Chapter 3 Stoichiometry of Formulas and Equations

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS 3.1 Cl Al 3.2 35.45 amu ≡ 35.45 g/mol Cl Mass Cl = (3 mol Cl) x (35.45 g Cl/l mol Cl) = 106.4 g Cl 26.98 amu ≡ 26.98 g/mol Al Mass Al = (2 mol Al) x (26.98 g Al/l mol Al) = 53.96 g Al Plan: The formulas are based on the mole ratios of the constituents. Avogadro’s number allows the change from moles to atoms. Solution:   12 mol C a) Moles of C atoms = (1 mol Sucrose )   = 12 mol C  1 mol C12 H 22 O11     6.022 x 1023 C atoms  12 mol C 24 b) C atoms = (1 mol C12 H 22 O11 )   = 7.226 x 10 C atoms   1 mol C H O 1 mol C 12 22 11     3.3 “1 mol of nitrogen” could be interpreted as a mole of nitrogen atoms or a mole of nitrogen molecules. Specify which to avoid confusion. The same problem is possible with other diatomic or polyatomic molecules, e.g., Cl2, Br2, S8, and P4. For these elements, as for nitrogen, it is not clear if atoms or molecules are being discussed. 3.4 The molecular mass is the sum of the atomic masses of the atoms or ions in a molecule. The molar mass is the mass of 1 mole of a chemical entity. Both will have the same numeric value for a given chemical substance but molecular mass will have the units of amu and molar mass will have the units of g/mol. 3.5 The mole maintains the same mass relationship between macroscopic samples as exist between individual chemical entities. It relates the number of chemical entities (atoms, molecules, ions, electrons) to the mass. 3.6 P4 molecules =   1 mol P4   Avogadro's Number P4 molecules   1 mol Ca 3 (PO 4 ) 2   2 mol P     1 mol P4  Molar Mass   1 mol Ca 3 (PO 4 ) 2   4 mol P    ( 2.5 g Ca 3 (PO4 )2 )  3.7 Plan: It is possible to relate the relative atomic masses by counting the number of atoms. Solution: a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yellow balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of six red balls means that they are that much lighter. Because the red ball is lighter, more red atoms are required to make 1 gram. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 gram. d) Both the left and right elements have the same number of atoms per mole. The number of atoms per mole (6.022 x 1023) is constant and so is the same for every element. 3.8 Plan: Locate each of the elements on the periodic table and record its atomic mass. The mass of the element times the number present in the formula gives the molar mass. 3-1 Solution: a) The molar mass, M, is the sum of the atomic weights, expressed in g/mol: Sr = = 87.62 g Sr/mol Sr(OH)2 O = (2 mol O) (16.00 g O/mol O) = 32.00 g O/mol Sr(OH)2 H = (2 mol H) (1.008 g H/mol H) = 2.016 g H/mol Sr(OH)2 = 121.64 g/mol of Sr(OH)2 b) M = (2 mol N) (14.01 g N/mol N) + (1 mol O) (16.00 g O/mol O) = 44.02 g/mol of N2O c) M = (1 mol Na) (22.99 g Na/mol Na) + (1 mol Cl) (35.45 g Cl/mol Cl) + (3 mol O) (16.00 g O/mol O) = 106.44 g/mol of NaClO3 d) M = (2 mol Cr) (52.00 g Cr/mol Cr) + (3 mol O) (16.00 g O/mol O) = 152.00 g/mol of Cr2O3 3.9 a) (NH4)3PO4 = 3(14.01) + 12(1.008) + 30.97 + 4(16.00) = 149.10 g/mol b) CH2Cl2 = 12.01 + 2(1.008) + 2(35.45) = 84.93 g/mol c) CuSO4·5H2O = 63.55 + 32.07 + 9(16.00) + 10(1.008) = 249.70 g/mol d) BrF5 = 79.90 + 5(19.00) = 174.90 g/mol 3.10 Plan: Locate each of the elements on the periodic table and record its atomic mass. The mass of the element times the number present in the formula gives the molar mass. Solution: a) M = (1 mol Sn) (118.7 g Sn/mol Sn) + (2 mol O) (16.00 g O/mol O) = 150.7 g/mol of SnO2 b) M = (1 mol Ba) (137.3 g Ba/mol Ba) + (2 mol F) (19.00 g F/mol F) = 175.3 g/mol of BaF2 c) M = (2 mol Al) (26.98 g Al/mol Al) + (3 mol S) (32.07 g S/mol S) + (12 mol O) (16.00 g O/mol O) = 342.17g/mol of Al2(SO4)3 d) M = (1 mol Mn) (54.94 g Mn/mol Mn) + (2 mol Cl) (35.45 g Cl/mol Cl) = 125.84 g/mol of MnCl2 3.11 a) N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol b) C8H10 = 8(12.01) + 10(1.008) = 106.16 g/mol c) MgSO4·7H2O = 24.31 + 32.07 + 11(16.00) + 14(1.008) = 246.49 g/mol d) Ca(C2H3O2)2 = 40.08 + 4(12.01) + 6(1.008) + 4(16.00) = 158.17 g/mol 3.12 Plan: The mass of a substance and its number of moles are related through the conversion factor of M, the molar mass expressed in g/mol. The moles of a substance and the number of entities per mole are related by the conversion factor, Avogadro’s number. Solution: a) M of KMnO4 = 39.10 + 54.94 + (4 x 16.00) = 158.04 g/mol of KMnO4  158.04 g KMnO 4  1 Mass of KMnO4 = ( 0.57 mol KMnO 4 )   = 90.08 = 9.0 x 10 g KMnO4 1 mol KMnO 4   b) M of Mg(NO3)2 = 24.31 + (2 x 14.01) + (6 x 16.00) = 148.33 g/mol Mg(NO3)2  1 mol Mg(NO3 ) 2   6 mol O atoms  Moles of O atoms = ( 8.18 g Mg(NO3 )2 )     148.33 g Mg(NO3 ) 2   1 mol Mg(NO3 ) 2  = 0.33088 = 0.331 mol O atoms c) M of CuSO4•5H2O = 63.55 + 32.07 + (4 x 16.00) + (5 x 18.016) = 249.70 g/mol (Note that the waters of hydration are included in the molar mass.)  1 mol Cu Cmpd  9 mol O atoms   6.022 x 1023 O atoms  O atoms = 8.1 x 10−3 g Cu Cmpd       249.70 g Cu Cmpd  1 mol Cu Cmpd   1 mol O atoms  = 1.7581 x 1020 = 1.8 x 1020 O atoms ( 3.13 )    46.01 g NO 2 1 mol NO2 a) Mass NO2 = 3.8 x1020 molecules NO2   6.022 x 1023 molecules NO   1 mol NO 2 2   –5 –5 = 2.9033 x 10 = 2.9 x 10 kg NO2 ( ) 3-2   1 kg    3    10 g   1 mol C2 H 4 Cl2  2 mol Cl atoms  b) Moles Cl atoms = ( 0.0425 g C2 H 4 Cl2 )     98.95 g C2 H 4 Cl2  1 mol C2 H 4 Cl2  = 8.5902 x 10–4 = 8.59 x 10–4 mol Cl atoms  1 mol SrH 2 c) Number of H– = ( 4.92 g SrH 2 )   89.64 g SrH 2 22 = 6.610495 x 10 = 6.61 x 1022 H– ions   2 mol H −     1 mol SrH 2  6.022 x 1023 H − ions     1 mol H −   3.14 Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. Solution: a) M of MnSO4 = (54.94 g Mn/mol Mn) + (32.07 g S/mol S) + [(4 mol O) (16.00 g O/mol O)] = 151.01 g/mol of MnSO4 Mass of MnSO4 = 0.64 mol MnSO4 x 151.01 g MnSO4/mol MnSO4 = 96.65 = 97 g MnSO4 b) M of Fe(ClO4)3 = (55.85 g Fe/mol Fe) + [(3 mol Cl) (35.45 g Cl/mol Cl)] + [(12 mol O) (16.00 g O/mol O)] = 354.20 g/mol of Fe(ClO4)3 Moles Fe(ClO4)3 = (15.8 g Fe(ClO4)3 ) / [(354.20 g Fe(ClO4)3/mol Fe(ClO4)3] = 0.044608 = 4.46 x 10–2 mol Fe(ClO4)3 c) M of NH4NO2 =[(2 mol N) (14.01 g N/mol N)] + [(4 mol H) (1.008 g H/mol H)] + [(2 mol O) (16.00 g O/mol O)] = 64.05 g/mol of NH4NO2   6.022 x 1023 N atoms   1 mol   2 mol N N atoms = 92.6 g NH 4 NO 2      1 mol N  64.05 g   1 mol NH 4 NO 2    3.15 FU = formula unit  1 mol CaF2   6.022 x 1023 FU CaF2 a) Ions = ( 38.1 g CaF2 )    1 mol CaF2  78.08 g CaF2   23 23 = 8.81550 x 10 = 8.82 x 10 ions   3 ions       1 FU CaF2   170.48 g CuCl2 • 2H 2 O   1 mg  b) Mass CuCl2•2H2O = ( 3.58 mol CuCl2 • 2 H 2 O )    −3   1 mol CuCl2 • 2H 2 O   10 g  = 610318.4 = 6.10 x 105 mg CuCl2•2H2O    485.1 g Bi(NO3 )3 • 5H 2 O   1 kg  1 mol c) Mass Bi(NO3)3•5H2O = 2.88 x 1022 FU  23  6.022 x 10 FU   1 FU Bi(NO ) • 5H O   103 g  3 3 2     = 0.0231997 = 0.0232 kg Bi(NO3)3•5H2O ( 3.16 ) Plan: The formula of each compound must be determined from its name. The molar mass for each formula comes from the formula and the periodic table entries. Avogadro’s number is also necessary. Solution: a) Carbonate is a polyatomic anion with the formula, CO32–. The correct formula for this ionic compound is Cu2CO3. M of Cu2CO3 = (2 x 63.55) + 12.01 + (3 x 16.00) = 187.11 g/mol  187.11 g Cu 2 CO3  3 Mass Cu2CO3 = ( 8.41 mol Cu 2 CO3 )   = 1573.595 = 1.57 x 10 g Cu2CO3 1 mol Cu CO 2 3   b) Dinitrogen pentaoxide has the formula N2O5. M of N2O5 = (2 x 14.01) + (5 x 16.00) = 108.02 g/mol    108.02 g N 2 O5  1 mol N 2 O5 Mass N2O5 = 2.04 x 1021 N 2 O5 molecules   6.022 x 1023 N O molecules   1 mol N O  2 5  2 5   = 0.365926 = 0.366 g N2O5 ( ) 3-3 c) The correct formula for this ionic compound is NaClO4. There are Avogadro’s number of entities (in this case, formula units) in a mole of this compound. M of NaClO4 = 22.99 + 35.45 + (4 x 16.00) = 122.44 g/mol  1 mol NaClO 4  Moles NaClO4 = ( 57.9 g NaClO 2 )   = 0.47288 = 0.473 mol NaClO4  122.44 g NaClO 4  FU = formula units  1 mol NaClO 4   6.022 x 1023 FU NaClO 4  FU NaClO4 = ( 57.9 g NaClO 2 )   =   1 mol NaClO 4  122.44 g NaClO 4    23 23 2.8477115 x 10 = 2.85 x 10 FU NaClO4 d) The number of ions or atoms is calculated from the formula units given in part c. Note the unrounded initially calculated value is used to avoid intermediate rounding.  1 Na + ion  = 2.85 x 1023 Na+ ions 2.8477115 x 1023 mol NaClO4   1 FU NaClO  4    1 ClO −4 ion 2.8477115 x 1023 mol NaClO4   1 FU NaClO 4   23 –  = 2.85 x 10 ClO4 ions   1 Cl atom  23 2.8477115 x 1023 mol NaClO4   = 2.85 x 10 Cl atoms 1 FU NaClO 4    4 O atoms  24 2.8477115 x 1023 mol NaClO4   = 1.14 x 10 O atoms 1 FU NaClO 4   3.17 a) The formula is Cr2(SO4)3•10H2O, and the molar mass is 572.4 g/mol.  572.4 g  Grams Cr2(SO4)3•10H2O = ( 3.52 mol Cr2 (SO 4 )3 • 10H 2 O )    mol  = 2014.848 = 2.01 x 103 g Cr2(SO4)3•10H2O b) The formula is Cl2O7, and the molar mass is 182.9 g/mol.    182.9 g Cl2 O7  1 mol Grams Cl2O7 = 9.64 x 1024 molecules Cl2 O7   23  6.022 x 10 molecules   1 mol    = 2927.858 = 2.93 x 103 g Cl2O7 c) The formula is Li2SO4, and the molar mass is 109.95 g/mol.  1 mol Li 2SO 4  Mol Li2SO4 = ( 56.2 g Li 2SO 4 )   = 0.5111414 = 0.511 mol Li2SO4  109.95 g Li 2SO 4  ( )  1 mol Li 2SO 4   6.022 x 10 23 FU  FU Li2SO4 = ( 56.2 g Li 2SO 4 )      109.95 g Li 2SO 4   1 mol Li 2SO 4  = 3.07809 x 1023 = 3.08 x 1023 FU Li2SO4 d) Note the unrounded initially calculated FU value is used to avoid intermediate rounding.  2 Li + ions  = 6.156187 x 1023 = 6.16 x 1023 Li+ ions 3.07809 x 1023 FU Li2SO4   1 FU Li SO  2 4    1 SO 4 2 − ion 3.07809 x 1023 FU Li2SO4   1 FU Li SO 2 4   23 23 2–  = 3.07809 x 10 = 3.08 x 10 SO4 ions   1 S atom  23 23 3.07809 x 1023 FU Li2SO4   = 3.07809 x 10 = 3.08 x 10 S atoms 1 FU Li SO 2 4   3-4  4 O atoms 3.07809 x 1023 FU Li2SO4   1 FU Li 2SO 4  24 24  = 1.231237 x 10 = 1.23 x 10 O atoms  3.18 Plan: Determine the formula and the molar mass of each compound. The formula gives the number of atoms of each type of element present. Masses come from the periodic table. Solution: a) Ammonium bicarbonate is an ionic compound consisting of ammonium ions, NH4+ and bicarbonate ions, HCO3–. The formula of the compound is NH4HCO3. M of NH4HCO3 = (14.01 g/mol) + (5 x 1.008 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 79.06 g/mol NH4HCO3 In 1 mole of ammonium bicarbonate, with a mass of 79.06 g, there are 5 H atoms with a mass of 5.040 g. (5 mol H) (1.008 g/mol H) x 100% = 6.374905 = 6.375% H 79.06 g/mol NH 4 HCO3 b) Sodium dihydrogen phosphate heptahydrate is a salt that consists of sodium ions, Na+, dihydrogen phosphate ions, H2PO4–, and seven waters of hydration. The formula is NaH2PO4•7H2O. Note that the waters of hydration are included in the molar mass. M of NaH2PO4•7H2O = (22.99 g/mol) + (16 x 1.008 g/mol) + (30.97g/mol) + (11 x 16.00 g/mol) = 246.09 g/mol NaH2PO4•7H2O In each mole of NaH2PO4.7H2O (with mass of 246.09 g), there are 11 x 16.00 g/mol or 176.00 g of oxygen. (11 mol O )(16.00 g/mol O ) x 100 % = 71.51855 = 71.52% O 246.09 g/mol 3.19 a) Sr(IO4)2 437.42 g/mol ( 2 mol I )(126.9 g/mol I ) x 100% = 54.0690 = 54.07% I 469.4 g/mol b) KMnO4 158.04 g/mol (1 mol Mn )( 54.94 g/mol Mn ) x 100% = 34.76335 = 34.76% Mn 158.04 g/mol 3.20 Plan: Mass fraction is related to the mass percentage, however the mass fraction is expressed in decimal rather than percentage form. Solution: a) Cesium acetate is an ionic compound consisting of Cs+ cations and C2H3O2– anions. (Note that the formula for acetate ions can be written as either C2H3O2– or CH3COO–.) The formula of the compound is CsC2H3O2. One mole of CsC2H3O2 weighs 191.9 g: M of CsC2H3O2 = 132.9 + (2 x 12.01) + (3 x 1.008) + (2 x 16.00) = 191.9 g/mol ( 2 mol C )(12.01 g/mol C ) Mass fraction of C = = 0.125169 = 0.1252 mass fraction C 191.9 g/mol b) The formula for this compound is UO2SO4•3H2O. M of UO2SO4•3H2O = 238.0 + (9 x 16.00) + 32.07 + (6 x 1.008) = 420.1 g/mol ( 9 mol O )(16.00 g/mol O ) = 0.3427755 = 0.3428 mass fraction O Mass fraction of O = 420.1 g/mol 3.21 a) The formula is Ca(ClO3)2, and the molar mass is 206.98 g/mol. ( 2 mol Cl ) ( 35.45 g/mol Cl ) = 0.342545 = 0.3425 mass fraction Cl Mass fraction of Cl = 206.98 g/mol 3-5 b) The formula is P4O6, and the molar mass is 219.88 g/mol. ( 4 mol P )( 30.97 g/mol P ) = 0.5633982 = 0.5634 mass fraction P Mass fraction of P = 219.88 g/mol 3.22 3.23  1 mol O 2   6.022 x 1023 O 2 molecules   2 O atoms  O atoms = ( 38.0 g O2 )       1 mol O 2  32.00 g O 2     1 O 2 molecule  = 1.430225 x 1024 = 1.43 x 1024 O atoms Plan: Determine the formula of cisplatin from the figure, and then calculate the molar mass from the formula. The molar mass is necessary for the subsequent calculations. Solution: The formula for cisplatin is Pt(Cl)2(NH3) 2 M of Pt(Cl)2(NH3) 2 = 195.1 + (2 x 35.45) + (2 x 14.01) + (6 x 1.008) = 300.1 g/mol  1 mol cisplatin  a) Moles cisplatin = ( 285.3 g cisplatin )   = 0.9506831 = 0.9507 mol cisplatin  300.1 g cisplatin   6 mol H   6.022 x 1023 H atoms  b) H atoms = ( 0.98 mol cisplatin )     1 mol H  1 mol cisplatin    = 3.540936 x 1024 = 3.5 x 1024 H atoms 3.24 The formula, from the figure, is (C3H5)2S, and the molar mass is 114.21 g/mol.  114.21 g allyl sulfide  a) Grams allyl sulfide = (1.63 mol allyl sulfide )    1 mol allyl sulfide  = 186.1623 = 186 g allyl sulfide  1 mol (C3 H5 ) 2 S     6.022 x 1023 C atoms  6 mol C b) C atoms = ( 4.77 g (C3 H 5 ) 2 S )      1 mol C  114.21 g (C3 H 5 ) 2 S   1 mol (C3 H 5 ) 2 S    = 1.5090591 x 1023 = 1.51 x 1023 C atoms 3.25 3.26 Plan: Determine the molar mass of rust. Use the molar mass to find the moles of rust. The moles of rust may be related to the grams of iron through the mole ratio. Solution: a) M of Fe2O3•4H2O = (2 x 55.85) + (3 x 16.00) + (4 x 18.016) = 231.76 g/mol  103 g   1 mol rust  Moles of compound = ( 65.2 kg rust )  = 281.3255 = 281 mol rust  1kg   231.76 g rust     b) The formula shows that there is one mole of Fe2O3 for every mole of rust, so there are also 281 mol of Fe2O3. c) Calculate grams of iron by determining the mole ratio and converting to grams.  