Solution Manual For Chemistry: An Atoms-Focused Approach, 3rd Edition

CHAPTER 2 | Atoms, Ions, and Molecules: The Building Blocks of Matter 2.1. Collect and Organize This question asks us to identify the element in Figure P2.1 that has the fewest protons in its nucleus. Analyze An element is defined by the number of protons in its nucleus; as we move to higher elements in the periodic table, the number of protons increases. Solve The lightest, lowest atomic number element highlighted is hydrogen (purple). Hydrogen therefore has the fewest protons (one) in its nucleus. Think About It The other elements highlighted have more protons in their nucleus: helium (blue) has 2, fluorine (green) has 9, sulfur (yellow) has 16, and arsenic (red) has 33. 2.2. Collect and Organize This question asks us to identify the element in Figure P2.1 that has the most neutrons in its nucleus. Analyze As we move to higher atomic numbers in the periodic table, the number of neutrons in the nucleus increases. Solve The largest atomic number element highlighted is arsenic. Arsenic (red), therefore, has the most neutrons in its nucleus. Think About It Arsenic is a monoisotopic element; its only stable isotope is 75As with 42 neutrons in its nucleus along with 33 protons. 2.3. Collect and Organize We are asked to identify which shaded element in Figure P2.1 is stable and yet has no neutrons in its nucleus. Analyze The presence of neutrons in nuclei helps to overcome the repulsive forces of more than one proton in the nucleus. Solve The only shaded element that does not have more than one proton in its nucleus and therefore would be stable without neutrons is the element shaded purple, hydrogen. Think About It Without neutrons, nuclei with more than one proton would fly apart. 2.4. Collect and Organize From the shaded elements in Figure P2.4, we are to choose the one for which no stable isotopes exist. Analyze When an element has no stable isotopes, the element is radioactive. Starting with bismuth (atomic number 83), all the elements are radioactive. Solve Lawrencium (shaded light orange) is the only shaded element with an atomic number greater than 83, so it is the unstable element in Figure P2.4. 42 Atoms, Ions, and Molecules | 43 Think About It The last naturally occurring element found on Earth is uranium (element 92); the rest, through the newest discovered elements, oganesson (Z = 118) in 2002 and tennessine (Z = 115) in 2010, are synthetic. 2.5. Collect and Organize From among the shaded elements in the periodic table shown in Figure P2.4, we are to identify a transition metal, an alkali metal, a halogen, and a chemically inert gas. Analyze Transition metals occupy the middle of the periodic table and reside in groups 3–12, alkali metals occupy the first column of the periodic table (group 1), and halogens occupy the second-to-last column on the right (group 17), next to the noble gases. Solve (a) The transition metal element is shaded green and is gold (Au). (b) The alkali metal is shaded blue and is sodium (Na). (c) The halogen is shaded lilac and is chlorine (Cl). (d) The chemically inert gas is shaded red and is neon (Ne). Think About It A few other regions of the periodic table also have names. Can you identify the alkaline earth metals, the noble gases, and the chalcogens? 2.6. Collect and Organize For each of the nuclides depicted, we are to write the nuclide symbols. Analyze Nuclide symbols AX are written to identify both the element X (from the number of protons in the nuclide, denoted as the atomic symbol) and the mass number A (the total number of protons and neutrons in the nuclide, denoted as a superscript to the upper left of the element symbol). Solve (a) This nuclide has 3 protons and 4 neutrons; it is Li with a mass number of 7: 7Li. (b) This nuclide has 3 protons and 3 neutrons; it is Li with a mass number of 6: 6Li. (c) This nuclide has 6 protons and 8 neutrons; it is C with a mass number of 14: 14C. (d) This nuclide has 7 protons and 7 neutrons; it is N with a mass number of 14: 14N. Think About It Carbon-14 is an unstable isotope of carbon with a half-life of 5,730 years and is used in radiocarbon dating 2.7. Collect and Organize Using Figure P2.7 and our knowledge of the charges and the masses of alpha (a) and beta (b) particles, we are to determine which arrow (red, blue, or green) represents each particle’s behavior as it moves through an electric field. Analyze Alpha particles are much more massive than beta particles and they have a positive charge, whereas beta particles have a negative charge. Solve Alpha particles, with their positive charges, will be deflected toward the negative side of the electric field. Their behavior is represented by the red arrow. Beta particles, however, are deflected toward the positive side of the electric field because they carry a negative charge; they are represented by the green arrow. Notice, too, that the alpha particles travel farther and with less arc in their deflection than the lighter beta particles. Because alpha 44 | Chapter 2 particles have greater mass, their momentum through the electric field is higher and it would take a stronger field to deflect them to the same degree. Think About It The blue arrow might represent either a neutral particle or electromagnetic radiation which are not deflected by an electric field. 2.8. Collect and Organize For this question we are asked to identify the subatomic particle that would curve in the same direction as the red arrow in Figure P2.7. Analyze The red arrow shows that the particle is deflected toward the negative plate, which means that the subatomic particle carries a positive charge. Solve Protons with their positive charge would curve in the same direction as alpha particles, the red arrow in Figure P2.7. Think About It Another subatomic particle represented in the diagram are neutrons (blue arrow). 2.9. Collect and Organize Using Figure P2.9, we are to determine whether the mass spectrum shown is for dichloromethane or cyclohexane. Analyze Dichloromethane has a molar mass of 84.93 g/mol and cyclohexane has a molar mass of 84.15 g/mol. Those masses are close to each other, so we will have to use other clues from the mass spectrum to help. The figure shows two obvious mass peaks at 86 and 88 u and another grouping around 49 u. Solve The molecular ion peaks showing up at 84, 86, and 88 u show that this compound has isotopes of significant abundance. Cyclohexane has only carbon and hydrogen atoms and the abundance of C and H isotopes are too low to show this significantly in the mass spectrum. The m/z peak at 84 u is due to the presence of two atoms of 35Cl (CH235Cl2), the m/z peak at 86 u is due to the presence of one atom of 35Cl and one atom of 37Cl (CH235Cl37Cl), and the m/z peak at 88 u is due to the presence of two atoms of 37Cl (CH237Cl2). Also, the m/z peak grouping around 50 u is consistent with dichloromethane, CH2Cl2, losing one 35Cl or one 37Cl atom. Think About It 35 Cl is about 76% naturally abundant and 37Cl is about 24% naturally abundant. The relative intensities of the isotopic peaks in the mass spectrum also reflect those differences in abundance. 2.10. Collect and Organize Given that krypton has six stable isotopes, we are to determine from the mass spectrum in Figure P2.10 the number of neutrons in the most abundant isotope and write the symbol for that isotope. Analyze The mass spectrum shows m/z peaks for masses of 78, 80, 82, 83, 84, and 86 u, of which the peak at 84 u has the highest relative intensity. Solve The most abundant isotope has the highest relative intensity; for krypton that is an m/z of 84 u. Krypton has 36 protons in its nucleus, and with the neutron having about the same mass as the proton that means that the most abundant isotope has 84 – 36 = 48 neutrons. The isotope symbol is 84Kr. Atoms, Ions, and Molecules | 45 Think About It The other isotopes have 42, 44, 46, 47, and 50 neutrons in their nuclei, respectively. 2.11. Collect and Organize This question asks us to correlate the position of an element in the periodic table with typical charges on the ions for the groups (or families) of elements. Analyze Figure 2.11 in the textbook shows the common charges on the elements used in forming compounds. That figure will help us determine which elements in monatomic form give the charges named in the question. Solve Highlighted elements in Figure P2.11 are K, Mg, Sc, Ag, O, and I. (a) Elements in group 1 form 1+ ions, so K will form K+ (dark blue). Silver (green) also typically forms a 1+ cation. (b) Elements in group 2 form 2+ ions, so Mg forms Mg2+ (gray). (c) Elements in group 3 form 3+ ions, so Sc forms Sc3+ (yellow). (d) Elements in group 17 (the halogens) form 1– ions, so I forms I– (purple). (e) Elements in group 16 form 2– ions, so O forms O2– (red). Think About It Elements on the left-hand side of the periodic table form cations, and elements on the right-hand side tend to form anions. 2.12. Collect and Organize Using the representations of molecules, atoms, and ions in Figure P2.12 and the atomic color palette, we are asked to match the representations to the descriptions. Analyze From the color palette we find what each color indicates: sulfur (yellow), oxygen (red), carbon (black), potassium (purple), iodine (lilac), neutron (dark gray), proton (white). We also note that [B], [D], [F], and [H] represent nuclei; [A] and [I] represent ionic solids; and [C], [E], and [G] represent molecules. Solve (a) Potassium iodide (KI) has a one-to-one ratio of K+ to I–; potassium oxide (K2O) has a two-to-one ratio of K+ to O2–. Representation [A] has ions in equal ratios and is therefore KI, whereas representation [I] has fewer red ions than purple ions and is therefore K2O. (b) When we count the gray particles in representations [B], [D], [F], and [H], the order by increasing number of neutrons is [H] (5 neutrons) < [D] (6 neutrons) < [F] (7 neutrons) 2p The mass number is also equal to the number of protons plus the number of neutrons (n), m=p+n Combining these expressions p + n > 2p and solving for n gives n > 2p – p n>p Therefore, the number of neutrons in this isotope is greater than the number of protons, and the neutron-to-proton ratio is greater than 1. Think About It We wouldn’t have had to express the relationships between the nuclear particles mathematically if the isotope had a mass number equal to twice the number of protons. Then the number of neutrons would have to be the same as the number of protons, giving a neutron-to-proton ratio of 1:1. 