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solutions chapter 3

solutions chapter 3 - CHAPTER 3 STOICHIOMETRY 43...

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Unformatted text preview: CHAPTER 3 STOICHIOMETRY 43 lmolSO2 >< 2molSO3 X 80.07gSO3 96.0 g SO; x 64.07 g SO2 2 mol SO2 mol SO3 = 120. g so3 10. Side reactions may occur. For example, in the combustion of CH4 (methane) to C02 and H20, some CO may also form. Also, reactions only go part way to completion, instead reaching a state of equilibrium where both reactants and products are present (see Ch. 13). Questions 19. isotope mass abundance 12c 12.0000amu 98.89% 13c 13.034 amu 1.11% average mass = 0.9889 (120000) + 0.0111(13034) = 12.01 amu From the relative abundances, there would be 9889 atoms of 12C and 111 atoms of 13C in the 10,000 atom sample. The average mass of carbon is independent of the sample size; it will always be 12.01 amu. total mass = 10,000 atoms x mu. : 1.201 x 105 amu atom For one mol of carbon (6.0221 x 1023 atoms C), the average mass would still be 12.01 amu. The number of 12C atoms would be 0.9889 (6.0221 x 1023) = 5.955 x 1023 atoms 12C and the number of ”C atoms would be 0.0111 (6.0221 x 1023) = 6.68 x 102‘ atoms l3c. total mass = 6.0221 x 1023 atoms x m = 7.233 x 1024 amu atom total mass m g = 6.0221 x 1023 atoms x M x ——J—,3—— = 12.01 g/mol atom 6.0221 x 10“ amu By using the carbon-12 standard to define the relative masses of all of the isotopes as well as to define the number of things in a mole, then each element’s average atomic mass in units of grams is the mass of a mole of that element as it is found in nature. 20. Consider a sample of glucose, C6H1206. The molar mass of glucose is 180.16 g/mol. The chemical formula allows one to convert from molecules of glucose to atoms of carbon, hydrogen, or oxygen present and vice versa. The chemical formula also gives the mole relationship in the formula. One mol of glucose contains 6 mol C, 12 mol H, and 6 mol 0. Thus, mole conversions between molecules and atoms are possible using the chemical for— mula. The molar mass allows one to convert between mass and moles of compound and Avogadro’s number (6.022 x 1023) allows one to convert between moles of compound and number of molecules. CHAPTER 3 STOICHIOMETRY 45 26. One method is to determine the actual mole ratio of XY to Y2 present and compare this ratio to the required 2:1 mole ratio from the balanced equation. Which ratio is larger will allow one to deduce the limiting reactant. Once the identity of the limiting reactant is known, then one can calculate the amount of product formed. A second method would be to pick one of the reactants and then calculate how much of the other reactant would be required to react with it all. How the answer compares to the actual amount of that reactant present allows one to deduce the identity of the limiting reactant. Once the identity is known, one would take the limiting reactant and convert it to mass of product formed. When each reactant is assumed limiting and the amount of product is calculated, there are two possible answers (assuming two reactants). The correct answer (the amount of product that could be produced) is always the smaller number. Even though there is enough of the other reactant to form more product, once the small quantity is reached, the limiting reactant runs out and the reaction cannot continue. Exercises Atomic Masses and the Mass Spectrometer 27. 