Atoms, Molecules, and Ions

CHAPTER 2 ATOMS, MOLECULES, AND IONS 2.7 First, convert 1 cm to picometers. 1 cm × 0.01 m 1 pm × = 1 × 1010 pm 1 cm 1 × 10−12 m ? He atoms = (1 × 1010 pm) × 2.8 1 He atom 2 1 × 10 pm Note that you are given information to set up the unit factor relating meters and miles. ratom = 104 rnucleus = 104 × 2.0 cm × 2.13 = 1 × 108 He atoms 1m 1 mi × = 0.12 mi 100 cm 1609 m For iron, the atomic number Z is 26. Therefore the mass number A is: A = 26 + 28 = 54 2.14 Strategy: The 239 in Pu-239 is the mass number. The mass number (A) is the total number of neutrons and protons present in the nucleus of an atom of an element. You can look up the atomic number (number of protons) on the periodic table. Solution: mass number = number of protons + number of neutrons number of neutrons = mass number − number of protons = 239 − 94 = 145 2.15 2.16 Isotope No. Protons No. Neutrons 3 2 He 4 2 He 24 12 Mg 25 12 Mg 48 22Ti 79 35 Br 195 78 Pt 2 1 2 2 12 12 12 13 22 26 35 44 78 117 Isotope No. Protons No. Neutrons No. Electrons 15 7N 33 16 S 63 29 Cu 84 38 Sr 130 56 Ba 186 74W 202 80 Hg 7 8 7 16 17 16 29 34 29 38 46 38 56 74 56 74 112 74 80 122 80 23 11 Na (b) 64 28 Ni 2.17 (a) 2.18 The accepted way to denote the atomic number and mass number of an element X is as follows: A ZX where, A = mass number Z = atomic number 24 CHAPTER 2: ATOMS, MOLECULES, AND IONS (a) 186 74W 201 80 Hg (b) 2.23 Helium and Selenium are nonmetals whose name ends with ium. (Tellerium is a metalloid whose name ends in ium.) 2.24 (a) Metallic character increases as you progress down a group of the periodic table. For example, moving down Group 4A, the nonmetal carbon is at the top and the metal lead is at the bottom of the group. (b) Metallic character decreases from the left side of the table (where the metals are located) to the right side of the table (where the nonmetals are located). 2.25 The following data were measured at 20°C. 3 3 H2O (0.98 g/cm ) 3 Hg (13.6 g/cm ) (a) Li (0.53 g/cm ) K (0.86 g/cm ) (b) Au (19.3 g/cm ) 3 Pt (21.4 g/cm ) (c) Os (22.6 g/cm ) (d) Te (6.24 g/cm ) 3 3 3 3 2.26 F and Cl are Group 7A elements; they should have similar chemical properties. Na and K are both Group 1A elements; they should have similar chemical properties. P and N are both Group 5A elements; they should have similar chemical properties. 2.31 (a) (b) (c) This is a polyatomic molecule that is an elemental form of the substance. It is not a compound. This is a polyatomic molecule that is a compound. This is a diatomic molecule that is a compound. 2.32 (a) (b) (c) This is a diatomic molecule that is a compound. This is a polyatomic molecule that is a compound. This is a polyatomic molecule that is the elemental form of the substance. It is not a compound. 2.33 Elements: Compounds: 2.34 There are more than two correct answers for each part of the problem. (a) (d) N 2 , S8 , H 2 NH3, NO, CO, CO2, SO2 H2 and F2 (b) H2O and C12H22O11 (sucrose) + Na 11 10 2+ Ca 20 18 HCl and CO 3+ Al 13 10 2+ Fe 26 24 (c) − I 53 54 − F 9 10 S8 and P4 2− S 16 18 2− O 8 10 3− 2.35 Ion No. protons No. electrons N 7 10 2.36 The atomic number (Z) is the number of protons in the nucleus of each atom of an element. You can find this on a periodic table. The number of electrons in an ion is equal to the number of protons minus the charge on the ion. number of electrons (ion) = number of protons − charge on the ion CHAPTER 2: ATOMS, MOLECULES, AND IONS + Ion No. protons No. electrons K 19 18 Mg 12 10 2+ 3+ Fe 26 23 − Br 35 36 Mn 25 23 2+ 4− C 6 10 Cu 29 27 25 2+ 2.43 (a) (b) (c) (d) Sodium ion has a +1 charge and oxide has a −2 charge. The correct formula is Na2O. The iron ion has a +2 charge and sulfide has a −2 charge. The correct formula is FeS. The correct formula is Co2(SO4)3 Barium ion has a +2 charge and fluoride has a −1 charge. The correct formula is BaF2. 