2.22. Collect and Organize For this question we are to compare the numbers of neutrons, protons, nucleons, and electrons in isotopes of oxygen. Analyze Atomic number gives the number of protons in the nucleus, and A, or the mass number, of the isotopes gives the number of nucleons (protons + neutrons in a nucleus). The number of neutrons can be calculated by subtracting the number of protons from the mass number. Finally, the number of electrons equals the number of protons in a neutral atom. Solve (a) The number of neutrons in each isotope is: 16O, 8; 17O, 9; and 18O, 10. (b) The number of protons in each isotope is: 16O, 8; 17O, 8; and 18O, 8. (c) The number of nucleons in each isotope is: 16O, 16; 17O, 17; and 18O, 18. (d) The number of electrons in each isotope is: 16O, 8; 17O, 8; and 18O, 8. Think About It Isotopes are defined as elements with different numbers of neutrons in their nuclei, but having the same number of protons. 2.23. Collect and Organize Given that most stable nuclides have at least as many neutrons in their nuclei as protons (and often more), we are to identify which element is an exception to that rule. Analyze Neutrons help stabilize the nucleus by counteracting the repulsive forces between protons. Solve Hydrogen, with only one proton, does not need neutrons to be stable and so 1H is the exception. Think About It Hydrogen, however, can have one (for deuterium, 2H) and even two (for tritium, 3H) neutrons in its nucleus. 2.24. Collect and Organize We are to explain why the mass number does not change when 19F becomes 19F–. 50 | Chapter 2 Analyze The mass number, A, gives the sum of protons and neutrons in the nucleus. It does not include the mass of the electrons outside of the nucleus. Electrons do not appreciably add to the mass of the atom. Solve In adding an electron to 19F to make the fluoride anion, we are not adding the electron to the nucleus, so that it would change the number or type of nucleons. Rather, an electron is added to the outside of the nucleus when an atom becomes a negative ion. Think About It Likewise, when an atom becomes a positive ion (cation) by losing an electron, the atomic number does not change. 2.25. Collect and Organize For each element in this question, we must look at the relationship of the neutrons, protons, and electrons. We need to determine the element’s atomic number from the periodic table and, from the mass number given, compute the number of neutrons to give the indicated isotope. Analyze An isotope is given by the symbol “!X, where X is the element’s symbol from the periodic table, Z is the atomic number (the number of protons in the nucleus), and A is the mass number (the number of protons and neutrons in the nucleus). Often, Z is omitted because the element symbol gives us the same information about the identity of the element. To determine the number of neutrons in the nucleus for each named isotope, we subtract Z (number of protons) from A (mass number). If the elements are neutral (no charge), the number of electrons equals the number of protons in the nucleus. Solve Mass Atomic Number = Atom Number Number of Protons (a) 14 (b) 59 (c) 90 (d) 210 Number of Neutrons = Mass Number of Electrons = Number – Atomic Number Number of Protons C 14 6 8 6 Fe 59 26 33 26 90 38 52 38 210 82 128 82 Sr Pb Think About It Isotopes of an element contain the same number of protons but a different number of neutrons. Thus, isotopes have different masses. 2.26. Collect and Organize For each element in this question, we must look at the relationship of the neutrons, protons, and electrons. We need to determine the element’s atomic number from the periodic table and, from the mass number given, compute the number of neutrons to give the indicated isotope. Analyze An isotope is given by the symbol “!X, where X is the element’s symbol from the periodic table, Z is the atomic number (the number of protons in the nucleus), and A is the mass number (the number of protons and neutrons in the nucleus). Often, Z is omitted because the element symbol gives us the same information about the identity of the element. To determine the number of neutrons in the nucleus for each named isotope, we subtract Z (number of protons) from A (mass number). If the elements are neutral (no charge), the number of electrons equals the number of protons in the nucleus. Atoms, Ions, and Molecules | 51 Solve Mass Atomic Number = Atom Number Number of Protons (a) 11 (b) 19 B Number of Neutrons = Mass Number of Electrons = Number – Atomic Number Number of Protons 11 5 6 5 F 19 9 10 9 (c) 131 I 131 53 78 53 (d) 222 Rn 222 86 136 86 Think About It As the atom gets more massive, the ratio of neutrons to protons gets larger. For fluorine the ratio is 10/9 = 1.1, but for radon the ratio is 136/86 = 1.6. We will see later that as the nucleus adds more protons, it needs more neutrons as “glue” to keep the nucleus stable. 2.27. Collect and Organize After calculating the neutron-to-proton ratio for 4He, 23Na, 59Co, and 197Au, we are to comment on how the ratio changes as Z increases. Analyze To calculate the neutron-to-proton ratios, we need to determine, from the mass number and the atomic number, the number of protons and neutrons for each element: 4 He, atomic number 2, has 2 protons and 2 neutrons 23 Na, atomic number 11, has 11 protons and 12 neutrons 59 Co, atomic number 27, has 27 protons and 32 neutrons 197 Au, atomic number 79, has 79 protons and 118 neutrons Solve The neutron-to-proton ratio for each element is (a) 4He, 2/2 = 1.00 (b) 23Na, 12/11 = 1.09 (c) 59Co, 32/27 = 1.19 (d) 197Au, 118/79 = 1.49 As the atomic number (Z) increases, the neutron-to-proton ratio increases. Think About It More neutrons are required to stabilize nuclei with more protons because of the strong repulsive forces between the positively charged protons. Neutrons bring added strong nuclear force to the nucleus to stabilize it. 2.28. Collect and Organize After calculating the neutron-to-proton ratios for group 15 nuclei 14N, 31P, 75As, 121Sb, and 123Sb, we are to comment on how the ratio changes as Z increases. Analyze To calculate the neutron-to-proton ratios, we need to determine, from the mass number and the atomic number, the number of protons and neutrons for each element: 14 N, atomic number 7, has 7 protons and 7 neutrons 31 P, atomic number 15, has 15 protons and 16 neutrons 75 As, atomic number 33, has 33 protons and 42 neutrons 121 Sb, atomic number 51, has 51 protons and 70 neutrons 123 Sb, atomic number 51, has 51 protons and 72 neutrons Solve The neutron-to-proton ratio for each element is (a) 14N, 7/7 = 1.00 (b) 31P, 16/15 = 1.07 52 | Chapter 2 (c) 75As, 42/33 = 1.27 (d) 121Sb, 70/51 = 1.37 (e) 123Sb, 72/51 = 1.41 As the atomic number (Z) increases, the neutron-to-proton ratio increases. Think About It Both isotopes of antimony, 121Sb and 123Sb, are present in nature in comparable abundances (57% and 43%, respectively). 2.29. Collect and Organize To fill in the table, we have to consider how the numbers of nuclear particles relate to one another. We also need to recall how the symbols for the isotopes are written. From the table, it is apparent that we have to work backward in some cases from the number of electrons or protons and mass number to get the element symbol. Analyze An isotope is given by the symbol “!X, where X is the element’s symbol from the periodic table, Z is the atomic number (the number of protons in the nucleus), and A is the mass number (the number of protons and neutrons in the nucleus). We can determine the number of neutrons in the nucleus for the isotopes by subtracting Z (number of protons) from A (mass number). If the elements are neutral (no charge), the number of electrons equals the number of protons in the nucleus. Except for the first element, we will assume that the rest of the elements are neutral; otherwise, there would be different possibilities for the numbers of protons and electrons. Solve Symbol Number of Protons Number of Neutrons Number of Electrons Mass Number 23 Na 11 12 11 23 89 Y 39 50 39 89 118 Sn 50 68 50 118 197 Au 79 118 79 197 Think About It Because the nuclear particles are all related to one another, either we can work from the isotope symbol to find the number of protons, neutrons, and electrons for a particular isotope, or we can work from the mass number and the number of electrons or protons to determine the number of neutrons and write the element symbol. 2.30. Collect and Organize To fill in the table, we have to consider how the numbers of nuclear particles relate to one another. We also need to recall how the symbols for the isotopes are written. From the table, it is apparent that we have to work backward in some cases from the number of electrons or protons and the mass number to get the element symbol. Analyze An isotope is given by the symbol “!X, where X is the element’s symbol from the periodic table, Z is the atomic number (the number of protons in the nucleus), and A is the mass number (the number of protons and neutrons in the nucleus). We can determine the number of neutrons in the nucleus for the isotopes by subtracting Z (number of protons) from A (mass number). If the elements are neutral (no charge), the number of electrons equals the number of protons in the nucleus. Except for the first element, we will assume that the rest of the elements are neutral, otherwise there would be different possibilities for the numbers of protons and electrons. Solve Symbol Number of Protons Number of Neutrons Number of Electrons Mass Number 27 Al 13 14 13 27 98 Mo 42 56 42 98 143 Nd 60 83 60 143 238 U 92 146 92 238 Atoms, Ions, and Molecules | 53 Think About It Because the nuclear particles are all related to one another, either we can work from the isotope symbol to find the number of protons, neutrons, and electrons for a particular isotope, or we can work from the mass number and the number of electrons or protons to determine the number of neutrons and write the element symbol. 