Chapter 3 Homework Solutions: Molecular Weights and Empirical Formulas - Prof. Michael P. , Assignments of Chemistry

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Chapter 3 Homework Solutions 9/9/08 5. a) C2H5NO2 b) MW = (2 atomsC) C C atom amu 0107.12 + (5 atomsH) H H atom amu 00794.1 + (1 atomN) N N atom amu 0067.14 + (2 atomsO) O O atom amu 9994.15 = 75.0666 amu c) %N = %) 100( amu 0666.75 atom amu 0067.14 )atom 1( N N N = 18.66 % 12. a) 6, 1, 2 c) 2, 2, 1, 4 e) 3, 2, 1, 6 g) 4, 9, 4, 10, 2 b) 1, 3, 2 d) 1, 6, 3, 2 f) 2, 1, 1, 2 14. a) SO3 (g) + H2O() H2SO4 (aq) b) B2S3 (s) + 6 H2O() 2 H3BO3 (aq) + 3 H2S(g) c) Pb(NO3)2 (aq) + 2 NaI PbI2 (s) + 2 NaNO3 (aq) d) 2 Hg(NO3)2 (s) 2 HgO(s) + 4 NO2 (g) + O2 (g) e) Cu(s) + 2 H2SO4 (aq) CuSO4 (aq) + SO2 (g) + 2 H2O() 16. a) 2 Ca(s) + O2 (g) 2 CaO(s) b) C3H6O() + 4 O2 (g) 3 CO2 (g) + 3 H2O() 18. a) 4 Al(s) + 3 O2 (g) 2 Al2O3 (s) b) Cu(OH)2 (s) CuO(s) + H2O(g) c) C7H16 () + 11 O2 (g) 7 CO2 (g) + 8 H2O() d) 2 C5H12O() + 15 O2 (g) 10 CO2 (g) + 12 H2O() 20. a) 2, 9, 6, 6; combustion d) 1, 3, 2; combination b) 1, 1, 2; decomposition e) 1, 1, 2; combination c) 1, 6, 5, 3; combustion 22. a) MWN2O = (2 atomsN) N N atom amu 0067.14 + (1 atomO) O O atom amu 9994.15 = 44.0128 amu b) MWBA = (7 atomsC) C C atom amu 0107.12 + (6 atomsH) 100794. amu atom H H + (2 atomsO) 15 9994. amu atom O O = 122.1213 amu c) MWMH = (1 atomsMg) Mg Mg atom amu 4.30502 + (2 atomsO) 15 9994. amu atom O O + (2 atomsH) 100794. amu atom H H = 58.3197 amu d) MWurea = (1 atomO O O atom amu 5.99941 + (1 atomsC) C C atom amu 2.01071 + (2 atomsN) 14 0067. amu atom N N + (4 atomsH) 100794. amu atom H H = 60.0553 amu e) MWIA = (7 atomsC) C C atom amu 0107.12 + (14 atomsH) 100794. amu atom H H + (2 atomsO) 15 9994. amu atom O O = 130.1849 amu 24. To save space on this problem, I am omitting the calculation of the molecular weights. a) %C = acetylene C c C g 26.0373 mol g 0107.12 mol 2 x 100% = 92.26%C [remember only report to the hundredths place] b) %H = acid ascorbic H H H g 176.1241 mol g 00794.1 mol 8 x 100% = 4.58%H c) %H = sulf. amm. H H H g 132.1395 mol g 00794.1 mol 8 x 100% = 6.10%H d) %Pt = platin-cis Pt Pt Pt g 300.045 mol g 078.195 mol 1 x 100% = 65.02%Pt e) %O = estradiol O O O g 272.3820 mol g 9994.15 mol 2 x 100% = 11.75%O f) %C = capsaiscin C C C g 305.4119 mol g 0107.12 mol 81 x 100% = 70.79%C formula of C10H14N2. 