Aqueous Solutions and Solubility Ch 2

Aqueous Solutions and Solubility Chapter 2 Problem 2.40. (a) We can use the information that the energy required to remove electrons from gaseous silver atoms to form gaseous silver cations is 731 kJ·mol–1 and that 296 kJ·mol–1 of energy is released when gaseous iodine atoms gain electrons to form gaseous iodide anions, to construct an energy diagram to find the energy change for the net reaction: Ag(g) + I(g) Ag+(g) + I–(g). The loss of an electron by each Ag(s) atom and gain of an electron by each I(g) atom is represented on the diagram by the slanting arrow showing that the electron lost by a silver atom is gained by an iodine atom, so there is no net loss or gain of electrons in the overall reaction The diagram shows that the net reaction energy is 435 kJ·mol–1. (b) Use the result from part (a) and the lattice energy for AgI(s) crystals, Elattice = 887 kJ·mol–1, to draw an energy diagram analogous to Figure 2.14 and use it to find Extal form for the formation of ionic crystals of AgI(s) from the gaseous atoms. Problem 2.41. In the energy diagrams we have seen in the textbook or constructed for the formation of solid ionic compounds, we have seen this relationship among the energies represented on the diagrams: Extal form = Eion form + (– Elattice) = Eion form – Elattice For KBr(s) crystals, the problem statement says that Elattice = 689 kJ·mol–1. The formation of separate gaseous atoms of potassium, K(g), and bromine, Br(g), from the ionic crystal requires 594 kJ·mol–1, so Extal form (energy change for the reverse of process, forming the ionic crystal from the gaseous atoms) = –594 kJ·mol–1. Substituting these values in the preceding equation, gives: –594 kJ·mol–1 = Eion form – (689 kJ·mol–1) Eion form = 95 kJ·mol–1 An energy diagram incorporating these data is shown here (not quite to scale, so the numeric values can be included with the arrows representing them): Problem 2.42. As in the solution for Problem 2.41, to find the lattice energy, Elattice, for magnesium fluoride, MgF2(s), we use the energy relationship: Extal form = Eion form + (– Elattice) = Eion form – Elattice Substituting the values given in the problem statement, Extal form = –1424 kJ·mol–1 and Eion form = 1533 kJ·mol–1, we get: –1424 kJ·mol–1 = (1533 kJ·mol–1) – Elattice Elattice = 2957 kJ·mol–1 An energy diagram incorporating these data is shown here: Problem 2.43. (a) When ammonium acetate, NH4C2H3O2 [= (NH4+)(C2H3O2–)], is dissolved in water, the mixture becomes quite cold. This observation means that the dissolution process is taking thermal energy from its surroundings, the molecules in the solution, the container, and your hand, if you are holding the container. A reaction that requires an input of thermal energy, Ereaction > 0, is endothermic. (b) The energy change for the process of dissolving ionic solutes in water can be broken into two parts. Lattice energy is required to break the coulombic electrical attractions between cations and anions in the lattice while hydration energy is released as water molecules surround these ions and are attracted to them by coulombic attractions. The difference between these two energies determines whether thermal energy will be absorbed (endothermic) or released (exothermic) by the dissolving process. For ammonium acetate, the dissolving process absorbs energy, so breaking the lattice attractions must take more energy than is gained back by hydration of the ions. An energy diagram that represents this case is: [Note that water is shown separately with the solid and gaseous ions as a reminder that the system as a whole contains the water into which the solid dissolves and which hydrates the ions. Sometimes, as in Figures 2.15 and 2.16 in the text and the solution to Problem 2.44 below, the water is omitted for simplicity, but this is probably not a good idea.] (c) The ions present in a solution of ammonium acetate are the same as the ions in the solid crystal, ammonium cation and acetate anion, which we write as NH4+(aq) and C2H3O2–(aq) to signify that the ions are hydrated, that is, surrounded by polar water molecules that are attracted by the ionic charges. (d) The molecular level interactions