103 g   1 mol rust   1 mol Fe 2 O3   2 mol Fe   55.85 g Fe  Grams of iron = ( 65.2 kg rust )          1 kg   231.76 g rust   1 mol rust   1 mol Fe 2 O3   1 mol Fe  = 31424.059 = 3.14 x 104 g Fe  1 mol C3 H8  a) Moles C3H8 = ( 75.3 g C3 H8 )   = 1.707870 = 1.71 mol C3H8  44.09 g C3 H8   1 mol C3 H8  3 mol C   12.01 g C  b) Grams C = ( 75.3 g C3 H8 )     = 61.534566 = 61.5 g C  44.09 g C3 H8  1 mol C3 H8   1 mol C  3-6 3.27 Plan: Determine the formulas for the compounds where needed. Determine the molar mass of each formula. Calculate the percent nitrogen by dividing the mass of nitrogen in a mole of compound by the molar mass of the compound, and multiply the result by 100%. Then rank the values. Solution: Name Formula Molar Mass (g/mol) Potassium nitrate KNO3 101.11 Ammonium nitrate NH4NO3 80.05 Ammonium sulfate (NH4)2SO4 132.12 60.06 Urea CO(NH2)2 Calculating the nitrogen percentages: (1 mol N ) (14.01 g/mol N ) x 100 = 13.856196 = 13.86% N Potassium nitrate 101.11 g/mol Ammonium nitrate ( 2 mol N )(14.01 g/mol N ) x100 = 35.003123 = 35.00% N Ammonium sulfate ( 2 mol N )(14.01 g/mol N ) Urea ( 2 mol N )(14.01 g/mol N ) 80.05 g/mol 132.12 g/mol 60.06 g/mol Rank is CO(NH2)2 > NH4NO3 > (NH4)2SO4 > KNO3 3.28 x 100 = 21.20799 = 21.21% N x 100 = 46.6533 = 46.65% N Plan: The volume must be converted from cubic feet to cubic centimeters (or vice versa). The volume and the density will give you mass, and the mass with the molar mass gives you moles. Part (b) requires a conversion from cubic decimeters, instead of cubic feet, to cubic centimeters. The density allows you to change these cubic centimeters to mass, the molar mass allows you to find moles, and finally Avogadro’s number allows you to make the last step. Solution: The molar mass of galena is 239.3 g/mol. 3  12 in   2.54 cm  a) Moles PbS = 1.00 ft PbS      1 ft   1 in  = 882.7566886 = 883 mol PbS ( ) 3 3 3  7.46 g PbS   1 mol PbS      3  1 cm   239.3 g PbS  3  0.1 m   1 cm   7.46 g PbS   1 mol PbS  1 mol Pb   6.022 x 1023 Pb atoms  b) 1.00 dm PbS      −2      3  1 mol Pb  1 dm   10 m   1 cm   239.3 g PbS  1 mol PbS    = 1.87731 x 1025 = 1.88 x 1025 Pb atoms ( 3 ) 3.29 Plan: If the molecular formula for hemoglobin (Hb) was known, the number of Fe2+ ions in a molecule of hemoglobin could be calculated. It is possible to calculate the mass of iron from the percentage of iron and the molar mass of the compound. From the mass of iron, the moles of iron per mole of hemoglobin may be found. Solution:  0.33% Fe   6.8 x 104 g   1 mol Fe  2+       = 4.0179 = 4.0 mol Fe /mol Hb 100% Hb mol 55.85 g Fe     Thus, there are 4 Fe2+/molecule Hb. 3.30 Determine the empirical formula from the mass % of the elements. Determine the E.F. from elemental composition. Determine the E.F. from combustion analysis. 3.31 a) No, you can obtain the empirical formula from the number of moles of each type of atom in a compound, but not the molecular formula. b) Yes, you can obtain the molecular formula from the mass percentages and the total number of atoms. 3-7 Solution: 1) Assume a 100.0 g sample and convert masses (from the mass % of each element) to moles using molar mass. 2) Identify the element with the lowest number of moles and use this number to divide into the number of moles for each element. You now have at least one elemental mole ratio (the one with the smallest number of moles) equal to 1.00 and the remaining mole ratios that are larger than one. 3) Examine the numbers to determine if they are whole numbers. If not, multiply each number by a whole number factor to get whole numbers for each element. You will have to use some judgment to decide when to round. 4) Write the empirical formula using the whole numbers from step 3. 5) Check the total number of atoms in the empirical formula. If it equals the total number of atoms given then the empirical formula is also the molecular formula. If not, then divide the total number of atoms given by the total number of atoms in the empirical formula. This should give a whole number. Multiply the number of atoms of each element in the empirical formula by this whole number to get the molecular formula. If you do not get a whole number when you divide, return to step 3 and revise how you multiplied and rounded to get whole numbers for each element. c) Yes, you can determine the molecular formula from the mass percent and the number of atoms of one element in a compound. Solution plan: 1) Follow steps 1-4 in part b. 2) Compare the number of atoms given for the one element to the number in the empirical formula. Determine the factor the number in the empirical formula must be multiplied by to obtain the given number of atoms for that element. Multiply the empirical formula by this number to get the molecular formula. d) No, the mass % will only lead to the empirical formula. e) Yes, a structural formula shows all the atoms in the compound. Solution plan: Count the number of atoms of each type of element and record as the number for the molecular formula. 3.32 MgCl2 is an empirical formula, since ionic compounds such as MgCl2 do not contain molecules. 3.33 Plan: Examine the number of atoms of each type in the compound. Divide all atom numbers by any common factor. The final answers must be the lowest whole-number values. Solution: a) C2H4 has a ratio of 2 carbon atoms to 4 hydrogen atoms, or 2:4. This ratio can be reduced to 1:2, so that the empirical formula is CH2. The empirical formula mass is 12.01 + 2(1.008) = 14.03 g/mol. b) The ratio of atoms is 2:6:2, or 1:3:1. The empirical formula is CH3O and its empirical formula mass is 12.01 + 3(1.008) + 16.00 = 31.03 g/mol c) Since, the ratio of elements cannot be further reduced, the molecular formula and empirical formula are the same, N2O5. The formula mass is 2(14.01) + 5(16.00) = 108.02 g/mol. d) The ratio of elements is 3 atoms of barium to 2 atoms of phosphorus to 8 atoms of oxygen, or 3:2:8. This ratio cannot be further reduced, so the empirical formula is also Ba3(PO4)2, with a formula mass of 3(137.3) + 2(30.97) + 8(16.00) = 601.8 g/mol. e) The empirical formula is TeI4, and the formula mass is 127.6 + 4(126.9) = 635.2 g/mol. 3.34 a) C4H8 = M.F. b) C3H6O3 = M.F. c) P4O10 = M.F. d) Ga2(SO4)3 = M.F. e) Al2Br6 = M.F. 3.35 Plan: Determine the molar mass of each empirical formula. The molar mass of each compound divided by its empirical formula mass gives the number of times the empirical formula is within the molecule. Multiply the empirical formula by the number of times the empirical formula appears to get the molecular formula. CH2 = E.F. CH2O = E.F. P2O5 = E.F. Ga2(SO4)3 = M.F. = E.F. AlBr3 = E.F. 3-8 E.F. mass = 14.03 g/mol E.F. mass = 30.03 g/mol E.F. mass = 141.94 g/mol E.F. mass = 427.6 g/mol E.F. mass = 266.7 g/mol Solution: Only approximate whole number values are needed. a) CH2 has empirical mass equal to 14.03 g/mol  42.08 g/mol    =3  14.03 g/mol  Multiplying the subscripts in CH2 by 3 gives C3H6 b) NH2 has empirical mass equal to 16.03 g/mol  32.05 g/mol   =2  16.03 g/mol  Multiplying the subscripts in NH2 by 2 gives N2H4 c) NO2 has empirical mass equal to 46.01 g/mol  92.02 g/mol   =2  46.01 g/mol  Multiplying the subscripts in NO2 by 2 gives N2O4 d) CHN has empirical mass equal to 27.03 g/mol  135.14 g/mol   = 5  27.03 g/mol  Multiplying the subscripts in CHN by 5 gives C5H5N5 3.36 a) CH has empirical mass equal to = 13.02 g/mol  78.11 g/mol  M.F. = 6 x E.F. = C6H6  = 6  13.02 g/mol  b) C3H6O2 has empirical mass equal to = 74.08 g/mol  74.08 g/mol  M.F. = 1 x E.F. = C3H6O2  =1  74.08 g/mol  c) HgCl has empirical mass equal to = 236.0 g/mol  472.1 g/mol  M.F. = 2 x E.F. = Hg2Cl2  = 2  236.0 g/mol  d) C7H4O2 has empirical mass equal to = 120.10 g/mol  240.20 g/mol  M.F. = 2 x E.F. = C14H8O4  = 2  120.10 g/mol  3.37 Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. All data must be converted to moles of an element. Using the smallest number of moles present, convert the mole ratios to whole numbers. Solution:  0.063 mol Cl   0.22 mol O  a)   =1   = 3.5  0.063 mol Cl   0.063 mol Cl  The formula is Cl1O3.5, which in whole numbers (x 2) is Cl2O7 3-9  1 mol Si  b) ( 2.45 g Si )   = 0.08722 mol Si  28.09 g Si   1 mol Cl   = 0.349788 mol Cl  35.45 g Cl  (12.4 g Cl )   0.08722 mol Si   0.349788 mol Cl    =1   =4 0.08722 mol Si    0.08722 mol Si  The empirical formula is SiCl4. c) Assume a 100 g sample and convert the masses to moles. 27.3% C   1 mol C  72.7% O   1 mol O  (100 g )  (100 g )    = 2.2731 mol C  = 4.5438 mol O    100%   12.01 g C   100%   16.00 g O   2.2731 mol C    =1  2.2731 mol C  The empirical formula is CO2. 3.38  4.5438 mol O    =2  2.2731 mol C   0.039 mol Fe   0.052 mol O  a)  =1   = 1.3333  0.039 mol Fe   0.039 mol Fe  The formula is Fe1O1.3333, which in whole numbers (x 3) is Fe3O4  1 mol P   1 mol Br  b) ( 0.903 g P )  ( 6.99 g Br )   = 0.029157 mol P  = 0.087484 mol Br  30.97 g P   79.90 g Br   0.029157 mol P   =1  0.029157 mol P  The empirical formula is PBr3.  0.087484 mol Br   =3  0.029157 mol P  c) Assume a 100 g sample and convert the masses to moles. 79.9% C   1 mol C  (100 g )   = 6.6528 mol C   100%   12.01 g C  (100 − 79.9)% H   1 mol H    1.008 g H  = 19.940 mol H 100%    (100 g )   6.6528 mol C   =1  6.6528 mol C   19.940 mol H   = 3  6.6528 mol C  The empirical formula is CH3. 3.39 Plan: The percent oxygen is 100% minus the percent nitrogen. Assume 100 grams of sample, and then the moles of each element may be found. Divide each of the moles by the smaller value, and convert to whole numbers to get the empirical formula. The empirical formula mass and the given mass lead to the molecular formula. Solution: a) % O = 100% - % N = 100% - 30.45% N = 69.55% O For a 100-gram sample, the mass, in grams, is numerically identical to the mass percent.  1 mol N  Moles N = ( 30.45 g N )   = 2.1734 mol N  14.01 g N   1 mol O  Moles O = ( 69.55 g O )   = 4.3469 mol O  16.00 g O   2.1734 mol N    =1  2.1734 mol N   4.3469 mol O    =2  2.1734 mol N  The empirical formula is NO2. 3-10 b) Formula mass of empirical formula = 14.01 g N/mol + 2(16.00 g O/mol) = 46.01 g/mol Molar mass / Empirical formula mass = (90 g/mol) / (46 g/mol) = 2 Thus, the molecular formula is twice the empirical formula. Note: only an approximate value is needed. The molecular formula is N2O4. 3.40 The percent silicon is 100% minus the percent chlorine. Assume 100 grams of sample, and then the moles of each element may be found. Divide each of the moles by the smaller value, and convert to whole numbers to get the empirical formula. The empirical formula mass and the given mass lead to the molecular formula. a) % Si = 100% - % Cl = 100% - 79.1% Cl = 20.9% Si For a 100-gram sample, the mass, in grams, is numerically identical to the mass percent.  1 mol Si  Moles Si = ( 20.9 g Si )   = 0.7440 mol Si  28.09 g Si   1 mol Cl  Moles Cl = ( 79.1 g Cl )   = 2.2313 mol Cl  35.45 g Cl   0.7440 mol Si   =1  0.7440 mol Si   2.2313 mol Cl   =3  0.7440 mol Si  The empirical formula is SiCl3. b) Formula mass of empirical formula = 28.09 g Si/mol + 3(35.45 g Cl/mol) = 134.44 g/mol Molar mass / Empirical formula mass = (269 g/mol) / (134.44 g/mol) = 2 Thus, the molecular formula is twice the empirical formula. Note: only an approximate value is needed. The molecular formula is Si2Cl6. 3.41 Plan: The balanced equation for this reaction is: M(s) + F2(g) → MF2(s) since fluorine, like other halogens, exists as a diatomic molecule. The moles of the metal are known, and the moles of everything else may be found from these moles using the balanced chemical equation. Solution: a) Determine the moles of fluorine.  2 mol F  Moles F = ( 0.600 mol M )   = 1.20 mol F  1 mol M  b) The grams of M are the grams of MF2 minus the grams of F present.  19.00 g F  Grams M = 46.8 g (M + F) - (1.20 mol F )   = 24.0 g M  1 mol F  c) The molar mass is needed to identify the element. Molar mass of M = 24.0 g M / 0.600 mol M = 40.0 g/mol The metal with the closest molar mass to 40.0 g/mol is calcium. 3.42  3 mol O  a) moles O = ( 0.370 mol M 2 O3 )   = 1.11 mol O  1 mol M 2 O3  b) The grams of M are the grams of M2O3 minus the grams of O present.  