, Sg/Bh The reverse order is necessary because the periodic table lists elements in order of increasing atomic number (protons in the nucleus) and not in order of increasing atomic masses. The Avogadro Constant and the Mole 55. (E) 6.022  1023 atoms Fe = 9.51  1024 atoms Fe 1 mol Fe 6.022  1023 atoms Ag (b) atoms of Ag = 0.000467 mol Ag  = 2.81  1020 atoms Ag 1 mol Ag 6.022  1023 atoms Na = 5.1  1013 atoms Na (c) atoms of Na = 8.5  1011 mol Na  1 mol Na (a) atoms of Fe = 15.8 mol Fe  56. (E) Since the molar mass of nitrogen is 14.0 g/mol, 25.0 g N is almost two moles (1.79 mol N), while 6.022 1023 Ni atoms is about one mole, and 52.0 g Cr (52.00 g/mol Cr) is also almost one mole. Finally, 10.0 cm 3 Fe (55.85 g/mol Fe) has a mass of about 79 g, and contains about 1.4 moles of atoms. Thus, 25.0 g N contains the greatest number of atoms. Note: Even if you take nitrogen as N2, the answer is the same. 57. (E) (a) moles of Zn = 415.0 g Zn  1 mol Zn = 6.347 mol Zn 65.38 g Zn (b) # of Cr atoms = 147, 400 g Cr  1 mol Cr 6.022  1023 atoms Cr  51.996 g Cr 1 mol Cr  1.707  1027 atoms Cr 1 mol Au 196.97 g Au   3.3  10 10 g Au (c) mass Au  1.0  1012 atoms Au  23 6.022  10 atoms Au 1 mol Au 1 mol F 18.998 g F (d) mass of F atom  1 atom F    3.1547  1023 g F 23 6.022  10 atoms F 1 mol F For exactly 1 F atom, the number of sig figs in the answer is determined by the least precise number in the calculation, namely the mass of fluorine. 44 M02_PETR5044_11_CSM_C02.pdf 18 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 58. (E) (a) number Kr atoms = 5.25 mg Kr  1 g Kr 1 mol Kr 6.022  1023 atoms Kr   1000 mg Kr 83.80 g Kr 1 mol Kr = 3.77  1019 atoms Kr (b) Molar mass is defined as the mass per mole of substance. mass = 2.09 g. This calculation requires that the number of moles be determined. 1 mol moles = 2.80 1022 atoms  = 0.0465 mol 6.022  1023 atoms mass 2.09 g molar mass = = = 44.9 g/mol The element is Sc, scandium. moles 0.0465 mol 1 mol Mg 1 mol P 30.9738 g P (c) mass P = 44.75 g Mg     57.03 g P 24.305 g Mg 1 mol Mg 1 mol P Note: The same answer is obtained if you assume phosphorus is P4 instead of P. 59. (E) Determine the mass of Cu in the jewelry, then convert to moles and finally to the number of atoms. If sterling silver is 92.5% by mass Ag, it is 100 – 92.5 = 7.5% by mass Cu. Conversion pathway approach: number of Cu atoms  33.24 g sterling  7.5 g Cu 1 mol Cu 6.022  1023 atoms Cu   100.0 g sterling 63.546 g Cu 1 mol Cu  2.4  1022 Cu atoms Stepwise approach: 7.5 g Cu  2.493 g Cu 100.0 g sterling 1 mol Cu  0.03923 mol Cu 2.493 g Cu  63.546 g Cu 33.24 g sterling  0.03923 mol Cu  60. 6.022  1023 atoms Cu = 2.4  1022 Cu atoms 1 mol Cu (E) We first need to determine the amount in moles of each metal. 9.4 g solder 67 g Pb 1 mol Pb amount of Pb = 50.0 cm3 solder    = 1.5 mol Pb 3 1 cm 100 g solder 207.2 g Pb 9.4 g solder 33 g Sn 1 mol Sn amount of Sn = 50.0 cm3 solder    = 1.3 mol Sn 3 1 cm 100 g solder 118.7 g Sn total atoms = 1.5 mol Pb +1.3 mol Sn   6.022  1023 atoms = 1.7  1024 atoms 1 mol 45 M02_PETR5044_11_CSM_C02.pdf 19 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 61. (E) We first need to determine the number of Pb atoms of all types in 215 mg of Pb, and then use the percent isotopic abundance to determine the number of 204 Pb atoms present. 1g 1 mol Pb 6.022  1023atoms 14 204 Pb atoms 204 Pb atoms  215 mg Pb     1000 mg 207.2 g Pb 1 mol Pb 1000 Pb atoms  8.7  1018 atoms 204 Pb 62. (E) mass of alloy = 7.25  1023 Cd atoms  1 mol Cd 112.4 g Cd 100.0 g alloy   23 6.022  10 Cd atoms 1 mol Cd 8.0 g Cd = 1.7  103 g alloy 63. (E) We will use the average atomic mass of lead, 207.2 g/mol, to answer this question. 