Chapter 3 Stoichiometry 83

CHAPTER THREESTOICHIOMETRYFor Review1. Counting by weighing utilizes the average mass of a particular unit of substance. For marbles, a large sample size will contain many different individual masses for the various marbles. However, the large sample size will have an average mass so that the marbles behave as if each individual marble has that average mass. This assumption is valid as long as the sample size is large. When a large sample of marbles is weighed, one divides the total mass of marbles by the average mass of a marble and this will equal the number of marbles present. For atoms, because we can’t count individual atoms, we “count” the atoms by weighing them and convert the mass in grams to the quantity of atoms in the sample. The mole scale of atoms is a huge number (6.022 × 1023 atoms = 1 mol), so the assumption that a weighable sample size behaves as a bunch of atoms each with an average mass is valid and very useful.2. The masses of all the isotopes are relative to a specific standard. The standard is one atom of the carbon-12 isotope weighing exactly 12.0000 amu. One can determine from experiment how much heavier or lighter any specific isotope is than 12C. From this information, we assign an atomic mass value to that isotope. For example, experiment tells one that 16O is about 4/3 heavier than 12C, so a mass of 4/3(12.00) = 16.00 amu is assigned to 16O.3. The two major isotopes of boron are 10B and 11B. The listed mass of 10.81 is the average mass of a very large number of boron atoms.4. There are several ways to do this. The three conversion factors to use are Avogadro’s number, the molar mass, and the chemical formula. Two ways to use these conversions to convert grams of aspirin to number of H atoms are given below: molar mass aspirin = 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol23 1 mol C9H8O4 8 mol H 6.022  10 atoms H 1.00 g C9H8O4 ×   180.15g C9H8O4 mol C9H8O4 mol H = 2.67 × 1022 H atoms or23 1 mol C9H8O4 6.022  10 molecules C9H8O4 1.00 g C9H8O4 ×  × 180.15g C9H8O4 mol C9H8O4 8 atoms H = 2.67 × 1022 H atoms molecule C9H8O440 41 CHAPTER 3 STOICHIOMETRYOf course, the answer is the same no matter the order of the conversion factors.5. CxHyOz + oxygen → x CO2 + y/2 H2O From the equation above, the only reactant that contains carbon is the unknown compound and the only product that contains carbon is CO2. From the mass of CO2 produced, one can calculate the mass of C present which is also the mass of C in C xHyOz. Similarly, all the hydrogen in the unknown compound ends up as hydrogen in water. From the mass of H2O produced, one can calculate the mass of H in CxHyOz. Once the mass of C and H are known, the remainder of the compound is oxygen. From the mass of C, H, and O in the compound, one can then go on to determine the empirical formula.6. The molecular formula tells us the actual number of atoms of each element in a molecule (or formula unit) of a compound. The empirical formula tells only the simplest whole number ratio of atoms of each element in a molecule. The molecular formula is a whole number multiple of the empirical formula. If that multiplier is one, the molecular and empirical formulas are the same. For example, both the molecular and empirical formulas of water areH2O. For hydrogen peroxide, the empirical formula is OH; the molecular formula is H2O2.7. The product of the reaction has two A atoms bonded to a B atom for a formula of A2B. The initial reaction mixture contains 4 A2 and 8 AB molecules and the final reaction mixture contains 8 A2B