28. 29. 30. 31. A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766) A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; From the periodic table, the element is Pb. A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu This is element Ti (titanium). Let A = mass of 185Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 — 117.0 2 0.3740(A) 69.2 0.3740 abundance 23Si 2 100.00 -(4.70 + 3.09) = 92.21%; From the periodic table, the average atomic mass of Si is 28.09 amu. A: = 185 amu (A = 184.95 amu without rounding to proper significant figures.) 28.09 = 0.9221(27.98) + 0.0470 (atomic mass 29Si) + 0.0309(29.97) ‘ atomic mass 29Si = 29.01 The mass of 29Si is actually a little less than 29 amu. There are other isotopes of silicon that are considered when determining the 28.09 amu average atomic mass of Si listed in the atomic table. There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br; molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 52 57. 58. CHAPTER 3 STOICHIOIVTETRY 32.04 g 9.27 x 10‘3 mol CH3OH x = 0.297 g CH3OH mol a. 14 mol c 12.01g +18 mol H 1'008 g + 2 mol N 14.01g + 5 mol 0 16mg mol C mol H mol N mol 0 = 294.30 g/mol b. 10.0 g aspartame x —1—m—°l— = 3.40 x 10‘2 mol 294.30 g c. 1.56 mol x 395395: 459 g , _ mol 23 d. 5.0 mg x 1g x—IEJI—XWfle—S: 1.0 x 1019 molecules 1000 mg 294.30 g mol 6. The chemical formula tells us that 1 molecule of aspartame contains two atoms of N. The chemical formula also says that 1 mol of aspartame contains two mol of N. 1 mol aspartame 2 mol N 6.02 ><1023 atoms N 1.2 g aspartame x ———-><——————><————————— 294.30 g aspartame mol aspartame mol N = 4.9 x 1021- atoms of nitrogen f. 1.0 x 109 molecules x 1““ xflg = 4.9 x 10‘13 g or 490 fg 6.02 X 1023 molecules mol lmol x 294.30 g —-———————————— = 4.887 x 10“22 g 6.022 X 1023 molecules mol g. 1 molecule aspartame x a. 2(12.01)+ 3(1.008)+ 3(35.45) + 2(16.00) = 165.39 g/mol ““01 = 3.023 mol c. 2.0 x 10'2mol x = 3.3 g b. 500.0 g x 165.39 g mol lmol x 6.02x1023 molecules X 3 atoms C1 165.39 g mol molecule d. 5.0 g C2H3C1302 X = 5.5 x 1022 atoms of chlorine 1mol Cl x mm C2H3C1302 X 165.39 g c2113c13o2 e. 1.0 g Cl x 35.45 g 3 mol Cl mol C2H3C1302 = 1.6 g chloral hydrate f. 500 molecules x ____1_n_13l___ x 165-39 g 6.022x1023 molecules mol = 1.373 x 10‘” g CHAPTER 3 STOICHIOMETRY 53 Percent Composition 59. a. C3H402: Molar mass = 302.01) + 40008) + 206.00) = 36.03 + 4.032 + 32.00 = 72.06 g/mol %C= “mx 100:50.00% C; %H= Mx 100 72.06 g compound 72.06 g compound = 5.595% H 32.00 g %o =100.00 — (50.00 + 5.595) = 44.41% 0 or %o = 72 06 x 100 = 44.41% o ~ g b. C4H602: Molar mass = 4(1201) + 60.008) + 206.