30 g Pb 1 dL 1 g Pb 1 mol Pb (a)   6   1.45  106 mol Pb / L 1 dL 0.1 L 10 g Pb 207.2 g 1.45  106 mol Pb 1L 6.022  1023 atoms (b)    8.7  1014 Pb atoms / mL L 1000 mL 1 mol (M) The concentration of Pb in air provides the principal conversion factor. Other conversion factors are needed to convert to and from its units, beginning with the 0.500 L volume, and ending with the number of atoms. 1 m3 3.11 g Pb 1 g Pb 1 mol Pb 6.022  1023 Pb atoms no. Pb atoms  0.500 L    6   1000 L 1 m3 10 g Pb 207.2 g Pb 1 mol Pb 64.  4.52  1012 Pb atoms 65. (M) To answer this question, we simply need to calculate the ratio of the mass (in grams) of each sample to its molar mass. Whichever elemental sample gives the largest ratio will be the one that has the greatest number of atoms. (a) Iron sample: 10 cm  10 cm  10 cm  7.86 g cm–3 = 7860 g Fe 1 mol Fe 7860 g Fe  = 141 mol of Fe atoms 55.845 g Fe 1.00  103g H 2 (b) Hydrogen sample:  1 mol H = 496 mol of H2 molecules = 2  (1.008 g H) 992 mol of H atoms 4 2.00  10 g S (c) Sulfur sample:  1 mol S = 624 mol of S atoms 32.06 g S 454 g Hg 1 mol Hg  = 172 mol of Hg atoms (d) Mercury sample: 76 lb Hg  1 lb Hg 200.6 g Hg Clearly, then, it is the 1.00 kg sample of hydrogen that contains the greatest number of atoms. 46 M02_PETR5044_11_CSM_C02.pdf 20 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 66. (M) (a) 23 g Na = 1 mol with a density ~ 1 g/cm3. 1 mole = 23 g, so volume of 25.5 mol ~ 600 cm3. (b) Liquid bromine occupies 725 mL or 725 cm3 (given). (c) 1.25 × 1025 atoms Cr is ~20 moles. At ~50 g/mol, this represents approximately 1000 g. Given the density of 9.4 g/cm3, this represents about 100 cm3 of volume. (d) 2150 g solder at 9.4 g/cm3 represents approximately 200 cm3. From this we can see that the liquid bromine would occupy the largest volume. INTEGRATIVE AND ADVANCED EXERCISES 67. (M) 0.9922 g = 2.50 g + 99.22 g =101.72 g 1mL 1.0085g mass of solution (20 C) = 100 mL × = 100.85g 1mL solid crystallized = total mass(40 C) – solution mass(20C) = 101.72 g – 100.85 g = 0.87 g (a) total mass(40 C) = 2.50 g + 100.0 mL × (b) The answer cannot be more precise because both the initial mass and the subtraction only allows the reporting of masses to + 0.01 g. For a more precise answer, more significant figures would be required for the initial mass (2.50 g) and the densities of the water and solution. 68. (M) We now recognize that the values of 24.3 u and 35.3 u for the masses of Mg and Cl represent a weighted-average that considers the mass and abundance of each isotope. The experimental mass of each isotope is very close to the natural number and therefore very close to an integer multiple of the mass of 1H, thus supporting Prout’s hypothesis. 69. (M) Each atom of 19 F contains 9 protons (1.0073 u each), 10 neutrons (1.0087 u each), and 9 electrons (0.0005486 u each). The mass of each atom should be the sum of the masses of these particles.  1.0073 u   1.0087 u   0.0005486 u  Total mass   9 protons   10 neutrons    9 electrons    1 proton   1 neutron   1 electron   = 9.0657 u + 10.087 u + 0.004937 u = 19.158 u This compares with a mass of 18.9984 u given in the periodic table. The difference, 0.160 u per atom, is called the mass defect and represents the energy that holds the nucleus together, the nuclear binding energy. This binding energy is released when 9 protons and 9 neutrons fuse to give a fluorine-19 nucleus. 70. (M) 4 volume of nucleus(single proton)   r 3  1.3333  3.14159  (0.5  10 13 cm)3  5  10 40 cm3 3 24 1.673  10 g density   3  1015 g/cm 3  40 3 5  10 cm 47 M02_PETR5044_11_CSM_C02.pdf 21 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 71. (M) This method of establishing the Avogadro constant uses fundamental constants. 