2.31. Collect and Organize An isotope of a monatomic ion can be given the symbol “!X # , where X is the element’s symbol from the periodic table, Z is the atomic number (the number of protons in the nucleus), A is the mass number (the number of protons and neutrons in the nucleus), and n is the charge on the species. Analyze If we are given the number of protons in the nucleus, the element can be identified from the periodic table. We can determine the mass number for the isotope by adding the protons and the neutrons in the nucleus. We can determine the number of neutrons or protons by subtracting Z (number of protons) or the number of neutrons, respectively, from A (mass number). We can account for the charge on the species by adding electrons (to form a negatively charged ion) or by subtracting electrons (to form a positively charged ion). Solve Symbol Number of Protons Number of Neutrons Number of Electrons Mass Number Cl– 17 20 18 37 37 23 Na+ 11 12 10 23 81 Br– 35 46 36 81 226 Ra2+ 88 138 86 226 Think About It To form a singly charged ion, there has to be one more electron (for a negative charge) or one fewer electron (for a positive charge) than the number of protons in the nucleus. For a doubly charged ion, we add or take away two electrons. 2.32. Collect and Organize An isotope of a monatomic ion can be given the symbol “!X # , where X is the symbol of the element from the periodic table, Z is the atomic number (the number of protons in the nucleus), A is the mass number (the number of protons and neutrons in the nucleus), and n is the charge on the species. Analyze If we are given the number of protons in the nucleus, the element can be identified from the periodic table. We can determine the mass number by adding the numbers of protons and neutrons. We can determine the number of neutrons or protons in the nucleus by subtracting Z (number of protons) or the number of neutrons, respectively, from A (mass number). We can account for the charge on the species by adding electrons (to form a negatively charged ion) or by subtracting electrons (to form a positively charged ion). Solve Symbol Number of Protons Number of Neutrons Number of Electrons Mass Number 137 Ba2+ 56 81 54 137 64 Zn2+ 30 34 28 64 32 2– S 16 16 18 32 90 Zr4+ 40 50 36 90 Think About It These are all multiply charged ions. For anions, we add electrons; for cations, we remove electrons. 2.33. Collect and Organize Knowing that Mendeleev labeled his groups on the left side of the periodic table on the basis of the formulas of the compounds they formed with oxygen, we are to assign his labels to groups 2, 3, and 4 on the modern periodic table. 54 | Chapter 2 Analyze Groups 2, 3, and 4 have cations with charges of 2+, 3+, and 4+, respectively. Oxygen forms an anion of 2– charge. The oxygen compounds that would form would balance the positive cation charge with the negative anion charge by combining the elements in that group with oxygen in whole-number ratios. Solve Group 2, with a 2+ charge, would form RO with oxygen; group 3, with a 3+ charge, would form an R2O3 compound with oxygen; and group 4, with a 4+ charge, would form an RO2 compound with oxygen. Think About It Mendeleev’s brilliant insight was to classify the elements based on similar chemical behaviors. 2.34. Collect and Organize Knowing that Mendeleev labeled his groups on the right side of the periodic table on the basis of the formulas of the compounds they formed with hydrogen, we are to assign modern periodic table labels to his groups of HR, H2R, and H3R. Analyze Because hydrogen has a charge of 1+, for those compounds we are looking for groups that have typical charges of their anions of 1–, 2–, and 3–. Solve Group 17, with a 1– charge, would form HR compounds with hydrogen; group 16, with a 2– charge, would form H2R compounds with hydrogen; and group 15, with a 3– charge, would form H3R compounds with hydrogen. Think About It Representative compounds with hydrogen from those groups are hydrochloric acid (HCl), water (H2O), and ammonia (NH3). 2.35. Collect and Organize We are asked why Mendeleev did not leave spaces for the noble gases in his periodic system. Analyze The noble gases are characterized by their remarkable unreactivity. Unreactive elements can be quite unnoticeable because they do not form compounds with other elements. The noble gases were discovered starting in 1894. Mendeleev proposed his periodic system in 1872. Solve The noble gases were not discovered until after Mendeleev put together his periodic table. He could not have predicted the existence of the noble gases at the time since (a) none of them were isolated and characterized on the basis of their reactivity (or lack thereof) to indicate their presence in nature and (b) he arranged the elements in order of increasing mass, not atomic number. If he had been aware of atomic numbers as characteristic of the elements, he would have noticed that the atomic numbers for the noble gases were missing as a column in his table. Think About It The noble gases are monatomic, are colorless and odorless, and have a remarkably narrow liquid range (their boiling points and melting points are close together). 2.36. Collect and Organize In this question we are asked to describe how the periodic table would look if the A- and B-labeled columns were stacked. Analyze For this, we will stack elements in the nA column above that of the nB column. Atoms, Ions, and Molecules | 55 Solve H Li Na K Cu Rb Ag Cs Au Fr Rg Be Mg Ca Zn Sr Cd Ba Hg Ra Cn B Al Sc Ga Y In La Tl Ac Nh C Si Ti Ge Zr Sn Hf Pb Rf Fl N P V As Nb Sb Ta Bi Db Mc O S Cr Se Mo Te W Po Sg Lv F Cl Mn Br Tc I Re At Bh Ts Ne Ar Fe Kr Ru Xe Os Rn Hs Og Co Ni Rh Pd Ir Pt Mt Ds This is nearly the same way that Mendeleev arranged his periodic table based on masses and similar properties of the elements. Think About It This other form loses a lot of information about what we now know about the periodic nature of the elements. In particular, it loses information about the electronic structure (how the electrons populate orbitals) which you will learn in Chapter 3. 2.37. Collect and Organize Knowing that the explosive TNT contains second-row elements in groups 14, 15, and 16 as well as hydrogen, we are to name those particular elements. Analyze The second-row elements start with lithium and end at neon. Solve The elements in TNT besides hydrogen are carbon (group 14), nitrogen (group 15), and oxygen (group 16). Think About It Many explosives have the same elements. The powerful C4 explosive is composed mainly of the explosive RDX2, which has the chemical formula C3H6N6O6. 2.38. Collect and Organize Knowing that the chemical weapon phosgene contains a second-row element in group 16 and a third-row element in group 17 as well as carbon, we are to name those particular elements and give their atomic numbers. Analyze The second-row elements start with lithium and end at neon. The third-row elements start at sodium and end at argon. Solve The elements in phosgene besides carbon are oxygen (group 16), with atomic number 8, and chlorine (group 17), with atomic number 17. Think About It Phosgene is a small molecule (COCl2) that is highly toxic and causes damage to the lungs, leading to suffocation when inhaled. 2.39. Collect and Organize Given information about the elements used in catalytic converters (their group numbers and relationship to one another in the periodic table), we are asked to name the elements. 56 | Chapter 2 Analyze In examining the clues in parts a–c, we see that these elements will all be transition metals located in the middle of the periodic table. Solve (a) The group 10 transition metal element in the fifth row of the periodic table (Rb–Xe) is palladium (Pd). (b) The transition metal element with one fewer proton than Pd is in group 9 and is rhodium (Rh). (c) The transition metal element with 32 more protons than Pd is platinum (Pt) with 78 protons. Think About It These metals are generally as expensive or more so than gold. In May, 2020 gold is priced at about $1700/oz, with platinum priced less (about $820/oz), but rhodium and palladium more expensive ($6500/oz and $1900/oz, respectively). 2.40. Collect and Organize We are asked to name the fourth-row halogen that can be used as an alternative to chlorine as a disinfectant. Analyze The fourth-row elements start at potassium and end at krypton. The halogens are the elements in group 17. Solve The fourth-row halogen element is bromine. Think About It Iodine, the fifth-row halogen, also can be used as a disinfectant or antiseptic. 2.41. Collect and Organize We are to count the metallic elements in the third row of the periodic table. Analyze The third row in the periodic table begins at sodium and ends at argon. The metalloids begin at silicon, and the nonmetals begin with phosphorus. Solve Three metallic elements are in the third row in the periodic table: sodium (Na), magnesium (Mg), and aluminum (Al). Think About It It’s easy to follow the color scheme in the periodic table of the elements on the inside front cover of the text: Metals are tan, nonmetals are blue, and metalloids are green. 2.42. Collect and Organize From the elements in the third row of the periodic table, we are to identify the element that has the chemical properties of a nonmetal but is more like a metal in its physical properties. Analyze The third row in the periodic table begins at sodium and ends at argon. The given characteristics mean that this particular element is between a metal and a nonmetal and is therefore one of the semimetals. Solve Silicon (Si) is the element described. Think About It As a metalloid, silicon has some, but not all, the properties of a metal but has a structure more like that of a nonmetal. Silicon is an important material in semiconducting computer chips. Atoms, Ions, and Molecules | 57 2.43. Collect and Organize We define weighted average for this question. Analyze An average is a number that expresses the middle of the data (here, for various masses of atoms or isotopes). Solve A weighted average takes into account the proportion of each value in the group of values to be averaged. For example, the average of 2, 2, 2, and 5 would be computed as (2 + 2 + 2 + 5)/4 = 2.75. That average shows the heavier weighting toward the values of 2. Think About It Because isotopes for any element are not equally present but have a range of natural abundances, the masses in the periodic table of the elements are calculated as weighted averages. 