molecules. The reaction is:8 AB(g) + 4 A2(g) → 8 A2B(g)Using the smallest whole numbers, the balanced reaction is:2 AB(g) + A2(g) → 2 A2B(g)2 mol A 2 B 2.50 mol A2 × = 5.00 mol A2B mol A 2The atomic mass of each A atom is 40.0/2 = 20.0 amu and the atomic mass of each B atom is30.0  20.0 = 10.0 amu. The mass of A2B = 2(20.0) + 10.0 = 50.0 amu.1 mol AB 1 mol A2 40.0 g A2 15.0 g AB ×   = 10.0 g A2 30.0 g AB 2 mol AB mol A2From the law of conservation of mass, the mass of product is:10.0 g A2 + 15.0 g AB = 25.0 g A2B or by stoichiometric calculation.1 mol AB 1 mol A2B 50.0 g A2B 15.0 g AB ×   = 25.0 g A2B 30.0 g AB mol AB mol A2B or CHAPTER 3 STOICHIOMETRY 421 mol A2 2 mol A2B 50.0 g A2B 10.0 g A2 ×   = 25.0 g A2B 40.0 g A2 mol A2 mol A2BGenerally, there are several ways to correctly do a stoichiometry problem. You should choose the method you like best.8. A limiting reactant problem gives you initial masses of at least two of the reactants and then asks for the amount of product that can form. Because one doesn’t know which reactant runs out first and hence determines the mass of product formed, this is a more difficult problem. The first step in solving the problem is to figure which reactant runs out first (is limiting).The strategy outlined in the text is to calculate the mole ratio of reactants actually present and compare this mole ratio to that required from the balanced reaction. Whichever ratio is larger allows one to deduce the identity of the limiting reactant and can, in turn, be used to calculate the amount of product formed. Another strategy is to pick one of the reactants and then calculate the mass of the other reactant required to react with it. By comparing the calculated mass to the actual mass present in the problem, one can deduce the identity of the limiting reactant and go on to solve the problem. A third common strategy is to assume each reactant is limiting and then calculate for each reactant the amount of product that could form. This gives two or more possible answers. The correct answer is the mass of product that is smallest. Even though there is enough of the other reactant to form more product, once the smaller amount of product is formed, the limiting reactant has run out.9. Balanced reaction: 2 SO2(g) + O2(g) → 2 SO3(g)We have 6 SO2 and 6 O2 molecules present. If all six of the SO2 molecules react, then 3 molecules of O2 will react producing 6 molecules of SO3. These numbers were determined using the balanced reaction. Since 6 molecules of O2 are present, and only 3 react when the SO2 reacts completely, SO2 is limiting. The product mixture will contain 6 – 6 = 0 SO2 molecules, 6 – 3 = 3 O2 molecules in excess, and 6 molecules of SO3 formed.1 mol SO2 1mol O2 32.00 g O2 96.0 g SO2 ×   = 24.0 g O2 64.07 g SO2 2 mol SO2 mol O2Because 32.0 g O2 are actually present, O2 is in excess and SO2 is the limiting reactant. Note that if the calculated amount of O2 was greater than 32.0 g, then we would have deduced that O2 is limiting. Solving the rest of the problem:1 mol SO2 2 mol SO3 80.07 g SO3 96.0 g SO2 ×   = 120. g SO3 64.07 g SO2 2 mol SO2 mol SO310. Side reactions may occur. For example, in the combustion of CH4 (methane) to CO2 and H2O, some CO may also form. Also, reactions only go part way to completion, instead reaching a state of equilibrium where both reactants and products are present (see Ch. 13).Questions19. isotope mass abundance 12C 12.0000 amu 98.89% 43 CHAPTER 3 STOICHIOMETRY13C 13.034 amu 1.11% average mass = 0.9889 (12.0000) + 0.0111(13.034) = 12.01 amuFrom the relative abundances, there