of hydrated ions are represented in Figure 2.9 by showing the negative oxygen end of water molecules oriented toward the positive cations and the positive hydrogen end of water molecules oriented toward the negative cations. We expect similar orientations and interactions for the NH4+ and C2H3O2– ions, but there is an added factor for these ions, because they can hydrogen bond with the water molecules and form even more directed interactions. Some of these H bonds are illustrated here: H O H H H H O H H N H H O H H O H H O H H H O O CH3 C O H O H H H O Problem 2.44. (a) Use the data in Table 2.3 for LiCl, Elattice = 861 kJ·mol–1 and Ehydration = –898 kJ·mol–1, to sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to find that Edissolve = –37 kJ·mol–1. Energy is released, so the dissolution reaction is exothermic. (b) Use the data in Table 2.3 for KBr, Elattice = 689 kJ·mol–1 and Ehydration = –670 kJ·mol–1, to sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to find that Edissolve = 19 kJ·mol–1. Energy is required, so the dissolution reaction is endothermic. Problem 2.45. We are asked to try to explain why lithium sulfate, Li2SO4, is quite soluble in water (261 g·L–1) while calcium sulfate, CaSO4, is essentially insoluble (4.9 mg·L–1). To see what factors might be responsible for such a difference, let’s compare the data for a pair calcium ionic compounds in Table 2.3: Compound Elattice kJ·mol–1 Ehydration kJ·mol–1 CaCl2(s) 2260 –2337 CaCO3(s) 2804 –2817 In CaCl2(s) and CaCO3(s), we have the cation with a 2+ charge and anions with a 1– charge and a 2– charge, respectively. CaCl2(s) is quite soluble in water, as you found in Investigate This 2.22, and CaCO3(s), marble or chalk, is quite insoluble. As a first approximation, these compounds are rather like Li2SO4(s) and CaSO4(s), where we have the anion with a 2– charge and cations with a 1+ charge and 2+ charge, respectively. The ion-ion interactions in the solids and ion-polar solvent interactions in aqueous solution should be roughly the same for an ionic compound with 1+ cations and a 2– anion as for a compound with a 2+ cation and 1– anions. If this supposition is correct, we see from the comparison in this table that the hydration energy should substantially outweigh the lattice energy, for a 2:1 ionic compound like Li2SO4(s) compared to a 2:2 ionic compound like CaSO4(s). As a first suggestion about the factor responsible for the high solubility of Li2SO4(s) compared to CaSO4(s), we would probably say that the hydration energy (due to solvation of the ions) favors dissolution of Li2SO4(s), Edissolve = –67 kJ·mol–1, more than CaSO4(s), Edissolve = –13 kJ·mol–1. This is a large factor in this case, but we have to be careful making to much of this argument, since we know that there are soluble ionic compounds whose dissolution is endothermic, so energetics cannot be the whole picture. Problem 2.46. We asked to describe or represent in three different ways what happens when sodium sulfate, Na2SO4(s), dissolves in water. First we describe the dissolution in words. Water molecules are attracted to the Na+ and SO42– ions in the solid crystal lattice, especially those at the edges and corners of the crystals. The negative oxygen ends of the water molecules are attracted to the Na+ and positive hydrogen ends in water are attracted to SO42-. These ion-dipole attractions compete with the ion-ion attractions, lattice energy, that holds Na+ and SO42- in the crystal. In the case of a soluble ionic compound like Na2SO4(s), the ion-dipole attractions are finally successful in the tug of war and the ions are broken away from the crystal. Once the ions are broken away from the crystal, more water molecules surround each ion creating a hydration layer. These hydration layers create a shield, making it difficult for hydrated ions that are oppositely charged to get too close to each other. In our example, hydrated Na+ and hydrated SO42- do not interact much with one another (until their concentrations become so high that they are forced close to one another because there are so many of them). An ionic equation that succinctly represents this process is usually written as: Na2SO4(s) 2Na+(aq) + SO42-(aq) Although