16.00 g O  Grams M = 55.4 g (M + O) - (1.11 mol O )   = 37.64 = 37.6 g M  1 mol O  c) First, the number of moles of M must be calculated.  2 mol M Moles M = ( 0.370 mol M 2 O3 )   1 mol M 2 O3   = 0.740 mol M  The molar mass is needed to identify the element. (Use the unrounded mass of M to avoid intermittent rounding errors.) Molar mass of M = 37.64 g M / 0.740 mol M = 50.86 g/mol The metal with the closest molar mass to 50.9 g/mol is vanadium. 3-11 3.43 Divide all the millimoles by the smallest value and convert to whole numbers to get the empirical formula. Since all the values are given in millimoles, there is no need to convert to another unit.  6.16 mmol C   8.56 mmol H   1.23 mmol N    =5   =7   =1  1.23 mmol N   1.23 mmol N   1.23 mmol N  All the ratios are close to whole numbers so no other adjustments are necessary. The empirical formula is C5H7N. 3.44 Plan: Assume 100 grams of cortisol so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. The empirical formula mass and the given molar mass will then relate the empirical formula to the molecular formula. Solution:  1 mol C  Moles C = ( 69.6 g C )   = 5.7952 mol C  12.01 g C   1 mol H  Moles H = ( 8.34 g H )   = 8.2738 mol H  1.008 g H   1 mol O  Moles O = ( 22.1 g O )   = 1.38125 mol O  16.00 g O   5.7952 mol C   8.2738 mol H   1.38125 mol O    = 4.20   = 6.00   = 1.00 1.38125 mol O 1.38125 mol O      1.38125 mol O  The carbon value is not close enough to a whole number to round the value. The smallest number that 4.20 may be multiplied by to get close to a whole number is 5. (You may wish to prove this to yourself.) All three ratios need to be multiplied by five to get the empirical formula of C21H30O5. The empirical formula mass is: 21 (12.01 g C/mol) + 30 (1.008 g H/mol) + 5 (16.00 g O/mol) = 362.45 g/mol The empirical formula mass and the molar mass given are the same, so the empirical and the molecular formulas are the same. The molecular formula is C21H30O5. 3.45 Determine the molecular formula from the figure, and the molar mass from the molecular formula. Divide the amount of each element by the molar mass and multiply the answer by 100%. Molecular formula = C8H9NO2 Molar mass = 8 (12.01 g C/mol) + 9 (1.008 g H/mol) + 1 (14.01 g/mol) + 2 (16.00 g O/mol) = 151.16 g/mol % C = [8(12.01g C/mol)/151.16 g/mol] x 100% = 63.5618 = 63.56% C % H = [9(1.008 g H/mol)/151.16 g/mol] x 100% = 6.00159 = 6.002% H % N = [14.01 g N/mol/151.16 g/mol] x 100% = 9.2683 = 9.268% N % O = [2(16.00 g O/mol)/151.16 g/mol] x 100% = 21.1696 = 21.17% O 3.46 Plan: In combustion analysis, finding the moles of carbon and hydrogen is relatively simple because all of the carbon present in the sample is found in the carbon of CO2, and all of the hydrogen present in the sample is found in the hydrogen of H2O. The moles of oxygen are more difficult to find, because additional O2 was added to cause the combustion reaction. The masses of CO2 and H2O are used to find both the mass of C and H and the moles of C and H. Subtracting the masses of C and H from the mass of the sample gives the mass of O. Convert the mass of O to moles of O. Take the moles of C, H, and O and divide by the smallest value, and convert to a whole number to get the empirical formula. Determine the empirical formula mass and compare it to the molar mass given in the problem to see how the empirical and molecular formulas are related. Finally, determine the molecular formula. Solution: (There is no intermediate rounding.) Initial mole determination:  1 mol CO 2   1 mol C  Moles C = ( 0.449 g CO 2 )    = 0.010202 mol C  44.01 g CO 2   1 mol CO 2  3-12  1 mol H 2 O   2 mol H  Moles H = ( 0.184 g H 2 O )    = 0.020422 mol H  18.02 g H 2 O   1 mol H 2 O  Now determine the masses of C and H:  12.01 g C  Grams C = ( 0.010202 mol C )   = 0.122526 g C  1 mol C   1.008 g H  Grams H = ( 0.020422 mol H )   = 0.020585 g H  1 mol H  Determine the mass and then the moles of O: 0.1595 g (C, H, and O) - 0.122526 g C - 0.020585 g H = 0.016389 g O  1 mol O  Moles O = ( 0.016389 g O )   = 0.0010243 mol O  16.00 g O  Divide by the smallest number of moles:  0.010202 mol C    = 10  0.0010243 mol O  (Rounding is acceptable for these answers.)  0.020422 mol H   0.0010243 mol O    = 20   =1  0.0010243 mol O   0.0010243 mol O  Empirical formula = C10H20O Empirical formula mass = 10 (12.01 g C/mol) + 20 (1.008 g H/mol) + 1 (16.00 g O/mol) = 156.26 g/mol The empirical formula mass matches the given molar mass so the empirical and molecular formulas are the same. The molecular formula is C10H20O. 3.47 A balanced chemical equation describes: a) The identities of the reactants and products b) The molar (and molecular) ratios by which reactants form products c) The physical states of all substances in the reaction 3.48 In a balanced equation, the total mass of the reactants is equal to the total mass of the products formed in the reaction. Thus, the law of conservation of mass is obeyed. 3.49 Students I and II are incorrect. Both students changed a given formula. Only coefficients should be changed when balancing. Student I failed to identify the product correctly, and Student II used atomic chlorine instead of molecular chlorine as a reactant. Student III followed the correct process. 3.50 Plan: Exam the diagram and label each formula. We will use A for red atoms and B for green atoms. Solution: The reaction shows A2 and B2 molecules forming AB molecules. Equal numbers of A2 and B2 combine to give twice as many molecules of AB. Thus, the reaction is A2 + B2 → 2 AB. This is the answer to part b. 3.51 Plan: Balancing is a trial and error procedure. Do one blank/one element at a time. Solution: a) 16 Cu(s) + __ S8(s) → 8 Cu2S(s) b) ___ P4O10(s) + 6 H2O(l) → 4 H3PO4(l) Hint: Balance the P first, because there is an obvious deficiency of P on the right side of the equation. Balance the H next, because H is present in only one reactant and only one product. Balance the O last, because it appears in both reactants and is harder to balance. c) __B2O3(s) + 6 NaOH(aq) → 2 Na3BO3(aq) + 3 H2O(l) Hint: Oxygen is again the hardest element to balance because it is present in more than one place on each side of the reaction. If you balance the easier elements first (B, Na, H), the oxygen will automatically be balanced. d) 2 CH3NH2(g) + 9/2 O2(g) → 2 CO2(g) + 5 H2O(g) + __N2(g) 4 CH3NH2(g) + 9 O2(g) → 4 CO2(g) + 10 H2O(g) + 2 N2(g) Hint: You should balance odd/even numbers of oxygen using the “half” method, and then multiply all coefficients by two. 3-13 3.52 a) Cu(NO3)2(aq) + 2 KOH(aq) → Cu(OH)2(s) + 2 KNO3(aq) b) BCl3(g) + 3 H2O(l) → H3BO3(s) + 3 HCl(g) c) CaSiO3(s) + 6 HF(g) → SiF4(g) + CaF2(s) + 3 H2O(l) d) (CN)2(g) + 4 H2O(1) → H2C2O4(aq) + 2 NH3(g) 3.53 Plan: The balancing is a trial and error procedure. Do one blank/one element at a time. Solution: a) 2 SO2(g) + __O2(g) → 2 SO3(g) b) __Sc2O3(s) + 3 H2O(l) → 2 Sc(OH)3(s) c) __H3PO4(aq) + 2 NaOH(aq) → __Na2HPO4(aq) + 2 H2O(l) d) C6H10O5(s) + 6 O2(g) → 6 CO2(g) + 5 H2O(g) 3.54 a) As4S6(s) + 9 O2(g) → As4O6(s) + 6 SO2(g) b) 2 Ca3(PO4)2(s) + 6 SiO2(s) + 10 C(s) → P4(g) + 6 CaSiO3(l) + 10 CO(g) c) 3 Fe(s) + 4 H2O(g) → Fe3O4(s) + 4 H2(g) d) 6 S2Cl2(l) + 16 NH3(g) → S4N4(s) + S8(s) + 12 NH4Cl(s) 3.55 Plan: The names must first be converted to chemical formulas. The balancing is a trial and error procedure. Do one blank/one element at a time. Solution: a) 4 Ga(s) + 3 O2(g) → 2 Ga2O3(s) b) 2 C6H14(l) + 19 O2(g) → 12 CO2(g) + 14 H2O(g) c) 3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCl(aq) 3.56 a) Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) b) Si2Cl6(l) + 4 H2O(l) → 2 SiO2(s) + 6 HCl(g) + H2(g) c) 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) 3.57 The stoichiometrically equivalent molar ratio is the ratio of the coefficients in the balanced equation. The molar ratio is equivalent to the mass ratios in mass units. This can be used to calculate masses of reactants or products in a chemical reaction. 3.58 First, write a balanced equation: a A + b B → c C    c Mol C   Molar Mass C  1 mol A Mass C = ( 5 g A )    =gC Mol C  Molar Mass A   a Mol A    3.59 Plan: First, write a balanced chemical equation. Determine if D or E is the limiting reagent. Use the limiting reagent to determine the amount of F formed. Solution: The balanced equation is: d D + e E → f F From the relationship moles = g/M, calculate the moles of D and E present. By comparing the ratio of moles present to that needed in the balanced equation, determine the limiting reactant (LR). Based on the number of moles of limiting reactant, proceed as follows.  1 mol LR   f Mol F  Molar Mass F  Mass F = ( g LR )    = g F Mol F  Molar Mass LR   Mol LR   3.60 The percent yield is the ratio of actual to theoretical value. Both yields can be expressed as a mass or mole comparison. The percent yield will be the same since mass and moles are directly proportional. 3.61 Plan: Use the mole ratio from the balanced chemical equation to determine the moles produced. Use the moles with the molar mass to determine the grams produced. 3-14 Solution:  1 mol Cl2  a) Moles Cl2 = (1.82 mol HCl )   = 0.455 mol Cl2  4 mol HCl   1 mol Cl2   70.90 g Cl2  b) Grams Cl2 = (1.82 mol HCl )   = 32.2595 = 32.3 g Cl2   4 mol HCl   1 mol Cl2  The beginning of the calculation is repeated to emphasize that the second part of the problem is simply an extension of the first part. There is no need to repeat the entire calculation, as only the final step times the answer of the first part will give the final answer to this part. Hint: Always check to see if the initial equation is balanced. If the equation is not balanced, it should be balanced before proceeding. 3.62 Bi2O3(s) + 3 C(s) → 2 Bi(s) + 3 CO(g)  1 mol Bi 2 O3 a) Moles Bi2O3 = ( 352 g Bi 2 O3 )   466.0 g Bi 2 O3   = 0.7553648 = 0.755 mol Bi2O3   1 mol Bi 2 O3   2 mol Bi  b) Moles Bi = ( 352 g Bi 2 O3 )    = 1.5107 = 1.51 mol Bi  466.0 g Bi 2 O3   1 mol Bi 2 O3  3.63 Plan: Convert the kilograms of oxygen to the moles of oxygen. Use the moles of oxygen and the mole ratios from the balanced chemical equation to determine the moles of KNO3. The moles of KNO3 and its molar mass will give the grams. Solution:  103 g   1 mol O 2   4 mol KNO3  = 2215 = 2.22 x 103 mol KNO3 a) Moles KNO3 = ( 88.6 kg O 2 )   1 kg   32.00 g O   5 mol O  2  2     103 g   1 mol O 2   4 mol KNO3   101.11 g KNO3  b) Grams KNO3 = ( 88.6 kg O 2 )   1 kg   32.00 g O   5 mol O   1 mol KNO  2  2  3    = 223958.65 = 2.24 x 105 g KNO3 The beginning of the calculation is repeated to emphasize that the second part of the problem is simply an extension of the first part. There is no need to repeat the entire calculation, as only the final step times the answer of the first part will give the final answer to this part. 3.64  1 mol Cr2S3   1 mol Cr2 O3 a) Moles Cr2O3 = ( 421 g Cr2S3 )    200.21 g Cr2S3   1 mol Cr2S3   = 2.10279 = 2.10 mol Cr2O3   1 mol Cr2S3   1 mol Cr2 O3   152.00 g Cr2 O3  b) Grams Cr2O3 = ( 421 g Cr2S3 )      200.21 g Cr2S3   1 mol Cr2S3   1 mol Cr2 O3  = 319.624 = 3.20 x 102 g Cr2O3 3.65 Plan: First, balance the equation. Convert the grams of diborane to moles of diborane using its molar mass. Use mole ratios from the balanced chemical equation to determine the moles of the products. Use the moles and molar mass of each product to determine the mass formed. Solution: The balanced equation is: B2H6(g) + 6 H2O(l) → 2 H3BO3(s) + 6 H2(g).  