would be 9889 atoms of 12C and 111 atoms of 13C in the 10,000 atom sample. The average mass of carbon is independent of the sample size; it will always be 12.01 amu.12.01amu total mass = 10,000 atoms × = 1.201 × 105 amu atomFor one mol of carbon (6.0221 × 1023 atoms C), the average mass would still be 12.01 amu. The number of 12C atoms would be 0.9889 (6.0221 × 1023) = 5.955 × 1023 atoms 12C and the number of 13C atoms would be 0.0111 (6.0221 × 1023) = 6.68 × 1021 atoms 13C. 12.01amu total mass = 6.0221 × 1023 atoms × = 7.233 × 1024 amu atom12.01amu 1g total mass in g = 6.0221 × 1023 atoms × × = 12.01 g/mol atom 6.0221  1023 amu By using the carbon-12 standard to define the relative masses of all of the isotopes as well as to define the number of things in a mole, then each element’s average atomic mass in units of grams is the mass of a mole of that element as it is found in nature.20. Consider a sample of glucose, C6H12O6. The molar mass of glucose is 180.16 g/mol. The chemical formula allows one to convert from molecules of glucose to atoms of carbon, hydrogen, or oxygen present and vice versa. The chemical formula also gives the mole relationship in the formula. One mol of glucose contains 6 mol C, 12 mol H, and 6 mol O. Thus, mole conversions between molecules and atoms are possible using the chemical formula. The molar mass allows one to convert between mass and moles of compound and Avogadro’s number (6.022 × 1023) allows one to convert between moles of compound and number of molecules.6.022  1023 dollars 21. Avogadro’s number of dollars = mol dollars 6.022  1023 dollars 1 mol dollars mol dollars = 1 × 1014 dollars/person 6  109 people1 trillion = 1,000,000,000,000 = 1 × 1012; Each person would have 100 trillion dollars.22. The molar mass is the mass of 1 mol of the compound. The empirical mass is the mass of 1 mol of the empirical formula. The molar mass is a whole number multiple of the empirical mass. The masses are the same when the molecular formula = empirical formula, and the masses are different when the two formulas are different. When different, the empirical mass CHAPTER 3 STOICHIOMETRY 44 must be multiplied by the same whole number used to convert the empirical formula to the molecular formula. For example, C6H12O6 is the molecular formula for glucose and CH2O is the empirical formula. The whole number multiplier is 6. This same factor of 6 is the multiplier used to equate the empirical mass (30 g/mol) of glucose to the molar mass (180 g/mol).23. The mass percent of a compound is a constant no matter what amount of substance is present. Compounds always have constant composition.24. A balanced reaction starts with the correct formulas of the reactants and products. The coefficients necessary to balance the reaction give molecule relationships as well as mole relationships between reactants and products. The state (phase) of the reactants and products is also given. Finally, special reaction conditions are sometimes listed above or below the arrow. These can include special catalysts used and/or special temperatures required for a reaction to occur.25. The specific information needed is mostly the coefficients in the balanced equation and the molar masses of the reactants and products. For percent yield, we would need the actual yield of the reaction and the amounts of reactants used.4 a. mass of CB produced = 1.00 × 10 molecules A2B2 ×1 mol A2B2 2 mol CB molar mass of CB 23   6.022  10 molecules A2B2 1 mol A2B2 mol CB 2 atoms A 4 b. atoms of A produced = 1.00 × 10 molecules A2B2 × 1 molecule A2B21 