2.44. Collect and Organize Given that the abundance of the two isotopes of an element are both 50%, we are to express the average atomic mass of the element if the mass of isotope X is mX and the mass of isotope Y is mY. Analyze Because each isotope is present in exactly 50% abundance, the average atomic mass will be the simple average of the two masses of isotopes X and Y. Solve average atomic mass = m X + mY 2 Think About It Some elements in nature have just one naturally occurring isotope. Can you find some examples in Appendix 3.3? 2.45. Collect and Organize For calculating the masses of binary ionic compounds (such as NaCl), we are to explain why we use the average masses of the neutral atoms rather than the masses of the ions. Analyze For a cation, the actual mass would be the mass of the neutral atom minus the mass of the one or more electrons lost; for an anion, the actual mass would be the atomic mass plus the mass of the electrons gained. Solve First, the mass of the electron is small in comparison with the mass of a neutral atom, so the mass of an ion is exceedingly close to the mass of the neutral atom. But, more importantly, in a neutral salt, the number of electrons gained to form the anion and the number of electrons lost to form the cation are equal, so no electrons have been lost or gained in total compared with the number of electrons in the neutral atoms. Think About It In NaCl, sodium (with 11 electrons) loses one electron to be Na+ (now with 10 electrons), and chlorine (with 17 electrons) gains one electron to be Cl– (with 18 electrons). The total count of electrons stays at 28. Can you do the electron count for MgCl2? 2.46. Collect and Organize We have to consider the concept of weighted average atomic mass to answer this question, and we’ll need to look up atomic masses for B, Li, and N from inside the front cover of the textbook. 58 | Chapter 2 Analyze We are asked to compare two isotopes and their weighted average mass. If the lighter isotope is more abundant, the average atomic mass will be less than the average if both isotopes were equally abundant. If the heavier isotope is more abundant, the average atomic mass will be greater than the simple average of the two isotopes. We are given the mass number for the isotopes as part of the isotope symbol, and we will take that as the mass of that isotope in atomic mass units. Solve (a) The simple average atomic mass for 10B and 11B would be 10.5 u. The actual average mass (10.811 u) is greater than this; therefore, 11B is more abundant. (b) The simple average atomic mass for 6Li and 7Li would be 6.5 u. The actual average mass (6.941 u) is greater than this; therefore, 7Li is more abundant. (c) The simple average atomic mass for 14N and 15N would be 14.5 u. The actual average mass (14.007 u) is less than this; therefore, 14N is more abundant. Think About It This is a quick question to answer for elements such as boron, lithium, and nitrogen that have the dominance of only two isotopes in terms of their abundance. Answering the same question for elements with more than two stable isotopes in relatively high abundances is a little harder. 2.47. Collect and Organize We are to determine which isotope of argon is most abundant, 36Ar, 38Ar, or 40Ar. Analyze To help here, it is useful to know from the periodic table that the average atomic mass of argon is 39.948 u. Solve Only if the highest mass isotope were most abundant would the average mass of argon be 39.948 u. Therefore, 40 Ar is the most abundant isotope of argon. Think About It Table A3.3 shows these natural abundances: 40Ar, 99.600%; 38Ar, 0.063%; and 36Ar, 0.337%. 2.48. Collect and Organize Given that the two most common isotopes of hydrogen have 0 or 1 neutron (i.e., are 1H and 2H) and those of oxygen have 8 or 10 neutrons (i.e., are 16O and 18O), we are to determine the possible values for the molecular weight (in u) of a single water molecule. We’ll need the masses for subatomic particles (Table 2.1). Analyze The possible water molecules are: 1H216O, 1H2H16O, 2H216O, 2H218O, 1H2H18O, and 1H218O. All of these have the same number of protons, 10, for which we can find the mass by multiplying by the mass of the proton (1.00728 u). Each of these neutral molecules has 10 electrons, for which we can find the mass by multiplying by the mass of the electron (5.48580 ´ 10–4 u). Finally, for each molecule we will calculate the total mass of neutrons using the mass of a neutron (1.00866 u). We will then sum these (mass of protons, electrons, and neutrons for each molecule) to get the atomic weight of each type of water molecule. Solve 1 H216O has 10 protons, 10 electrons, and 8 neutrons: ⎛ 1.00728 u ⎞ ⎛ 5.48580 × 10 –4 u ⎞ ⎛ 1.00866 u ⎞ ⎜⎝ 10 protons × proton ⎟⎠ + ⎜⎝ 10 electrons × ⎟⎠ + ⎜⎝ 8 neutrons × proton ⎟⎠ = 18.1476 u electron 1 H2H16O: ⎛ 1.00728 u ⎞ ⎛ 5.48580 × 10 –4 u ⎞ ⎛ 1.00866 u ⎞ 10 protons × + 10 electrons × + ⎜ 9 neutrons × = 19.1562 u ⎜⎝ ⎟ ⎜ ⎟ proton ⎠ ⎝ electron proton ⎟⎠ ⎠ ⎝ Atoms, Ions, and Molecules | 59 2 H216O: ⎛ 1.00728 u ⎞ ⎛ 5.48580 × 10 –4 u ⎞ ⎛ 1.00866 u ⎞ 10 protons × + 10 electrons × ⎜⎝ ⎟⎠ + ⎜⎝ 10 neutrons × proton ⎟⎠ = 20.1649 u proton ⎟⎠ ⎜⎝ electron 2 H218O: ⎛ 1.00728 u ⎞ ⎛ 5.48580 × 10 –4 u ⎞ ⎛ 1.00866 u ⎞ 10 protons × + 10 electrons × + ⎜ 12 neutrons × = 22.1822 u ⎜⎝ ⎟ ⎜ ⎟ proton ⎠ ⎝ electron proton ⎟⎠ ⎠ ⎝ 1 H2H18O: ⎛ 1.00728 u ⎞ ⎛ 5.48580 × 10 –4 u ⎞ ⎛ 1.00866 u ⎞ 10 protons × + 10 electrons × + ⎜ 11 neutrons × = 21.1735 u ⎜⎝ ⎟ ⎜ ⎟ proton ⎠ ⎝ electron proton ⎟⎠ ⎠ ⎝ 1 H218O: ⎛ 1.00728 u ⎞ ⎛ 5.48580 × 10 –4 u ⎞ ⎛ 1.00866 u ⎞ 10 protons × + 10 electrons × + ⎜ 10 neutrons × = 20.1649 u ⎜⎝ ⎟ ⎜ ⎟ proton ⎠ ⎝ electron proton ⎟⎠ ⎠ ⎝ Think About It Hydrogen has three common isotopes and because each has a very different mass, they have special names: 1H is protium, 2H is deuterium, and 3H is tritium. 2.49. Collect and Organize In this question we are asked to explain the observation that the average atomic mass of platinum is 195.08 u, whereas the natural abundance of 195Pt is only 33.8%. Analyze The average atomic mass on the periodic table is the weighted average of all the masses of the naturally occurring isotopes for that element. Only if an element has one naturally occurring element will the average atomic mass match the mass of that isotope. Solve We are given that the 195Pt isotope is not 100% abundant. Therefore, we must conclude that the other isotopes with masses greater than 195 u have natural abundances in equal proportion to those isotopes with masses lower than 195 so that the weighted average atomic mass calculates to 195.08 u. Think About It For example, platinum might have three isotopes, each in 33% abundance: 194Pt, 195Pt, and 196Pt. 2.50. Collect and Organize In this question we are asked to determine whether we can conclude that europium has only one stable isotope on the basis of the information that its atomic mass is only 0.04 u off from a whole number. Analyze The average atomic mass on the periodic table is the weighted average of all the masses of the naturally occurring isotopes for that element. Only if an element has one naturally occurring element will the average atomic mass match the mass of that isotope. The mass of an isotope is the sum of the masses of all its protons, neutrons, and electrons. Solve No, we cannot conclude from this information that only one stable isotope of europium exists. Heavier and lighter isotopes might “cancel” each other in the calculation of the average atomic mass, or other stable isotopes might be in very small abundances and so they do not contribute significantly to the weighted atomic mass. 60 | Chapter 2 Think About It In fact, europium has two naturally occurring isotopes: 151Eu, which is 52% abundant, and 153Eu, which is about 48% abundant. 2.51. Collect and Organize For a sample of hydrogen gas (H2) in which 1H is 99.99745% abundant and 2H is 0.00255% abundant, we are to calculate the average atomic mass of hydrogen and compare it to the mass of H in the periodic table in the textbook. Analyze To calculate the average atomic mass, we have to consider the relative abundances according to the following formula: mX = a1m1 + a2m2 + a3m3 + . . . where an refers to the abundance of isotope n and mn refers to the mass of isotope n. If the relative abundances are given as percentages, the value we use for an in the formula is the percentage divided by 100. The masses of 1 H and 2H are given in the statement of the problem. Solve mH = (0.9999745 ´ 1.007825 u) + (0.0000255 ´ 2.014102 u) = 1.007851 u This value matches that given in the periodic table (1.0079 u) to five significant figures. Think About It Because the 1H isotope is in much, much greater abundance than 2H, we expect the average atomic mass of hydrogen in any sample of H2 to be very close to 1 u rather than 2 u. 2.52. Collect and Organize For a boron-containing sample of water in which 11B is 81.07% abundant and 10B is 18.93% abundant, we are to calculate the average atomic mass of boron and compare it to the mass of B in the periodic table in the textbook. Analyze To calculate the average atomic mass, we have to consider the relative abundances according to the following formula: mX = a1m1 + a2m2 + a3m3 + . . . where an refers to the abundance of isotope n and mn refers to the mass of isotope n. If the relative abundances are given as percentages, the value we use for an in the formula is the percentage divided by 100. The masses of 11 B and 10B are given in the statement of the problem. Solve mB = (0.8107 ´ 11.0093 u) + (0.1893 ´ 10.0129 u) = 10.82 u This value nearly matches that on the periodic table (10.811 u) to four significant figures. Think About It Because the 11B isotope is in much greater abundance than the 10B isotope, we expect the average atomic mass to be closer to 11 u rather than 10 u. 2.53. Collect and Organize Here we are asked to find out whether the mass of magnesium on Mars is the same as here on Earth. We are given the masses of each of the three isotopes of Mg in the Martian sample. Once we calculate the weighted average for Mg for the Martian sample, we can compare it with the average mass for Mg found on Earth. Analyze To calculate the average atomic mass, we have to consider the relative abundances according to the following formula: mX = a1m1 + a2m2 + a3m3 + . . . Atoms, Ions, and Molecules | 61 where an refers to the abundance of isotope n and mn refers to the mass of isotope n. If the relative abundances are given as percentages, the value we use for an in the formula is the percentage divided by 100. Solve For the average atomic mass of magnesium in the Martian sample mMg = (0.7870 ´ 23.9850 u) + (0.1013 ´ 24.9858 u) + (0.1117 ´ 25.9826 u) = 24.31 u The average mass of Mg on Mars is the same as here on Earth. Think About It The mass of Mg on Mars should be close to the same value as on Earth; the magnesium on both planets arrived in the solar system via the same ancient stardust. 