1 mol B2 H 6   2 mol H3 BO3   61.83 g H3 BO3  Mass H3BO3 = ( 33.61 g B2 H 6 )      27.67 g B2 H 6   1 mol B2 H 6   1 mol H3 BO3  = 150.206 = 150.2 g H3BO3 3-15  1 mol B2 H 6   6 mol H 2   2.016 g H 2  Mass H2 = ( 33.61 g B2 H 6 )      27.67 g B2 H 6   1 mol B2 H 6   1 mol H 2  = 14.69268 = 14.69 g H2 3.66 First, balance the equation: Ag2S(s) + 2 HCl(aq) → 2 AgCl(s) + H2S(g)  1 mol Ag 2S   2 mol AgCl   143.4 g AgCl  Grams AgCl = 174 g Ag 2S     = 201.304 = 201 g AgCl  247.9 g Ag 2S   1 mol Ag 2S   1 mol AgCl   1 mol Ag 2S   1 mol H 2S  34.09 g H 2S  Grams H2S = 174 g Ag 2S     = 23.9276 = 23.9 g H2S  247.9 g Ag 2S   1 mol Ag 2S  1 mol H 2S  3.67 Plan: Write the balanced equation by first writing the formulas for the reactants and products. Reactants: formula for phosphorus is given as P4 and formula for chlorine gas is Cl2 (chlorine occurs as a diatomic molecule). Products: formula for phosphorus pentachloride — the name indicates one phosphorus atom and five chlorine atoms to give the formula PCl5. Convert the mass of phosphorus to grams, use the mole ratio from the balanced chemical equation, and finally use the molar mass of chlorine to get the mass of chlorine. Solution: Formulas give the equation: P4 + Cl2 → PCl5 Balancing the equation: P4 + 10 Cl2 → 4 PCl5  1 mol P4   10 mol Cl2   70.90 g Cl2  3 Grams Cl2 = ( 355 g P4 )     = 2031.76 = 2.03 x 10 g Cl2 123.88 g P 1 mol P 1 mol Cl 4  4  2   3.68 First, balance the equation: S8(s) + 24 F2(g) → 8 SF6(s)  1 mol S8   24 mol F2   38.00 g F2  Grams F2 = (17.8 g S8 )     = 63.27409 = 63.3 g F2  256.56 g S8   1 mol S8   1 mol F2  3.69 Plan: Begin by writing the chemical formulas of the reactants and products in each step. Next, balance each of the equations. Combine the equations for the separate steps by adjusting the equations so the intermediate (iodine monochloride) cancels. Finally, change the mass of product to mole and use the mole ratio and molar mass of iodine to determine the mass of iodine. Solution: a) First step: I2(s) + Cl2(g) → 2 ICl(s) Second step: ICl(s) + Cl2(g) → ICl3(s) b) Multiply the coefficients of the second equation by 2, so that ICl(s), an intermediate product, can be eliminated from the overall equation. I2(s) + Cl2(g) → 2 ICl(s) 2 ICl(s) + 2 Cl2(g) → 2 ICl3(s) I2(s) + Cl2(g) + 2 ICl(s) + 2 Cl2(g) → 2 ICl(s) + 2 ICl3(s) Overall equation: I2(s) + 3 Cl2(g) → 2 ICl3(s)  103 g   1 mol ICl3   1 mol I 2   253.8 g I 2  c) Grams I2 = ( 31.4 kg ICl3 )   1 kg   233.2 g ICl   2 mol ICl   1 mol I  3  3  2    4 = 17086.88 = 1.71 x 10 g I2 3.70 ∆ a) 2 PbS(s) + 3 O2(g) → 2 PbO(s) + 2 SO2(g) ∆ 2 PbO(s) + PbS(s) → 3 Pb(l) + SO2(g) ∆ b) PbS(s) + O2(g) → Pb(l) + SO2(g) 3-16 c) 1 metric ton = tonne = 1000 kg  103 kg   103 g   1 mol Pb   1 mol SO 2   64.07 g SO 2 Tonne SO2 = ( tonne Pb )   1 tonne   1 kg   207.2 g Pb   1 mol Pb   1 mol SO   2    = 0.309218 = 0.3092 tonne SO2 3.71   1 kg  1 tonne     3   3   10 g  10 kg  Plan: Convert the given masses to moles and use the mole ratio from the balanced chemical equation to find the moles of CaO that will form. The reactant that produces the least moles of CaO is the limiting reactant. Convert the moles of CaO from the limiting reactant to grams using the molar mass. Solution:  1 mol Ca   2 mol CaO  a) Moles CaO from Ca = ( 4.20 g Ca )    = 0.104790 = 0.105 mol CaO  40.08 g Ca   2 mol Ca   1 mol O 2   2 mol CaO  b) Moles CaO from O2 = ( 2.80 g O 2 )    = 0.17500 = 0.175 mol CaO  32.00 g O 2   1 mol O 2  c) Calcium is the limiting reactant since it will form less calcium oxide.  1 mol Ca   2 mol CaO  56.08 g CaO  d) Grams CaO = ( 4.20 g Ca )     = 5.8766 = 5.88 g CaO  40.08 g Ca   2 mol Ca  1 mol CaO  3.72 SrH2(s) + 2 H2O(l) → Sr(OH)2(s) + 2 H2(g)  1 mol SrH 2   2 mol H 2  a) Moles H2 from SrH2 = ( 5.63 g SrH 2 )    = 0.125614 = 0.126 mol H2  89.64 g SrH 2   1 mol SrH 2   1 mol H 2 O   2 mol H 2  b) Moles H2 from H2O = ( 4.80 g H 2 O )    = 0.26637 = 0.266 mol H2  18.02 g H 2 O   2 mol H 2 O  c) SrH2 is the limiting reagent since it will yield fewer moles of hydrogen gas.  1 mol SrH 2   2 mol H 2   2.016 g H 2  d) Grams H2 = ( 5.63 g SrH 2 )     = 0.2532369 = 0.253 g H2  89.64 g SrH 2   1 mol SrH 2   1 mol H 2  3.73 Plan: First, balance the chemical equation. Determine which of the reactants is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of material formed and the amount of the other reactant used. The difference between the amount of reactant used and the initial reactant supplied gives the amount of excess reactant remaining. Solution: The balanced chemical equation for this reaction is: 2 ICl3 + 3 H2O → ICl + HIO3 + 5 HCl Hint: Balance the equation by starting with oxygen. The other elements are in multiple reactants and/or products and are harder to balance initially. Next, find the limiting reactant by using the molar ratio to find the smaller number of moles of HIO3 that can be produced from each reactant given and excess of the other:  1 mol ICl3   1 mol HIO3  Moles HIO3 from ICl3 = ( 685 g ICl3 )    = 1.468696 = 1.47 mol HIO3  233.2 g ICl3   2 mol ICl3   1 mol H 2 O  1 mol HIO3  Moles HIO3 from H2O = (117.4 g H 2 O )    = 2.17166 = 2.17 mol HIO3  18.02 g H 2 O  3 mol H 2 O  ICl3 is the limiting reagent and will produce 1.47 mol HIO3. Use the limiting reagent to find the grams of HIO3 formed.  1 mol ICl3  1 mol HIO3   175.9 g HIO3  Grams HIO3 = ( 685 g ICl3 )     = 258.355 = 258 g HIO3  233.2 g ICl3  2 mol ICl3   1 mol HIO3  3-17 The remaining mass of the excess reagent can be calculated from the amount of H2O combining with the limiting reagent.  1 mol ICl3   3 mol H 2 O   18.02 g H 2 O  Remaining H2O = 117.4 g H2O - ( 685 g ICl3 )      233.2 g ICl3   2 mol ICl3   1 mol H 2 O  = 38.002 = 38.0 g H2O 3.74 First balance the equation: Al2S3 + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g) Determine the limiting reactant:  1 mol Al2S3   3 mol H 2S  Moles H2S from Al2S3 = (158 g Al2S3 )    = 3.15642 = 3.16 mol H2S  150.17 g Al 2S3   1 mol Al2S3   1 mol H 2 O   3 mol H 2S  Moles H2S from H2O = (131 g H 2 O )    = 3.63485 = 3.63 mol H2S  18.02 g H 2 O   6 mol H 2 O  Al2S3 is the limiting reagent and 3.16 mol of H2S will form. Using the limiting reagent, we can calculate the mass of H2S that will form.  1 mol Al2S3   3 mol H 2S   34.09 g H 2S  Grams H2S = (158 g Al2S3 )     = 107.602 = 108 g H2S  150.17 g Al2S3   1 mol Al2S3   1 mol H 2S  The remaining mass of the excess reagent can be calculated from the amount of H2O combining with the limiting reagent.  1 mol Al2S3   6 mol H 2 O   18.02 g H 2 O  Remaining H2O = 131 g H2O - (158 g Al2S3 )      150.17 g Al 2S3   1 mol Al2S3   1 mol H 2 O  = 17.24 = 17 g H2O 3.75 Plan: Write the balanced equation: formula for carbon is C, formula for oxygen is O2 and formula for carbon dioxide is CO2. Determine the limiting reagent by seeing which reactant will yield the smaller amount of product. The limiting reactant is used for all subsequent calculations. Solution: C(s) + O2(g) → CO2(g)  1 mol CO 2  Moles CO2 from C = ( 0.100 mol C )   = 0.100 mol CO2  1 mol C   1 mol O 2   1 mol CO 2  Moles CO2 from O2 = ( 8.00 g O 2 )    = 2.5000 = 2.50 mol CO2  32.00 g O 2   1 mol O 2  The C is the limiting reactant and will be used to determine the amount of CO2 that will form.  1 mol CO 2   44.01 g CO 2  Grams CO2 = ( 0.100 mol C )   = 4.401 = 4.40 g CO2   1 mol C  1 mol CO 2  Since the C is limiting, the O2 is in excess. The amount remaining depends on how much combines with the limiting reagent.  1 mol O 2   32.00 g O 2  Remaining O2 = 8.00 g O2 - ( 0.100 mol C )   = 4.8000 = 4.80 g O2   1 mol C   1 mol O 2  3-18 3.76 First, balance the equation: 2 H2(g) + O2(g) → 2 H2O(l) Determine the limiting reagent:  1 mol H 2   2 mol H 2 O  Mole H2O from H2 = ( 0.0359 g H 2 )    = 0.0178075 = 0.0178 mol H2O  2.016 g H 2   2 mol H 2   2 mol H 2 O  Mole H2O from O2 = ( 0.0175 mol O 2 )   = 0.0350 mol H2O  1 mol O 2  The hydrogen is the limiting reactant, and will be used to determine the amount of water that will form.  1 mol H 2   2 mol H 2 O   18.02 g H 2 O  Grams H2O = ( 0.0359 g H 2 )     = 0.32089 = 0.321 g H2O  2.016 g H 2   2 mol H 2   1 mol H 2 O  Since the hydrogen is limiting; the oxygen must be excess reactant. The amount of excess reactant is determined from the limiting reactant.  32.00 mol O 2   1 mol H 2   1 mol O 2  32.00 g O 2  Remaining O2 = ( 0.0175 mol O 2 )   - ( 0.0359 g H 2 )      1 mol O 2   2.016 g H 2   2 mol H 2  1 mol O 2  = 0.275079 = 0.28 g O2 3.77 Plan: The question asks for the mass of each substance present at the end of the reaction. “Substance” refers to both reactants and products. Solve this problem using multiple steps. Recognizing that this is a limiting reactant problem, first write a balanced chemical equation. Using the molar relationships from the balanced equation, determine which reactant is limiting. Any product can be used to predict the limiting reactant; in this case, AlCl3 is used. Additional significant figures are retained until the last step. Solution: The balanced chemical equation is: Al(NO2)3(aq) + 3 NH4Cl(aq) → AlCl3(aq) + 3 N2(g) + 6 H2O(l) Now determine the limiting reagent. We will use the moles of AlCl3 produced to determine which is limiting.  1 mol Al(NO 2 )3   1 mol AlCl3  Mole AlCl3 from Al(NO2)3 = ( 62.5 g Al(NO 2 )3 )     165.01 g Al(NO 2 )3   1 mol Al(NO 2 )3  = 0.37876 = 0.379 mol AlCl3  1 mol NH 4 Cl   1 mol AlCl3  Mole AlCl3 from NH4Cl = ( 54.6 g NH 4 Cl )    = 0.34025 = 0.340 mol AlCl3  53.49 g NH 4 Cl   3 mol NH 4 Cl  Ammonium chloride is the limiting reactant, and it is important for all subsequent calculations. Mass of substances after the reaction: Al(NO2)3:  1 mol NH 4 Cl   1 mol Al(NO 2 )3   165.01 g Al(NO 2 )3  62.5 g Al(NO2)3 - ( 54.6 g NH 4 Cl )      53.49 g NH 4 Cl   3 mol NH 4 Cl   1 mol Al(NO 2 )3  = 6.35526 = 6.4 g Al(NO2)3 (It is the limiting reactant.) AlCl3:  1 mol NH 4 Cl   1 mol AlCl3   133.33 g AlCl3  ( 54.6 g NH 4 Cl )   = 45.3656 = 45.4 g AlCl3    53.49 g NH 4 Cl   3 mol NH 4 Cl   1 mol AlCl3  N2:  1 mol NH 4 Cl   3 mol N 2   28.02 g N 2  ( 54.6 g NH 4 Cl )     = 28.601 = 28.6 g N2  53.49 g NH 4 Cl   3 mol NH 4 Cl   1 mol N 2  H2O:  1 mol NH 4 Cl   6 mol H 2 O   18.02 g H 2 O     = 36.7879 = 36.8 g H2O  53.49 g NH 4 Cl   3 mol NH 4 Cl   1 mol H 2 O  ( 54.6 g NH 4 Cl )  3-19 3.78 The balanced chemical equation is: Ca(NO3)2(s) + 2 NH4F(s) → CaF2(s) + 2 N2O(g) + 4 H2O(g) Now determine the limiting reagent. We will use the moles of CaF2 produced to determine which is limiting.  1 mol Ca(NO3 ) 2  1 mol CaF2  Mole CaF2 from Ca(NO3)2 = (16.8 g Ca(NO3 ) 2 )     164.10 g Ca(NO3 ) 2  1 mol Ca(NO3 ) 2  = 0.1023766 = 0.102 mol CaF2  1 mol NH 4 F   1 mol CaF2  Mole CaF2 from NH4F = (17.50 g NH 4 F )    = 0.236231 = 0.236 mol CaF2  37.04 g NH 4 F   2 mol NH 4 F  Calcium nitrate is the limiting reactant, and it is important for all subsequent calculations. Mass of substances after the reaction: Ca(NO3)2: None (It is the limiting reactant.) NH4F:  1 mol Ca(NO3 ) 2  2 mol NH 4 F   37.04 g NH 4 F  17.50 g NH4F - (16.8 g Ca(NO3 ) 2 )      164.10 g Ca(NO3 ) 2  1 mol Ca(NO3 ) 2   1 mol NH 4 F  CaF2: = 9.9159 = 9.92 g NH4F  1 mol Ca(NO3 ) 2   1 mol CaF2   78.08 g CaF2    164.10 g Ca(NO3 ) 2   1 mol Ca(NO3 ) 2   1 mol CaF2 (16.8 g Ca(NO3 )2 )    = 7.99356 = 7.99 g CaF2  N2O:  1 mol Ca(NO3 ) 2  2 mol N 2 O   44.02 g N 2 O     = 9.0132 = 9.01 g N2O  164.10 g Ca(NO3 ) 2  1 mol Ca(NO3 )2   1 mol N 2 O  (16.8 g Ca(NO3 )2 )  H2O:  1 mol Ca(NO3 ) 2  4 mol H 2 O   18.02 g H 2 O     = 7.3793 = 7.38 g H2O  164.10 g Ca(NO3 ) 2  1 mol Ca(NO3 ) 2   1 mol H 2 O  (16.8 g Ca(NO3 )2 )  3.79 Plan: Multiply the yield of the first step by that of the second step to get the overall yield. Solution: It is simpler to use the decimal equivalents of the percent yields, and then convert to percent using 100%. (0.82) (0.65) (100%) = 53.3 = 53% 3.80 It is simpler to use the decimal equivalents of the percent yields, and then convert to percent using 100%. (0.48) (0.73) (100%) = 35.04 = 35% 3.81 Plan: Balance the chemical equation using the formulas of the substances. Determine the yield (theoretical yield) for the reaction from the mass of tungsten(VI) oxide. The actual yield divided by the yield just calculated (with the result multiplied by 100%) gives the percent yield. Solution: (Rounding to the correct number of significant figures will be postponed until the final result.) The balanced chemical equation is: WO3(s) + 3 H2(g) → W(s) + 3 H2O(l) Theoretical yield of H2O:  1 mol WO3   3 mol H 2 O   18.02 g H 2 O  ( 41.5 g WO3 )     = 9.67439 g H2O  231.9 g WO3   1 mol WO3   1 mol H 2 O  Actual yield, in grams, of H2O:  1.00 g H 2 O  ( 9.50 mL H 2 O )   = 9.50 g H2O  1 mL H 2 O  3-20 Calculate the percent yield:  9.50 g H 2 O   Actual Yield   x 100% = 98.1974 = 98.2%   x 100% =   Theoretical Yield   9.67439 g H 2 O  3.82 Balance the chemical equation using the formulas of the substances. Determine the yield (theoretical yield) for the reaction from the mass of tungsten(VI) oxide. The actual yield divided by the yield just calculated (with the result multiplied by 100%) gives the percent yield. (Rounding to the correct number of significant figures will be postponed until the final result.) The balanced chemical equation is: PCl3(l) + 3 H2O(l) → H3PO3(aq) + 3 HCl(g) Theoretical yield of H2O:  1 mol PCl3  3 mol HCl   36.46 g HCl  ( 200. g PCl3 )     = 159.3067 g HCl (Theoretical yield)  137.32 g PCl3  1 mol PCl3   1 mol HCl  The actual yield is given (128 g HCl). Calculate the percent yield:  Actual Yield   128 g HCl    x 100% =   x 100% = 83.3481586 = 83.3%  Theoretical Yield   159.3067 g HCl  3.83 Plan: Write the balanced chemical equation. Since quantities of reactants are present, we must determine which is limiting. Only 80.0% of the calculated amounts of products will form. (Rounding to the correct number of significant figures will be postponed until the final result.) Solution: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)  1 mol CH 4   1 mol CH3Cl  Mole HCl from CH4 = (18.5 g CH 4 )    = 1.153367 mol CH3Cl  16.04 g CH 4   1 mol CH 4   1 mol Cl2   1 mol CH3Cl  Mole HCl from Cl2 = ( 43.0 g Cl2 )    = 0.606488 mol CH3Cl  70.90 g Cl2   1 mol Cl2  Chlorine is the limiting reactant.  1 mol Cl2   1 mol CH3Cl   50.48 g CH 3Cl   80.0%  Grams CH3Cl = ( 43.0 g Cl2 )       70.90 g Cl2   1 mol Cl2   1 mol CH3Cl   100%  = 24.4924 = 24.5 g CH3Cl The beginning of the calculation is repeated to emphasize that the second part of the problem is simply an extension of the first part. There is no need to repeat the entire calculation as only the final step(s) times the answer of the first part will give the final answer to this part. 3.84 First, balance the chemical equation: 3 Ca(s) + N2(g) → Ca3N2(s)  1 mol Ca   1 mol Ca 3 N 2  Mole Ca3N2 from Ca = ( 56.6 g Ca )    = 0.470725 mol Ca3N2  40.08 g Ca   3 mol Ca   1 mol N 2   1 mol Ca 3 N 2  Mole Ca3N2 from N2 = ( 30.5 g N 2 )    = 1.0885 mol CH3Cl  28.02 g N 2   1 mol N 2  Ca is the limiting reactant.  