1 mol 12 C 6.022  1023 12 C atoms 12 u 12   12  6.022  1023 u 1.000 g C  12 12 12 g C 1 mol C 1 C atom 72. (M) Let Z = # of protons, N = # of neutrons, E = # of electrons, and A = # of nucleons = Z + N. (a) Z + N = 234 The mass number is 234 and the species is an atom. N = 1.600 Z The atom has 60.0% more neutrons than protons. Next we will substitute the second expression into the first and solve for Z. Z  N  234  Z  1.600 Z  2.600 Z 234  90 protons Z 2.600 Thus this is an atom of the isotope 234 Th. (b) Z = E + 2 The ion has a charge of +2. Z = 1.100 E There are 10.0% more protons than electrons. By equating these two expressions and solving for E, we can find the number of electrons. E  2  1.100 E 2 E  20 electrons Z  20  2  22, (titanium). 2  1.100 E  E  0.100 E 0.100 The ion is Ti 2  . There is not enough information to determine the mass number. (c) Z + N = 110 The mass number is 110. Z=E+2 The species is a cation with a charge of +2. N = 1.25 E Thus, there are 25.0% more neutrons than electrons. By substituting the second and third expressions into the first, we can solve for E, the number of electrons. 108 E  48 ( E  2)  1.25 E  110  2.25 E  2 108  2.25 E 2.25 Then Z  48  2  50, (the element is Sn) N  1.25  48  60 Thus, it is 110Sn 2 . 73. (E) Because the net ionic charge (+ 2) is one-tenth of its the nuclear charge, the nuclear charge is + 20. This is also the atomic number of the nuclide, which means the element is calcium. The number of electrons is 20 for a neutral calcium atom, but only 18 for a calcium ion with a 2 net 2+ charge. Four more than 18 is 22 neutrons. The ion described is 42 20 Ca . 74. (M) A = Z + N = 2.50 Z The mass number is 2.50 times the atomic number. The neutron number of selenium-82 equals 82 – 34 = 48, since Z = 34 for Se. The neutron number of isotope Y also equals 48, which equals 1.33 times the atomic number of isotope Y. 48  36 Thus 48  1.33  Z Y Z Y  1.33 The mass number of isotope Y = 48 + 36 = 84 = the atomic number of E, and thus, the element is Po. Thus, from the relationship in the first line, the mass number of E  2.50 Z  2.50  84  210 The isotope E is 210 Po . 48 M02_PETR5044_11_CSM_C02.pdf 22 1/29/16 10:03:57 PM Chapter 2: Atoms and the Atomic Theory 75. As a result of the redefinition, all masses will decrease by a factor of 35.00000/35.453 = 0.98722. (a) atomic mass of He atomic mass of Na atomic mass of I 4.00260  0.98722  3.9515 22.9898  0.98722  22.696 126.905  0.98722  125.28 (b) These three elements have nearly integral atomic masses based on C-12 because these three elements and C-12 all consist mainly of one stable isotope, rather than a mixture of two or more stable isotopes, with each being present in significant amounts (10% or more), as is the case with chlorine. 76. (M) To solve this question, represent the fractional abundance of 14 N by x and that of 14 N by (1 – x). Then use the expression for determining average atomic mass. 14.0067  14.0031 x  15.0001(1  x) 14.0067  15.0001  14.0031 x  15.0001 x OR 0.9934  0.9970 x 0.9934  100%  99.64%  percent isotopic abundance of 14 N. 0.9970 Thus, 0.36%  percent isotopic abundance of 15 N. x 77. (D) In this case, we will use the expression for determining average atomic mass—the sum of products of nuclidic mass times fractional abundances (from Figure 2-14)—to answer the question. 196 Hg: 195.9658 u  0.00146  0.286 u 198 199 197.9668 u  0.1002  19.84 u 198.9683 u  0.1684  33.51 u Hg : Hg : 200 201 200.9703 u  0.1322  26.57 u Hg : 199.9683 u  0.2313  46.25 u Hg : 202 204 201.9706 u  0.2980  60.19 u 203.9735 u  0.0685  14.0 u Hg : Hg : Atomic weight = 0.286 u + 19.84 u + 33.51 u + 46.25 u + 26.57 u + 60.19 u + 14.0 u = 200.6 u 78. (D) The sum of the percent abundances of the two minor isotopes equals 100.00% – 84.68% = 15.32%. Thus, we denote the fractional abundance of 73Ge as x, and the other as (0.1532 – x). These fractions are then used in the expression for average atomic mass. atomic mass  72.64 u  (69.92425 u  0.2085)  (71.92208 u  0.2754)  (73.92118 u  0.3629)  (72.92346 u  x) + [75.92140 u  (0.1532  x)] 72.64 u  (14.5792 u)  (19.8073 u)  (26.8260 u)  (72.92346 u x)  [11.6312 u  75.92140 u x)] 0.2037  2.99794x Hence: x  0.068. 