mol A B 4 2 2 c. mol of C reacted = 1.00 × 10 molecules A2B2 × 23 × 6.022  10 molecules A2B2 2 mol C1 mol A2B2 actual mass d. % yield = × 100; The theoretical mass of CB produced was calculated theoretical mass in part a. If the actual mass of CB produced is given, then the percent yield can be determined for the reaction using the % yield equation.26. One method is to determine the actual mole ratio of XY to Y2 present and compare this ratio to the required 2:1 mole ratio from the balanced equation. Which ratio is larger will allow one to deduce the limiting reactant. Once the identity of the limiting reactant is known, then one can calculate the amount of product formed. A second method would be to pick one of the reactants and then calculate how much of the other reactant would be required to react with it all. How the answer compares to the actual amount of that reactant present allows one to deduce the identity of the limiting reactant. Once the identity is known, one would take the limiting reactant and convert it to mass of product formed.When each reactant is assumed limiting and the amount of product is calculated, there are two possible answers (assuming two reactants). The correct answer (the amount of product that could be produced) is always the smaller number. Even though there is enough of the 45 CHAPTER 3 STOICHIOMETRY other reactant to form more product, once the small quantity is reached, the limiting reactant runs out and the reaction cannot continue.ExercisesAtomic Masses and the Mass Spectrometer27. A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766)A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; From the periodic table, the element is Pb.28. A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amuThis is element Ti (titanium).29. Let A = mass of 185Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207  117.0 = 0.3740(A) 69.2 A = = 185 amu (A = 184.95 amu without rounding to proper significant figures.) 0.3740 30. abundance 28Si = 100.00 (4.70 + 3.09) = 92.21%; From the periodic table, the average atomic mass of Si is 28.09 amu.28.09 = 0.9221(27.98) + 0.0470 (atomic mass 29Si) + 0.0309(29.97) atomic mass 29Si = 29.01The mass of 29Si is actually a little less than 29 amu. There are other isotopes of silicon that are considered when determining the 28.09 amu average atomic mass of Si listed in the atomic table.31. There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br 2 molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2 or 78.92. This corresponds to 79Br. The second isotope is 81Br with mass equal to 79 79 81 81 161.84/2 = 80.92. The peaks in the mass spectrum correspond to Br2, Br Br, and Br2 in order of increasing mass. The intensities of the highest and lowest mass tell us the two isotopes are present in about equal abundance. The actual abundance is 50.69% 79Br and 49.31% 81Br. The calculation of the abundance from the mass spectrum is beyond the scope of this text.32. GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have 2 peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60:40 or 3:2.144 146 CHAPTER 3 STOICHIOMETRY 4669 69 71 71 Ga2As2 can be Ga2As2, Ga GaAs2, or Ga2As2. The mass spectrum will have 3 peaks at 288, 290, and 292 with intensities in the ratio of 36:48:16 or 9:12:4. We get this ratio from the following probability table:69Ga (0.60) 71Ga (0.40) 69Ga (0.60) 0.36 0.24 71Ga (0.40) 0.24 0.16288 290 292Moles and Molar Masses33. When more than one conversion factor is necessary to determine the answer, we will usually put all the conversion factors into one calculation instead of determining intermediate answers. This method reduces round-off error and is a time saver.1 mol Fe 55.85 g Fe 500. atoms Fe   = 4.64 × 1020 g Fe 6.0221023 atoms Fe mol Fe1 mol Fe 34. 