water is obviously involved in the process, it is usually left out of such equations, because its stoichiometry is not easily represented. If we just add H2O(l) to the reactant side of the equation, it might be interpreted as one molecule of water for each formula unit of Na2SO4 dissolved, and that would be misleading. We just have to remember that water is an active reactant in the dissolution process, not an inert by-stander. A molecular level representation of the dissolution process is shown in this very rough sketch showing a two-dimensional crystal interacting with solvent (water): S + Na+ cation 2– SO4 2– anion S solvent, water molecules + + + 2– + + 2– 2– + S + S S S S S 2– S + S + + S S S S S 2– S 2– + S + S S S S For simplicity, the many water molecules that are not interacting with the ions are omitted and the polar ends of the water molecules are not represented. See Figure 2.9(b) for a representation of the orientation of the water molecules with respect to the cations and anions. Problem 2.47. (a) The mixing process described in this problem is represented in this table, modeled after Table 2.4 in the text: Before Mixing Positive ion(s) Negative ion(s) Conductivity? Precipitate? Na3PO4 solution Na+ (aq) PO4-(aq) yes CaBr2 solution Ca2+(aq) Br–(aq) yes After Mixing Na3PO4 and CaBr2 Na+(aq)+ Ca2+(aq) PO4- (aq)+ Br– (aq) yes yes (b) After mixing, two new combinations of cations and anions are possible: NaBr and Ca3(PO4)2. Our solubility rules say that ionic compounds of alkali metal cations and halide anions are soluble, so sodium bromide, NaBr, is likely to be a soluble compound. Calcium phosphate, Ca3(PO4)2, must be the precipitate and it fits our solubility rule that says that ionic compounds of multiply-charged cations and anions are likely to be insoluble. (c) This diagram is a simple molecular level representation of this mixing and reaction (precipitate formation) with only enough of each ion shown to represent the stoichiometry of the reaction: Note that each of the individual solutions that are mixed contains ions and there are also ions in the solution remaining after the solid has precipitated. The presence of the ions in all three solutions explains why they are all conduct an electric current, as shown in the table above. (d) The complete ionic equation that represents the precipitation reaction involves all four ions: 6Na+(aq) + 2PO4–(aq) + 3Ca2+(aq) + 6 Br–(aq) 6Na+(aq) + 6Br--(aq) + Ca3(PO4)2(s) (e) The net ionic equation that represents the precipitation reaction involves only the two ions that react to form the precipitate (the spectator ions are omitted): 3Ca2+(aq) + 2PO4–(aq) Ca3(PO4)2(s) Problem 2.48. (a) When aqueous solutions of potassium chloride, KCl, and sodium bromide, NaBr, are mixed, the solution contains all four ions, Na+(aq), K+(aq), Cl–(aq), and Br–(aq), and no precipitate is formed. We can conclude that both NaCl(s) and KBr(s) are water soluble, since the solids could have formed from this mixture, but did not. We also note that all of these ionic compounds are combinations of an alkali metal cation and a halide anion, which our solubility rules predict will be soluble. (b) This molecular level representation, similar to Figure 2.17, illustrates the result when the two solutions are mixed. (c) The complete ionic equation representing what happens when the two solutions are mixed involves all four ions: K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) The ions are present in two separate solutions on the left-hand (reactant) side of this equation and together in a single solution on the right-hand (product) side. The separation on the left is not represented in this standard ionic equation, but we can amend it slightly to suggest the separation by bracketing the separate solution components on the left: [K+(aq) + Cl-(aq)] + [Na+(aq) + Br-(aq)] K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) Since there is no net reaction (no apparent reaction of any kind), there is no net ionic equation for this mixing. In a sense, all the ions are spectator ions. Problem 2.49. The solubility rules indicate that multiple charged cations and anions tend to form insoluble ionic compounds. Barium