54. This problem is solved in a manner that is similar to 46, 50, and 52 except that we can treat the magnesium sulfate and water as units without subdividing them into atoms. masswater = 5.061 g – 2.472 g = 2.589 g molMgSO4 = (2.472 gMgSO4) MgSO4 MgSO4 g 3676.120 mol 1 = 0.02054 molMgSO4 molH2O = (2.589 gH2O) H2O H2O g 0153.18 mol 1 = 0.1437 molH2O MgSO4 H2O g 02054.0 mol 1437.0 = 6.997 molH2O per molMgSO4 the formula is MgSO4•7H2O 58. a) molCO2 = (0.400 molC6H12O6) C6H12O6 CO2 mol 1 mol 2 = 0.800 molCO2 b) massC6H12O6 = (7.50 gC2H5OH) C6H12O6 C6H12O6 C2H5OH C6H12O6 C2H5OH C2H5OH mol g 1559.180 mol 2 mol 1 g 46.0684 mol 1 = 14.7 gC6H12O6 c) massCO2 = (7.50 gC2H5OH) CO2 CO2 C2H5OH CO2 C2H5OH C2H5OH mol g 4.00954 mol 2 mol 2 g 46.0684 mol 1 = 7.16 gCO2 60. a) Fe2O3 (s) + 3 CO(g) 2 Fe(s) + 3 CO2 (g) b) massCO = (0.150 kgFe2O3) CO CO Fe2O3 CO Fe2O3 Fe2O3 Fe2O3 Fe2O3 mol 1 g 28.0101 mol 1 mol 3 g 688.159 mol 1 kg 1 g 1000 = 78.9 gCO c) As in (b), massFe = 105 gFe; massCO2 = 124 gCO2 d) massreactants = 150 g + 78.9 g = 229 g massproducts = 105 g + 124 g = 229 g massreactants = massproducts 72. a) To work this problem, assume you use up all of one of the reagents, then calculate how much of the other would be required to accomplish this. molAl(OH)3 (required) = (0.500 molH2SO4) H2SO4 Al(OH)3 mol 3 mol 2 = 0.333 molAl(OH)3 This means that if all of the sulfuric acid were consumed, then 0.333 mol of aluminum hydroxide would be consumed, but you have 0.500 moles of aluminum hydroxide, which is more than this. Thus, if all of the sulfuric acid were used up, you’d have some aluminum hydroxide left over, hence the sulfuric acid is the limiting reagent. b) molAl2(SO4)3 = (0.500 molH2SO4) H2SO4 Al2(SO4)3 mol 3 mol 1 = 0.167 molAl2(SO4)3 c) molAl(OH)3 = 0.500 molAl(OH)3 – 0.333 molAl(OH)3 = 0.167 molAl(OH)3 74. Determine the limiting reagent by first calculating the number of moles of each substance present. molNH3 = (1.50 g) NH3 NH3 g 0305.17 mol 1 = 0.0881 molNH3 molO2 = (2.75 g) 1 mol 31.9988 g = 0.0859 molO2 Now determine how much oxygen is required to use up all of the ammonia. molO2(required) = (0.0881 molNH3) 5 mol 4 mol O2 NH3 = 0.101 molO2 But you only have 0.0859 molO2 to begin with, so there isn’t enough oxygen to consume all of the ammonia. Thus, oxygen is the limiting reagent. massNO (formed) = (2.75 gO2) NO NO O2 NO O2 O2 mol 1 g 0061.30 mol 5 mol 4 g 31.9988 mol 1 = 2.06 gNO massH2O (formed) = (2.75 gO2) H2O H2O O2 H2O O2 O2 mol 1 g 8.01531 mol 5 mol 6 g 31.9988 mol 1 = 1.86 gH2O massNH3 (used) = (2.75 gO2) 1 4 5 17.0305 g 1 mol 31.9988 g mol mol mol O2 O2 NH3 O2 NH3 NH3 = 1.17 gNH3 massNH3 (remaining) = 1.50 g – 1.17 g = 0.33 gNH3 starting mass = 1.50 g + 2.75 g = 4.25 g final mass = 0.33 g + 2.06 g + 1.86 g = 4.25 g 76. H2SO4 (aq) + Pb(C2H3O2)2 (aq) 2 HC2H3O2 (aq) + PbSO4 (s) molH2SO4 = (7.50 gH2SO4) H2SO4 H2SO4 g 98.0875 mol 1 = 0.764 molH2SO4 molPb(C2H3O2)2 = (7.50 gPb(OAc)2) 2Pb(C2H3O2) 2Pb(C2H3O2) g 325.3 mol 1 = 0.0230 molPb(C2H3O2)2 molPb(C2H3O2)2(required) = (0.764 molH2SO4) H2SO4 2Pb(C2H3O2) mol 1 mol 1 = 0.764 mol Pb(C2H3O2)2 But you only have 0.0230 molPb(C2H3O2)2 to begin with, so there isn’t enough lead(II) acetate to consume all of the sulfuric acid. Thus, lead(II) acetate is the limiting reagent. mass Pb(C2H3O2)2 = 0 g (since it is the limiting reagent, it’s all used up) massH2SO4 (used) = (7.50 gPb(OAc)2) H2SO4 H2SO4 2Pb(C2H3O2) H2SO4 2Pb(C2H3O2) 2Pb(C2H3O2) mol 1 g 0875.98 mol 1 mol 1 g 325.3 mol 1 = 2.26 gH2SO4 massH2SO4(remaining) = 7.50 g – 2.26 g = 5.24 g massHC2H3O2 (formed) = (7.50 gPb(OAc)2) HC2H3O2 HC2H3O2 2Pb(C2H3O2) HC2H3O2 2Pb(C2H3O2) 2Pb(C2H3O2) mol 1 g 0.05206 mol 1 mol 2 g 325.3 mol 1 = 2.77 gHC2H3O2 massPbSO4 (formed) = (7.50 gPb(OAc)2) PbSO4 PbSO4 2Pb(C2H3O2) PbSO4 2Pb(C2H3O2) 2Pb(C2H3O2) mol 1 g 03.33 mol 1 mol 1 g 325.3 mol 1 = 6.99 gPbSO4 90. If you can’t do this one, it’s ok, but this is an interesting problem. The key is to figure out how to begin. You are given the percent bromine in the sample. If you assume 100 g of compound, this gives you the mass of bromine (52.92 g). From this you can determine the number of moles of bromine. The formula tells you the molar ratio of potassium to bromine (1:1). Knowing the number of moles of potassium you can calculate the mass of potassium. Knowing the masses of bromine & potassium allows you to calculate the mass of oxygen. At this point, the problem is solved just like all of the other empirical formula problems. The actual math follows below. molBr = (52.92 gBr) Br Br g 904.79 mol 1 = 0.6623 molBr molK = molBr = 0.6623 molK massK = (0.6623 molK) K K mol 1 g 0983.39 = 25.89 gK massO = 100.00 g – 52.92 g – 25.89 g = 21.19 gO molO = (21.19 gO) O O g 15.9994 mol 1 = 1.324 molO molar ratio: Br O mol 6623.0 mol 324.1 = 2.0 molO per molBr formula is KBrO2 103. First begin with a balanced equation: 2 C8H18 () + 25 O2 (g) 18 H2O() + 16 CO2 (g) Now calculate the volume of gasoline needed: Vgas = (225 mi) mi 20.5 gal 1 = 11.0 galgas From this you can calculate the mass of the gasoline and from that the mass of CO2 produced. massCO2 = (11.0 galgas) CO2 CO2 C8H18 CO2 C8H18 C8H18 mol 1 g 0144 mol 2 mol 16 g 2114 mol 1 mL g 690 L mL 1000 qt 1.057 L 1 gal qt 4 . . . = 88,500 gCO2 (or 88.5 kg)