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol %C = 4804 g x 100 = 55.80% c; %H = 6'048 g x 100 = 7.025% H 86.09 g 86.09 g %0 = 100.00 - (55.80 + 7.025) = 37.18% 0 c. C3H3N: Molar mass = 302.01) + 3(1.008) + 10401) = 36.03 + 3.024 + 14.01 = 53.06 g/mol %C = 36.03g x 100 = 67.90% C; %H = 3024 g x 100 = 5.699% H 53.06g 53.06g 14.01 g %N = x 100 = 26.40% N or %N = 100.00 — (67.90 + 5.699) = 26.40% N 53.06 g 60. molar mass = 2002.01) + 290.008) + 19.00 + 306.00) = 336.43 g/mol %c = MEE— x100=71.40% C 336.43 g compound %H = W— x 100 = 8.689% H 336.43 g compound %F= JOK— x100=5.648%F 336.43 g compound %0 = 100.00 - (71.40 + 8.689 + 5.648) = 14.26% 0 or: %o _ 3(16.00)g o _ ————————————— x 100: 14.27% 0 336.43 g compound 14.01 g N 30.01 g NO 61. a. NO: %N = x 100 = 46.68% N 54 CHAPTER 3 STOICHIOIVIETRY b. N02: %N = w x 100 = 30.45% N 46.01 g NO2 c. N204: %N = Mx 100 = 30.45% N 92.02 g N204 d. N20: %N = M x 100 = 63.65% N 44.02 g N20 The order from lowest to highest mass percentage of nitrogen is: N0; = N204 < NO < N20. 62. C8H10N402: molar mass = 802.01) + 100.008) + 404.01) + 206.00) = 194.20 g/mol %c = w x100 = 96'08 x100 = 49.47% C 194.20g CgHmN40Z 194.20 C12 H220“: molar mass = 1202.01) + 220.008) + 1106.00): 342.30 g/mol %C = M x 100 = 42.10% C 342.30 g C12H22O11 C2H5OH: molar mass = 202.01) + 60.008) + 106.00) = 46.07 g/mol %C = Mic—4 100 =52.14% c 46.07 g CZHSOH The order from lowest to highest mass percentage of carbon is: sucrose (CnszOn) < caffeine (CgHmN4Oz) < ethanol (CZHSOH) 63. There are many valid methods to solve this problem. We will assume 100.00 g of compound, then determine from the information in the problem how many mol of compound equals 100.00 g of compound. From this information, we can determine the mass of one mol of compound (the molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin: mol cyanocobalamin = 4.34 g Co x ______1mol CO x—-—-————————————1 mol cyanocobalamm 58.93 g Co mol Co = 7.36 x 10’2 mol cyanocobalamin W2— : fl ’ x = molar mass = 1360 g/mo] 1 mol cyanocobalamln 7.36 ><10‘2 mol 64. There are 0.390 g Cu for every 100.00 g of fungal laccase. Assuming 100.00 g fungal laccase: CHAPTER 3 STOICHIOMETRY 55 mol fungal laccase = 0.390 g Cu x flxw = 1.53 x 10‘3 mol 63.55 g Cu 4 mol Cu x g fungal laccase _ 100.00 g _ “mu-T, x = molar mass = 6.54 x 104 g/mol 1 mol fungal laccase 1.53 x 10 mol Empirical and Molecular Formulas 65. a. Molar mass of CHZO = 1 mol C (2952) + 2 mol H {M} mol C mol H + 1 mol 0 11095—9 = 30.03 g/mol mol 0 %c = flLgE—x 100 = 39.99% c; %H = JQEEELX 100 = 6.713% H 30.03 g (31120 30.03 g C1120 %o = ——1—6'30—g—0— x 100 = 53.28% 0 or %o = 100.00 — (39.99 + 6.713): 53.30% 30.03 g CHZO b. Molar Mass of C6H1206 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol 76“ g C x 100 = 40.00%; %H = W %C = —— 180.16gC6H1206 180.16g x 100 = 6.714% %O = 100.00 — (40.00 + 6.714) = 53.29% c. Molar mass of HC2H302 = 2(12.0l) + 4(1.008) + 2(16.00) = 60.05 g/mol %C = x 100 = 6.714% 24.02 g x 100 = 40.00%; %H = 4032 g 60.05 g 60.05 g %0 = 100.00 - (40.00 + 6.714) = 53.29% 66. All three compounds have the same empirical formula, CHZO, and different molecular formulas. The composition of all three in mass percent is also the same (within rounding differences). Therefore, elemental analysis will give us only the empirical formula. 67. a. The molecular formula is N204. The smallest whole number ratio of the atoms (the empirical formula) is N02. b. Molecular formula: C3H6; empirical formula = CH2 c. Molecular formula: P4010; empirical formula = P205 d. Molecular formula: C6H1206; empirical formula 2 CHZO 56 CHAPTER 3 STOICHIOMETRY 68. a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g 188.35 g = 4000; So the molecular formula is (SNH)4 or S4N4H.1. 