500.0 g Fe × = 8.953 mol Fe 55.85 g Fe6.022  1023 atoms Fe 8.953 mol Fe × = 5.391 × 1024 atoms Fe mol Fe 0.200 g C 1 mol C 6.022  1023 atoms C 35. 1.00 carat ×   = 1.00 × 1022 atoms C carat 12.01g C mol C1 mol C 36. 5.0 × 1021 atoms C × = 8.3 × 103 mol C 6.022 1023 atoms C12.01g C 8.3 × 3 mol C × = 0.10 g C 10 mol C 47 CHAPTER 3 STOICHIOMETRY37. Al2O3: 2(26.98) + 3(16.00) = 101.96 g/molNa3AlF6: 3(22.99) + 1(26.98) + 6(19.00) = 209.95 g/mol38. HFC  134a, CH2FCF3: 2(12.01) + 2(1.008) + 4(19.00) = 102.04 g/molHCFC 124, CHClFCF3: 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 g/mol39. a. The formula is NH3. 14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol b. The formula is N2H4. 2(14.01) + 4(1.008) = 32.05 g/mol c. (NH4)2Cr2O7: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 g/mol40. a. The formula is P4O6. 4(30.97 g/mol) + 6(16.00 g/mol) = 219.88 g/mol b. Ca3(PO4)2: 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol c. Na2HPO4: 2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 g/mol1 mol NH3 41. a. 1.00 g NH3 × = 0.0587 mol NH3 17.03 g NH31 mol N 2 H 4 b. 1.00 g N2H4 × = 0.0312 mol N2H4 32.05 g N 2 H 4 1mol (NH ) Cr O 4 2 2 7 3 c. 1.00 g (NH4)2Cr2O7 × = 3.97 × 10 mol (NH4)2Cr2O7 252.08 g (NH4 )2 Cr2O71 mol P 4 O6 42. a. 1.00 g P O × = 4.55 × 3 mol P O 4 6 219.88 g 10 4 6 1 mol Ca (PO ) 3 4 2 3 b. 1.00 g Ca3(PO4)2 × = 3.22 × 10 mol Ca3(PO4)2 310.18 g1 mol Na HPO 2 4 3 c. 1.00 g Na2HPO4 × = 7.04 × 10 mol Na2HPO4 141.96 g17.03 g NH 4 43. a. 5.00 mol NH3 × = 85.2 g NH3 mol NH332.05 g N 2 H 4 b. 5.00 mol N2H4 × = 160. g N2H4 mol N 2 H 4252.08 g (NH 4 ) 2 Cr2O7 c. 5.00 mol (NH4)2Cr2O7 × = 1260 g (NH4)2Cr2O7 1mol (NH 4 ) 2 Cr2O 7 CHAPTER 3 STOICHIOMETRY 48219.88 g 3 44. a. 5.00 mol P4O6 × = 1.10 × 10 g P4O6 1 mol P 4 O6310.18 g 3 b. 5.00 mol Ca3(PO4)2 × = 1.55 × 10 g Ca3(PO4)2 mol Ca3 (PO4 )2 141.96 g 2 c. 5.00 mol Na2HPO4 × = 7.10 × 10 g Na2HPO4 mol Na 2 HPO 445. Chemical formulas give atom ratios as well as mol ratios.1 mol N 14.01g N a. 5.00 mol NH3 ×  = 70.1 g N mol NH3 mol N2 mol N 14.01g N b. 5.00 mol N2H4 ×  = 140. g N mol N2H4 mol N 2 mol N 14.01g N c. 5.00 mol (NH4)2Cr2O7 ×  = 140. g N mol (NH4 )2 Cr2O7 mol N4 mol P 30.97 g P 46. a. 5.00 mol P4O6 ×  = 619 g P mol P4O 6 mol P2 mol P 30.97 g P b. 5.00 mol Ca3(PO4)2 ×  = 310. g P mol Ca3 (PO4 )2 mol P1 mol P 30.97 g P c. 5.00 mol Na2HPO4 ×  = 155 g P mol Na 2 HPO 4 mol P23 1 mol NH3 6.022 10 molecules NH3 22 47. a. 1.00 g NH3 ×  = 3.54 × 10 molecules NH3 17.03 g NH3 mol NH323 1 mol N 2 H 4 6.022 10 molecules N 2 H 4 b. 1.00 g N2H4 ×  32.05 g N 2 H 4 mol N 2 H 4 22 = 1.88 × 10 molecules N2H41mol (NH4 )2 Cr2O7 c. 1.00 g (NH4)2Cr2O7 × 252.08 g (NH4 )2 Cr2O723 6.022 10 formula units (NH4 )2 Cr2O7 21  = 2.39 × 10 formula units (NH4)2Cr2O7 mol (NH4 )2 Cr2O71 mol P O 23 4 6 6.022 10 molecules 21 48. a. 1.00 g P4O6 ×  = 2.74 × 10 molecules P4O6 219.88 g mol P4O 6 49 CHAPTER 3 STOICHIOMETRY23 1 mol Ca 3 (PO 4 ) 2 6.022 10 formula units b. 1.00 g Ca3(PO4)2 ×  310.18 g mol Ca 3 (PO 4 ) 221 = 1.94 × 10 formula units Ca3(PO4)223 1 mol Na 2 HPO 4 6.022 10 formula units c. 1.00 g Na2HPO4 ×  141.96 g mol Na 2 HPO 421 = 4.24 × 10 formula units Na2HPO449. Using answers from Exercise 47:1atom N 22 22 a. 3.54 × 10 molecules NH3 × = 3.54 × 10 atoms N molecule NH3 2 atoms N 22 22 b. 1.88 × 10 molecules N2H4 × = 3.76 × 10 atoms N molecule N 2 H 4 2 atoms