2.44 (a) (b) (c) (d) The copper ion has a +1 charge and bromide has a −1 charge. The correct formula is CuBr. The manganese ion has a +3 charge and oxide has a −2 charge. The correct formula is Mn2O3. 2+ − We have the Hg2 ion and iodide (I ). The correct formula is Hg2I2. Magnesium ion has a +2 charge and phosphate has a −3 charge. The correct formula is Mg3(PO4)2. 2.45 (a) CN 2.46 Strategy: An empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms. Can you divide the subscripts in the formula by some factor to end up with smaller wholenumber subscripts? (b) CH (c) C9H20 (d) P2 O 5 (e) BH3 Solution: (a) (b) (c) (d) Dividing both subscripts by 2, the simplest whole number ratio of the atoms in Al2Br6 is AlBr3. Dividing all subscripts by 2, the simplest whole number ratio of the atoms in Na2S2O4 is NaSO2. The molecular formula as written, N2O5, contains the simplest whole number ratio of the atoms present. In this case, the molecular formula and the empirical formula are the same. The molecular formula as written, K2Cr2O7, contains the simplest whole number ratio of the atoms present. In this case, the molecular formula and the empirical formula are the same. 2.47 The molecular formula of glycine is C2H5NO2. 2.48 The molecular formula of ethanol is C2H6O. 2.49 Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually molecular. Ionic: Molecular: 2.50 LiF, BaCl2, KCl SiCl4, B2H6, C2H4 Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually molecular. Ionic: Molecular: NaBr, BaF2, CsCl. CH4, CCl4, ICl, NF3 26 CHAPTER 2: ATOMS, MOLECULES, AND IONS 2.57 (a) (b) (c) (d) (e) (f) (g) 2.58 Strategy: When naming ionic compounds, our reference for the names of cations and anions is Table 2.3 of the text. Keep in mind that if a metal can form cations of different charges, we need to use the Stock system. In the Stock system, Roman numerals are used to specify the charge of the cation. The metals that + have only one charge in ionic compounds are the alkali metals (+1), the alkaline earth metals (+2), Ag , 2+ 2+ 3+ Zn , Cd , and Al . potassium dihydrogen phosphate potassium hydrogen phosphate hydrogen bromide (molecular compound) hydrobromic acid lithium carbonate potassium dichromate ammonium nitrite (h) (i) (j) (k) (l) (m) iodic acid phosphorus pentafluoride tetraphosphorus hexoxide cadmium iodide strontium sulfate aluminum hydroxide When naming acids, binary acids are named differently than oxoacids. For binary acids, the name is based on the nonmetal. For oxoacids, the name is based on the polyatomic anion. For more detail, see Section 2.7 of the text. Solution: + (a) This is an ionic compound in which the metal cation (K ) has only one charge. The correct name is potassium hypochlorite. Hypochlorite is a polyatomic ion with one less O atom than the chlorite ion, − ClO2 (b) silver carbonate (c) This is an oxoacid that contains the nitrite ion, NO2 . The “-ite” suffix is changed to “-ous”. The correct name is nitrous acid. (d) potassium permanganate sulfate (g) This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Fe ion. Since the oxide ion has a −2 charge, the Fe ion has a +2 charge. The correct name is iron(II) oxide. (h) iron(III) oxide (i) This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Ti ion. Since each of the four chloride ions has a −1 charge (total of −4), the Ti ion has a +4 charge. The correct name is titanium(IV) chloride. (j) sodium hydride − (e) (k) cesium chlorate (f) lithium nitride (l) potassium ammonium sodium oxide + 2− (m) This is an ionic compound in which the metal cation (Na ) has only one charge. The O2 ion is called the peroxide ion. Each oxygen has a −1 charge. You can determine that each oxygen only has a −1 charge, because each of the two Na ions has a +1 charge. Compare this to sodium oxide in part (l). The correct name is sodium peroxide. 