sulfate, BaSO4, falls into this category. The mixture that doctors use to x-ray the gastrointestinal (GI) tract is a suspension of insoluble solid barium sulfate in water. The solubility of the solid ionic compound is so low that the suspension contains only a tiny amount of Ba2+ cation. Its concentration is below the toxic level of Ba2+, so the patient is not harmed (except by having to swallow a substantial amount of chalky suspension). Problem 2.50. We are asked to suggest a sequence of selective precipitation reactions to separate Ag+, Ba2+, and Fe3+ from solution in which all three cations are present. The information we have to work with is in this table: Cation Test Solution Ag+(aq) Ba2+(aq) Fe3+(aq) NaCl ppt no ppt no ppt NaOH ppt no ppt ppt Na2SO4 no ppt ppt no ppt The objective of our sequence of additions of the test solutions (which we might also call the separation reagents) is to precipitate the cations one at a time without forming any mixtures of precipitates that contain more than one cation. For example, if we were to add the NaOH test solution to the original mixture of cations, both the Ag+(aq) and Fe3+(aq) cations would react to form precipitates, so the solid product would be a mixture of solids and we would not have separated the cations from one another. However, if the Ag+(aq) cation had already been removed from the solution, addition of the NaOH test solution would result in a precipitate containing the Fe3+(aq) cations and we would have the Fe3+(aq) cations as a separate precipitate. Our strategy is to look for a test solution that will precipitate only one of the cations, use it to precipitate that cation and then use a second test solution that will precipitate one or the other of the remaining cations. There are two ways to start our sequence. (1) We can add the NaCl test solution to precipitate the Ag+(aq) cation as AgCl(s), leaving behind a solution containing the Ba2+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added with the test solution]. Or, (2) we can begin by adding the Na2SO4 test solution to precipitate the Ba2+(aq) cations as BaSO4(s), leaving behind a solution containing the Ag+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added with the test solution]. To pick the next step in sequence (1), consider this table that tells us how our test solutions react with the remaining cations: Cation 2+ Test Solution Ba (aq) Fe3+(aq) NaOH no ppt ppt Na2SO4 ppt no ppt We have two choices, both of which will separate these two cations from one another. In the pathway that we will label (1a), we add NaOH test solution to the solution we get after removing the AgCl(s) in order to precipitate the Fe3+(aq) as Fe(OH)3(s). The solution now contains the Ba2+(aq) [and the Na+(aq) cations added with the test solutions]. The cations are now separated as two precipitates, AgCl(s) and Fe(OH)3(s), with the remaining cation, Ba2+(aq), in solution. If we need to have the Ba2+ separated from all the Na+(aq) also in the solution, we can add the Na2SO4 test solution to precipitate it as BaSO4(s). The net ionic reactions for the three steps of pathway/sequence (1a) [and the alternative (1b)] are: Pathway (1a) Ag+(aq) + Cl–(aq) 3+ – Fe (aq) + 3OH (aq) AgCl(s) Fe(OH)3(s) Pathway (1b) Ag+(aq) + Cl–(aq) 2+ 2– Ba (aq) + SO4 (aq) AgCl(s) BaSO4(s) Ba2+(aq) + SO42–(aq) BaSO4(s) Fe3+(aq) + 3OH–(aq) Fe(OH)3(s) The net ionic reactions for the three steps of pathway (2) are given here. You can reason through the logic that leads to this sequence and the reason why there is no alternative possibility for this sequence. Pathway (2) Ba2+(aq) + SO42–(aq) BaSO4(s) Ag+(aq) + Cl–(aq) 3+ – Fe (aq) + 3OH (aq) AgCl(s) Fe(OH)3(s) Problem 2.51. (a) When an aluminum nitrate, Al(NO3)3, solution is mixed with a sodium oxalate, Na2C2O4, solution the solution formed contains aluminum, Al3+(aq), and sodium, Na+(aq), cations, and nitrate, NO3–(aq), and oxalate, C2O42–(aq), anions. A precipitate forms from this mixed solution. The possibilities for this solid ionic compound are Al2(C2O4)3(s) and NaNO3(s). Our solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be soluble, so the precipitate is not NaNO3(s). The precipitate, Al2(C2O4)3(s), is