N 21 c. 2.39 × 10 formula units (NH4)2Cr2O7 × formula unit (NH 4 ) 2 Cr2O7 = 4.78 × 1021 atoms N 50. Using answers from Exercise 48:4 atoms P 21 22 a. 2.74 × 10 molecules P4O6 × = 1.10 × 10 atoms P molecule P4O6 2 atoms P 21 21 b. 1.94 × 10 formula units Ca3(PO4)2 × = 3.88 × 10 atoms P formula unit Ca 3 (PO 4 ) 2 1atom P 21 21 c. 4.24 × 10 formula units Na2HPO4 × = 4.24 × 10 atoms P formula unit Na 2 HPO 451. Molar mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol1g 1mol 500.0 mg ×  = 2.839 × 10 3 mol 1000 mg 176.12 g6.022 1023 molecules 2.839 × 10-3 mol × = 1.710 × 1021 molecules mol52. a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol1g 1mol b. 500. mg ×  = 2.78 × 3 mol 1000 mg 180.15 g 10 6.022 1023 molecules 2.78 × 103 mol × = 1.67 × 1021 molecules mol CHAPTER 3 STOICHIOMETRY 501 mol 53. a. 150.0 g Fe O × = 0.9393 mol Fe O 2 3 159.70 g 2 31g 1 mol 4 b. 10.0 mg NO2 ×  = 2.17 × 10 mol NO2 1000 mg 46.01g1mol 16 8 c. 1.5 × 10 molecules BF3 × = 2.5 × 10 mol BF3 6.02 1023 molecules1g 1mol 4 54. a. 20.0 mg C8H10N4O2 ×  = 1.03 × 10 mol C8H10N4O2 1000 mg 194.20 g 1 mol b. 2.72 × 1021 molecules C H OH × 2 5 6.022 1023 molecules 3 = 4.52 × 10 mol C2H5OH 1 mol c. 1.50 g CO × = 3.41 × 2 mol CO 2 44.01g 10 255. a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to show how these conversion factors can be used.Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol23 1 mol C 2 H 5O 2 N 6.022 10 molecules C 2 H 5O 2 N 5.00 g C2H5O2N ×  × 75.07 g C 2 H 5O 2 N mol C 2 H 5O 2 N 1atom N = 4.01 × 1022 atoms N molecule C 2 H 5O 2 N b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol23 1mol Mg 3 N 2 6.022 10 formula units Mg 3 N 2 2 atoms N 5.00 g Mg3N2 ×   100.95 g Mg 3 N 2 mol Mg 3 N 2 mol Mg 3 N 2= 5.97 × 1022 atoms N c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol23 1mol Ca(NO3 ) 2 2 mols N 6.022  10 atoms N 5.00 g Ca(NO3)2 ×   164.10 g Ca(NO3 ) 2 mol Ca(NO3 ) 2 mol N= 3.67 × 1022 atoms N d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 51 CHAPTER 3 STOICHIOMETRY23 1 mol N2O4 2 mol N 6.022  10 atoms N 5.00 g N2O4 ×   92.02 g N2O4 mol N2O4 mol N = 6.54 × 1022 atoms N1 mol 56. 4.24 g C H × = 5.43 × 2 mol C H 6 6 78.11g 10 6 6 23 2 6.022  10 molecules 22 5.43 × 10 mol C6H6 × = 3.27 × 10 molecules C6H6 molEach molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 total atoms.22 12 atoms total 23 3.27 × 10 molecules C6H6 × = 3.92 × 10 atoms total molecule18.02 g 0.224 mol H2O × = 4.04 g H2O mol23 6.022  10 molecules 23 0.224 mol H2O × = 1.35 × 10 molecules H2O mol23 3 atoms total 23 1.35 × 10 molecules H2O × = 4.05 × 10 atoms total molecule1 mol 22 2 2.71 × 10 molecules CO2 × = 4.50 × 10 mol CO2 6.022  1023 molecules2 44.01g 4.50 × 10 mol CO2 × = 1.98 g CO2 mol3 atoms total 22 22 2.71 × 10 molecules CO2 × = 8.13 × 10 atoms total molecule CO21 molecule 3.35 × 1022 atoms total × = 5.58 × 1021 molecules CH OH 6 atoms total 31 mol 21 3 5.58 × 10 molecules CH3OH × = 9.27 × 10 mol CH3OH 6.022  1023 molecules3 32.04 g 9.27 × 10 mol CH3OH × = 0.297 g CH3OH mol12.01g  1.008 g  14.01g  16.00 g  57. a. 14 mol C   + 18 mol H   + 2 mol N   + 5 mol O    mol C   mol H   mol N   mol O = 294.30 g/mol CHAPTER 3 STOICHIOMETRY 521 mol b. 10.0 g aspartame × = 3.40 × 102 mol 294.30 g294.30 g c. 1.56 mol × = 459 g mol 1g 1mol 6.02 1023 molecules d. 5.0 mg ×   = 1.0 × 1019 molecules 1000 mg 294.30 g mol e. The chemical formula tells us that 1 molecule of aspartame contains two atoms of N. The chemical formula also says that 1 mol of aspartame contains two mol of N.1 mol aspartame 2 mol N 6.02 1023 atoms N 1.2 g aspartame ×   294.30 g aspartame mol aspartame mol N = 4.9 × 1021 atoms of nitrogen1mol 294.30 g f. 1.0 × 109 molecules ×  = 4.9 × 1013 g or 490 fg 6.02 1023 molecules mol1 mol 294.30 g g. 1 molecule aspartame ×  = 4.887 × 10 22 g 6.022 1023 molecules mol58. a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol1 mol 165.39 g b. 500.0 g × = 3.023 mol c. 2.0 × 102 mol × = 3.3 g 165.39 g mol1 mol 6.02 1023 molecules 3 atoms Cl d. 5.0 g C H Cl O ×   2 3 3 2 165.39 g mol molecule = 5.5 × 1022 atoms of chlorine1 mol Cl 1 mol C H Cl O 165.39 g C H Cl O e. 1.0 g Cl ×  2 3 3 2  2 3 3 2 = 1.6 g chloral hydrate 35.45 g 3 mol Cl mol C2H3Cl3O21 mol 165.39 g f. 500 molecules ×  = 1.373 × 10 19 g 6.022 1023 molecules molPercent Composition59. a. C3H4O2: Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06 g/mol 53 CHAPTER 3 STOICHIOMETRY36.03 g C 4.032 g H %C = × 100 = 50.00% C; %H = × 100 72.06 g compound 72.06 g compound = 5.595% H 32.00 g %O = 100.00 - (50.00 + 5.595) = 44.41% O or %O = × 100 = 44.41% O 72.06 g b. C4H6O2: Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol 48.04 g 6.048 g %C = × 100 = 55.80% C; %H = × 100 = 7.025% H 86.09 g 86.09 g%O = 100.00 - (55.80 + 7.025) = 37.18% O c. C3H3N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01 = 53.06 g/mol 36.03 g 3.024 g %C = × 100 = 67.90% C; %H = × 100 = 5.699% H 53.06 g 53.06 g14.01g %N = × 100 = 26.40% N or %N = 100.00  (67.90 + 5.699) = 26.40% N 53.06 g60. molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol20(12.01) g C %C = × 100 = 71.40% C 336.43 g compound29(1.008) g H %H = × 100 = 8.689% H 336.43 g compound19.00 g F %F = × 100 = 5.648% F 336.43 g compound%O = 100.00  (71.40 + 8.689 + 5.648) = 14.26% O or:3(16.00) g O %O = × 100 = 14.27% O 336.43 g compound14.01g N 61. a. NO: %N = × 100 = 46.68% N 30.01g NO 14.01g N b. NO2: %N = × 100 = 30.45% N 46.01g NO228.02 g N c. N2O4: %N = × 100 = 30.45% N 92.02 g N 2O 4 CHAPTER 3 STOICHIOMETRY 5428.02 g N d. N2O: %N = × 100 = 63.65% N 44.02 g N2OThe order from lowest to highest mass percentage of nitrogen is: NO2 = N2O4 < NO < N2O.62. C8H10N4O2: molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol8(12.01) g C 96.08 %C = × 100 = × 100 = 49.47% C 194.20 g C8H10 N 4O 2 194.20C12 H22O11: molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol12(12.01) g C %C = × 100 = 42.10% C 342.30 g C12 H 22O11C2H5OH: molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol2(12.01) g C %C = × 100 = 52.14% C 46.07 g C 2 H 5OHThe order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH)63. There are many valid methods to solve this problem. We will assume 100.00 g of compound, then determine from the information in the problem how many mol of compound equals 100.00 g of compound. From this information, we can determine the mass of one mol of compound (the molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin:1mol Co 1mol cyanocobalamin mol cyanocobalamin = 4.34 g Co ×  58.93 g Co mol Co= 7.36 × 10 2 mol cyanocobalamin x g cyanocobalamin 100.00 g = , x = molar mass = 1360 g/mol 1 mol cyanocobalamin 7.36 102 mol64. There are 0.390 g Cu for every 100.00 g of fungal laccase. Assuming 100.00 g fungal laccase:1 mol Cu 1 mol fungal laccase mol fungal laccase = 0.390 g Cu ×  = 1.53 × 103 mol 63.55 g Cu 4 mol Cu x g fungallaccase 100.00 g = , x = molar mass = 6.54 × 104 g/mol 1 mol fungal laccase 1.53103 mol 55 CHAPTER 3 STOICHIOMETRYEmpirical and Molecular Formulas12.01g C  1.008 g H      65. a. Molar mass of CH2O = 1 mol C   + 2 mol H    mol C   mol H 16.00 g O  + 1 mol O   = 30.03 g/mol  mol O 12.01g C 2.016 g H %C = × 100 = 39.99% C; %H = × 100 = 6.713% H 30.03 g CH 2O 30.03 g CH 2O16.00 g O %O = × 100 = 53.28% O or %O = 100.00 - (39.99 + 6.713) = 53.30% 30.03 g CH 2O b. Molar Mass of C6H12O6 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol76.06 g C 12.