2.59 (a) (f) RbNO2 KH2PO4 (b) (g) K2S IF7 (c) (h) NaHS (NH4)2SO4 (d) (i) Mg3(PO4)2 AgClO4 (e) (j) CaHPO4 BCl3 2.60 Strategy: When writing formulas of molecular compounds, the prefixes specify the number of each type of atom in the compound. When writing formulas of ionic compounds, the subscript of the cation is numerically equal to the charge of the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges of the CHAPTER 2: ATOMS, MOLECULES, AND IONS 27 cation and anion are numerically equal, then no subscripts are necessary. Charges of common cations and anions are listed in Table 2.3 of the text. Keep in mind that Roman numerals specify the charge of the cation, not the number of metal atoms. Remember that a Roman numeral is not needed for some metal cations, + because the charge is known. These metals are the alkali metals (+1), the alkaline earth metals (+2), Ag , 2+ 2+ 3+ Zn , Cd , and Al . When writing formulas of oxoacids, you must know the names and formulas of polyatomic anions (see Table 2.3 of the text). Solution: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) The Roman numeral I tells you that the Cu cation has a +1 charge. Cyanide has a −1 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is CuCN. − Strontium is an alkaline earth metal. It only forms a +2 cation. The polyatomic ion chlorite, ClO2 , has a −1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. The correct formula is Sr(ClO2)2. − Perbromic tells you that the anion of this oxoacid is perbromate, BrO4 . The correct formula is HBrO4(aq). Remember that (aq) means that the substance is dissolved in water. − Hydroiodic tells you that the anion of this binary acid is iodide, I . The correct formula is HI(aq). + Na is an alkali metal. It only forms a +1 cation. The polyatomic ion ammonium, NH4 , has a +1 3− charge and the polyatomic ion phosphate, PO4 , has a −3 charge. To balance the charge, you need 2 + Na cations. The correct formula is Na2(NH4)PO4. The Roman numeral II tells you that the Pb cation has a +2 charge. The polyatomic ion carbonate, 2− CO3 , has a −2 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is PbCO3. The Roman numeral II tells you that the Sn cation has a +2 charge. Fluoride has a −1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. The correct formula is SnF2. This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the molecule. The correct formula is P4S10. The Roman numeral II tells you that the Hg cation has a +2 charge. Oxide has a −2 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is HgO. The Roman numeral I tells you that the Hg cation has a +1 charge. However, this cation exists as 2+ 2+ Hg2 . Iodide has a −1 charge. You need two iodide ion to balance the +2 charge of Hg2 . The correct formula is Hg2I2. This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the molecule. The correct formula is SeF6. 2.61 Uranium is radioactive. It loses mass because it constantly emits alpha (α) particles. 2.62 Changing the electrical charge of an atom usually has a major effect on its chemical properties. The two electrically neutral carbon isotopes should have nearly identical chemical properties. 2.63 The number of protons = 65 − 35 = 30. The element that contains 30 protons is zinc, Zn. There are two 2+ fewer electrons than protons, so the charge of the cation is +2. The symbol for this cation is Zn . 28 CHAPTER 2: ATOMS, MOLECULES, AND IONS 2.64 Atomic number = 127 − 74 = 53. This anion has 53 protons, so it is an iodide ion. Since there is one more − electron than protons, the ion has a −1 charge. The correct symbol is I . 2.65 (a) (b) (c) Species with the same number of protons and electrons will be neutral. A, F, G. Species with more electrons than protons will have a negative charge. B, E. Species with more protons than electrons will have a positive charge. C, D. (d) A: 10 5B B: 14 3− 7N C: 39 + 19 K D: 66 2+ 30 Zn E: 81 − 35 Br F: 11 5B G: 19 9F 2.66 NaCl is an ionic compound; it doesn’t form molecules. 