1 mol Ca   1 mol Ca 3 N 2   148.26 g Ca 3 N 2   93.0%  Grams Ca3N2 = ( 56.6 g Ca )       40.08 g Ca   3 mol Ca   1 mol Ca 3 N 2   100%  = 64.90444 = 64.9 g Ca3N2 3-21 3.85 Plan: The first step is to determine the chemical formulas so a balanced chemical equation can be written. The limiting reactant must be determined. Finally, the mass of CF4 is determined from the limiting reactant. Solution: (Rounding to the correct number of significant figures will be postponed until the final result.) The balanced chemical equation is: (CN)2(g) + 7 F2(g) → 2 CF4(g) + 2 NF3(g)  1 mol (CN) 2   2 mol CF4  Mole CF4 from (CN)2 = ( 80.0 g (CN) 2 )    = 3.074558 mol CF4  52.04 g (CN) 2   1 mol (CN) 2   1 mol F2   2 mol CF4  Mole CF4 from F2 = ( 80.0 g F2 )    = 0.60150 mol CF4  38.00 g F2   7 mol F2  F2 is the limiting reactant, and will be used to calculate the yield.  1 mol F2  2 mol CF4   88.01 g CF4  Grams CF4 = ( 80.0 g F2 )     = 52.9383 = 52.9 g CF4  38.00 g F2  7 mol F2   1 mol CF4  3.86 Start by deriving the balanced chemical equation: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Determine the limiting reactant. (Rounding to the correct number of significant figures will be postponed until the final result.)  1 mol NH3   4 mol NO  Mole NO from NH3 = ( 466 g NH 3 )    = 27.36348 mol NO  17.03 g NH3   4 mol NH 3   1 mol O 2   4 mol NO  Mole NO from O2 = ( 812 g O 2 )    = 20.3 mol NO  32.00 g O 2   5 mol O 2  O2 is the limiting reactant, and will be used to calculate the yield.  1 mol O 2  4 mol NO   30.01 g NO  Grams NO = ( 812 g O 2 )     = 609.203 = 609 g NO  32.00 g O 2  5 mol O 2   1 mol NO  3.87 Determine the balanced chemical equation: 2 C4H10(g) + 13 O2(g) → 8 CO2 (g) + 10 H2O(g)  0.579 g   1 mol C 4 H10   13 mol O 2   32.00 g O 2  a) Grams O2 = ( 6.50 mL C 4 H10 )       mL   58.12 g C 4 H10   2 mol C 4 H10   1 mol O 2  = 13.4688 = 13.5 g O2  0.579 g   1 mol C 4 H10   8 mol CO 2  b) Moles CO2 = ( 6.50 mL C4 H10 )      mL   58.12 g C 4 H10   2 mol C 4 H10  = 0.259016 = 0.259 mol CO2  0.579 g   1 mol C 4 H10   8 mol CO 2  c) Total moles = ( 6.50 mL C4 H10 )   +   mL   58.12 g C 4 H10   2 mol C 4 H10  0.579 g   1 mo l C4 H10   10 mol H 2 O  ( 6.50 mL C4 H10 )    = 0.5827858 mol   mL   58.12 g C 4 H10   2 mol C 4 H10   6.002 x 1023 molecules  Total molecules = ( 0.5827858 mol )    1 mol   = 3.5095 x 1023 = 3.51 x 1023 molecules 3-22 3.88 Balancing the chemical equation gives: 2 NaH(s) + B2H6(g) → 2 NaBH4(s) Determine the limiting reactant. (Rounding to the correct number of significant figures will be postponed until the final result.)  1 mol NaH   2 mol NaBH 4  Mole NaBH4 from NaH = ( 7.88 g NaH )    = 0.32833 mol NaBH4  24.00 g NaH   2 mol NaH   1 mol B2 H 6   2 mol NaBH 4  Mole NaBH4 from B2H6 = ( 8.12 g B2 H 6 )  = 0.585647 mol NaBH4  27.73 g B2 H 6   1 mol B2 H 6     NaH is the limiting reactant, and will be used to calculate the yield.  1 mol NaH   2 mol NaBH 4   37.83 g NaBH 4   95.5%  ( 7.88 g NaH )      = 11.8619 = 11.9 g NaBH4  24.00 g NaH   2 mol NaH   1 mol NaBH 4   100%  3.89 Plan: The spheres represent particles of solute and the amount of solute per given volume of solution determines its concentration. Solution: a) Box C has more solute added because it contains 2 more spheres than Box A contains. b) Box B has more solvent because solvent molecules have displaced two solute molecules. c) Box C has a higher molarity, because it has more moles of solute per volume of solution. d) Box B has a lower concentration (and molarity), because it has fewer moles of solute per volume of solution. 3.90 Molarity = 3.91 Volumes may not be additive when two different solutions are mixed, so the final volume may be slightly different from 1000.0 mL. The correct method would state, “Take 100.0 mL of the 10.0 M solution and add water until the total volume is 1000 mL.” 3.92 Plan: In all cases, the definition of molarity (moles/liters) will be important. The molar mass is important in some cases. The chemical formulas are necessary. Solution:  10−3 L   0.207 mol Ca(C 2 H3O 2 ) 2   158.17 g Ca(C2 H3O 2 ) 2  a) Grams Ca(C2H3O2)2 = (175.8 mL )     1 mL   1L   1 mol Ca(C 2 H3O 2 ) 2    = 5.7559 = 5.76 g Ca(C2H3O2)2  21.1 g KI   1 mL   1 mol KI  b) Molarity KI =    −3    = 0.254217 = 0.254 M KI  500. mL   10 L   166.0 g KI  moles Solute Liters Solution The amount of solute = concentration x volume of solution Moles solute = molarity x liters of solution Mass solute = molarity x liters of solution x molar mass  0.850 mol NaCN  c) Moles NaCN =   (145.6 L ) = 123.76 = 124 mol NaCN 1L   3.93 Molarity is a convenient way of expressing concentration; and, it is useful to use the definition (mole/L) in place of M.    1 mol KOH  1L a) Liters KOH solution =    8.42 g KOH  2.26 mol KOH   56.11 g KOH  = 0.066399 = 0.0664 L KOH solution 3-23  6.022 x 1023 Cu 2 + ions    1 mol Cu 2 +     1 mol CuCl  ( 52 L )  1 mol Cu 2 +  2    25 25 2+ = 7.2023 x 10 = 7.2 x 10 Cu ions  135 mmol glucose  c) M glucose =   = 0.490909 = 0.491 M glucose 275 mL    2.3 mol CuCl2 b) Number Cu2+ ions =  L  3.94 Plan: It will help to rewrite M as its definition (mole/L). (a) You will need to convert milliliters to liters, and determine the molar mass of potassium sulfate. (b) The simplest way will be to convert the milligrams to millimoles. Molarity may not only be expressed as moles/L, but also as mmoles/mL. (c) Convert the milliliters to liters, and find the moles of solute. It will be necessary to use Avogadro’s number to determine the number of ions present. Solution:  5.62 x 10−2 mol K 2SO 4  10−3 L   174.27 g K 2SO 4  a) Grams potassium sulfate =    ( 475 mL )      L  1 mol K 2SO 4     1 mL  = 4.6521 = 4.65 g K2SO4  6.55 mg CaCl2   1 mmol CaCl2  b) Molarity calcium chloride =   = 0.05902 = 0.0590 M CaCl2  1 mL    110.98 mg CaCl2  If you believe that molarity must be moles/liters then the calculation becomes:  6.55 mg CaCl2   10−3 g   1 mL   1 mol CaCl2    −3    = 0.05902 = 0.0590 M CaCl2    1 mL    1 mg   10 L   110.98 g CaCl 2  Notice that the two central terms cancel each other.  1 mol Mg 2 +  6.022 x 1023 Mg 2 + ions   0.184 mol MgBr2   10−3 L  c) Number of Mg2+ ions =  mL ( )         1L 1 mol Mg 2 +    1 mL   1 mol MgBr2   = 1.1080 x 1020 = 1.11 x 1020 Mg2+ ions 3.95  46.0 g AgNO3 a) Molarity of silver nitrate =  335 mL    1 mL   1 mol AgNO3   = 0.80820 = 0.808 M AgNO3   −3     10 L   169.9 g AgNO3   1 mol MnSO 4   1L b) Liters manganese(II) sulfate solution = ( 57.0 g MnSO 4 )     151.01 g MnSO 4  0.385 mol MnSO 4  = 0.98041 = 0.980 L MnSO4 solution    10−3 mol ATP   1 mL  1L 1.68 mmol ATP )  −3  c) Milliliters ATP solution =   6.44 x 10−2 mol ATP   1 mmol ATP  (  10 L       = 26.0867 = 26.1 mL ATP solution This problem can be simplified by using M = mmol/mL instead of M = mol/L:   1 mL   (1.68 mmol ATP ) = 26.0867 = 26.1 mL ATP solution −2  6.44 x 10 mmol ATP  3.96 Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new volume, however, it is much easier to use M1V1 = M2V2. Part (c) may be done as two dilution problems or as a mole problem. The dilution equation does not require a volume in liters; it only requires that the volume units match. Solution: a) M1 = 0.250 M KCl V1 = 37.00 mL M2 = ? V2 = 150.00 mL M2 = M1V1 / V2 = (0.250 M KCl) (37.00 mL) / (150.00 mL) = 0.061667 = 0.0617 M KCl 3-24 b) M1 = 0.0706 M (NH4)2SO4 V1 = 25.71 mL M2 = ? V2 = 500.00 mL M2 = M1V1 / V2 = (0.0706 M (NH4)2SO4) (25.71 mL) / (500.00 mL) = 0.003630 = 0.00363 M (NH4)2SO4 c) When working this as a mole problem it is necessary to find the individual number of moles of sodium ions in each separate solution. (Rounding to the proper number of significant figures will only be done for the final answer.)  1 mol Na +   0.288 mol NaCl   10−3 L  Moles Na+ from NaCl solution =   ( 3.58 mL )     1L    1 mL   1 mol NaCl  = 0.00103104 mol Na+  6.51 x10−3 mol Na 2SO 4   10−3 L   2 mol Na +  Moles Na+ from Na2SO4 solution =  500. mL )  (     1 mol Na SO     1 mL  1L 2 4      + = 0.006510 mol Na  ( 0.00103104 + 0.006510 ) mol Na +   1 mL    −3  = 0.01497486 = 0.0150 M Na+ ions Molarity of Na+ =     10 L  + 3.58 500. mL ( )   3.97 These are dilution problems using the equation M1V1 = M2V2. V1 = ? M2 = 0.8543 M Cu(NO3)2 V2 = 750.0 mL a) M1 = 2.050 M Cu(NO3)2 V1 = M2V2 / M1 = (0.8543 M) (750.0 mL) / (2.050 M) = 312.5488 = 312.5 mL  1.03 mol CaCl2   2 mol Cl−  – b) 1.03 M CaCl2 gives: M Cl– =   = 2.06 M Cl ions   1 L 1 mol CaCl   2  M1 = 2.06 M Cl– V1 = ? M2 = 2.66 x 10–2 M Cl– ions V2 = 350. mL V1 = M2V2 / M1 = (2.66 x 10–2 M) (350. mL) / (2.06 M) = 4.5194 = 4.52 mL M2 = 0.0700 M Li2CO3 V2 = ? c) M1 = 0.155 M Li2CO3 V1 = 18.0 mL V2 = M1V1 / M2 = (0.155 M) (18.0 mL) / (0.0700 M) = 39.8571 = 39.9 mL 3.98 Plan: You will need to pay particular attention to the units throughout this problem. The 70.0% by mass translates to 70.0 g solute/100 g solution. The molar mass of nitric acid is also needed. Solution:  1.41 g Solution   1 mL   70.0 g HNO3  a) Mass HNO3 per liter =    −3    = 987 g HNO3 / L 1 mL    10 L   100 g Solution   1.41 g Solution   1 mL   70.0 g HNO3   1 mol HNO3  b) Molarity of HNO3 =     −3    1 mL    10 L   100 g Solution   63.02 g HNO3  = 15.6617 = 15.7 M HNO3 3.99 3.100  18.3 mol H 2SO 4   10−3 L  –2 a) moles per milliliter =   = 1.83 x 10 mol H2SO4 / mL   1 L 1 mL    b) To calculate the mass percent, the mass of H2SO4 in each milliliter (1.84 g) of solution is needed.  18.3 mol H 2SO 4   10−3 L   98.09 g H 2SO 4    1mL Mass percent =     (100% )     1L    1 mL   1 mol H 2SO 4   1.84 g Solution  = 97.5569 = 97.6% H2SO4 by mass Plan: Convert the mass of calcium carbonate to moles, and use the balanced chemical equation to find the moles of hydrochloric acid required. The moles of acid along with the molarity of the acid will give the volume required. The molarity of the solution is given in the calculation as mol/L. 3-25 Solution: 2 HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)  1 mol CaCO3   2 mol HCl     1 mL  1L Volume required = (16.2 g CaCO3 )      −3   100.09 g CaCO3   1 mol CaCO3   0.383 mol HCl   10 L  = 845.1923 = 845 mL HCl solution 3.101 Convert the molarity of sodium hydroxide to moles, and use the balanced chemical equation to find the moles of sodium dihydrogen phosphate required. The moles of sodium dihydrogen phosphate, along with its molar mass, will give the grams required. The molarity of the solution is given in the calculation as mol/L. NaH2PO4(s) + 2 NaOH(aq) → Na3PO4(aq) + 2 H2O(l) Mass of sodium dihydrogen phosphate =  10−3 L   0.275 mol NaOH   1 mol NaH 2 PO 4   119.98 g NaH 2 PO 4  ( 38.74 mL )       1L   2 mol NaOH  1 mol NaH 2 PO 4   1 mL   = 0.639103 = 0.639 g NaH2PO4 3.102 Plan: The first step is to write and balance the chemical equation for the reaction. Use the molarity and volume of each of the reactants to determine the moles of each as a prelude to determining which is the limiting reactant. Use the limiting reactant to determine the mass of barium sulfate that will form. Solution: The balanced chemical equation is: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) The mole and limiting reactant calculations are:  10−3 L   0.160 mol BaCl2   1 mol BaSO 4  Moles BaSO4 from BaCl2 = ( 25.0 mL )   = 0.00400 mol BaSO4   1 mL   1L  1 mol BaCl2     10−3 L   0.055 mol Na 2SO4   1 mol BaSO4  Moles BaSO4 from Na2SO4 = ( 68.0 mL )   = 0.00374 mol BaSO4   1 mL   1L  1 mol Na 2SO4    Sodium sulfate is the limiting reactant.  10−3 L   0.055 mol Na 2SO4   1 mol BaSO4  233.4 g BaSO4  Grams BaSO4 = ( 68.0 mL )      1 mL   1L  1 mol Na 2SO4  1 mol BaSO4    = 0.872916 = 0.87 g BaSO4 3.103 The first step is to write and balance the chemical equation for the reaction. Use the molarity and volume of each of the reactants to determine the moles of each as a prelude to determining which is the limiting reactant. Use the limiting reactant to determine the mass of the other substance consumed. H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) We can use either product to determine the limiting reactant. We will use sodium sulfate.  10−3 L   0.210 mol H 2SO 4   1 mol Na 2SO 4  Moles Na2SO4 from H2SO4 = ( 350.0 mL )     1 mL   1L  1 mol H 2SO 4    = 0.0735 mol Na2SO4  0.196 mol NaOH   1 mol Na 2SO 4  Moles Na2SO4 from NaOH = ( 0.500 L )    = 0.0490 mol Na2SO4 1L    2 mol NaOH  NaOH is the limiting reactant. Finish the problem using the limiting reactant. Moles H2SO4 remaining = Initial moles - Moles reacting with NaOH  10−3 L   0.210 mol H 2SO 4   0.196 mol NaOH   1 mol H 2SO 4  - ( 0.500 L )  = ( 350.0 mL )        1 mL  1L 1L     2 mol NaOH    = 0.0245 mol H2SO4 3-26 3.104 3.105  20.5 g NaClO   1 mL   1 mol NaClO  Molarity of sodium hypochlorite =    −3     375 mL   10 L   74.44 g NaClO  = 0.73437 = 0.734 M NaClO Plan: The first part of the problem is a simple dilution problem (M1V1 = M2V2). The second part requires the molar mass of the HCl along with the molarity. Solution: M2 = 3.5 M V2 = 5.0 gal a) M1 = 11.7 M V1 = ? V1 = M2V2 / M1 = (3.5 M) (5.0 gal) / (11.7 M) = 1.4957 gallons (unrounded) Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 3.0 gallons of water into the container. Add slowly and with mixing 1.5 gallon of concentrated HCl into the water. Dilute to 5.0 gallons with water.  