73 Ge has 6.8% isotopic abundance and 76Ge has 8.5% isotopic abundance. From the calculations, we can see that the number of significant figures drops from four in the percent isotopic abundances supplied to only two significant figures owing to the imprecision in the supplied values of the percent isotopic abundance. 49 M02_PETR5044_11_CSM_C02.pdf 23 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 79. (D) First, it must be understood that because we don’t know the exact percent abundance of 84 Kr, all the percent abundances for the other isotopes will also be approximate. From the question, we may initially infer the following:  Assume percent abundance of 84Kr ~ 55% as a start (somewhat more than 50)  Let percent abundance of 82Kr = x%; percent abundance 83Kr ~ 82Kr = x%  86Kr = 1.50(percent abundance of 82Kr) = 1.50(x%)  80Kr = 0.196(percent abundance of 82Kr) = 0.196(x%)  78Kr = 0.030(percent abundance of 82Kr) = 0.030(x%) 100% = %78Kr + %80Kr + %82Kr + %83Kr + %84Kr + %86Kr 100% = 0.030(x%) + 0.196(x%) + x% + x% + %84Kr + 1.50(x%) 100% = 3.726(x%) + %84Kr Assuming percent abundance of 84Kr is 55%, solving for x gives a value of 12.1% for percent abundance of 82Kr, from which the remaining abundances can be calculated based on the above relationships, as shown below: 78 Kr: 0.03 × 12.1 = 0.363%; 80Kr: 0.196 × 12.1 = 2.37%; 83Kr: same as 82Kr; 86Kr: 1.5 × 12.1 = 18.15%. The weighted-average isotopic mass calculated from the above abundances is as follows: Weighted-average isotopic mass = 0.030(12.1%)(77.9204 u) + 0.196(12.1%)(79.9164 u) + 12.1%(81.9135 u) + 12.1%(82.9141 u) + 55%(83.9115 u) + 1.50(12.1%)(85.9106 u) = 83.8064 u As stated above, the problem here is the inaccuracy of the percent abundance for 84Kr, which is crudely estimated to be ~ 55%. If we vary this percentage, we vary the relative abundance of all other isotopes accordingly. Since we know the weighted-average atomic mass of Kr is 83.80, we can try different values for 84Kr abundance and figure out which gives us the closest value to the given weighted-average isotopic mass: Percent Weighted-Average Abundance 84Kr Isotopic Mass 50% 83.793 51% 83.796 52% 83.799 53% 83.801 54% 83.803 55% 83.806 From this table, we can see that the answer is somewhere between 52% and 53%. 50 M02_PETR5044_11_CSM_C02.pdf 24 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 80. (D) Four molecules are possible, given below with their calculated molecular masses. 35 Cl 79Br mass = 34.9689 u + 78.9183 u = 113.8872 u 35 Cl 81Br mass = 34.9689 u + 80.9163 u = 115.8852 u 37 Cl 79Br mass = 36.9658 u + 78.9183 u = 115.8841 u 37 Cl 81Br mass = 36.9658 u + 80.9163 u = 117.8821 u Each molecule has a different intensity pattern (relative number of molecules), based on the isotopic abundance of the isotopes making up each molecule. If we divide all of the values by the lowest ratio, we can get a better idea of the relative ratio of each molecule. 35 Cl 79Br Intensity = (0.7577) × (0.5069 ) = 0.3841  0.1195 = 3.214 35 Cl 81Br Intensity = (0.7577) × (0.4931 ) = 0.3736  0.1195 = 3.127 37 Cl 79Br Intensity = (0.2423) × (0.5069 ) = 0.1228  0.1195 = 1.028 37 Cl 81Br Intensity = (0.2423) × (0.4931 ) = 0.1195  0.1195 = 1.000 A plot of intensity versus molecular mass reveals the following pattern under ideal circumstances (high resolution mass spectrometry). 81. (M) Let’s begin by finding the volume of copper metal. 2.54 cm = 0.08118 cm wire diameter (cm) = 0.03196 in.  1 in. The radius is 0.08118 cm  1/2 = 0.04059 cm The volume of Cu(cm3) =  r2 h = 3.1416  (0.04059 cm)2  150 cm = 0.7764 cm3 8.92 g Cu So, the mass of Cu = 0 cm3  = 6.93 g Cu 1 cm3 1 mol Cu The number of moles of Cu = 6.93 g Cu  = 0.109 mol Cu 63.546 g Cu 6.022  1023 atoms Cu Cu atoms in the wire = 0.109 mol Cu  = 6.56  1022 atoms 1 mol Cu 51 M02_PETR5044_11_CSM_C02.pdf 25 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 82. (D) volume = l ´ w´ h – p r 2 h 2 æ 2.50 cm ÷ö = (15.0 cm ´12.5 cm ´ 0.300 cm) – (3.1416) ´çç ´(0.300 cm) çè 2 ÷÷ø = (56.25 cm -1.47 cm) = 54.8 cm3 mass of object = 54.8 cm3 ´ 8.80 g = 482 g Monel metal 1 cm3 Then determine the number of silicon atoms in this quantity of alloy. 