2.67 Yes. The law of multiple proportions requires that the masses of sulfur combining with phosphorus must be in the ratios of small whole numbers. For the three compounds shown, four phosphorus atoms combine with three, seven, and ten sulfur atoms, respectively. If the atom ratios are in small whole number ratios, then the mass ratios must also be in small whole number ratios. 2.68 The species and their identification are as follows: (a) (b) (c) (d) (e) (f) SO2 S8 Cs N2O5 O O2 molecule and compound element and molecule element molecule and compound element element and molecule 2.69 (a) (b) (c) (d) This is an ionic compound. Prefixes are not used. The correct name is barium chloride. Iron has a +3 charge in this compound. The correct name is iron(III) oxide. − NO2 is the nitrite ion. The correct name is cesium nitrite. Magnesium is an alkaline earth metal, which always has a +2 charge in ionic compounds. The roman numeral is not necessary. The correct name is magnesium bicarbonate. 2.70 (a) (b) (c) (d) Ammonium is NH4 , not NH3 . The formula should be (NH4)2CO3. Calcium has a +2 charge and hydroxide has a −1 charge. The formula should be Ca(OH)2. 2− 2− Sulfide is S , not SO3 . The correct formula is CdS. 2− 2− Dichromate is Cr2O7 , not Cr2O4 . The correct formula is ZnCr2O7. 2.71 Symbol Protons Neutrons Electrons Net Charge + (g) (h) (i) (j) (k) (l) O3 CH4 KBr S P4 LiF element and molecule molecule and compound compound element element and molecule compound + 11 5B 54 2+ 26 Fe 31 3− 15 P 196 79 Au 222 86 Rn 5 6 5 0 26 28 24 +2 15 16 18 −3 79 117 79 0 86 136 86 0 2.72 (a) (b) Ionic compounds are typically formed between metallic and nonmetallic elements. In general the transition metals, the actinides and lanthanides have variable charges. 2.73 (a) (b) (c) Li , alkali metals always have a +1 charge in ionic compounds 2− S − I , halogens have a −1 charge in ionic compounds + CHAPTER 2: ATOMS, MOLECULES, AND IONS (d) (e) (f) (g) 29 3− N 3+ Al , aluminum always has a +3 charge in ionic compounds + Cs , alkali metals always have a +1 charge in ionic compounds 2+ Mg , alkaline earth metals always have a +2 charge in ionic compounds. 23 2.74 The symbol Na provides more information than 11Na. The mass number plus the chemical symbol identifies a specific isotope of Na (sodium) while combining the atomic number with the chemical symbol tells you nothing new. Can other isotopes of sodium have different atomic numbers? 2.75 The binary Group 7A element acids are: HF, hydrofluoric acid; HCl, hydrochloric acid; HBr, hydrobromic acid; HI, hydroiodic acid. Oxoacids containing Group 7A elements (using the specific examples for chlorine) are: HClO4, perchloric acid; HClO3, chloric acid; HClO2, chlorous acid: HClO, hypochlorous acid. Examples of oxoacids containing other Group A-block elements are: H3BO3, boric acid (Group 3A); H2CO3, carbonic acid (Group 4A); HNO3, nitric acid and H3PO4, phosphoric acid (Group 5A); and H2SO4, sulfuric acid (Group 6A). Hydrosulfuric acid, H2S, is an example of a binary Group 6A acid while HCN, hydrocyanic acid, contains both a Group 4A and 5A element. 2.76 Mercury (Hg) and bromine (Br2) 2.77 (a) (b) Isotope No. Protons No. Neutrons 4 2 He 20 10 Ne 40 18 Ar 84 36 Kr 132 54 Xe 2 2 10 10 18 22 36 48 54 78 neutron/proton ratio 1.00 1.00 1.22 1.33 1.44 The neutron/proton ratio increases with increasing atomic number. 2.78 H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn 2.79 Cu, Ag, and Au are fairly chemically unreactive. This makes them specially suitable for making coins and jewelry, that you want to last a very long time. 2.80 They do not have a strong tendency to form compounds. Helium, neon, and argon are chemically inert. 