2.54. Collect and Organize Given the exact mass for each isotope of lithium and its abundance, we are to calculate the average mass for lithium and compare that value with the mass found in the periodic table. Analyze The weighted atomic mass is computed according to the formula mX = a1m1 + a2m2 + a3m3 + . . . where an refers to the abundance of isotope n and mn refers to the mass of isotope n. If the relative abundances are given as percentages, the value we use for an in the formula is the percentage divided by 100. Solve mLi = (0.0773 ´ 6.015123 u) + (0.9227 ´ 7.016003 u) = 6.939 u The atomic mass of Li in the periodic table is 6.941 u. These values are close. Think About It We expect the average mass computed here to closely match the value in the periodic table because the periodic table value also is the weighted average for the atomic mass. 2.55. Collect and Organize In this problem, we again use the concept of weighted average atomic mass, but here we have to work backward from the average mass to find the exact mass of the 48Ti isotope, one of the five stable Ti isotopes. Analyze We can use the formula for finding the weighted average atomic mass, but this time our unknown quantity is one of the isotope masses. Here, mTi = a 46 Ti m 46 Ti + a 47 Ti m 47 Ti + a 48 Ti m 48 Ti + a 49 Ti m 49 Ti + a 50 Ti m50 Ti Solve 47.867 u = (0.0825 ´ 45.9526 u) + (0.0744 ´ 46.9518 u) + (0.7372 ´ m 48 ) Ti + (0.0541 ´ 48.94787 u) + (0.0518 ´ 49.94479 u) m 48 = 47.95 u Ti Think About It That answer makes sense because the exact mass of 48Ti should be close to 48 u. 2.56. Collect and Organize In this problem, we again use the concept of weighted average atomic mass, but here we are asked to work backward from the average mass to find the exact mass of the 92Zr isotope. 62 | Chapter 2 Analyze We can use the formula for finding the weighted average atomic mass, only this time our unknown quantity is one of the isotope masses. Here, mZr = a 90 Zr m90 Zr + a 91Zr m91Zr + a 92 Zr m92 Zr + a 94 Zr m94 Zr + a 96 Zr m96 Zr Solve 91.224 u = (0.5145 ´ 89.905 u) + (0.1122 ´ 90.906 u) + (0.1715 ´ m 92 m92 Zr Zr ) + (0.1738 ´ 93.906 u) + (0.0280 ´ 95.908) = 91.906 u Think About It That answer makes sense because the exact mass of 92Ti should be close to 92 u. 2.57. Collect and Organize From the formulas for four ionic compounds, NaI, CaS, Al2O3, and NH4Cl, we are to calculate the masses of the formula units. Analyze For each formula we will sum the masses of the elements, making sure that we also account for the number of a particular element present in the formula. Solve (a) NaI: 22.990 u + 126.90 u = 149.89 u (b) CaS: 40.078 u + 32.065 u = 72.143 u (c) Al2O3: 2(26.982 u) + 3(15.999 u) = 101.961 u (d) NH4Cl: 14.007 u + 4(1.0079 u) + 35.453 u = 53.492 u Think About It In determining the formula mass, use as many significant figures in your calculation as listed in the periodic table. Resist the temptation to round up or down, which would make the calculation for the mass less accurate. 2.58. Collect and Organize From the formulas for four ionic compounds, LiCl, TiO2, MgCO3, and CaHPO4, we are to calculate the masses of the formula units. Analyze For each formula we will sum the masses of the elements, making sure that we also account for the number of a particular element present in the formula. Solve (a) LiCl: 6.941 u + 35.453 u = 42.394 u (b) TiO2: 47.867 u + 2(15.999 u) = 79.865 u (c) MgCO3: 24.305 u + 12.011 u + 3(15.999 u) = 84.313 u (d) CaHPO4: 40.078 u + 1.0079 u + 30.974 u + 4(15.999 u) = 136.056 u Think About It In determining the formula mass, use as many significant figures in your calculation as listed in the periodic table. Resist the temptation to round up or down, which would make the calculation for the mass less accurate. 2.59. Collect and Organize For each given formula, we are to determine the number of carbon atoms in each molecule. Atoms, Ions, and Molecules | 63 Analyze Each C in the formula stands for a C atom, and any subscript after the C atom is the number of carbons in that part of the molecule. Solve (a) CO2 has 1 C atom per molecule (b) HOCH2CH2OH has 2 atoms per molecule (c) CH3COOH has 2 atoms per molecule (d) C2H5OH has 2 atoms per molecule Think About It Those molecules are all organic molecules, and writing their fully condensed formulas as CaHbNcOd followed by other elements, if present, is customary. Formulas like HOCH2CH2OH or C2H5OH show how atoms are connected in the molecule. 2.60. Collect and Organize For each given formula in Problem 2.59, we are to determine the number of oxygen atoms in each. Analyze The subscript in each formula after the O atom is the number of oxygens in the molecule. Solve (a) CO2, has 2 atoms per molecule (b) HOCH2CH2OH, has 2 atoms per molecule (c) CH3COOH, has 2 atoms per molecule (d) C2H5OH, has 1 atom per molecule Think About It Those molecules are all organic molecules, and writing their formula as CaHbNcOd followed by other elements, if present, is customary. 2.61. Collect and Organize For a list of five compounds, we are to rank them in order of increasing mass. First, we have to determine their molecular masses. Analyze When we use the masses on the periodic table to calculate the molecular masses, we obtain the following: (a) CS2 = 76.14 u (b) HI = 127.91 u (c) NO2 = 46.01 u (d) HClO = 52.46 u (e) C4H10 = 58.12 u Solve In order of increasing molecular mass: (c) NO2 < (d) HClO < (e) C4H10 < (a) CS2 (e) BF3 > (c) N2O > (b) CO2 > (a) H2O. Think About It In this problem we needed five significant figures for the mass to discriminate the order between CO2 and N2O. 2.63. Collect and Organize For describing a collection of atoms or molecules, we are asked why using the unit dozen to express the number of atoms or molecules we have might not be a good idea. Analyze A dozen is 12 objects and therefore a relatively small group. Solve Although a dozen is a convenient and recognizable unit for donuts and eggs, it is too small a unit to express the very large number of atoms, ions, or molecules present in a mole. 6.022 ´ 1023 atoms 1 dozen ´ = 5.02 ´ 1022 dozen/mole mole 12 atoms Think About It The mole (6.022 ´ 1023) is a much more convenient unit to express the number of atoms or molecules in a sample. 2.64. Collect and Organize We are asked whether equal masses of two isotopes of an element contain the same number of atoms. Analyze The number of atoms present for a given mass is dependent on the molar mass of the substance. Isotopes with higher molar masses will have fewer atoms for a given mass of the element than those with a lower isotopic mass. Solve No, given samples of equal masses, the isotope with the higher molar mass will contain fewer atoms than the isotope of lower molar mass. Think About It That difference may not be significant, however, since many isotopic molar masses are close to each other. 2.65. Collect and Organize In this exercise, we convert the given number of atoms or molecules of each gas to moles. Analyze To convert the number of atoms or molecules to moles, we divide by the Avogadro constant. Solve 14 (a) 4.4 ´ 10 23atoms of Ne = 7.3 ´ 10-10 mol Ne 6.022 ´ 10 atoms/mol (c) 2.5 ´ 1012 molecules of O3 = 4.2 ´ 10-12 mol O3 6.022 ´ 1023 molecules/mol Atoms, Ions, and Molecules | 65 4.9 ´ 10 molecules of NO2 4.2 ´ 1013 molecules of CH4 = 8.1 ´ 10-15 mol NO2 = 7.0 ´ 10-11 mol CH 4 (d) 23 23 6.022 ´ 10 molecules/mol 6.022 ´ 10 molecules/mol 9 (b) Think About It The trace gas with the most atoms or molecules present also has the most moles present. In that sample of air, the amount of the trace gases decreases in the order Ne > CH4 > O3 > NO2. 2.66. Collect and Organize In this exercise, we convert the number of molecules of each gas found in the sample to moles. Analyze To convert the number of molecules to moles, we divide by the Avogadro constant. Solve 13 molecules of H2 (a) 1.4 ´ 10 23 = 2.3 ´ 10-11 mol H2 6.022 ´ 10 molecules/mol 14 atoms of He (b) 1.5 ´ 10 23 = 2.5 ´ 10-10 mol He 6.022 ´ 10 atoms/mol (c) (d) 7.7 ´ 1012 molecules of N2 O = 1.3 ´ 10-11 mol N 2 O 6.022 ´ 1023 molecules/mol 3.0 ´ 1012 molecules of CO = 5.0 ´ 10-12 mol CO 6.022 ´ 1023 molecules/mol Think About It The trace gas with the most atoms or molecules present also has the most moles present. In that sample of air, the amount of the trace gases decreases in the order He > H2 > N2O > CO. 2.67. Collect and Organize From the chemical formulas for various nitrogen compounds with oxygen and hydrogen, we are asked to determine how many moles of nitrogen are in 1 mol of each substance. Analyze The chemical formula reflects the molar ratios of the elements in the compound. If one atom of nitrogen is in the compound’s chemical formula, then 1 mol of nitrogen is in 1 mol of the compound. Likewise, if three atoms of nitrogen are in the chemical formula, 3 mol of nitrogen is present in 1 mol of the substance. Solve (a) One atom of nitrogen is in NO; therefore, 1 mol of NO contains 1 mol of nitrogen. (b) One atom of nitrogen is in NO2; therefore, 1 mol of NO2 contains 1 mol of nitrogen. (c) Two atoms of nitrogen are in NH4NO3; therefore, 1 mol of NH4NO contains 2 mol of nitrogen. (d) Two atoms of nitrogen are in N2O5; therefore, 1 mol of N2O5 contains 2 mol of nitrogen. Think About It If the question had asked how many moles of oxygen were present in 1 mol of N2O5, the answer would be 5 mol of oxygen. 2.68. Collect and Organize From the chemical formulas for various potassium salts, we are asked to determine how many moles of K+ ions are in 1 mol of each substance. Analyze The chemical formula reflects the molar ratios of the elements in the compound. If one atom of potassium is in the compound’s chemical formula, then 1 mol of K+ is in 1 mol of the compound. Likewise, if two atoms of potassium are in the chemical formula, 2 mol of K+ is present in 1 mol of the substance. Solve (a) One atom of potassium is in KBr; therefore, 1 mol of KBr contains 1 mol of K+. (b) Two atoms of potassium are in K2SO4; therefore, 1 mol of K2SO4 contains 2 mol of K+. 