2.2´10-4 g Si 1 mol Si 6.022´1023 Si atoms 482 g Monel metal´ ´ ´ = 2.3´1021 Si atoms 1.000 g metal 28.05 g Si 1 mol Si Finally, determine the number of 30 Si atoms in this quantity of silicon. æ 3.10 30 Si atoms ö÷ ÷ = 7.1´1019 30 Si number of Si atoms = (2.3´10 Si atoms)´çç çè 100 Si atoms ø÷÷ 30 83. 21 (M) The percent isotopic abundance of deuterium of 0.015% means that, in a sample of 100,000 H atoms, only 15 2H are present. mass H 2 = 2.50×10 21 atoms 2 H × 100,000 H atoms 2 15 atoms H × 1 mol H 23 6.022×10 H atoms × 1 mol H 2 2 mol H × 2.016 g H 2 1 mol H 2 mass H 2 = 27.9 g H 2 is required such that we have 2.50×1021 atoms 2 H. 84. (M) The numbers sum to 21 (= 10 + 6 + 5). Thus, in one mole of the alloy there is 10 21 mol Bi, 6 21 mol Pb, and 5 21 mol Sn. The mass of this mole of material is figured in a similar fashion to computing a weighted-average atomic mass from isotopic masses.  10 209.0 g   6 207.2 g   5 118.7     mol Pb     mol Sn   mass of alloy   mol Bi  1 mol Bi   21 1 mol Pb   21 1 mol Sn   21  99.52 g Bi  59.20 g Pb  28.26 g Sn  186.98 g alloy 85. (M) The atom ratios are of course, the same as the mole ratios. We first determine the mass of alloy that contains 5.00 mol Ag, 4.00 mol Cu, and 1.00 mol Zn. 107.87 g Ag 63.546 g Cu 65.38g Zn  4.00 mol Cu   1.00 mol Zn  1mol Ag 1mol Cu 1mol Zn  539.4 g Ag  254.2 g Cu  65.38g Zn  859 g alloy mass  5.00 mol Ag  52 M02_PETR5044_11_CSM_C02.pdf 26 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory Then, for 1.00 kg of the alloy, (1000 g), we need 1000/859 moles. (859 g is the alloy’s “molar mass.”) 539 g Ag mass of Ag  1000 g alloy   627 g Ag 859 g alloy 254 g Cu mass of Cu  1000 g alloy   296 g Cu 859 g alloy 65.4 g Zn mass of Zn  1000 g alloy   76.1 g Zn 859 g alloy 86. (M) The relative masses of Sn and Pb are 207.2 g Pb (assume one mole of Pb) to (2.73 × 118.71 g/mol Sn =) 324 g Sn. Then the mass of cadmium, on the same scale, is 207.2/1.78 = 116 g Cd. 324 g Sn 324 g Sn % Sn   100%   100%  50.1%Sn 207.2  324  116 g alloy 647 g alloy % Pb  207.2 g Pb  100%  32.0% Pb 647 g alloy % Cd  116 g Cd  100%  17.9% Cd 647 g alloy 87. (M) We need to apply the law of conservation of mass and convert volumes to masses: Calculate the mass of zinc: 125 cm3 × 7.13 g/cm3 = 891 g Calculate the mass of iodine: 125 cm3 × 4.93 g/cm3 = 616 g Calculate the mass of zinc iodide: 164 cm3 × 4.74 g/cm3 = 777 g (891 + 616 – 777) g = 730 g Calculate the mass of zinc unreacted: Calculate the volume of zinc unreacted: 730 g × 1cm3 / 7.13 g = 102 cm3 88. (D) First, calculate the total number of Ag atoms in a 1 cm3 crystal:. 1 mol Ag 6.02  1023 Ag atoms 10.5 g Ag    5.86  1022 atoms of Ag 107.87 g Ag 1 mol Ag The actual volume taken up by the Ag atoms (considering that there is 26% empty space in the crystal) is: 1 cm3 × 0.74 = 0.74 cm3 Therefore, the volume of each atom is: 0.74 cm3 = 1.263 × 10–23 cm3 5.86  1022 atoms Volume of a sphere is expressed as V = 1.263 × 10–23 cm3 = (4/3) π r3 Solving for r, we get 1.44 × 10–8 cm or 144 pm. 53 M02_PETR5044_11_CSM_C02.pdf 27 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory FEATURE PROBLEMS 89. (M) The product mass differs from that of the reactants by  5.62  2.50 =  3.12 grains. In order to determine the percent gain in mass, we need to convert the reactant mass to grains. 