47.09 g b. NPClg: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol 347.64 g = 3.0000; Molecular formula is (NPC12)3 or N3P3C16. 115.88 g c. C0C404: 58.93 + 4(12.01) + 4(1600) = 170.97 g/mol 41. 4 _3____9_g = 2.0000; Molecular formula: CongOg 170.97 g 184.32 g d. SN: 32.07 + 14.01 = 46.08 g/mol; —— = 4.000; Molecular formula: S4N4 46.08 g 69. Out of 100.00 g of adrenaline, there are: 56.79 g C x Jm—Ol—C— = 4.729 mol C; 6.56 g H x infli- = 6.51 mol H 12.01 g C 1.008 g H 28.37 g o x 33131—9— = 1.773 mol 0; 8.28 g N x 3391i: 0.591 mol N 16.00 g 0 14.01 g N Dividing each mol value by the smallest number: 4.729 = 800; 6.51 = 11.0; 1.773 = 300; 0.591 = 1.00 0.591 0.591 0.591 0.591 This gives adrenaline an empirical formula of C3H1103N. 70. Assuming 100.00 g of nylon-6: 63.68 g C x 2‘39— = 5.302 mol C; 12.38 g N x Emili— = 0.8837 mol N 12.01 g C 14.01 g N 9.80 g H x _1_r_n_cfl§_ = 9.72 mol H; 14.14 g o x Ego— = 0.8838 mol 0 1.008 gH 16.00gO Dividing each mol value by the smallest number: 5.302 =6.000; 9.72 :1“) 0.8837 0.8837 The empirical formula for nylon-6 is C6H11NO 60 CHAPTER 3 STOICHIOMETRY 79. 80. The combustion data allow determination of the amount of hydrogen in cumene. One way to determine the amount of carbon in cumene is to determine the mass percent of hydrogen in the compound from the data in the problem; then determine the mass percent of carbon by difference (100.0 - mass %H = mass %C). 42.8mgHsz 1g x 2.016gH X1000mg = 4.79 mg H 1000 mg 18.02 g H20 g 4.79 mg H 47.6 mg cumene %H: x 100 = 10.1% H; %C = 100.0 — 10.1 = 89.9% C Now solve this empirical formula problem. Out of 100.0 g cumene, we have: 899ng —1"—‘°]—C— =7.49 mol C; 10.1ng-flfl— = 10.0m01H 12.01g c 1.008g H 10.0 77—9 = 1.34 2 %, i.e., mol H to mol C are in a 4:3 ratio. Empirical formula = C3H4 Empirical formula mass 3 3(12) + 4(1) = 40 g/mol The molecular formula is (C3H4)3 or C9le since the molar mass will be between 115 and 125 g/mol (molar mass A“ 3 x 40 g/mol = 120 g/mol). First, we will determine composition by mass percent: 16.01 mg co2 x ix—lflii x m: 4.369 mg C 1000 mg 44.01 g CO2 g %c = 4369 mg C x 100 = 40.91% c 10.68 mg compound 1000 mg 18.02 g H20 g 0.489 mg %H = x 100 = 4.58% H; %O = 100.00 - (40.91 + 4.58): 54.51% 0 10.68 mg So, in 100.00 g of the compound, we have: 40.91 g c x—lm—OIC—= 3.406 mol 0, 4.58 g H x—IM— = 4.54 mol H 12.01gC 1.008g H 54.51 g o x i999— : 3.407 mol 0 16.00g o 4.54 4 Dividing by the smallest number: = 1.33 z ;the em irical formula is C O . 3.406 p 3H“ 3 CHAPTER 3 STOICHIOMETRY 61 The empirical formula mass of C311403 is z 3(12) + 4(1) + 3(16) = 88 g. Because 176.1 = 2.0, the molecular formula is C6H306. Balancing Chemical Equations 81. 82. 83. When balancing reactions, start with elements that appear in only one of the reactants and one of the products, then go on to balance the remaining elements. a. One of the most important parts to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. a. b. C6H1206(S) + 02(g) -+ C02(g) + H20(g) Balance C atoms: C6H1206 + Oz —-> 6 C0; + H20 Balance H atoms: (361—11206 + 02 "9 6 C02 + 6 H20 Lastly, balance 0 atoms: C6H1206(S) + 6 02(g) —> 6 C02(g) + 6 H20(g) 136-483(3) + HCl(g) —> FeC13(s) + HZS(g) Balance Fe atoms: F6283 + HCl —» 2 FeCl3 + HZS Balance S atoms: F6283 + HCl —> 2 FeCl3 + 3 H28 There are 6 H and 6 Cl on