66 | Chapter 2 (c) One atom of potassium is in KH2PO4; therefore, 1 mol of KH2PO contains 1 mol of K+. (d) Two atoms of potassium are in K2O; therefore, 1 mol of K2O contains 2 mol of K+. Think About It The number of potassium ions appearing in those chemical formulas balances the charge on the anions in the salts. For example, two potassium cations are needed to produce the neutral salt of SO42–. 2.69. Collect and Organize Given the formulas and the moles of each substance in a pair, we are asked to decide which compound contains more moles of oxygen. Analyze To answer this question, we have to take into account the moles of oxygen present in the substance formulas as well as the initial number of moles specified for each substance. Solve (a) One mole of MnO2 contains 2 mol of oxygen, and 1 mol of Fe2O3 contains 3 mol of oxygen. Therefore, the one mole of Fe2O3 contains more moles of oxygen. (b) One and one-half moles of CO2 contains 3 mol of oxygen, and 1 mol of N2O5 contains 5 mol of oxygen. Therefore, the 1 mol of N2O5 contains more moles of oxygen. (c) Three moles of NO contain 3 mol of oxygen, and 2 mol of SO2 contains 4 mol of oxygen. Therefore, the 2 mol of SO2 contains more oxygen. Think About It We cannot decide which substance has more moles of oxygen by comparing only the amounts of the substances present. If that were the case, we would have incorrectly concluded that 3 mol of NO contains more moles of oxygen than 2 mol of SO2. 2.70. Collect and Organize Given the formulas and the moles of each substance in a pair, we are asked to decide which compound contains more moles of oxygen. Analyze To answer this question, we have to take into account the moles of oxygen present in the substance formulas as well as the initial number of moles specified for each substance. Solve (a) One mole of N2O contains 1 mol of oxygen, and 1 mol of NO2 contains 2 mol of oxygen. Therefore, the 1 mol of NO2 contains more oxygen. (b) One mole of H2SO4 contains 4 mol of oxygen, and 1.5 mol of H2SO3 contains 4.5 mol of oxygen. Therefore, the 1.5 mol of H2SO3 contain more moles of oxygen. (c) One mole of Ca3(PO4)2 contains 8 mol of oxygen, and 1.5 mol of Ca(HCO3)2 contains 9 mol of oxygen. Therefore, the 1.5 mol of Ca(HCO3)2 contains more oxygen. Think About It We cannot decide which substance has more moles of oxygen by comparing only the amounts of substances present. 2.71. Collect and Organize This exercise has us compute the molar masses of various molecular compounds of oxygen. Analyze We can find the molar mass of each gas by adding the molar mass of each element from the periodic table, taking into account the number of moles of each atom present in 1 mol of the substance. Atoms, Ions, and Molecules | 67 Solve (a) SO2: 32.065 + 2(15.999) = 64.063 g/mol (b) O3: 3(15.999) = 47.997 g/mol (c) CO2: 12.011 + 2(15.999) = 44.009 g/mol (d) N2O5: 2(14.007) + 5(15.999) = 108.009 g/mol Think About It The three gases SO2, O3, and CO2 have three atoms in their chemical formulas, but each molecule has a different molar mass because they have different constituent atoms. 2.72. Collect and Organize This exercise has us compute the molar masses of various minerals. Analyze We can find the molar mass of each mineral by adding the molar mass of each element from the periodic table, taking into account the number of moles of each atom present in 1 mol of the mineral. Solve (a) Rhodonite, MnSiO3: 54.938 + 28.086 + 3(15.999) = 131.021 g/mol (b) Scheelite, CaWO4: 40.078 + 183.84 + 4(15.999) = 287.91 g/mol (c) Ilmenite, FeTiO3: 55.845 + 47.867 + 3(15.999) = 151.709 g/mol (d) Magnesite, MgCO3: 24.305 + 12.011 + 3(15.999) = 84.313 g/mol Think About It Scheelite, CaWO4, has the highest molar mass of those minerals; therefore, if we had 100 g of each mineral, that mass of scheelite would have the smallest number of moles. 2.73. Collect and Organize This exercise has us compute the molar mass of various flavorings. Analyze We can find the molar mass of each flavoring by adding the molar mass of each element from the periodic table, taking into account the number of moles of each atom present in 1 mol of the flavoring. Each flavoring contains only carbon (12.011 g/mol), hydrogen (1.0079 g/mol), and oxygen (15.999 g/mol). Solve (a) Vanillin, C8H8O3: 8(12.011) + 8(1.0079) + 3(15.999) = 152.148 g/mol (b) Oil of cloves, C10H12O2: 10(12.011) + 12(1.0079) + 2(15.999) = 164.203 g/mol (c) Anise oil, C10H12O: 10(12.011) + 12(1.0079) + 15.999 = 148.204 g/mol (d) Oil of cinnamon, C9H8O: 9(12.011) + 8(1.0079) + 15.999 = 132.161 g/mol Think About It Each flavoring has a distinctive odor and flavor due in part to its different chemical formula. Another factor, however, in differentiating these flavorings is their chemical structure, or the arrangement in which the atoms are attached, as shown by the structures of those flavorings: H H CH3 O H O H H H H H H OCH3 OCH3 OH Vanillin OH Oil of cloves OCH3 Anise oil Oil of cinnamon 68 | Chapter 2 2.74. Collect and Organize This exercise has us compute the molar masses of various sweeteners. Analyze We can find the molar mass of each sweetener by adding the molar mass of each element from the periodic table, taking into account the number of moles of each atom present in 1 mol of the sweetener. These flavorings contain carbon (12.011 g/mol), hydrogen (1.0079 g/mol), oxygen (15.999 g/mol), nitrogen (14.007 g/mol), and sulfur (32.065 g/mol). Solve (a) Sucrose, C12H22O11: 12(12.011) + 22(1.0079) + 11(15.999) = 342.295 g/mol (b) Saccharin, C7H5O3NS: 7(12.011) + 5(1.0079) + 3(15.999) + 14.007 + 32.065 = 183.186 g/mol (c) Aspartame, C14H18N2O5: 14(12.011) +18(1.0079) + 2(14.007) + 5(15.999) = 294.305 g/mol (d) Fructose, C6H12O6: 6(12.011) + 12(1.0079) + 6(15.999) = 180.155 g/mol Think About It Two of those sweeteners are natural (fructose and sucrose), and two are artificial (saccharin and aspartame). 2.75. Collect and Organize We are asked to convert a mass of SiO2 in quartz in grams to moles. Analyze We need the mass of 1 mol of SiO2 to compute the number of moles in the 7.474 g quartz sample. From the periodic table, we calculate the molar mass to be 60.084 g/mol. Solve 7.474 g × 1 mol = 0.1244 mol 60.084 g Think About It Because the mass of the sample was much lower than the molar mass, the moles of quartz in this sample is quite less than a mole. 2.76. Collect and Organize We are asked to convert a mass of aluminum in kg to moles. Analyze We need the mass of 1 mol of aluminum to compute the number of moles of Al in the 2.8 kg sample. From the periodic table, we see that the molar mass of aluminum is 26.982 g/mol. We also must convert the mass of the aluminum in kg to the mass in grams, using the conversion 1 kg = 1000 g. Solve 2.8 kg × 1000 g 1 mol × = 104 mol or 100 mol (2 significant figures) 1 kg 26.982 g Think About It The number of moles calculated for aluminum is large because we have a large mass of a light element. 2.77. Collect and Organize Given a molar amount of calcium titanate, we are asked to determine the number of moles and mass (in g) of Ca2+ ions in the substance. Atoms, Ions, and Molecules | 69 Analyze The formula for calcium titanate gives us the number of moles of Ca in the compound. Because the mass of the two missing electrons in the Ca2+ cation is negligible, the molar mass of the Ca2+ ion is taken to be the same as the molar mass of Ca. We can use that value to determine the mass of Ca2+ in the 0.25 mol of calcium titanate. Solve Because calcium titanate contains one atom of Ca in its formula, 0.25 mol of CaTiO3 contains 0.25 mol of Ca2+ ions. The mass of Ca2+ ions in the sample, therefore, is 0.25 mol Ca 2+ × 40.078 g = 10 g Ca 2+ 1 mol Think About It Our answer makes sense. One-quarter of a mole of Ca2+ should give us 40/4, or about 10 g, of Ca in the 0.25 mol of calcium titanate. 2.78. Collect and Organize Given a molar amount of aluminum oxide, we are asked to determine the number of moles and mass of O2– ions in the substance. Analyze Aluminum oxide, Al2O3, contains 3 mol of oxygen ions per 1 mol as given by its formula. Because the mass of the two additional electrons in the O2– anion is negligible, the molar mass of the O2– anion is taken to be the same as the molar mass of O. We can use that value to determine the mass of O2– in the 0.55 mol of aluminum oxide. Solve Because aluminum oxide contains three atoms of O in its formula, 0.55 mol of Al2O3 contains 0.55 ´ 3 = 1.65 mol of O2– ions (note that the final answer should be expressed in two significant figures). The mass of O2– ions in the sample, therefore, is 15.999 g 1.65 mol O 2– × = 26 g 1 mol Think About It Be careful to take into account the number of atoms in the formula of a substance. Here we might have wrongly thought that only 0.55 mol of O2– ions was in the substance and computed a mass that would be one-third the true mass present in 0.55 mol of aluminum oxide. 2.79. Collect and Organize Between two balloons filled with 10.0 g of different gases, we are to choose which balloon has more particles. Analyze The balloon with more particles has more moles. The greater number of moles contained in 10.0 g of a gas is for the gas with the lowest molar mass. A gas with a lower molar mass contains more moles in a 10.0 g mass and, therefore, has more moles than a 10.0 g mass of a higher molar mass gas. Solve (a) The molar mass of CO2 is about 44 g/mol, and the molar mass of NO is about 30 g/mol. Therefore, the balloon containing NO has more particles. (b) The molar mass of CO2 is 44 g/mol, and the molar mass of SO2 is 64 g/mol. Therefore, the balloon containing CO2 has more particles. (c) The molar mass of O2 is 32 g/mol, and the molar mass of Ar is 40 g/mol. Therefore, the balloon containing O2 has more particles. Think About It Although we could numerically determine the number of moles of gas in each balloon to make the comparisons in this problem, doing so is unnecessary because we know the relationship between moles and molar mass. 70 | Chapter 2 2.80. Collect and Organize If we have equal masses of two salts with different formula weights, which contains more ions? Analyze To determine the salt with more ions, we have to take into account not only the formula weight difference between the salts but also the number of ions in the chemical formula for the salt. If both salts contain only one cation and one anion, then the salt sample with the lowest formula mass contains more ions. If the salts contain different numbers of ions, then we have to compute the moles of ions in each case, by assuming a certain amount (in grams) of the salts. Solve (a) The formula mass of NaBr is 102.9 g/mol, and that of KCl is 74.6 g/mol. Because each salt has two ions in its formula, the one with the lowest formula mass has more ions for a given mass of salt, KCl. (b) The molar mass of NaCl is 58.44 g/mol, and the molar mass of MgCl2 is 95.21 g/mol. Because NaCl has two ions in its formula and MgCl2 has three, which has more ions is not immediately obvious. To determine that, assume a convenient number of grams (10 g), convert the grams to moles of ions for each salt, and compare: 1 mol 2 ions × = 0.34 mol ions in NaCl 58.44 g NaCl formula 1 mol 3 ions 10 g MgCl2 ´ ´ = 0.32 mol ions in MgCl2 95.21 g MgCl2 formula Comparing those results, we see that, for a given mass, more ions are in NaCl than in MgCl2. (c) The formula mass for CrCl3 is 158.355 g/mol, and the formula mass for Na2S is 78.045 g/mol. Because CrCl3 has four ions in its formula and Na2S has three, which has more ions is not immediately obvious. To determine that, assume a convenient number of grams (10 g), convert the grams to moles of ions for each salt, and compare: 10 g NaCl × 1 mol 4 ions × = 0.25 mol ions in CrCl3 158.355 g CrCl3 formula unit 1 mol 3 ions 10 g Na 2S × × = 0.38 mol ions in Na 2S 78.045 g Na 2S formula unit Comparing those results, we see that, for a given mass, more ions are in Na2S than in CrCl3. 