8 gros 72 grains 13 onces  = 104 gros  104 + 2  gros  = 7632 grains 1 once 1 gros 3.12 grains increase % mass increase   100%  0.0409% mass increase 7632 + 2.50 grains original The sensitivity of Lavoisier’s balance can be as little as 0.01 grain, which seems to be the limit of the readability of the balance; alternatively, it can be as large as 3.12 grains, which assumes that all of the error in the experiment is due to the (in)sensitivity of the balance. Let us convert 0.01 grains to a mass in grams. 1 gros 1 once 1 livre 30.59 g     3  105 g  0.03 mg minimum error  0.01 gr  72 gr 8 gros 16 once 1 livre 3  10 5 g maximum error  3.12 gr   9  10 3 g  9 mg 0.01 gr The maximum error is close to that of a common modern laboratory balance, which has a sensitivity of 1 mg. The minimum error is approximated by a good quality analytical balance. Thus we conclude that Lavoisier’s results conform closely to the law of conservation of mass. 90. (D) One way to determine the common factor of which all 13 numbers are multiples is to first divide all of them by the smallest number in the set. The ratios thus obtained may be either integers or rational numbers whose decimal equivalents are easy to recognize. 1 2 3 4 5 6 7 8 9 10 11 12 13 Obs. Quan. 19.6 24.60 29.62 34.47 39.38 44.42 49.41 53.91 59.12 63.68 68.65 78.34 83.22 Ratio 1.00 1.251 1.507 1.753 2.003 2.259 2.513 2.742 3.007 3.239 3.492 3.984 4.233 Mult. 4.00 5.005 6.026 7.013 8.012 9.038 10.05 10.97 12.03 12.96 13.97 15.94 16.93 Int. 4 5 6 7 8 9 10 11 12 13 14 16 17 The row labeled “Mult.” is obtained by multiplying the row “Ratio” by 4.000. In the row labeled “Int.” we give the integer closest to each of these multipliers. It is obvious that each of the 13 measurements is exceedingly close to a common quantity multiplied by an integer. 54 M02_PETR5044_11_CSM_C02.pdf 28 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 91. (M) In a 60-year-old chemistry textbook, the atomic mass for oxygen would be exactly 16 u because chemists assigned precisely 16 u as the atomic mass of the naturally occurring mixture of oxygen isotopes. This value is slightly higher than the value of 15.9994 in modern chemistry textbooks. Thus, we would expect all other atomic masses to be slightly higher as well in the older textbooks. 92. (D) We begin with the amount of reparations and obtain the volume in cubic kilometers with a series of conversion factors. Conversion pathway approach: 1troy oz Au 31.103g Au 1mol Au 6.022  1023 atoms Au 9 V  $28.8  10     $21.25 1troy oz Au 196.97 g Au 1mol Au 1ton seawater 2000 lb seawater 453.6 g seawater 1cm 3 seawater     4.67  1017 Au atoms 1ton seawater 1 lb seawater 1.03g seawater 3  1m 1km     2.43  105 km3   100 cm 1000 m  Stepwise approach: $28.8 109  1troy oz Au 31.103g Au   4.22  1010g Au $21.25 1troy oz Au 4.22 1010 g Au  1mol Au 6.022  1023 atoms Au   1.29  1032 atoms Au 196.97 g Au 1mol Au 1.29 1032 atoms Au  1ton seawater 2000 lb seawater  5.52 1017 lb seawater  17 4.67 10 Au atoms 1ton seawater 5.52 1017 lb seawater  453.6 g seawater 1cm 3 seawater   2.43 1020cm3 seawater 1 lb seawater 1.03g seawater 3  1m 1km  5 3  2.43 10 cm seawater     2.43 10 km 100 cm 1000 m   20 3 93. (D) We start by using the percent isotopic abundances for 87Rb and 85Rb along with the data in the “spiked” mass spectrum to find the total mass of Rb in the sample. Then, we calculate the Rb content in the rock sample in ppm by mass by dividing the mass of Rb by the total mass of the rock sample, and then multiplying the result by 106 to convert to ppm. 87 Rb = 27.83% isotopic abundance 85Rb = 72.17% isotopic abundance 87 Rb(isotopic) 27.83% = = 0.3856 Therefore, 85 Rb(isotopic) 72.17% For the 87Rb(spiked) sample, the 87Rb peak in the mass spectrum is 1.12 times as tall as the 87 Rb(isotopic)  87 Rb(spiked) 85 Rb peak. Thus, for this sample = 1.12 85 Rb(isotopic) 55 M02_PETR5044_11_CSM_C02.pdf 29 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory Using this relationship, we can now find the masses of both 85Rb and 87Rb in the sample. 