right, so balance with 6 HCl on left: F6233“) + 6 HCl(g) —+ 2 FeCl3(s) + 3 HZS(g). (332(1) + NH3(g) -+ H25(g) + NH43CN(S) -Wmumb.wm N. .. C and S balanced; balance N: CS; + 2 NH3 —> H28 + NH48CN H is also balanced. So: CSz(l) + 2 NH3(g) ——> H28(g) + NH48CN(s) 3 Pb(N03)2(aq) + 2 Na3PO4(aq) —-> Pb3(PO4)3(S) + 6 NaN03(aq) Zn(s) + 2 HCl(aq) —> ZnC12(aq) + H2(g) Sr(OH)2(aq) + 2 HBr(aq) —-> 2H20(1) + SrBr2(aq) 3 Ca(OH)2(aq) + 2 H3PO4(aq) -—> 6 H200) + Ca3(PO4)2(s) mwwmmhwu a. Mr» l i i g z i 62 CHAPTER 3 STOICHIOMETRY W 84. 85. 86. b. C. a. Al(OH)3(s) + 3 HCl(aq) —+ A]Cl;(aq) + 3 1120(1) 2 AgN03(aq) + HZSO4(aq) 44 AngO4(s) + 2 HNO3(aq) 2 K02“) + 2 H200) .4 2 KOH(aq) + 02(g) + H202(aq) or 4 K02(s) + 6 H200) —> 4 KOH(aq) + 02(g) + 4 H202(aq) Fe203(s) + 6 HN03(aq) —» 2 Fe(N03)3(aq) + 3 H200) 4 NH3(g) + 5 02(g) —+ 4 NO(g) + 6 H20(g) PC15(1) + 4 H200) —> H3PO4(aq) + 5 HCl(g) 2 CaO(s) + 5 C(s) —) 2 CaC2(s) + C02(g) 2 MoSz(s) + 7 02(g) ~+ 2 M003(s) + 4 802(g) FeC03(s) +H2CO3(aq) —> Fe(HC03)2(aq) The formulas of the reactants and products are C6H6(l) + 02(g) —> C02(g) + H20(g). To balance this combustion reaction, notice that all of the carbon in C6H§ has to end up as carbon in C02 and all of the hydrogen in C6}16 has to end up as hydrogen in H20. To balance C and H, we need 6 C02 molecules and 3 H20 molecules for every 1 molecule of C6H6. We do oxygen last. Because we have 15 oxygen atoms in 6 C02 molecules and 3 H20 molecules, we need 15/2 02 molecules in order to have 15 oxygen atoms on the reactant side. C6H6(l) +—1—2-5— 02(g) ——+ 6 C02(g) + 3 H20(g); Multiply by two to give whole numbers. 2 C6H6(l) + 15 02(g) —-> 12 C02(g) + 6 H20(g) The formulas of the reactants and products are C4H10(g) + 02(g) -——> C02(g) + H20(g). C4H10(g) + l2202(g) -—> 4 C02(g) + 5 H20(g); Multiply by two to give whole numbers. 2 C4H10(g) + 13 02(g) -* 8 C02(g) + 10 H20(g) C12H22011(S) + 12 02(g) —} 12 C02(g) + 11 H20(g) 2 Fe(s) + g02(g) —> FeZO3(s); For whole numbers: 4 Fe(s) + 3 02(g) —~> 2 Fe203(s) 2 FeO(s) + % 02(g) ——> Fe203(s); For whole numbers, multiply by two. 4 FeO(s) + 02(g) —> 2 Fe203(s) 16 Cr(s) + 3 53(3) ——> 8 CrZS3(s) 2 NaHCO3(s) —» NaZCO3(s) + C02(g) + H20(g) 64 CHAPTER 3 STOICHIOMETRY 1000 g Al X lmol A1 3 mol NHACIO‘. X 117.49 g NH4CIO4 91. 1.000 kg Al x x kg Al 26.98 g A] 3 mol A1 mol NH4CIO4 = 4355 g 92. a. Ba(OH)2 0 8H20(s) + 2 NILSCN(S) —-> Ba(SCN)2(s) + 10 H200) + 2 NH3(g) 1molBa(OH)2 0 8HZO 315.4g b. 6.5 g Ba(OH)2 o 8H20 x = 0.0206 mol = 0.021 mol 2 mol NH4SCN x 76.13 g NH4SCN 1molBa(OH)2 0 8H20 mol NH4SCN = 3.2 g NI-LSCN 3.0 g NH: x 1000g x lmol NH4+ ><1molC5H702N x 100 kg waste kg 18.04 g NH: 55 mol NH; 0.021 mol Ba(OH)2 o 8H20 x 93. 1.0 ><104 kg waste x W = 3.4 x 104 g tissue if all NH;r converted mol C5H702N Since only 95% of the N111” ions react: mass of tissue = (0.95) (3.4 x 104 g) = 3.2 x 104 g or 32 kg bacterial tissue 75gCa3(PO4)2 X lmolCa3(PO4)2 X 94. 1.0 x 103 g phosphorite x _ 100 g phosphorlte 310.18 g Ca3(P04)2 lmolP 4 X123.88gP4 =150gP4 2 mol Ca3(PO4)2 mol P4 1molC7H6O3 x lmolC4H6O3 ><102.09gC4H603 138.12gC7H603 lmolC—,H603 lrnolC4H603 = 73.9 g C4H603 95. a. 1.00x102gC7H603x 1molC7H603 x lmol €911.304 X 180.15 g (:gHsp4 b. 1.00 x 102 g €711.03 x 138.12 g C7H603 1molC7H603 mol C9H304 = 1.30 x 102 g aspirin 96. 