10 g CrCl3 × Think About It We can easily imagine in part b that the salt with more ions, not the one with the lowest formula mass, will have more ions. That is not necessarily the case, however, so we must compare them numerically. An example is for MgF2 (formula weight = 62.3 g/mol, 0.48 mol ions in 10 g) compared with NaCl (formula weight = 58.4 g/mol, 0.34 mol ions in 10 g). 2.81. Collect and Organize Given different amounts of different aluminum-containing substances we are asked to rank them in order of increasing amounts of aluminum. Analyze To make the comparisons, we can conveniently convert all the amounts to moles of aluminum. (a) One mole of Al2O3 will contain 2 mol Al. (b) 76 g of Al2O3 will contain 76 g / (101.96 g/mol) ´ 2 mol Al = 1.49 mol Al. (c) 27 g of Al will contain 27 g / (26.982 g/mol) = 1.00 mol Al. (d) 1.1 ´ 1024 atoms of Al will contain 1.1 ´ 1024 atoms / (6.022 ´ 1024 atoms/mol) = 1.8 mol Al. (e) 1.2 mol of AlCl3 will contain 1.2 mol of Al. Solve The order of these substance in increasing amounts of aluminum: (c) 27 g of Al < (e) 1.2 mol AlCl3 < (b) 76 g Al2O3 < (d) 1.1 ´ 1024 atoms of Al < (a) 1 mol Al2O3 Atoms, Ions, and Molecules | 71 Think About It It’s easy to forget in (b) that for every mole of Al2O3 calculated from the molar mass, there are two moles of Al. 2.82. Collect and Organize We are to compare the size of a block of 1 mol of aluminum with the size of a block of 1 mol of strontium. Analyze Starting with 1 mol of each element, we can obtain the mass by multiplying 1 mol by the element’s molar mass (Al = 26.982 g/mol, Sr = 87.62 g/mol). We can find the volume that 1 mol of the element occupies by multiplying the mass by the inverse of the given densities (Al = 1 cm3/2.70 g, Sr = 1 cm3/2.64 g). The length of a side of the cube for each element is the cube root of the volume. Solve 1 mol Al × 26.982 g 1 cm 3 × = 9.99 cm 3 1 mol 2.70 g Length of side of Al cube = 3 9.99 cm 3 = 2.15 cm 1 mol Sr × 87.62 g 1 cm 3 × = 33.2 cm 3 1 mol 2.64 g Length of side of Sr cube = 3 33.2 cm 3 = 3.21 cm From those results, the aluminum cube is smaller, with dimensions of 2.15 cm ´ 2.15 cm ´ 2.15 cm. Think About It The volume of the Sr mole is more than three times greater than that of the Al mole. The volume that 1 mol of a substance takes up is called its molar volume. We have calculated that elemental property here for aluminum and strontium. 2.83. Collect and Organize In this question we are asked how mass spectrometry provides information about a molecule. Analyze Mass spectrometry plots the mass-to-charge ratio (m/z) versus relative intensity or relative abundance of the peaks. Solve In mass spectrometry molecules of the compound are ionized with high-energy electrons to form +1 cations. When a molecule loses one electron, the ion that forms is called the molecular ion, M+. This and other cations produced when the molecule fragments are separated based on their mass-to-charge ratios (m/z). The m/z ratio of a molecular ion with a charge of 1+ corresponds to the molecular mass of the compound. Often, but not always, this is the peak that appears at the highest m/z ratio. Each fragment ion is also counted by the detector giving relative intensities of each fragment ion. Think About It Remember that it is the m/z ratio that is determined in mass spectrometry. If a cation has a +2 charge—that is, it loses two electrons in the ionization step—it will appear at an m/z value that is half of its mass. 2.84. Collect and Organize We are to explain the possible origin of the small peak at m/z = 15 u in the mass spectrum of CH4, which shows a molecular ion peak at m/z = 16 u. Analyze Mass spectrometry plots the mass-to-charge ratio (m/z) versus relative intensity or relative abundance of the peaks. Peaks in the mass spectrum may be due to the presence of isotopes of the elements composing the substance or from fragmentation of the molecule. 72 | Chapter 2 Solve The isotopes that might be present in CH4 (with mass of 16 u) would be present in the mass spectrum at higher m/z values (because the next most abundant isotopes for carbon and hydrogen would be 13C and 2H). Therefore, the m/z peak at 15 must be due to fragmentation; it must be due to the loss of an H atom from CH4. Think About It If 2H and 13C were more abundant or our mass spectrometer were very sensitive, we would expect to see m/z peaks at 17 u (CH32H or 13CH4), 18 u (CH22H2 or 13CH32H), 19 u (13CH22H2 or CH2H3), 20 u (C2H4 or 13CH32H), and 21 u (13C2H4). 2.85. Collect and Organize We are asked to consider whether the mass spectrum of CO2 and that of C3H8 would show the same molecular ion peak. Analyze Mass spectrometry plots the mass-to-charge ratio (m/z) versus relative intensity or relative abundance of the peaks. Molecules with the same molecular mass will show the same m/z ratio for their molecular ions. Solve To the nearest unit, the molecular mass of CO2 is 44 u and for C3H8 the molecular mass also is 44 u. Therefore, both molecules will show the same molecular ion peak at m/z 44. Think About It That result does not mean that the mass spectra of these two compounds will be the same, however. The two molecules will probably show different fragmentation patterns in their mass spectra. 2.86. Collect and Organize We are asked to consider whether the mass spectrum of CO2 and that of C3H8 would show the same pattern. Analyze Mass spectrometry plots the mass-to-charge ratio (m/z) versus relative intensity or relative abundance of the peaks. Molecules with the same molecular mass will show the same m/z ratio for their molecular ions. In the mass spectrometer, the molecules may also (and often do) fragment in distinctive patterns. Solve Because CO2 and C3H8 have very different compositions, we would expect that, despite their having molecular ion peaks at the same m/z value due to their identical molecular masses, their fragmentation patterns would be different. Think About It The fragmentation pattern of a molecule in the mass spectrum can be used as a fingerprint to identify a molecule. 2.87. Collect and Organize For the explosive materials given we are to calculate the masses of their molecular ion peaks in the mass spectrum. Analyze The molecular ion is formed through the loss of one electron from the molecule. The mass of an electron is negligible, so the mass of the molecular ion peak is that of the molecule. We can compute the mass of a molecule using masses of the elements from the periodic table and their ratios in the molecular formula. Solve (a) C3H6N6O6: (3 ´ 12 u) + (6 ´ 1 u) + (6 ´ 14 u) + (6 ´ 16 u) = 222 u (b) C4H8N8O8: (4 ´ 12 u) + (8 ´ 1 u) + (8 ´ 14 u) + (8 ´ 16 u) = 296 u (c) C5H8N4O12: (5 ´ 12 u) + (8 ´ 1 u) + (4 ´ 14 u) + (12 ´ 16 u) = 316 u (d) C14H6N6O12: (14 ´ 12 u) + (6 ´ 1 u) + (6 ´ 14 u) + (12 ´ 16 u) = 450 u Atoms, Ions, and Molecules | 73 Think About It These explosives have the following structures. What common features do you see? (a) O O N (b) O N N N O N O O O N O N N O O O O N O O (c) N N N N O (d) N N O N O O O O O N O O O O NO2 N O O2 N O2 N NO2 NO2 O2 N 2.88. Collect and Organize For the gases emitted from landfills, CH4, C2H6S, and C2H2Cl2, we are to calculate the masses of their molecular ion peaks in the mass spectrum. Analyze The molecular ion is formed through the loss of one electron from the molecule. The mass of an electron is negligible, so the mass of the molecular ion peak is that of the molecule. We can compute the mass of a molecule by using masses of the elements from the periodic table and their ratios in the molecular formula. Solve CH4: (1 ´ 12 u) + (4 ´ 1 u) = 16 u C2H6S: (2 ´ 12 u) + (6 ´ 1 u) + (1 ´ 32 u) = 62 u C2H2Cl2: (2 ´ 12 u) + (2 ´ 1 u) + (2 ´ 35.5 u) = 97 u Think About It Chlorine has two relatively abundant isotopes, 35Cl (78% abundant) and 37Cl (24% abundant), so we would expect to see three molecular ions peaks for C2H2Cl2: one for C2H235Cl2, one for C2H235Cl37Cl, and one for C2H237Cl2. At what m/z values would you see each of those peaks? 2.89. Collect and Organize From the mass spectrum for H2S shown in Figure P2.89, we are asked to explain how the spectrum shows the sequential loss of H from the molecule. The low-intensity peaks at m/z ratios above the molecular ion peak suggests the presence of isotopes. We may need the mass numbers and natural abundances of the naturally occurring isotopes of sulfur (Table A3.3). Analyze The possible combinations of the sulfur isotopes in H2S are H232S (34 u), H233S (35 u), H234S (36 u) and H235S (37 u). Loss of hydrogen from these would decrease m/z by 1 u. Solve The most intense peak, at 34 u, is the molecular ion H232S+ (containing the most abundant isotope of sulfur). The peaks at 33 and 32 u are produced when electron bombardment of H232S+ ions fragments them, forming ions H32S+ and 32S+ atoms. The source of the small peak at 36 u is probably the molecular ion of H234S. The even smaller peak at 35 u is probably a combination of mostly H34S and a very little H233S. 74 | Chapter 2 Think About It Fragmentation patterns such as this for H2S can be important in using mass spectrometry to identify compounds. 