2.81 Magnesium and strontium are also alkaline earth metals. You should expect the charge of the metal to be the same (+2). MgO and SrO. 2.82 All isotopes of radium are radioactive. It is a radioactive decay product of uranium-238. Radium itself does not occur naturally on Earth. 2.83 (a) (b) (c) 2.84 Berkelium (Berkeley, CA); Europium (Europe); Francium (France); Scandium (Scandinavia); Ytterbium (Ytterby, Sweden); Yttrium (Ytterby, Sweden). Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie); Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence). Arsenic, Cesium, Chlorine, Chromium, Iodine. Argentina is named after silver (argentum, Ag). 30 CHAPTER 2: ATOMS, MOLECULES, AND IONS 2.85 The mass of fluorine reacting with hydrogen and deuterium would be the same. The ratio of F atom to hydrogen (or deuterium) is 1:1 in both compounds. This does not violate the law of definite proportions. When the law of definite proportions was formulated, scientists did not know of the existence of isotopes. 2.86 (a) (d) NaH, sodium hydride AlF3, aluminum fluoride 2.87 (a) Br 2.88 All of these are molecular compounds. We use prefixes to express the number of each atom in the molecule. The names are nitrogen trifluoride (NF3), phosphorus pentabromide (PBr5), and sulfur dichloride (SCl2). 2.89 (b) alkali metals Rn (b) (e) (c) Se B2O3, diboron trioxide OF2, oxygen difluoride (d) alkaline earth metals Rb (c) (f) (e) Pb noble gases halogens 1A Na2S, sodium sulfide SrCl2, strontium chloride 8A 2A 3A 4A 5A 6A 7A The metalloids are shown in gray. 2.90 Cation 2+ Mg Anion − HCO3 Sr Cl 2+ 3+ Fe Mn Sn 2+ 4+ Co 2+ Hg2 Cu 2+ + + Li 3+ Al 2.91 (a) (b) (c) (d) (e) − NO2 − ClO3 − − Br PO4 I 3− − CO3 N S 3− 2− 2− Formula Mg(HCO3)2 SrCl2 Name Magnesium bicarbonate Strontium chloride Fe(NO2)3 Iron(III) nitrite Mn(ClO3)2 SnBr4 Manganese(II) chlorate Tin(IV) bromide Co3(PO4)2 Cobalt(II) phosphate Hg2I2 Mercury(I) iodide Cu2CO3 Li3N Copper(I) carbonate Lithium nitride Al2S3 Aluminum sulfide CO2(s), solid carbon dioxide NaCl, sodium chloride N2O, nitrous oxide CaCO3, calcium carbonate CaO, calcium oxide (f) (g) (h) (i) (j) Ca(OH)2, calcium hydroxide NaHCO3, sodium bicarbonate Na2CO3⋅10H2O, sodium carbonate decahydrate CaSO4⋅2H2O, calcium sulfate dihydrate Mg(OH)2, magnesium hydroxide CHAPTER 2: ATOMS, MOLECULES, AND IONS 2.92 31 The change in energy is equal to the energy released. We call this ΔE. Similarly, Δm is the change in mass. E , we have Because m = c2 1000 J (1.715 × 103 kJ) × ΔE 1 kJ Δm = = = 1.91 × 10−11 kg = 1.91 × 10−8 g 2 8 2 (3.00 × 10 m/s) c ⎛ 1 kg ⋅ m 2 ⎞ Note that we need to convert kJ to J so that we end up with units of kg for the mass. ⎜1 J = ⎟ ⎜ ⎟ s2 ⎝ ⎠ We can add together the masses of hydrogen and oxygen to calculate the mass of water that should be formed. 12.096 g + 96.000 = 108.096 g −8 The predicted change (loss) in mass is only 1.91 × 10 g which is too small a quantity to measure. Therefore, for all practical purposes, the law of conservation of mass is assumed to hold for ordinary chemical processes. 2.93 CH4, C2H6, and C3H8 each only have one structural formula. H H C H H H H H C C H H H H C4H10 has two structural formulas. H H H H H C C C C H H H H H H C C C H H H H H HH H C H H H C C H HH C H H C5H12 has three structural formulas. H H H 2.94 H H H H H C C C C C H H H H H H HH H C C C HH H C C H H H H H HH H C H H C C C HH C H H H H (a) Rutherford’s experiment is described in detail in Section 2.2 of the text. From the average magnitude of scattering, Rutherford estimated the number of protons (based on electrostatic interactions) in the nucleus. (b) Assuming that the nucleus is spherical, the volume of the nucleus is: V = 4 3 4 πr = π(3.04 × 10−13 cm)3 = 1.177 × 10−37 cm3 3 3 32 CHAPTER 2: ATOMS, MOLECULES, AND IONS The density of the nucleus can now be calculated. d = m 3.82 × 10−23 g = = 3.25 × 1014 g/cm 3 V 1.177 × 10−37 cm3 To calculate the density of the space occupied by the electrons, we need both the mass of 11 electrons, and