an ionic compound formed from a multiply-charged cation and a multiply-charged anion, which our rules suggest is likely to be insoluble. (b) The net ionic equation for the reaction that forms Al2(C2O4)3(s) has the aqueous ions with appropriate stoichiometric coefficients as reactants and the solid as product: 2Al3+(aq) + 3C2O42-(aq) Al2(C2O4)3(s) Problem 2.52. When a solution of lithium nitrate, LiNO3, is mixed with a solution of sodium phosphate, Na3PO4, the solution contains lithium, Li+(aq), and sodium, Na+(aq), cations and nitrate, NO3– (aq), and phosphate, PO43–(aq), anions. A white precipitate is observed to form from this mixture. The possibilities for this solid ionic compound are Li3PO4(s) and NaNO3(s). Our solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be soluble, so the precipitate is not NaNO3(s). The precipitate must be Li3PO4(s). Although ionic compounds with alkali metal cations are generally expected to be soluble, note that, in Worked Example 2.34, lithium is discussed as an exception to the general rule. (b) The net ionic equation for the precipitation reaction is: 3Li+(aq) + PO43–(aq) Li3PO4(s) Problem 2.53. [Note that there seems to be no symbol for the equilibrium double arrow that is common to both PC and Macintosh computer platforms, so the symbol “ ” is used in these notes to make them cross-platform compatible.] When the reactants and products are separated by a forward arrow over a backward arrow, “ ” in a reaction equation, it means that the reaction can or is going in both directions. That is, the reactant species are combining to yield products and the product species are combining to give reactants and both processes are occurring simultaneously. Problem 2.54. (a) When a solution of cadmium chloride, CdCl2, is mixed with a solution of ammonium sulfide, (NH4)2S, the solution contains cadmium, Cd2+(aq), and ammonium, NH4+(aq), cations and chloride, Cl–(aq), and sulfide, S2–(aq), anions. A yellow-orange precipitate is observed to form from this mixture. The possibilities for this solid ionic compound are CdS(s) and NH4Cl(s). Our solubility rules indicate that ionic compounds with a singly-charged cation and halide anion are likely to be soluble, so the precipitate is not NH4Cl(s). The precipitate must be CdS(s). (Cadmium sulfide -- cadmium yellow -- is used as a pigment in paints.) (b) The net ionic equation for the precipitation reaction is: Cd2+(aq) + S2-(aq) CdS(s) Problem 2.55. The objective of this problem is to use our solubility rules, including the few exceptions we have noted, to predict the products of mixing four pairs of aqueous solutions and to write complete and net ionic equations for those reactions that produce a precipitate. (a) barium chloride(aq) + sodium sulfate(aq) The mixed solution contains barium, Ba2+(aq), and sodium, Na+(aq), cations and chloride, Cl– (aq), and sulfate, SO42–(aq), anions. Our solubility rules indicate that an ionic compound with a multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that BaSO4(s) is insoluble and write this complete and net ionic reaction equation: Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq) Ba+2(aq) + SO42-(aq) BaSO4(s) (b) silver nitrate(aq) + magnesium chloride(aq) The mixed solution contains silver, Ag+(aq), and magnesium, Mg2+(aq), cations and nitrate, NO3–(aq), and chloride, Cl–(aq), anions. Our solubility rules indicate that ionic compounds with a halide or nitrate anion are likely to be soluble, so the initial instinct is to predict that no precipitate will form. However, when we recall that the silver ion is a notable exception to the rules, we predict that silver chloride, AgCl(s), is insoluble and write this complete and net ionic reaction equation: 2Ag+(aq) + 2NO3–(aq) + Mg2+(aq) + 2Cl–(aq) 2AgCl(s) + Mg2+(aq) + 2NO3–(aq) 2Ag+(aq) + 2Cl–(aq) 2AgCl(s) (c) strontium nitrate(aq) + potassium nitrate(aq) The mixed solution contains strontium, Sr2+(aq), and potassium, K+(aq), cations and nitrate, NO3–(aq), anions. No new products are possible, because there is only one anion that is common to both soluble starting compounds. Thus we can write: Sr2+(aq) + 3NO3-(aq) + K+(aq) + NO3-(aq) NO APPARENT REACTION (d) ammonium