1 mol HCl    1 mL  1L b) Volume needed = ( 9.55 g HCl )     −3   36.46 g HCl  11.7 mol HCl   10 L  = 22.38725 = 22.4 mL muriatic acid solution 3.106 Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)  0.750 mol HCl  Moles of HCl reacting with Mg =   ( 0.100 L ) - 0.0125 mol HCl = 0.0625 mol HCl 1L    1 mol Mg   24.31 g Mg  Grams of Mg reacting = ( 0.0625 mol HCl )    = 0.7596875 g Mg  2 mol HCl   1 mol Mg   0.7596875 g Mg  Mass percent Mg =   (100% ) = 57.552 = 57.6% Mg  1.32 g Sample  3.107 . 3.108 a) An atomic mass unit (amu) is exactly 1/12 the mass of a 12C atom.  1 atom 12 C   12 g 12 C  1 mol 12 C Mass in grams = (1 amu )    12 amu  6.022 x 1023 atoms 12 C   1mol 12 C      = 1.6605778 x 10–24 = 1.661 x 10–24 g Remember that all the values in this calculation except Avogadro’s number are exact numbers b) Either use 1.661 x 10–24 g / amu or 1g / 6.022 x 1023 amu. First, determine the empirical formula:  1 mol S  ( 2.288 g S)   = 0.0713439 mol S  32.07 g S   1 mol N   = 0.0713776 mol N  14.01 g N  (1.000 g N )  Dividing both moles by the smaller value shows the empirical formula to be SN (empirical formula mass = 46.08 g/mol). The ratio of the given molar mass to the empirical formula mass is (184.27 g/mol) / (46.08 g/mol) = 4 The molecular formula is four times the empirical formula, giving (SN)4 or S4N4. 3.109 Plan: The moles of narceine and the moles of water are required. We can assume any mass of narceine hydrate (we will use 100 g), and use this mass to determine the mass of water present and convert the mass to moles of the hydrate. The mass of water will be converted to moles. Finally, the ratio of the moles of hydrate to moles of water will give the amount of water present. 3-27 Solution:  1 mol narceine hydrate  Moles narceine hydrate = (100 g narceine hydrate )    499.52 g narceine hydrate  = 0.20019 mol narceine hydrate    1 mol H 2 O  10.8% H 2 O Moles H2O = (100 g narceine hydrate )     100% narceine hydrate  18.02 g H 2 O  = 0.59933 mol H2O The ratio of water to hydrate is: (0.59933 mol) / (0.20019 mol) = 3 Thus, there are three water molecules per mole of hydrate. The formula for narceine hydrate is narceine•3H2O 3.110 Plan: Determine the formula, then the molar mass of each compound. Determine the mass of hydrogen in each formula. The mass of hydrogen divided by the molar mass of the compound (with the result multiplied by 100%) will give the mass percent hydrogen. Ranking, based on the percents, is easy. Solution: Name Chemical Molar mass Mass percent H formula (g/mol) [(mass H) / (molar mass)] x 100% 30.07 [(6 x 1.008) / (30.07)] x 100% = 20.11% H Ethane C2H6 44.09 [(8 x 1.008) / (44.09)] x 100% = 18.29% H Propane C3H8 480.83 [(64 x 1.008) / (480.83)] x 100% = 13.42% H Cetyl palmitate C32H64O2 46.07 [(6 x 1.008) / (46.07)] x 100% = 13.13% H Ethanol C2H5OH 78.11 [(6 x 1.008) / (78.11)] x 100% = 7.743% H Benzene C6H6 The hydrogen percentage decreases in the following order: Ethane > Propane > Cetyl palmitate > Ethanol > Benzene 3.111 First, determine the empirical formula and empirical formula mass. The empirical formula mass and the molar mass give the molecular formula for the compound. Assume 100 grams of sample, thus the percentages are the mass, in grams, of each element.  1 mol C  Mole C = ( 68.2 g C )   = 5.6786 mol C  12.01 g C   1 mol H  Mole H = ( 6.86 g H )   = 6.8056 mol H  1.008 g H   1 mol N  Mole N = (15.9 g N )   = 1.1349 mol N  14.01 g N   1 mol O  Mole O = ( 9.08 g O )   = 0.5675 mol O  16.00 g O  Divide each mole value by the smallest value (mole O). C: 5.6786 mol / 0.5675 mol = 10.0 H: 6.8056 mol / 0.5675 mol = 12.0 N: 1.1349 mol / 0.5675 mol = 2.00 O: 0.5675 mol / 0.5675 mol = 1.00 This gives the empirical formula: C10H12N2O with a formula mass = 10(12.01g C/mol) + 12 (1.008 g H/mol) + 2(14.01 g N/mol) + 1 (16.00 g O/mol) = 176.22 g/mol The empirical formula mass and the molar mass are the same, thus, the molecular and empirical formulas are the same. Molecular formula: C10H12N2O 3-28 3.112 3.113 ∆ a) 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ∆ b) 4 KClO3(s) → KCl(s) + 3 KClO4(s) c) 3 H2(g) + Fe2O3(s) → 2 Fe(s) + 3 H2O(g) ∆ d) 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) e) 2 FeCl2(s) + 2 ClF3(g) → 2 FeF3(s) + 3 Cl2(g) Isobutylene + O2 → CO2 + H2O  1 mol CO 2   1 mol C  Mole C = ( 2.657 g CO 2 )    = 0.06037 mol C  44.01 g CO 2   1 mol CO 2   1 mol H 2 O   2 mol H  Mole H = (1.089 g H 2 O )    = 0.1209 mol H  18.02 g H 2 O   1 mol H 2 O  Divide each mole value by the smaller value (mole C). C: 0.06037 mol / 0.06037 mol = 1.00 H: 0.1209 mol / 0.06037 mol = 2.00 This gives an empirical formula of CH2. 3.114 C7H8 + 9 O2(g) → 7 CO2(g) + 4 H2O(g)  0.867 g C7 H8   1 mol C7 H8  Moles C7H8 = (15.0 mL )   = 0.14116 mol C7H8 (unrounded)  1 mL    92.13 g C7 H8   9 mol O 2   32.00 g O 2  a) Grams oxygen = ( 0.14116 mol C7 H8 )    = 40.654 = 40.7 g O2  1 mol C7 H8   1 mol O 2   11 mol gas  b) Total moles of gas = ( 0.14116 mol C7 H8 )   = 1.55276 = 1.55 mol of gas  1 mol C7 H8  The 11 mol of gas is an exact, not measured, number, so it does not affect the significant figures.  4 mol H 2 O   6.022 x 1023 H 2 O molecules  c) Molecules of water = ( 0.14116 mol C7 H8 )     1 mol H 2 O  1 mol C7 H8    = 3.40026 x 1023 = 3.40 x 1023 molecules H2O 3.115 Plan: The key to solving this problem is determining the overall balanced equation. Each individual step must be set up and balanced first. The separate equations can then be combined to get the overall equation. The mass of iron may be converted to moles of iron, and the mole ratio and molar mass of carbon monoxide used to determine the mass of carbon monoxide. Solution: a) In the first step, ferric oxide (ferric denotes Fe3+) reacts with carbon monoxide to form Fe3O4 and carbon dioxide: 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) (1) In the second step, Fe3O4 reacts with more carbon monoxide to form ferrous oxide: Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) (2) In the third step, ferrous oxide reacts with more carbon monoxide to form molten iron: FeO(s) + CO(g) → Fe(l) + CO2(g) (3) Common factors are needed to allow these equations to be combined. The intermediate products are Fe3O4 and FeO, so multiply equation (2) by 2 to cancel Fe3O4 and equation (3) by 6 to cancel FeO: 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) 2 Fe3O4(s) + 2 CO(g) → 6 FeO(s) + 2 CO2(g) 6 FeO(s) + 6 CO(g) → 6 Fe(l) + 6 CO2(g) 3 Fe2O3(s) + 9 CO(g) → 6 Fe(s) + 9 CO2(g) 3-29 Then divide by 3 to obtain the smallest integer coefficients: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) b) A metric ton is a tonne, and is equal to 1000 kg.  103 kg   103 g   1 mol Fe   3 mol CO   28.01 g CO  Grams carbon monoxide = ( 40.0 tonnes Fe )   1 tonne   1 kg   55.85 g Fe   2 mol Fe   1 mol CO        = 3.00913 x 107 = 3.01 x 107 g CO 3.116 Plan: If 100.0 g of dinitrogen tetroxide reacts with 100.0 g of hydrazine (N2H4), what is the theoretical yield of nitrogen if no side reaction takes place? First, we need to identify the limiting reactant. The limiting reactant can be used to calculate the theoretical yield. Determine the amount of limiting reactant required to produce 10.0 grams of NO. Reduce the amount of limiting reactant by the amount used to produce NO. The reduced amount of limiting reactant is then used to calculate an “actual yield.” The “actual” and theoretical yields will give the maximum percent yield. Solution: Determining the limiting reactant:  1 mol N 2 O 4   3 mol N 2  N2 from N2O4 = (100.0 g N 2 O 4 )    = 3.26016 mol N2  92.02 g N 2 O 4   1 mol N 2 O 4  N2O4 is the limiting reactant.  1 mol N 2 O 4   3 mol N 2   28.02 g N 2  Theoretical yield of N2 = (100.0 g N 2 O 4 )      92.02 g N 2 O 4   1 mol N 2 O 4   1 mol N 2  = 91.3497 g N2 (unrounded) How much limiting reactant used to produce 100.0 g NO?  1 mol NO   2 mol N 2 O 4   92.02 g N 2 O 4  Grams N2O4 used = (10.0 g NO )   = 10.221 g N2O4 (unrounded)    30.01 g NO   6 mol NO   1 mol N 2 O 4  Determine the “actual yield.”  1 mol N 2 O 4   3 mol N 2   28.02 g N 2  “Actual yield” of N2 = ( (100.0 − 10.221) g N 2 O 4 )      92.02 g N 2 O 4   1 mol N 2 O 4   1 mol N 2  = 82.01285 g N2 (unrounded) Theoretical yield = [“Actual yield” / theoretical yield] x 100% [(82.01285 g N2) / (91.3497 g N2)] x 100% = 89.7790 = 89.8% 3.117 M = molarity V = volume usually in liters dil = diluted solution conc = concentrated solution The equation works because the quantity (moles) of solute remains the same. 3.118 Plan: Count the number of each type of molecule in the reactant box and in the product box. Subtract any molecules of excess reagent (molecules appearing in both boxes). The remaining material is the overall equation. This will need to be simplified if there is a common factor among the substances in the equation. The balanced chemical equation is necessary for the remainder of the problem. Solution: a) The contents of the boxes give: 6 AB2 + 5 B2 → 6 AB3 + 2 B2 B2 is in excess, so two molecules need to be removed from each side. This gives: 6 AB2 + 3 B2 → 6 AB3 Three is a common factor among the coefficients, and all coefficients need to be divided by this value to give the final balanced equation: 2 AB2 + B2 → 2 AB3 b) B2 was in excess, thus AB2 is the limiting reactant. c) From 3.0 mol of B2, it is possible to make 6.0 mol of AB3 (3.0 x 2/1). From 5.0 mol of AB2 it is possible to make 5.0 mol of AB3 (5.0 x 2/2). Thus, AB2 is limiting and the maximum is 5.0 mol of AB3. 3-30 d) A total of 5.0 mol AB3 would react with 5.0 x 1/2 = 2.5 mol of B2. The un-reacted B2 is 3.0 mol - 2.5 mol = 0.5 mol B2. 3.119 These are all dilution problems and the equation MconcVconc = MdilVdil. In parts c and d further calculations are necessary. Mdil = 0.309 M Vdil = 2.00L a) Mconc = 18.0 M Vconc = ? (18.0 M) (Vconc) = (0.309 M) (2.00L) ( 0.309 M )( 2.00 L ) = 0.034333 = 0.0343 L Vconc = (18.0 M ) b) Mconc = 0.225 M Vconc = 80.6 mL Mdil = ? Vdil = 0.250 L (0.225 M) (80.6 mL) = (Mdil)0.250 L) ( 0.225 M )(80.6 mL )  10−3 L  Mdil =   = 0.072540 = 0.0725 M ( 0.250 L )  1 mL  c) Mconc = 0.0262 M Vconc = 0.150 L Mdil = 0.0100 M Vdil = ? (0.0262 M) (0.150 L) = (0.0100 M) (Vdil) ( 0.0262 M )( 0.150 L ) = 0.393 L Vdil = ( 0.0100 M ) Final volume - initial volume = volume added. 0.393 L − 0.150 L = 0.243 L of water Vconc = 64.0 mL Mdil = ? Vdil = 0.100 L d) Mconc = 0.745 M (0.745 M) (64.0 mL) = (Mdil) (0.100 L) ( 0.745 M )( 64.0 mL )  10−3 L  Mdil =   = 0.4768 M (unrounded) ( 0.100 L )  1 mL  The concentration must now be converted to grams per milliliter.  0.4768 mol Ca(NO3 ) 2   10−3 L   164.10 g Ca(NO3 ) 2     1 mL   1 mol Ca(NO )  = 0.07824 = 0.0782 g Ca(NO3)2/mL L   3 2   3.120 Plan: The molar mass of the compound must be calculated. The molar mass and the numbers in the problem (with a few unit conversions) are then placed in the definition of molarity. Solution: Molar Mass C12H11N5 = 12 (12.01 g C/mol) + 11 (1.008 g H/mol) + 5 (14.01 g N/mol) = 225.26 g/mol  0.030 mg C12 H11 N5   10−3 g   1 mL   1 mol C12 H11 N5  M =    −3      150. mL    1 mg   10 L   225.26 g C12 H11 N5  = 8.878629 x 10–7 = 8.9 x 10–7 M C12H11N5 Note: By setting up the central conversions as shown, it is obvious that they are irrelevant. 3.121 The formula and molar mass are needed in addition to the information in the problem. The formula is: C4H6O2, and the molar mass is 86.09 g/mol.  3.74 x 108 lb C 4 H 6 O 2   1 yr   1 kg   103 g   1 mol C4 H 6 O 2  Moles/month =            1 yr    12 months   2.205 lb   1 kg   86.09 g C4 H 6 O 2  = 1.64183 x 108 = 1.64 x 108 mol C4H6O2 / month 3.122  85% NaCl  a) ( 4.0% ions )   = 3.4% NaCl  100% ions  3-31  22.99 g Na +  = 1.3375 = 1.3% Na+ ions b)% Na+ ions = ( 3.4% NaCl )   58.44 g NaCl     35.45 g Cl−  = 2.062 = 2.1% Cl– ions % Cl– ions = ( 3.4% NaCl )   58.44 g NaCl     3.4 g NaCl   1.025 g Seawater   1 mL   1 mol NaCl  c) M NaCl =     −3    = 0.596338 = 0.60 M NaCl 1 mL  100 g Seawater     10 L   58.44 g NaCl  3.123 a) False, a mole of one substance has the same number of units as a mole of any other substance. b) True c) False, a limiting-reactant problem is presented when the quantity of available material is given for more than one reactant. d) False, to prepare 1.00 L of 3.00 M NaCl, weigh 175.5 g of NaCl and add enough water to make 1.00 L e) True 3.124 Plan: Count the total number of spheres in each box. The number in box A divided by the volume change in each part will give the number we are looking for and allow us to match boxes. Solution: The number in each box is: A = 12, B = 6, C = 4, and D = 3. a) When the volume is tripled, there should be 12/3 = 4 spheres in a box. This is box C. b) When the volume is doubled, there should be 12/2 = 6 spheres in a box. This is box B. c) When the volume is quadrupled, there should be 12/4 = 3 spheres in a box. This is box D. 3.125 a) Equal b) O3 c) SO2 3.126 P4S3(s) + 8 O2(g) → P4O10(s) + 3 SO2(g) a) 1 molecule of P4S3 reacts with 8 molecules of O2 to produce 1 molecule of P4O10 and 3 molecules of SO2. b) 1 mol P4S3 reacts with 8 mol O2 to produce 1 mol P4O10 and 3 mol SO3. c) 220.09 g of P4S3 react with 256.00 g of O2 to produce 383.33 g of P4O10 and 192.21 g of SO3. 