2 LiOH(s) + C02(g) —> LizCO3(aq) + H200) The total volume of air exhaled each minute for the 7 astronauts is 7 x 20. = 140 L/min. lmolLiOH lmolCO2 X 44.01gCO2 x lOOgair 25,000 g LiOH x . x _ 23.95 g LlOH 2 molL10H mol CO2 4.0 g CO2 X lmLair 1L 1min x 1hr ————————— x —— >< 0.0010g air 1000mL 140Lair 60 min = 68 hr = 2.8 days 68 CHAPTER 3 STOICHIOMETRY 104. a. 1142 g C6H5Cl x M = 10.1 mol C6H5C1 112.55 g C6H5Cl 485 g C2H0C13 x w = 3.29 mol C2H0C13 147.38 g CZHOCI3 From the balanced equation, the required mole ratio is M = 2. The actual lmol CZHOCl3 10.1 moi C6H5Cl 3.29 moi C2H0C13 the required mole ratio, so the denominator of actual mole ratio (CZHOCI3) is limiting. mole ratio present is = 3.07. The actual mole ratio is greater than imoi C14H9C15 x 354.46 g CMHgCls 3.29 mol C2H0C13 x mol C2H0C13 moi CMH9Ci5 = 1170 g C14H9C15 (DDT) b. C2H0C13 is limiting and C6H5Cl is in excess. 2 moi C6H5C1 X 112.55 g C6H5Cl c. 3.29 moi C2H0C13 x moi C2H0C13 moi C6H5Cl = 741 g C6H5C1 reacted 1142 g — 741 g = 401 g C6H5Cl in excess d. % yield = W x100 =17.1% 1170 g DDT 105. 2.50 metric tons CU3FCS3 x ———————1000 kg x——1000 g X________1mol 01313683 ><—-————-———~-—3 mol CU x——-——6355 g metric ton kg 342.71 g 1 mol Cu3FeS3 mol Cu = 1.39 x 106 g Cu (theoretical) 863 g C" (adual) = 1.20 x 106 g Cu = 1.20 x 103 kg Cu 1.39 x 106 g Cu (theoretical) x ————-—————,—— 100. g Cu (theoretical) = 1.20 metric tons Cu (actual) 106. P4(s) + 6 F2(g) -—> 4 PF3(g); The theoretical yield of PF3 is: 100.0 g PF3 (theoretical) 120. PF actual x = 154 PF theoretical g 3( ) 78.1g PF3 (actual) g 3( ) 154gPF3>< imolPF3 6molF2 X 3800ng :99};ng x 87.97 g PF3 4 mol PF3 mol F2 99.8 g F2 are needed to produce an actual PF3 yield of 78.1%. 78 CHAPTER 3 STOICHIONIETRY MM 135. 136. 137. % yield = M x 100 = 75% 4mol><M b. The product of the percent yields of the individual steps must equal the overall yield, 75%. (0.87) (0.98) (x) = 0.75, x = 0.88; Step III has a % yield = 88%. 10.00 g XClg + excess C1; -—> 12.55 g XCl4; 2.55 g Cl reacted with XClz to form XC14. XC14 contains 2.55 g Cl and 10.00 g XClz. From mol ratios, 10.00 g XClz must also contain 2.55 g Cl; mass X in XC12 = 10.00 - 2.55 = 7.45 g X. 2.55 g c1 x ——————l ““01 C‘ x————1 mOIXCIZ x————-——-—l mOIX = 3.60 x 10‘2moi X 35.45 g C] 2 mol Cl molXCl2 So, 3.60 x 10'2 mol X has a mass equal to 7.45 g X. The molar mass of X is: 7.45 g X ———-—~————2—-————— = 207 g/mol X; Atomic mass = 207 amu so X is Pb. 3.60 XIO‘ mol X 4.000 g M2S3 ———> 3.723 g M02 There must be twice as many mol of M02 as mol of M283 in order to balance M in the reaction. Setting up an equation for 2 mol M02 = mol M2S3 where A = molar mass M: 2( 4.000g ]_ 3.723g 8.000 3.723 2A+3(32.07) _ A +2(16.00) ’ 2A +9621 _ A +3200 8.000 A + 256.0 = 7.446 A + 358.2, 0.554 A = 102.2, A = 184 g/mol; atomic mass = 184 amu Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms 02' anions. The simplest compound of the two elements is A1203. Similarly, we would expect the formula of any group 6A element with A1 to be A12X3. Assuming this, out of 100.00 g of compound there are 18.56 g Al and 81.44 g of the unknown element, X. Let’s use this information to determine the molar mass of X which will allow us to identify X from the periodic table. 1 mol Al 3 mol X x = ...
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