phosphate(aq) + calcium bromide(aq) The mixed solution contains barium, NH4+(aq), and calcium, Ca2+(aq), cations and phosphate, PO43–(aq), and bromide, Br–(aq), anions. Our solubility rules indicate that an ionic compound with a multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that Ca3(PO4)2(s) is insoluble and write this complete and net ionic reaction equation: 6NH4+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Br-(aq) Ca3(PO4)2(s) + 6NH4+(aq) + 6Br-(aq) 3Ca2+(aq) + 2PO43-(aq) Ca3(PO4)2(s) Problem 2.56. The objective of this problem is to use your knowledge of the solubility rules to propose a way to prepare (synthesize) several solid ionic compounds by precipitation reactions from mixtures of soluble ionic compounds. (a) The net ionic reaction for preparing BaSO4(s) is: Ba2+(aq) + SO42–(aq) BaSO4(s) Mixing aqueous solutions of the soluble ionic compounds Ba(NO3)2 (nitrates are soluble) and Na2SO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction. (b) The net ionic reaction for preparing AgCl(s) is: Ag+(aq) + Cl–(aq) AgCl(s) Mixing aqueous solutions of the soluble ionic compounds AgNO3 (nitrates are soluble) and KCl (alkali metal halide salts are soluble) will provide the reactant ions necessary for this reaction. (c) The net ionic reaction for preparing Ca3(PO4)2(s) is: 3Ca2+(aq) + 2PO43- (aq) Ca3(PO4)2 (s) Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are soluble) and K3PO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction. (d) The net ionic reaction for preparing CaC2O4(s) is: Ca2+ (aq) + C2O42-(aq) CaC2O4 (s) Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are soluble) and K2C2O4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction. Problem 2.57. Since the concentration of a solution (assuming it is well mixed after preparation) is uniform throughout the entire volume, spilling some of it will not change the concentration of the remaining solution. Concentration is an intensive variable (like density or temperature) that does not depend upon the amount of solution you have. Problem 2.58. (a) If it is properly labeled, the solution in a bottle labeled, "0.5 M CaCl2" contains 0.5 moles of CaCl2(aq) per liter of solution. There are 0.5 moles of Ca2+(aq) per liter of solution and 1 mole of Cl–(aq) per liter of solution. (b) If a 0.5-L bottle is about half full, then it contains approximately 0.25 L of 0.5 M CaCl2 solution. Use the definition of molarity, moles of solute per liter of solution, to determine the number of moles of the solute in about one-quarter liter of solution: mol CaCl2 in 0.25 L = (0.25 L) 0.5 mol 1L = 0.13 mol CaCl2 (c) Use the molar formula mass of CaCl2, 111 g·mol–1, to convert the number of moles of CaCl2 from part (b) to mass: 0.13 mol CaCl2 = (0.13 mol CaCl2) 111 g 1 mol CaCl 2 = 14 g CaCl2 Problem 2.59. (a) Counting the number of atoms of each element in the molecular structure for vitamin C in Problem 2.15 gives the molecular formula C6H8O6. (b) Find the molar mass of vitamin C by summing the masses of each element in a mole of the compound: 12.01 g = 72.06 g (6 mol C) 1 mol C 1.008 g (8 mol H) = 8.06 g 1 mol H 16.00 g (6 mol O) = 96.00 g 1 mol O molar mass vitamin C = 176.12 g = 176 g (accurate enough for the rest of the data) (c) To find the number of moles of vitamin C in a tablet that contains 500-mg of the vitamin, we use the molar mass to convert this mass to moles. The actual mass of vitamin in a vitamin tablet is only accurate to a few percent, so we will assume that the mass of vitamin C in the tablet is about 0.50 g [= (500 mg) 1 g 1000 mg ], with an implied accuracy of 1 part in 50 or about 2%. 