87 87 Rb(isotopic) Rb(isotopic) 85 So, = 0.3856; Rb(isotopic) = 85 Rb(isotopic) 0.3856 1.12  87 Rb (isotopic) = 2.905 87Rb(isotopic) 0.3856 87 Rb(spiked) = 1.905 87Rb(isotopic) 87 87 Rb(isotopic)  87 Rb(spiked) Rb(isotopic)  87 Rb(spiked) = and = 1.12 87 85 Rb(isotopic) Rb(isotopic) 0.3856 Since the mass of 87Rb(spiked) is equal to 29.45 g, the mass of 87Rb(isotopic) must be 29.45 g = 15.46 g of 87Rb(isotopic) 1.905 15.46 g of 87 Rb(isotopic) = 40.09 g of 85Rb(isotopic) So, the mass of 85Rb(isotopic) = 0.3856 Therefore, the total mass of Rb in the sample = 15.46 g of 87Rb(isotopic) + 40.09 g of 85 Rb(isotopic) = 55.55 g of Rb. Convert to grams: 1 g Rb = 5.555  10–5 g Rb = 55.55 g of Rb  1  106 μg Rb 87 Rb(isotopic) + 87Rb(spiked) = Rb content (ppm) = 5.555  10 5 g Rb  106 = 159 ppm Rb 0.350 g of rock SELF-ASSESSMENT EXERCISES 94. (E) (a) A Z E : The element “E” with the atomic number of Z (i.e., Z protons in the nucleus) and atomic mass of A (i.e., total of protons and neutrons equals A). (b) β particle: An electron produced as a result of the decay of a neutron (c) Isotope: Nuclei that share the same atomic number but have different atomic masses (d) 16 (e) Molar mass: Mass of one mole of a substance O: An oxygen nucleus containing 8 neutrons 56 M02_PETR5044_11_CSM_C02.pdf 30 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 95. (E) (a) The total mass of substances present after the chemical reaction is the same as the total mass of substances before the chemical reaction. More universally, mass is neither created nor destroyed, but converts from one form to another. (b) Rutherford’s model of the atom postulates the existence of positively charged fundamental particles at the nucleus of the atom. (c) An average value used to express the atomic mass of an element by taking into account the atomic masses and relative abundances of all the naturally occurring isotopes of the element. (d) A spectrum showing the mass/charge ratio of various atoms in a matrix 96. (E) (a) Cathode rays are beams of electrons being generated from a negatively charged surface (cathode) moving to a positively charged surface (anode). X-rays are high energy photons, which are typically generated when the high energy beam of electrons impinges on the anode. (b) Protons and neutrons are both fundamental particles that make up an atom’s nucleus. Protons are positively charged and neutrons have no charge. (c) Nuclear charge is determined by the numbers of protons in the nucleus. Ionic charge is determined by the difference between the number of protons and electrons in the atom. (d) Periods are horizontal rows in the periodic table, while groups are vertical columns. (e) Metals are generally characterized by their malleability, ductility, and ability to conduct electricity and heat well. Nonmetals are generally brittle and nonconductive. (f) The Avogadro constant is the number of elementary entities (atoms, molecules, etc.) in one mole of a substance. A mole is the amount of a substance that contains the same number of elementary entities as there are atoms in exactly 12 g of pure carbon-12. 97. 1  4  (E)  × 10.013 u     11.009 u   10.810 u. Therefore, the element is boron. 5  5  98. (E) The answer is (b). If all of the zinc reacts and the total amount of the product (zinc sulfide) is 14.9 g, then 4.9 g of S must have reacted with zinc. Therefore, 3.1 g of S remain. 99. (E) The answer is (d). It should be remembered that atoms combine in ratios of whole numbers. Therefore: (a) 16 g O × (1 mol O/16 g O) = 1 mol O, and 85.5 g Rb × (1 mol Rb/85.5 g Rb) = 1 mol Rb Therefore, the O: Rb ratio is 1:1. (b) Same calculations as above give an O: Rb ratio of 0.5:0.5, or 1:1. (c) Same type calculation gives an O: Rb ratio of 2:1. Because all of the above combine in O and Rb in whole number ratios, they are all at least theoretically possible. 