3.127 2 H2(g) + O2(g) → 2 H2O(g)  103 g   1 mol H 2 O  2 mol H 2   2.016 g H 2 Mass H2 = ( 85.0 kg H 2 O )        1 kg   18.02 g H 2 O  2 mol H 2 O   1 mol H 2 = 9.62493 x 103 = 9.62 x 103 g H2 Equal numbers of particles 0.4 mol O3 has a greater mass (4.0 g N2O4) (1mol/92.02 g N2O4) = 0.043 mol N2O4 (3.3 g SO2) (1 mol/64.07 g SO2) = 0.052 mol SO2) (0.6 mol C2H4) (28.05 g/1 mol C2H4) = 17g d) F2 (0.6 mol F2) (38.00 g/mol F2) = 23 g Note that if each of these values is properly rounded to one significant figure, the answers are identical. (2.2 mol MgCl2) (3 ions/mol MgCl2) = 6.6 mol e) MgCl2 (2.3 mol NaClO3) (2 ions/mol NaClO3) = 4.6 The compound with the lower molar mass will have more molecules in a given mass. f) H2O (0.500 mol NaBr/L) (0.500L) (1 mol Na+/mol NaBr) = 0.250 mol Na+ g) Equal (0.0146 kg NaCl) (103 g/1kg) (1mol NaCl/58.44 g) (1 mol Na+/mol NaCl) = 0.250 mol Na+ Multiply the moles by Avogadro’s number to find the number of ions. This step is unnecessary since the moles are equal. The heavier atoms will give a greater total mass. h) 238U 3-32   100%      98.8%  3.128 Plan: This problem may be done as two dilution problems with the two final molarities added, or, as done here, it may be done by calculating, then adding the moles and dividing by the total volume. Solution: Total Moles KBr Moles KBr from Solution 1 + Moles KBr from Solution 2 M KBr = = Total Volume Volume Solution 1 + Volume Solution 2  0.053 mol KBr   0.078 mol KBr    ( 0.200 L ) +   ( 0.550 L ) 1 L 1L     M KBr = = 0.071333 = 0.071 M KBr 0.200 L + 0.550 L 3.129  3.0 kg   60 s  60 min   24 hr   365.25 d  9 a) Mass of atmosphere =      50 x 10 y   s   1min  1 hr   1 d   1 y  = 4.73364 x 1018 = 4.7 x 1018 kg atmosphere b) Begin with the unrounded total mass of the atmosphere. 3  75.5%   10 g   1 mol N 2  Moles N2 = 4.73364 x 1018 kg Air       100%   1 kg   28.02 g N 2  = 1.27548 x 1020 = 1.3 x 1020 mol N2 ( ( 3.130 ) )  1 mol NH 4 Br  a) Moles NH4Br = ( 0.588 g NH 4 Br )   = 0.0060037 = 0.00600 mol NH4Br  97.94 g NH 4 Br   1 mol KNO3   1 mol K +   6.022 x 1023 K + ions  b) Number of K+ ions = ( 68.5 g KNO3 )       1 mol K +  101.11 g KNO3   1 mol KNO3    = 4.07978 x 1023 = 4.08 x 1023 K+ ions  92.09 g C3 H8 O3  c) Mass C3H8O3 = ( 5.85 mol C3 H8 O3 )   = 538.7265 = 539 g C3H8O3  1 mol C3 H8 O3   119.37 g CHCl3   mL d) Volume CHCl3 = ( 2.55 mol CHCl3 )    = 205.671 = 206 mL CHCl3  1 mol CHCl3  1.48 g CHCl3   2 mol Na +   6.022 x 1023 Na + ions  e) Number of Na+ = ( 2.11 mol Na 2 CO3 )    1 mol Na CO   1 mol Na + 2 3    24 24 + = 2.54128 x 10 = 2.54 x 10 Na ions  10−6 g   1 mol Cd   6.022 x 1023 Cd atoms  f) Number of Cd atoms = (10.0 µg Cd )    1 µg   112.4 g Cd   1 mol Cd     = 5.35765 x 1016 = 5.36 x 1016 Cd atoms  2 mol F   6.022 x 1023 F atoms  g) Number of F atoms = ( 0.0015 mol F2 )     1 mol F  1 mol F2    = 1.8066 x 1021 = 1.8 x 1021 F atoms 3.131 Neither A nor B has any XY3 molecules. Both C and D have XY3 molecules. D forms both XY3 and XY molecules. Only C has a single XY3 product, thus the answer is C. 3.132 Plan: Deal with the methane and propane separately, and combine the results. Balanced equations are needed for each hydrocarbon. The total mass and the percentages will give the mass of each hydrocarbon. The mass of each hydrocarbon is changed to moles, and through the balanced chemical equation the amount of CO2 produced by each gas may be found. Summing the amounts of CO2 gives the total from the mixture. 3-33 Solution: The balanced chemical equations are: Methane: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Propane: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) Mass of CO2 from each:  25.0%   1 mol CH 4   1 mol CO 2   44.01 g CO 2  Methane: ( 200.g Mixture )     = 137.188 g CO2   100%   16.04 g CH 4   1 mol CH 4   1 mol CO 2   75.0%   1 mol C3 H8   3 mol CO 2   44.01 g CO 2  Propane: ( 200.g Mixture )     = 449.183 g CO2   100%   44.09 g C3 H8   1 mol C3 H8   1 mol CO 2  Total CO2 = 137.188 g + 449.183 g = 586.318 = 586 g CO2 3.133  0.893 mol HCl  Total moles of HCl in final solution =   ( (1.20 + 3.37 ) L ) = 4.08101 mol HCl (unrounded) 1L    0.325 mol HCl  Moles HCl in solution of known concentration =   (1.20 L ) = 0.390 mol HCl 1L   Moles HCl added from solution with unknown concentration = Total moles - moles from known solution = (4.08101 - 0.390) mol HCl = 3.69101 mol HCl (unrounded) Molarity of unknown solution = (3.69101 mol HCl) / (3.37 L) = 1.095255 = 1.10 M HCl 3.134 Plan: If we assume a 100-gram sample of fertilizer, then the 30:10:10 percentages become the masses, in grams, of N, P2O5, and K2O. These masses may be changed to moles of substance, and then to moles of each element. To get the desired x:y:1.0 ratio, divide the moles of each element by the moles of potassium. Solution: A 100-gram sample of 30:10:10 fertilizer contains 30 g N, 10 g P2O5, and 10 g K2O.  1 mol N  Moles N = ( 30 g N )   = 2.1413 mol N (unrounded)  14.01 g N   1 mol P2 O5   2 mol P  Moles P = (10 g P2 O5 )    = 0.14090 mol P (unrounded)  141.94 g P2 O5   1 mol P2 O5   1 mol K 2 O   2 mol K  Moles K = (10 g K 2 O )    = 0.21231 mol K (unrounded)  94.20 g K 2 O   1 mol K 2 O  This gives a ratio of 2.1413:0.14090:0.21231 The ratio must be divided by the moles of K and rounded. (2.1413/0.21231):(0.14090/0.21231):(0.21231/0.21231) 10.086:0.66365:1.000 10:0.66:1.0 3.135 SrX2(aq) + H2SO4(aq) → SrSO4(s) + 2 HX(aq) 0.652 g 0.755 g Let M = the atomic mass of X.  1 mol SrSO 4  1 mol SrX 2  Moles SrX2 = ( 0.755 g SrSO 4 )    = 0.004110185 mol SrX2 (unrounded)  183.69 g SrSO 4  1 mol SrSO 4   (87.62 + 2M)g SrX 2   =0.652 g SrX2 1 mol SrX 2   ( 0.004110185 mol SrX 2 )  ( 0.004110185 (87.62 + 2M ) ) =0.652 0.360134 + 0.00822037 M = 0.652 M = 35.505 = 35.5 g/mol = Cl The original halide formula is SrCl2. 3-34 3.136 Assuming 100 g of mixture, we have 40.0 g CH4 + 60.0 g C2H6.  1 mol CH 4  1 mol C   12.01 g C  C from CH4 = ( 40.0 g CH 4 )     = 29.950 g C (unrounded)  16.04 g CH 4  1 mol CH 4   1 mol C   1 mol C2 H 6   2 mol C C from C2H6 = ( 60.0 g C2 H 6 )    30.07 g C2 H 6   1 mol C 2 H 6   12.01 g C    = 47.928 g C (unrounded)   1 mol C   ( 29.950 + 47.928 ) g  Mass percent C =   x 100% = 77.878 = 77.9% C  100 g Sample   3.137 Plan: Assume 100 grams of mixture. This means the mass of each compound, in grams, is the same as its percentage. Solution:  1 mol CO   1 mol C   12.01 g C  C from CO = ( 35.0 g CO )     = 15.007 g C (unrounded)  28.01 g CO   1 mol CO   1 mol C   1 mol CO 2   1 mol C   12.01 g C  C from CO2 = ( 65.0 g CO 2 )     = 17.738 g C (unrounded)  44.01 g CO 2   1 mol CO 2   1 mol C   (15.007 + 17.738 ) g  Mass percent C =   x 100% = 32.745 = 32.7% C  100 g Sample   3.138 Ferrocene + ? O2(g) → CO2 + H2O 0.9437 g 2.233 g 0.457 g  1 mol CO 2  1 mol C  Moles C = ( 2.233 g CO 2 )    = 0.050738 mol C (unrounded)  44.01 g CO 2  1 mol CO 2   1 mol H 2 O   2 mol H  Moles H = ( 0.457 g H 2 O )    = 0.050721 mol H (unrounded)  18.02 g H 2 O   1 mol H 2 O  Mass of Fe = mass of sample - mass of C - mass of H = 0.9437 g Fe, C, H - (0.050738 mol C) (12.01 g C/mol C) - (0.050721 mol H) (1.008 g H/mol H) = (0.9437 - 0.60936 - 0.05113) g = 0.28321 g Fe (unrounded) Moles Fe = (0.28321 g Fe) (1 mol Fe/ 55.85 g Fe) = 0.0050709 mol Fe Divide all moles by the smallest (Fe): C: (0.050738 mol / 0.0050709 mol) = 10.0 H: (0.050721 mol / 0.0050709 mol) = 10.0 Fe: (0.0050709 mol / 0.0050709 mol) = 1.0 Thus, the empirical formula is: C10H10Fe 3.139 Plan: Determine the molecular formula from the figure. Once the molecular formula is known, use the periodic table to determine the molar mass. With the molar mass and the additional information in part (b), the moles can be found. Solution: a) The formula of citric acid obtained by counting the number of carbon atoms, oxygen atoms, and hydrogen atoms is C6H8O7. Molar mass = (6 x 12.01) + (8 x 1.008) + (7 x 16.00) = 192.12 g/mol b) Determine the mass of citric acid in the lemon juice, and then use the molar mass to find the moles.  1 L   1 mL   1.09 g   6.82%   1 mol C6 H8 O7  Moles C6H8O7 = (1.50 qt )    −3       1.057 qt   10 L   mL   100%   192.12 g acid  = 0.549104 = 0.549 mol C6H8O7 3-35 3.140 Plan: Determine the formulas of each reactant and product, then balance the individual equations. Combine the three smaller equations to give the overall equation, where some substances serve as intermediates and will cancel. The amount of nitrogen plus the balanced overall chemical equation will give the amount of nitric acid formed. Solution: a) Derive the formulas and balance the equations. Nitrogen and oxygen combine to form nitrogen monoxide: N2(g) + O2(g) → 2 NO(g) Nitrogen monoxide reacts with oxygen to form nitrogen dioxide: 2 NO(g) + O2(g) → 2 NO2(g) Nitrogen dioxide combines with water to form nitric acid and nitrogen monoxide: 3 NO2(g) + H2O(l) → 2 HNO3(g) + NO(g) b) Combining the reactions may involve adjusting the equations in various ways to cancel out as many materials as possible other than the reactants added and the desired products. 2 x (N2(g) + O2(g) → 2 NO(g)) 3 x (2 NO(g) + O2(g) → 2 NO2(g)) 2 x (3 NO2(g) + H2O(l) → 2 HNO3(g) + NO(g)) Multiplying the above equations, summing the results, and canceling any substance appearing on both sides gives: 2 N2(g) + 5 O2(g) + 2 H2O(l) → 4 HNO3(g) c) Metric tons HNO3 =  103 kg   103 g   1 mol N 2   4 mol HNO3   63.02 g HNO3   1 kg   1t  1.25 x 103 t N 2         3   3     1t   1 kg   28.02 g N 2   2 mol N 2   1 mol HNO3   10 g   10 kg  = 5.622769 x 103 = 5.62 x 103 metric tons HNO3 ( 3.141 ) a) The moles of anhydrous alum and of water are needed. Dividing the moles of water by the moles of the anhydrous salt gives the value of x. Mass of water driven off on heating = 0.5404 g - 0.2941 g = 0.2463 g H2O  1 mol H 2 O  Moles water = ( 0.2463 g H 2 O )   = 0.013668 mol H2O (unrounded)  18.02 g H 2 O   1 mol KAl(SO 4 ) 2  Moles anhydrous = ( 0.2941 g KAl(SO 4 ) 2 )    258.22 g KAl(SO 4 ) 2  = 0.00113895 mol anhydrous (unrounded) x = (0.013668 mol H2O) / (0.00113895 mol anhydrous) = 12.00 = x Complete formula: KAl(SO4)2•12H2O b) There is a 1:1 mole ratio relating the Al to alum. Using the formula determined in part (a), the molar mass of alum is 474.46 g/mol. The theoretical yield (unrounded) is:  1 mol Al   1 mol Alum   474.46 g Alum  Grams Alum = ( 0.7500 g Al )     = 13.18921 g Alum  26.98 g Al   1 mol Al   1 mol Alum   Actual Yield  Percent yield =   x 100% =  Theoretical Yield  3.142  8.500 g    x 100% = 64.4466 = 64.45% yield  13.18921g   1 mol CO 2   1 mol C  Moles C = ( 2.838 g CO 2 )    = 0.0644853 mol C (unrounded)  44.01 g CO 2   1 mol CO 2   1 mol H 2 O   2 mol H  Moles H = ( 0.8122 g H 2 O )    = 0.090144 mol H (unrounded)  18.02 g H 2 O   1 mol H 2 O  The iron calculation needs an adjustment because of the differing sample size. This adjustment involves multiplying the iron by the sample size of the other sample, and dividing by the size of the iron sample as: (1.5173g / 0.3355 g). 