0.50 g vit C = (0.50 g vit C) 1 mol vit C 176 g = 0.0028 mol vit C = 2.8 10–3 mol vit C (d) To find the number of molecules of vitamin C in a tablet that contains 500-mg of the vitamin, we use Avogadro’s number to convert number of moles from part (c) to number of molecules: 2.8 –3 10 mol vit C = (2.8 = 1.7 6.02 10 23 molec 10 mol vit C) 1 mol –3 1021 molec vit C Problem 2.60. (a) To convert moles of aspartame to mass, we need the molar mass of aspartame, C14H18N2O5: 12.01 g = 168.14 g (14 mol C) 1 mol C 1.008 g = 18.14 g (18 mol H) 1 mol H 14.01 g 1 mol N 16.00 g (5 mol O) 1 mol O (2 mol N) = 28.02 g = 80.00 g molar mass aspartame = 294.30 g = 294 g (accurate enough for the rest of the data) The mass of aspartame in 2.5 mol of aspartame is: 294 g = 735 g = 7.4 2.5 mol = (2.5 mol) 1 mol aspartame 102 g = 0.74 kg (b) To convert moles of aspirin to mass, we need the molar mass of aspirin, C9H8O4: 12.01 g = 108.09 g (9 mol C) 1 mol C 1.008 g = 8.06 g (8 mol H) 1 mol H 16.00 g = 64.00 g (4 mol O) 1 mol O molar mass aspirin = 180.15 g = 180 g (accurate enough for the rest of the data) The mass of aspirin in 0.040 mol of aspirin is: 0.040 mol = (0.040 mol) 180 g = 7.2 g 1 mol aspirin (c) To convert number of molecules of cholesterol, C27H46O, to mass, we first need to convert number of molecules to number of moles (using Avogadro’s number) and then number of moles to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in 2.5 1023 molecules of cholesterol is: 2.5 1023 molec = (2.5 1023 molec) 1 mol 23 6.02 10 molec = 0.42 mol The molar mass of cholesterol is: 12.01 g = 324.27 g (27 mol C) 1 mol C 1.008 g = 46.37 g (46 mol H) 1 mol H 16.00 g = 16.00 g (1 mol O) 1 mol O molar mass cholesterol = 386.64 g = 387 g (accurate enough for the rest of the data) The mass of cholesterol in 0.42 mol cholesterol is: 387 g = 163 g = 1.6 0.42 mol = (0.42 mol) 1 mol cholesterol 102 g = 0.16 kg (d) To convert number of molecules of caffeine, C8H10N4O2, to mass, we first need to convert number of molecules to number of moles (using Avogadro’s number) and then number of moles to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in 1.2 1022 molecules of caffeine is: 1.2 1022 molec = (1.2 1022 molec) 1 mol 6.02 10 23 molec = 0.020 mol The molar mass of caffeine is: 12.01 g = 96.08 g (8 mol C) 1 mol C 1.008 g = 10.08 g (10 mol H) 1 mol H 14.01 g = 56.04 g (4 mol N) 1 mol N 16.00 g = 32.00 g (2 mol O) 1 mol O molar mass caffeine = 194.20 g = 194 g (accurate enough for the rest of the data) The mass of caffeine in 0.020 mol of caffeine is: 194 g = 3.9 g 0.020 mol = (0.020 mol) 1 mol caffeine Problem 2.61. To find the number of atoms of carbon in 5 mg of niacin, we need a strategy to get from the mass of a substance to the number of atoms of an element in that mass of the substance. This problem involves understanding and applying the concept of the mole. One way to plan your work is to reason backward from the desired answer, using the information given in the problem, the mass of niacin, 5 mg, and its molecular structure: H H H C C C N O C C C H N H H The molecular formula for niacin, from its structure, is C6H6N2O, so each molecule of niacin contains six atoms of carbon. (Note that you might write the formula a different way, ON2C6H6 for example, and that is OK. The format we used, C6H6N2O, is the conventional way chemists write formulas for carbon-containing compounds: carbon first, hydrogen second, and then all other elements in alphabetical order. Any format that shows the correct number of each atom in the molecule is fine for stoichiometric problems like the one here.) If we know the number of molecules of niacin in 5 mg of niacin, there are six times as many atoms of carbon in the sample. If we know the number of moles of niacin in the sample, we use Avogadro’s number to get the number of molecules in the sample. Finally, use the mass (in grams) of the sample of niacin and its molar mass (from the formula) to get the number of moles of niacin. Going forward through the solution, our strategy is summarized in this sequence: mass moles molecules atoms To find the number of moles of niacin in a 0.005 g [= (5 mg) 1 g 1000 mg ] sample, we need the molar mass of niacin: 12.01 g = 72.06 g (6 mol C) 1 mol C 1.008 g = 6.05 g (6 mol H) 1 mol H 14.01 g = 28.02 g (2 mol N) 1 mol N 16.00 g = 16.00 g (1 mol O) 1 mol O molar mass niacin = 122.13 g = 122 g (accurate enough for the rest of the data) The number of moles of niacin is: 0.005 g niacin = (0.005 g niacin) 1 mol niacin =4 122 g 10–5 mol niacin The number of molecules of niacin is: 4 –5 10 mol niacin = (4 = 2.4 6.02 10 23 molec 10 mol niacin) 1 mol –5 1019 molec niacin The number on atoms of carbon in the sample is: 2.4 1019 molec niacin = (2.4 = 1.4 1019 molec niacin) 6 atoms carbon 1 molec niacin 