57 M02_PETR5044_11_CSM_C02.pdf 31 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 100. (E) If the compound contains 46.7% X, then the percent of oxygen would be 100.0% − 46.7 = 53.3%. Assuming the sample is 100.0 g, there would be 53.3 g of oxygen present.  1 mol O  53.3 g O    3.331 mol O  16.00 g O  The formula of the compound is XO; therefore, X and O are in a 1:1 mole ratio. 46.7 g X  14.0 g/mol X 3.331 mol X 101. (E) The answer is (c). Cathode rays are beams of electrons, and as such have identical properties to β particles, although they may not have the same energy. 102. (E) The answer is (a), that the greatest portion of the mass of an atom is concentrated in a small but positively charged nucleus. 103. (E) The answer is (d). A hydrogen atom has one proton and one electron, so its charge is zero. A neutron has the same charge as a proton, but is neutral. Since most of the mass of the atom is at the nucleus, a neutron has nearly the same mass as a hydrogen atom. 104. (E) All the choices in this question are fundamental particles (e). 105. (E) Dalton (d) is correct. 106. (E) There are no particles that have the same mass as the hydrogen atom and a negative charge. The correct answer is (e). 35 107. (E) 17 Cl 108. (E) The answer is (d), calcium, because they are in the same group. 109. (E) (a) Group 18, (b) Group 17, (c) Group 13 and Group 1 (d) Group 18 110. (E) (d) and (f) 111. (E) (c), because it is not close to being a whole number 112. (M) The answer is (d). Even with the mass scale being redefined based on 84Xe, the mass ratio between 12C and 84Xe will remain the same. Using 12C as the original mass scale, the mass ratio of 12C : 84Xe is 12 u/83.9115 u = 0.1430. Therefore, redefining the mass scale by assigning the exact mass of 84 u to 84Xe, the relative mass of 12C becomes 84 × 0.14301 = 12.0127 u. 58 M02_PETR5044_11_CSM_C02.pdf 32 1/29/16 10:03:58 PM Chapter 2: Atoms and the Atomic Theory 113. (M) The answer is (b) 1000 g 1 mol Fe   1.000  10 2 mol Fe 5.585 kg Fe  1 kg 55.85 g Fe 1 mol C  50.01 mol C 600.6 g C  12.01 g C Therefore, 100 moles of Fe has twice as many atoms as 50 moles of C. 114. (E) (d) is correct. mol 6.022  1023 atoms 22 electrons 91.84 g     2.542  10 25 electrons 47.867 g mol Ti atom 115. (M) 1 mol Fe  0.0417 mol Fe 55.8452 g Fe 1 mol O  0.0625 mol O 1.000 g O  15.999 g O Dividing the two mole values to obtain the mole ratio, we get: 0.0625/0.0417 = 1.50. That is, 1.50 moles (or atoms) of O per 1 mole of Fe, or 3 moles of O per 2 moles of Fe (Fe2O3). Performing the above calculations for a compound with 2.618 g of Fe to 1.000 g of O yields 0.0469 mol of Fe and 0.0625 mol of O, or a mole ratio of 1.333, or a 4:3 ratio (Fe3O4). 2.327 g Fe  116. (D) The weighted-average atomic mass of Sr is expressed as follows: atomic mass of Sr = 87.62 amu = 83.9134(0.0056) + 85.9093x + 86.9089[1 – (0.0056 + 0.8258 + x)] + 87.9056(0.8258) Rearrange the above equation and solve for x, which is 0.095 or 9.5%, which is the relative abundance of 86Sr. Therefore, the relative abundance of 87Sr is 0.0735 or 7.3%. The reason for the imprecision is the low number of significant figures for 84Sr. 117. (M) This problem lends itself well to the conversion pathway: 0.15 mg Au 1 ton seawater 1 kg 1.03 g seawater 250 mL     1 ton seawater 1000 kg 1000 mg 1 mL seawater sample 1 g Au 1 mol Au 6.02  1023 atoms     1.2  1014 atoms of Au 1000 mg Au 196.97 g Au 1 mol Au 118. (M) In sections 2-7 and 2-8, the simplest concept is the concept of mole. Mole is defined by the number of atoms in 12 g of 12C. Other topics emanate from this basic concept. Molar mass (weight) is defined in terms of moles, as is mole ratios. Percent isotopic abundance is another topic defined directly by the concept of moles. 59 M02_PETR5044_11_CSM_C02.pdf 33 1/29/16 10:03:58 PM