3-36  1 mol Fe 2 O3   2 mol Fe   1.5173 g  Moles Fe = ( 0.0758 g Fe 2 O3 )     = 0.004293 mol Fe(unrounded)  159.70 g Fe 2 O3   1 mol Fe 2 O3   0.3355 g  To determine the amount of oxygen the above moles need to be converted to grams of each element and subtracted from the original sample mass (1.5173 g). Mass C = (0.0644853 mol C) (12.01 g C/mol) = 0.774468 g C Mass H = (0.090144 mol H) (1.008 g H/mol) = 0.090865 g H Mass Fe = (0.004293 mol Fe) (55.85 g Fe/mol) = 0.239764 g Fe Mass of O = (1.5173 g Fe, C, H, O) - (0.774468 g C) - (0.090865 g H) - (0.239764 g Fe) = 0.412203 g O Moles of O = (0.412203 g O (1 mol O / 16.00 g O)) = 0.0257627 mol O Divide each of the moles by the smallest value (Fe): C: (0.0644853 mol) / (0.004293 mol) = 15.0 H: (0.090144 mol) / (0.004293 mol) = 21.0 Fe: (0.004293 mol) / (0.004293 mol) = 1.0 O: (0.0257627 mol) / (0.004293 mol) = 6.0 These values give an empirical formula = C15H21O6Fe 3.143 a) Moles of each element are needed. These may be determined from the mass of each element. The mass of Pt is given and the mass of F is (0.519 - 0.327) g F = 0.192 g F. Calculating the moles of each element gives: Moles Pt = (0.327 g Pt) (1 mol Pt / 195.1 g Pt) = 0.001676 mol Pt (unrounded) Moles F = (0.192 g F) (1 mol F / 19.00 g F) = 0.010105 mol F (unrounded) Dividing each by the smaller value (Pt) gives: Pt: (0.001676 mol) / (0.001676 mol) = 1.0 F: (0.010105 mol) / (0.001676 mol) = 6.0 Empirical formula = PtF6 b) Determine the moles of Xe and of PtF6. Moles PtF6 = (0.265 g PtF6) (1 mol PtF6 / 309.1 g PtF6) = 0.0008573 mol PtF6 (unrounded) Moles Xe = [(0.378 - 0.265)g Xe] (1 mol Xe / 131.3 g Xe) = 0.0008606 mol Xe (unrounded) The moles are essentially identical so dividing by the smaller value gives 1 in each case. The empirical formula is: XePtF6. c) This problem can be solved as a system of two equations and two unknowns. The two equations are: The two unknowns are: x = mol XeF4 produced Xe(g) + 2 F2(g) → XeF4(s) y = mol XeF6 produced Xe(g) + 3 F2(g) → XeF6(s) Moles of Xe consumed = 1.85 x 10–4 mol present - 9.00 x 10–6 mol excess = 1.76 x 10–4 mol Xe Then x + y = 1.76 x 10–4 mol Xe consumed 2 x + 3 y = 5.00 x 10–4 mol F2 consumed Solve for x using the first equation and substitute the value of x into the second equation: x = 1.76 x 10–4 - y 2 (1.76 x 10–4 - y) + 3 y = 5.00 x 10–4 (3.52 x 10–4) - 2 y + 3 y = 5.00 x 10–4 y = (5.00 x 10–4) - (3.52 x 10–4) = 1.48 x 10–4 mol XeF6 x = (1.76 x 10–4) - (1.48 x 10–4) =2.8 x 10–5 mol XeF4 Convert moles of each product to grams using the molar masses (unrounded values): Mass XeF4 = (2.8 x 10–5 mol XeF4) x (207.3 g/mol) = 5.8044 x 10–3 g XeF4 Mass XeF6 = (1.48 x 10–4 mol XeF6) x (245.3 g/mol) = 3.63044 x 10–2 g XeF6 Calculate the percent of each compound using the total weight of the products: (5.8044 x 10–3 + 3.63044 x 10–2) g = 0.0421088 g % XeF4 = [(5.8044 x 10–3 g) / (0.0421088 g)] x 100% = 13.784 = 14% XeF4 % XeF6 = [(3.63044 x 10–2 g) / (0.0421088 g)] x 100% = 86.2157 = 86.2% XeF6 3-37 3.144 Plan: This problem involves running through a series of unit conversions. Solution:   6.0% heme a) Grams of heme = ( 0.45 g hemoglobin )   = 0.0270 = 0.027 g heme  100% hemoglobin   1 mol heme  –5 –5 b) Mole of heme = ( 0.027 g heme )   = 4.37966 x 10 = 4.4 x 10 mol heme 616.49 g heme    1 mol Fe   55.85 g Fe  c) Grams of Fe = 4.37966 x 10−5 mol heme     1 mol heme   1 mol Fe  = 2.44604 x 10–3 = 2.4 x 10–3 g Fe  1 mol hemin  651.94 g hemin  d) Grams of hemin = 4.37966 x 10−5 mol Heme     1 mol heme  1 mol hemin  ( ) ( ) = 2.85528 x 10–2 = 2.9 x 10–2 g hemin 3.145 Mn:O ratio In sample: 1.00: 1.42 or 0.704 In braunite: 2.00: 3.00 or 0.667 In manganosite: 1.00:100 or 1.00 a) The total amount of ore is equal to the amount of braunite (B) + the amount of manganosite (M). B + M = 1.00 M = 1.00 - B The amount of Mn is dependent on the sample’s composition. M(1.00) + B(0.667) = 0.704 (1.00 - B) (1.00) + B(0.667) = 0.704 B = 0.888889 mol braunite M = 0.111111 mol manganosite  157.88 g  Mass of braunite = 0.888889 mol   = 140.338 g braunite  1 mol   70.94 g  Mass of manganosite (g) = 0.111111 mol   = 7.88221 g manganosite  1 mol  Mass percent braunite = [140.338 g / (140.338 g + 7.88221 g)] x 100% = 94.6821% Mass percent manganosite = [7.88221 g / (140.338 g + 7.88221 g)] x 100% = 5.3179% Mass of braunite = (542.3 g) (94.6821%/ 100%) = 513.461 = 514 g braunite Mass of manganosite = (542.3 g) (5.3179% / 100%) = 28.839 = 28.8 g manganosite 2 ( 0.88889 ) b) Mn3+: Mn2+ = = 16.000 = 16.0 ( 0.11111) 3.146 First, a balanced chemical equation is needed: 4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)  103 g   1 mol SO 2   4 mol FeS2   119.99 g FeS2  Grams FeS2 = (1.00 kg SO 2 )   1 kg   64.07 g SO   8 mol SO   1 mol FeS  2  2  2    = 936.3977 = 936 g FeS2  103 g   1 mol SO 2  11 mol O 2   32.00 g O 2  = 686.749 = 687 g O2 Grams O2 = (1.00 kg SO 2 )   1 kg   64.07 g SO  8 mol SO   1 mol O  2  2  2     103 g   1 mol SO 2  2 mol Fe 2 O3   159.70 g Fe 2 O3  Grams Fe2O3 = (1.00 kg SO 2 )   1 kg   64.07 g SO  8 mol SO   1 mol Fe O  2  2  2 3    = 623.147 = 623 g Fe2O3 3-38 3.147 Plan: Determine the molecular formula and the molar mass of each of the compounds. From the amount of nitrogen present and the molar mass, the percent nitrogen may be determined. The moles need to be determined for part (b). Solution: a) To find mass percent of nitrogen, first determine molecular formula, then the molar mass of each compound. Mass percent is then calculated from the mass of nitrogen in the compound divided by the molar mass of the compound, and multiply by 100%. Urea: CH4N2O, M = 60.06 g/mol % N = [(2 x 14.01 g/mol N) / 60.06 g/mol CH4N2O] x 100% = 46.6580 = 46.66% N in urea Arginine: C6H15N4O2, M = 175.22 g/mol % N = [(4 x 14.01 g/mol N) / 175.22 g/mol C6H15N4O2] x 100% = 31.98265 = 31.98% N in arginine Ornithine: C5H13N2O2, M = 133.17 g/mol % N = [(2 x 14.01 g/mol N) / 133.17 g/mol C5H13N2O2)] x 100% = 21.04077 = 21.04% N in ornithine b) Grams of N =  1 mol C5 H13 N 2 O 2  1 mol CH 4 N 2 O     14.01 g N  2 mol N (143.2 g C5 H13 N 2 O2 )       133.17 g C5 H13 N 2 O 2  1 mol C5 H13 N 2 O 2   1 mol CH 4 N 2 O   1 mol N  = 30.130390 = 30.13 g N 3.148  1 mol C7 H 6 O3   1 mol C9 H8 O 4  a) Mole aspirin from C7H6O3 = ( 3.027 g C7 H 6 O3 )     138.12 g C7 H 6 O3   1 mol C7 H 6 O3  = 0.0219157 mol C9H8O4 (unrounded)  1.080 g   1 mol C4 H 6 O3  1 mol C9 H8 O 4  Mole aspirin from C4H6O3 = ( 6.00 mL C4 H 6 O3 )      mL   102.09 g C4 H 6 O3  1 mol C 4 H 6 O3  = 0.064734 mol C9H8O4 (unrounded) The limiting reactant is C7H6O3 b) First, calculate the theoretical yield from the limiting reagent:  1 mol C7 H 6 O3   1 mol C9 H8 O 4   180.15 g C9 H8 O 4  Grams C9H8O4 = ( 3.027 g C7 H 6 O3 )      138.12 g C7 H 6 O3   1 mol C7 H 6 O3   1 mol C9 H8 O 4  = 3.9481 g C9H8O4 (unrounded)  Actual Yield  Percent yield =   x 100% =  Theoretical Yield   3.261 g    x 100% = 82.5967 = 82.60% yield  3.9481 g  ( Number of Moles )( Molar Mass of Desired Product ) c) % atom economy = x 100% (Sum of Number of Moles )( Molar Mass for all Products ) = 3.149 (1 mol )(180.15 g / mol ) (180.15 + 60.05) g / mol x 100% = 75.00% atom economy Plan: Determine the molar mass of each product and plug into atom percent equation. Solution: NaCl: 58.44 g/mol H2O: 18.02 g/mol Molar masses of product: N2H4: 32.05 g/mol ( Number of Moles )( Molar Mass of Desired Product ) % atom economy = x 100% (Sum of Number of Moles )( Molar Mass for all Products ) (1mol )( 32.05 g / mol ) ( ( 32.05 + 58.44 + 18.02 ) g / mol ) x 100% = 29.5364 = 29.54% atom economy 3-39 3.150 mass FeCr2O4 =  103 g   0.511%   1 mol PbCrO 4   1 mol Cr      kg   100%   323.2 g PbCrO 4   1 mol PbCrO 4 = 1.769606 = 1.77 g FeCr2O4 ( kg Paint )  3.151   1 mol FeCr2 O 4    2 mol Cr   223.85 g FeCr2 O 4      1 mol FeCr2 O 4  Plan: Convert the mass of ethanol to moles, and use moles to determine the theoretical yield of diethyl ether. This theoretical yield, along with the given (actual) yield, is used to determine the percent yield. The difference between the actual and theoretical yields is related to the quantity of ethanol that did not produce diethyl ether. Fifty percent of the mass difference is converted to moles of diethyl ether, then to moles of ethanol and ethylene, and finally to the mass of ethylene. Solution: a) The determination of the theoretical yield: Grams diethyl ether =  1 mol CH3CH2OH  1 mol CH3CH2OCH2CH3  74.12 g CH3CH2OCH2CH3  ( 50.0 g CH3CH2OH )     2 mol CH3CH2 OH  46.07 g CH3CH2OH   1 mol CH3CH2OCH2CH3  = 40.2214 g diethyl ether (unrounded) Determining the percent yield:  Actual Yield   33.9 g  Percent yield =   x 100% =   x 100% = 84.2835 = 84.3% yield  Theoretical Yield   40.2214 g  b) To determine the amount of ethanol not producing diethyl ether, we will use the difference between the theoretical yield and actual yield (unrounded) to determine the amount of diethyl ether that did not form and hence, the amount of ethanol that did not produce the desired product. Fifty percent of this amount will be used to determine the amount of ethylene formed. Mass difference = 40.2214 g - 33.9 g = 6.3214 g diethyl ether that did not form Grams ethylene =  1 mol (C2 H5 ) 2 O  2 mol C2 H 6 O  1 mol C2 H 4   28.05 g C2 H 4   50.0%  ( 6.3214 g (C2 H5 )2 O )        74.12 g (C2 H 5 ) 2 O  1 mol (C2 H 5 ) 2 O  1 mol C 2 H 6 O   1 mol C2 H 4   100%  = 2.39227 = 2.39 g ethylene 3.152 First, balance the given equation: 8 Zn(s) + S8(s) → 8 ZnS(s) The reactions with oxygen are: 2 Zn(s) + O2(g) → 2 ZnO(s) S8(s) + 8 O2(g) → 8 SO2(g) a) Since the amounts of both starting materials are given, we must first find the limiting reagent:  1 mol Zn   8 mol ZnS  Moles of ZnS from Zn = ( 85.2 g Zn )    = 1.30295 mol ZnS (unrounded)  65.39 g Zn   8 mol Zn   1 mol S8   8 mol ZnS  Moles of ZnS from S8 = ( 52.4 g S8 )    = 1.6339 mol ZnS (unrounded)  256.56 g S8   1 mol S8  The zinc will produce less zinc sulfide, thus, zinc is the limiting reactant and will first be used to determine the theoretical yield and then the percent yield.  1 mol Zn   8 mol ZnS   97.46 g ZnS  Theoretical yield = ( 85.2 g Zn )     = 126.986 g ZnS (unrounded)  65.39 g Zn   8 mol Zn   1 mol ZnS   Actual Yield  Percent yield =   x 100% =  Theoretical Yield  3-40  105.4 g    x 100% = 83.001 = 83.0% yield  126.986 g  b) The theoretical yield indicates that 83.0% of the zinc produced zinc sulfide so (100 - 83.0)%= 17.0% of the zinc became zinc oxide. This allows the calculation of the amount of zinc oxide formed.  17.0%   1 mol Zn   2 mol ZnO  81.39 g ZnO  Grams zinc oxide = ( 85.2 g Zn )       100%   65.39 g Zn   2 mol Zn  1 mol ZnO  = 18.0280 = 18.0 g ZnO The calculation is slightly different for the sulfur. We need to determine the amount of sulfur not in zinc sulfide. The sulfur not in the zinc sulfide must be in sulfur dioxide. The amount of sulfur not in zinc sulfide will be converted to the mass of sulfur dioxide.   1 mol S8   1 mol ZnS   1 mol S8    8 mol SO 2   64.07 g SO 2  Mass SO2 = 52.4 g S8   − (105.4 g ZnS)        97.46 g ZnS   8 mol ZnS    1 mol S8   1 mol SO 2   256.56 g S8   = 35.3959 = 35 g SO2 The result of the subtraction in the calculation left only two significant figures. 3.153 Plan: Careful tracking of the units will carry you through this problem. Solution: 3 −3  2.50 kg Salt   10 g   10 L  a) Dissolved salt =      ( 50.0 mL ) = 125 g salt  L    1 kg   1 mL  b) Convert the grams from part (a) to moles of salt, then to moles of cocaine, and finally to grams of cocaine. A calculation similar to that in part (a) will give the volume required.  1 mol Salt  1 mol Cocaine  303.35 g Cocaine  Mass cocaine = (125 g Salt )      339.81 g Salt  1 mol Salt  1 mol Cocaine  = 111.588 g cocaine (unrounded)  10 −3 L    1L  Additional water needed = (111.588 g Cocaine )   - (50.0 mL )  1.70 g Cocaine   1 mL  = 65.59 = 65.6 L H2O 3.154 a) x = 0 La2Sr0CuO4 = 2 (138.9 g La/mol) + 0 (87.62 g Sr/mol) + 1 (63.55 g Cu/mol) + 4 (16.00 g O/mol) = 405.4 g/mol x=1 La1Sr1CuO4 = 1 (138.9 g La/mol) + 1 (87.62 g Sr/mol) + 1 (63.55 g Cu/mol) + 4 (16.00 g O/mol) = 354.1 g/mol x = 0.163 La(2-0.163)Sr0.163CuO4 = (2 -0.163) (138.9g La/mol) + (0.163) (87.62g Sr/mol) + 1 (63.55g Cu/mol) + 4 (16.00g O/mol) = 397.0 g/mol b) Assuming x grams to be the “equal” mass leads to:  1 mol BaCO3   2 mol Product  Moles product from BaCO3 = ( x g BaCO3 )    = 0.002534x mol product  197.3 g BaCO3   4 mol BaCO3   1 mol CuO   2 mol Product  Moles product from CuO = ( x g CuO )    = 0.004190x mol product  79.55 g CuO   6 mol CuO   1 mol Y2 O3  2 mol Product  Moles product from Y2O3 = ( x g Y2 O3 )    = 0.008857x mol product  225.82 g Y2 O3  1 mol Y2 O3  BaCO3 is the limiting reactant. 3-41 c) These calculations are based on the limiting reactant. BaCO3 remaining = 0% (Limiting reagent)  1 mol BaCO3  6 mol CuO   79.55 g CuO  CuO remaining = x g CuO - ( x g BaCO3 )      197.3 g BaCO3  4 mol BaCO3   1 mol CuO  = 0.39521x g CuO (unrounded)  0.39521 x g  Percent CuO =   x 100% = 39.521 = 39.52% CuO remaining xg    1 mol BaCO3  1 mol Y2 O3   225.82 g Y2 O3  Y2O3 remaining = x g Y2O3 - ( x g BaCO3 )      197.3 g BaCO3  4 mol BaCO3   1 mol Y2 O3  = 0.713862x g Y2O3 (unrounded)  0.713862 x g  Percent Y2O3 =   x 100% = 71.3862 = 71.38% Y2O3 remaining xg   3.155 Plan: The balanced chemical equation is needed. From the balanced chemical equation and the masses we can calculate the limiting reagent. The limiting reagent will be used to calculate the theoretical yield, and finally the percent yield. Solution: Determine the balanced chemical equation: ZrOCl2•8H2O(s) + 4 H2C2O4•2H2O(s) + 4 KOH(aq) → K2Zr(C2O4)3(H2C2O4)•H2O(s) + 2 KCl(aq) + 14 H2O(l) Determine the limiting reactant:  1 mol Zr cmpd  1 mol Product  Moles product from ZrOCl2•8H2O = (1.60 gZr cmpd )     322.25 g Zr cmpd  1 mol Zr cmpd  = 0.0049651 mol product (unrounded) Moles product from H2C2O4•2H2O =  1 mol H 2 C2 O 4 • 2H 2 O    1 mol Product ( 5.20 g H 2 C2 O4 • 2H 2 O )     126.07 g H 2 C2 O 4 • 2H 2 O   4 mol H 2 C 2 O 4 • 2H 2 O  = 0.0103117 mol product (unrounded) Moles product from KOH = irrelevant because KOH is stated to be in excess. The ZrOCl2•8H2O is limiting, and will be used to calculate the theoretical yield:  1 mol Zr cmpd   1 mol Product   541.53 g Product  Grams product = (1.60 g Zr cmpd )      322.25 g Zr cmpd   1 mol Zr cmpd   1 mol Product  = 2.68874 g product (unrounded) Finally, calculating the percent yield:  Actual Yield   1.20 g  Percent yield =   x100% =   x 100% = 44.630 = 44.6% yield  Theoretical Yield   2.68874 g  3.156 Convert the mass of carbon to moles and carry through the steps using these moles.  1 mol C   6.022 x 1023 C atoms  19 19 a) Number of atoms = ( 0.001 g C )   = 5.01415 x 10 = 5 x 10 C atoms   12.01 g C 1 mol C     1 mol C  3 mol PbO  223.2 g PbO  b) Mass PbO = ( 0.001 g C )     = 0.05575 = 0.06 g PbO  12.01 g C  1 mol C  1 mol PbO  3-42 3.157 Alcohols with C4H10O: H H H H H C C C C H H H H O H H H H H C C C C H H O H H H H H H C H C C H H H H H H C H O H H H C H H C H C C H O H 3-43 H H