1020 atoms carbon Problem 2.62. In order to determine the molar concentration, mol·L–1, of DNA in the given bacterium, we need to know the number of moles of DNA and the volume of the bacterium in which it is found. We are told that there is one molecule of DNA in the bacterium, so we can use Avogadro’s number to find the number of moles of DNA: 1 mol 1 molec DNA = (1 molec DNA) = 1.7 10–24 mol 23 6.02 10 molec We are told that the bacterium is spherical and has a diameter, d, of 1 10–6 m. The formula for the volume of a sphere is 4 3 r3 = 1 6 d3 (where radius, r, = d/2). We want the volume in liters and one way to get it is to recall that 1 L = 1 dm3. Therefore, if we express the diameter of the bacterium in decimeters, 1 dm = 10–1 m, the volume we get will be in dm3 (= liters). (Another approach is to calculate the volume in cubic centimeters, which are equal to milliliters, and thence to liters.) The required diameter of the bacterium is: 1 dm d = 1 10–6 m = (1 10–6 m) = 1 10–5 dm 10 –1 m The volume, V, of the bacterium is: V = 1 6 d3 = 1 6 (1 10–5 dm)3 = 5 10–16 dm3 = 5 10–16 L The DNA concentration is: mol DNA 1.7 10 –24 mol = = 3 10–9 M [DNA] = 5 10 –16 L V Problem 2.63. To find the number of sodium ions in 50 mL of blood serum, we need a strategy to get from the volume of a solution to the number of ions of an element in that volume of the solution. One way to plan your work is to reason backward from the desired answer, using the information given in the problem, the volume of solution (serum), 50 mL, and its molarity, 0.14 M in NaCl. If we know the number of moles of sodium ion in the sample, we use Avogadro’s number to get the number of sodium ions in the sample. We know that every mole of NaCl dissolved in the serum produces a mole of sodium ions, Na+(aq). Finally, use the molarity of the serum sample and its volume (in liters) to get the number of moles of NaCl dissolved. Going forward through the solution, our strategy is summarized in this sequence: volume solution moles NaCl moles ions number ions The volume of the serum sample is: volume = 50 mL = (50 mL) 1L 1000 mL = 0.050 L The number of moles of NaCl in the serum sample is: 0.14 mol 0.050 L = ( 0.050 L) = 7.0 10–3 mol 1L The number of moles of sodium ions in the serum sample is: 1 mol Na + –3 –3 7.0 10 mol NaCl = (7.0 10 mol NaCl) = 7.0 1 mol NaCl The number of molecules of sodium ions in the serum sample is: 6.02 10 23 molec –3 + –3 + 7.0 10 mol Na = (7.0 10 mol Na ) 1 mol 10–3 mol Na+ = 4.2 1021 Na+ Problem 2.64. For each solution in this problem we are asked to calculate the mass of solute present in a known volume of solution of a known molar concentration. The route map shown in Worked Example 2.50 (read from bottom to top) shows the sequence of steps required to get these masses: volume solution moles solute mass solute These steps can be combined: mol solute g solute 1 L solution 1 mol solute In each case, we will need the molar mass of the solute and will calculate this first. mass solute, g = (volume solution, L) (a) The molar mass of K2Cr2O7 is: 39.10 g (2 mol K) = 78.20 g 1 mol K 52.00 g = 104.00 g (2 mol Cr) 1 mol Cr 16.00 g = 112.00 g (7 mol O) 1 mol O molar mass K2Cr2O7 = 294.20 g = 294 g (accurate enough for the rest of the data) The mass of K2Cr2O7 in 350 mL of 0.105 M K2Cr2O7 is: mass K2Cr2O7 = (0.350 L) 0.105 mol K2 Cr2O 7 1 L solution 294 g 1 mol K2 Cr2O 7 = 10.8 g (b) The molar mass of FeCl3·6H2O is: 55.85 g (1 mol Fe) = 55.85 g 1 mol Fe 35.45 g = 106.35 g (3 mol Cl) 1 mol Cl 1.008 g = 12.10 g (12 mol H) 1 mol H 16.00 g = 112.00 g (6 mol O) 1 mol O molar mass FeCl3·6H2O = 286.30 g = 286 g (accurate enough for the rest of the data) The mass of FeCl3·6H2O in 50 mL of 1.0 M FeCl3·6H2O is: mass FeCl3·6H2O = (0.050 L) 1.0 mol FeCl3 ·6H2 O 1 L solution 286 g 1 mol FeCl3 ·6H2 O = 14.3 g (c) The molar mass of KCl is: 39.10 g (1 mol K) = 39.10 g 1 mol K 35.45 g = 35.45 g (1 mol Cl) 1 mol Cl molar mass KCl = 74.55 g = 74.6 g (accurate enough for the rest of the data) The mass of KCl in 0.3 L of 1.70 M KCl is: 1.70 mol KCl mass KCl = (0.3 L) 1 L solution 74.6 g 1 mol KCl = 38.0 g Or we get about 4 102 g, if the volume is really only known to about 1 part in 3 ( 33%).