the volume occupied by these electrons. The mass of 11 electrons is: 11 electrons × 9.1095 × 10−28 g = 1.00205 × 10−26 g 1 electron The volume occupied by the electrons will be the difference between the volume of the atom and the volume of the nucleus. The volume of the nucleus was calculated above. The volume of the atom is calculated as follows: 186 pm × Vatom = 1 × 10−12 m 1 cm × = 1.86 × 10−8 cm −2 1 pm 1 × 10 m 4 3 4 πr = π(1.86 × 10−8 cm)3 = 2.695 × 10−23 cm3 3 3 Velectrons = Vatom − Vnucleus = (2.695 × 10 −23 3 cm ) − (1.177 × 10 −37 3 cm ) = 2.695 × 10 −23 3 cm As you can see, the volume occupied by the nucleus is insignificant compared to the space occupied by the electrons. The density of the space occupied by the electrons can now be calculated. d = m 1.00205 × 10−26 g = = 3.72 × 10−4 g/cm 3 V 2.695 × 10−23 cm3 The above results do support Rutherford's model. Comparing the space occupied by the electrons to the volume of the nucleus, it is clear that most of the atom is empty space. Rutherford also proposed that the nucleus was a dense central core with most of the mass of the atom concentrated in it. Comparing the density of the nucleus with the density of the space occupied by the electrons also supports Rutherford's model. 2.95 Two different structural formulas for the molecular formula C2H6O are: H H H C C H H H O H H C H H O C H H In the second hypothesis of Dalton’s Atomic Theory, he states that in any compound, the ratio of the number of atoms of any two of the elements present is either an integer or simple fraction. In the above two compounds, the ratio of atoms is the same. This does not necessarily contradict Dalton’s hypothesis, but Dalton was not aware of chemical bond formation and structural formulas. CHAPTER 2: ATOMS, MOLECULES, AND IONS 2.96 (a) Ethane 2.65 g C 0.665 g H 33 Acetylene 4.56 g C 0.383 g H Let’s compare the ratio of the hydrogen masses in the two compounds. To do this, we need to start with the same mass of carbon. If we were to start with 4.56 g of C in ethane, how much hydrogen would combine with 4.56 g of carbon? 4.56 g C 0.665 g H × = 1.14 g H 2.65 g C We can calculate the ratio of H in the two compounds. 1.14 g ≈ 3 0.383 g This is consistent with the Law of Multiple Proportions which states that if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. In this case, the ratio of the masses of hydrogen in the two compounds is 3:1. (b) For a given amount of carbon, there is 3 times the amount of hydrogen in ethane compared to acetylene. Reasonable formulas would be: Ethane CH3 C2H6 2.97 (a) Acetylene CH C2H2 The following strategy can be used to convert from the volume of the Pt cube to the number of Pt atoms. 3 cm → grams → atoms 1.0 cm3 × (b) 21.45 g Pt 1 cm × 3 1 atom Pt 3.240 × 10−22 g Pt = 6.6 × 1022 Pt atoms Since 74 percent of the available space is taken up by Pt atoms, 6.6 × 10 volume: 3 22 atoms occupy the following 3 0.74 × 1.0 cm = 0.74 cm We are trying to calculate the radius of a single Pt atom, so we need the volume occupied by a single Pt atom. 0.74 cm3 = 1.12 × 10−23 cm3 /Pt atom volume Pt atom = 22 6.6 × 10 Pt atoms 4 3 πr . Solving for the radius: 3 4 V = 1.12 × 10−23 cm3 = πr 3 3 The volume of a sphere is 3 r = 2.67 × 10 r = 1.4 × 10 −8 −24 cm 3 cm 34 CHAPTER 2: ATOMS, MOLECULES, AND IONS Converting to picometers: radius Pt atom = (1.4 × 10−8 cm) × 2.98 0.01 m 1 pm × = 1.4 × 102 pm −12 1 cm 1 × 10 m The mass number is the sum of the number of protons and neutrons in the nucleus. Mass number = number of protons + number of neutrons Let the atomic number (number of protons) equal A. The number of neutrons will be 1.2A. Plug into the above equation and solve for A. 55 = A + 1.2A A = 25 The element with atomic number 25 is manganese, Mn. 2.99 2.100 S N B I The acids, from left to right, are chloric acid, nitrous acid, hydrocyanic acid, and sulfuric acid.