2.90. Collect and Organize From the mass spectrum for AsH3 shown in Figure P2.90 and given that arsenic has only one stable isotope, we are to assign a formula to each peak in the mass spectrum. Analyze Arsine, AsH3, has a molecular mass of 78 u. Solve The molecular ion peak AsH3+ is assigned to m/z 78 u. The other three peaks are due to the loss of H atoms with AsH2+ at 77 u, AsH+ at 76 u, and As+ at 75 u. Think About It The AsH+ peak is the most intense peak, not the molecular ion peak, in this mass spectrum—showing that the molecular ion is not always the most intense peak in the spectrum. 2.91. Collect and Organize Thomson’s experiment revealed the electron and its behavior in magnetic and electric fields. In this question, we examine his experiment. Analyze Thomson showed that a cathode ray was deflected by a magnetic field in one direction and by an electric field in the other direction. He saw the deflection of the cathode ray when the ray hit a fluorescent plate at the end of his experimental apparatus, as shown in the textbook in Figure 2.2. The cathode ray was deflected by the electrically charged plates as shown. We can imagine the experiment proceeding from no voltage across the charged plates to low voltages and then to higher voltages. From this thought experiment, the ray must be deflected more by an increase in the voltage across the charged plates. Thomson reasoned that the cathode ray was composed of tiny charged particles, which he called “corpuscles.” Solve (a) Today we call the particles that make up the cathode rays electrons. (b) The beam of electrons was deflected between the charged plates because they were attracted to the oppositely charged plate and repelled by the negatively charged plate as the beam passed through the electric field. Indeed, in Figure 2.2b we see the beam deflected up toward the (+) plate. (c) If the polarities of the plates were switched, the electron would still be deflected toward the positively charged plate, which would now be at the bottom of the tube, and the light spot would appear on the bottom of the screen instead of at the top as in Figure 2.2b. Think About It That experiment was key to the discovery of subatomic particles, which until then in the atomic theory were not known to exist. It was believed before that time that the atom was the smallest indivisible component of matter. 2.92. Collect and Organize In this problem we are given the masses of the three isotopes of magnesium, from which we can figure out the nuclides (24Mg = 23.9850 u, 25Mg = 24.9858 u, and 26Mg = 25.9826 u). We are also given that the abundance of 24 Mg is 78.99%. From that information and the average (weighted) atomic mass of magnesium from the periodic table (24.305 u), we must calculate the abundances of the other two isotopes, 25Mg and 26Mg. Analyze The average atomic mass is derived from a weighted average of the isotopes’ atomic masses. If x = abundance of 25 Mg and y = abundance of 26Mg, the weighted average of magnesium is (0.7899 ´ 23.9850) + 24.9858x + 25.9826y = 24.305 Atoms, Ions, and Molecules | 75 Because the sum of the abundances of the isotopes must add up to 1.00, 0.7899 + x + y = 1.00 So x = 1.00 – 0.7899 – y = 0.2101 – y Substituting this expression for x in the weighted average mass equation gives (0.7899 ´ 23.9850) + 24.9858(0.2101 – y) + 25.9826y = 24.305 Solve 18.94575 + 5.249517 – 24.9858y + 25.9826y = 24.305 0.9968y = 0.1097 y = 0.1101 So x = 0.2101 – 0.1101 = 0.1000 The abundance of 25Mg is x ´ 100 = 10.00%, and the abundance of 26Mg is y ´ 100 = 11.01%. Think About It Although the abundances of 25Mg and 26Mg are nearly equal to each other at the end of that calculation, we cannot assume that in setting up the equation. We have to solve the problem algebraically by setting up two equations with two unknowns. 2.93. Collect and Organize Using the periodic table in Figure P2.93, we are to assign atomic numbers to the highlighted elements. Analyze The atomic number increases by 1 as we move from hydrogen in the upper left-hand corner of the periodic table to the right and then move down to the next row. Solve The atomic numbers for the highlighted elements are as follows: for blue, 9; for yellow, 16; for green, 25; for red, 33; and for lilac, 73. Think About It Be careful in identifying the atomic number for the lilac element. The lanthanide elements occur in the periodic table after La and continue through Lu (atomic numbers 58–71) 2.94. Collect and Organize Given the diameter of spherical silver nanoparticles and the number of atoms in one nanoparticle, we are to calculate the number of nanoparticles in 1.00 g. Analyze If we divide the mass (1.00 g) of silver by the molar mass of silver and then multiply the result by the Avogadro constant, we will obtain the number of silver atoms in 1.00 g of nanoparticles. If we then divide that result by the number of atoms of silver in one nanoparticle, we will obtain the number of nanoparticles in the 1.00 g sample. Solve 1.00 g × 1 mol 6.022 × 10 23 atoms 1 nanoparticle × × = 1.2 × 1014 nanoparticles 107.87 g mol 4.8 × 10 7 atoms Think About It We did not need the information given in this problem pertaining to the diameter of the nanoparticles. 76 | Chapter 2 2.95. Collect and Organize For CdS and CdSe quantum dots we are asked to calculate their formula masses, find how many grams of Se are in a sample of CdSe that contains 2.7 ´ 107 atoms of Cd, and calculate the masses of Cd and S in 4.3 ´ 10–15 g of CdS. Analyze The ratio of Cd to both Se and S in these quantum dots is 1:1. We can use the masses of the elements on the periodic table for Cd, Se, and S to calculate the formula masses of the CdSe and CdS quantum dots. Then we can use the formula mass of CdS with the masses of Cd and S to calculate the masses of Cd and S in a 4.3 ´ 10–15 g sample of CdS. Solve (a) Formula mass of CdS: 112.41 + 32.065 = 144.48 u Formula mass of CdSe: 112.41 + 78.96 = 191.37 u (b) 2.7 × 10 7 atoms Cd × 1 atom Se = 2.7 × 10 7 atoms Se 1 atom Cd (c) 1 formula unit CdS 1 Cd atom 112.41 u × × = 3.3 × 10 −15 g Cd 144.48 u 1 formula unit CdS 1 Cd atom 1 formula unit CdS 1 S atom 32.065 u 4.3 × 10 −15 g CdS × × × = 9.5 × 10 −16 g S 144.48 u 1 formula unit CdS 1 S atom 4.3 × 10 −15 g CdS × Think About It Which do you think would contain more formula units: 1.0 g of CdSe or 1.0 g of CdS? 2.96. Collect and Organize Given that the average molar mass of air is 28.8 g/mol and that each mole of air had 411.31 ´ 10–6 mol of CO2, we are to calculate how many micrograms of CO2 that sample of air contains. Analyze This is a problem in unit analysis. If we multiply the moles of CO2 per mole of air by the molar mass of air and then multiply that result by the molar mass of CO2, we will obtain the mass of CO2 per gram of air. Solve 411.31× 10 −6 mol CO 2 mol air 44.0 g CO 2 1 µg CO 2 × × × = 628 µg CO 2 / g air mol air 28.8 g air mol CO 2 1× 10 –6 g CO 2 Think About It Always label your values with units and make sure that they cancel correctly to arrive at your final answer. 2.97. Collect and Organize From the mass spectrum of cocaine shown in Figure P2.97, we are to determine the molar mass. Analyze The molar mass can be determined from a mass spectrum from the molecular ion peak. The molecular ion peak is (usually) the peak at the largest m/z value. Solve The largest m/z value is at 303 u. Therefore, the molar mass of cocaine is 303 g/mol. Think About It The molecular ion peak is not the most intense peak in this mass spectrum. Atoms, Ions, and Molecules | 77 2.98. Collect and Organize From the mass spectrum shown in Figure P2.98, we are to identify the molecular ion peak and show that it is consistent with the formula C21H28O2. Analyze The molecular ion peak is (usually) the peak at the largest m/z value. We can compute the expected mass of the compound by using values from the periodic table for C, H, and O. Solve The molecular ion peak is at 312 u, and that is consistent with the mass of C21H28O2, which is (21 ´ 12 u) + (28 ´ 1 u) + (2 ´ 16 u) = 312 u. Think About It The molecular ion peak is not the most intense peak, but the largest mass peak, in a mass spectrum. 2.99. Collect and Organize To determine the number of moles and atoms of carbon in the Hope Diamond, we have to first convert its given mass of 45.52 carats to mass in grams. Analyze To convert the mass of the diamond to grams, we use the relationship 1 carat = 200 mg. The moles of carbon atoms is equal to the mass of the diamond divided by the molar mass of carbon (12.011 g/mol); the number of carbon atoms in the diamond is the number of moles of carbon multiplied by the Avogadro constant. Solve (a) The mass of the diamond in grams is 200 mg 1g ´ = 9.104 g 1 carat 1000 mg The number of moles of carbon atoms in the diamond is 1 mol 9.104 g × = 0.7580 mol of C 12.011 g (b) The number of carbon atoms in the diamond is 45.52 carats ´ 0.7580 mol ´ 6.022 ´ 1023 atoms = 4.565 ´ 1023 atoms of C 1 mol Think About It Diamond is the hardest natural substance and is an electrical insulator while being an excellent conductor of heat. 2.100. Collect and Organize Given the molecular formulas for three compounds that compose the minerals called feldspars, we are to rank them in order of increasing percentage Al by mass. Analyze The percent by mass for each of the compounds is determined by dividing the mass of the aluminum in the formula (mol of Al ´ 26.982 u) by the mass of the formula unit of the compounds, then multiplying by 100 for the percent. Using the periodic table, we can calculate the masses of the compounds: (a) NaAlSi3O8 = 22.990 u + 26.982 u + 3(28.086 u) + 8(15.999 u) = 262.22 u (b) KAlSi3O8 = 39.098 u + 26.982 u + 3(28.086 u) + 8(15.999 u) =278.33 u (c) CaAl2Si2O8= 40.078 u + 2(26.982 u) + 2(28.086 u) + 8(15.999 u) = 278.21 u. Solve (a) NaAlSi3O8 26.982 u × 100 = 10.290% 262.22 u 78 | Chapter 2 (b) KAlSi3O8 26.982 u × 100 = 9.6942% 278.33 u (c) CaAl2Si2O8 2 × 26.982 u × 100 = 19.397% 278.21 u In order of increasing percent by mass of aluminum: (b) KAlSi3O8 < (a) NaAlSi3O8 < (c) CaAl2Si2O8 Think About It Remember that in multiplication or division we use the rule of the number of the least significant figures; therefore, the percentages here have 5 significant figures.