(1.008) g %C = × 100 = 40.00%; %H = × 100 = 6.714% 180.16 g C6 H12O6 180.16 g %O = 100.00 - (40.00 + 6.714) = 53.29% c. Molar mass of HC2H3O2 = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol24.02 g 4.032 g %C = × 100 = 40.00%; %H = × 100 = 6.714% 60.05 g 60.05 g %O = 100.00 - (40.00 + 6.714) = 53.29%66. All three compounds have the same empirical formula, CH2O, and different molecular formulas. The composition of all three in mass percent is also the same (within rounding differences). Therefore, elemental analysis will give us only the empirical formula.67. a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical formula) is NO2. b. Molecular formula: C3H6; empirical formula = CH2 c. Molecular formula: P4O10; empirical formula = P2O5 d. Molecular formula: C6H12O6; empirical formula = CH2O 68. a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g188.35 g = 4.000; So the molecular formula is (SNH) or S N H . 47.09 g 4 4 4 4 b. NPCl2: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol CHAPTER 3 STOICHIOMETRY 56347.64 g = 3.0000; Molecular formula is (NPCl ) or N P Cl . 115.88 g 2 3 3 3 6 c. CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol341.94 g = 2.0000; Molecular formula: Co C O 170.97 g 2 8 8 184.32 g d. SN: 32.07 + 14.01 = 46.08 g/mol; = 4.000; Molecular formula: S N 46.08 g 4 469. Out of 100.00 g of adrenaline, there are:1 mol C 1 mol H 56.79 g C × = 4.729 mol C; 6.56 g H × = 6.51 mol H 12.01g C 1.008 g H1 mol O 1mol N 28.37 g O × = 1.773 mol O; 8.28 g N × = 0.591 mol N 16.00 g O 14.01g NDividing each mol value by the smallest number:4.729 6.51 1.773 0.591 = 8.00; = 11.0; = 3.00; = 1.00 0.591 0.591 0.591 0.591This gives adrenaline an empirical formula of C8H11O3N.70. Assuming 100.00 g of nylon-6:1 mol C 1mol N 63.68 g C × = 5.302 mol C; 12.38 g N × = 0.8837 mol N 12.01g C 14.01g N1 mol H 1 mol O 9.80 g H × = 9.72 mol H; 14.14 g O × = 0.8838 mol O 1.008 g H 16.00 g ODividing each mol value by the smallest number:5.302 9.72 = 6.000; = 11.0 0.8837 0.8837The empirical formula for nylon-6 is C6H11NO71. Compound I: mass O = 0.6498 g HgxOy  0.6018 g Hg = 0.0480 g O1 mol Hg 0.6018 g Hg × = 3.000 × 3 mol Hg 200.6 g Hg 10 57 CHAPTER 3 STOICHIOMETRY1 mol O 0.0480 g O × = 3.00 × 103 mol O 16.00 g OThe mol ratio between Hg and O is 1:1, so the empirical formula of compound I is HgO.Compound II: mass Hg = 0.4172 g HgxOy  0.016 g O = 0.401 g Hg1 mol Hg 1 mol O 0.401 g Hg × = 2.00 × 103 mol Hg; 0.016 g O × 200.6 g Hg 16.00 g O = 1.0 × 103 mol OThe mol ratio between Hg and O is 2:1, so the empirical formula is Hg2O.1mol N 1 mol H 72. 1.121 g N × = 8.001 × 102 mol N; 0.161 g H × = 1.60 × 101 mol H 14.01g N 1.008 g H1 mol C 1 mol O 0.480 g C × = 4.00 × 2 mol C; 0.640 g O × = 4.00 × 2 mol O 12.01g C 10 16.00 g O 10 Dividing all mol values by the smallest number:8.001102 1.60 101 4.00 102 = 2.00; = 4.00; = 1.00 4.00 102 4.00 102 4.00 102Empirical formula = N2H4CO73. Out of 100.0 g, there are:1 mol S 1mol N 69.6 g S × = 2.17 mol S; 30.4 g N × = 2.17 mol N 32.07 g S 14.01g NEmpirical formula is SN since mol values are in a 1:1 mol ratio. The empirical formula mass of SN is ~ 46 g. Because 184/46 = 4.0, the molecular formula isS4N4.74. Assuming 100.0 g of compound:1 mol P 1mol N 26.7 g P × = 0.862 mol P; 12.1 g N × = 0.864 mol N 30.97 g P 14.01g N1mol Cl 61.2 g Cl × = 1.73 mol Cl 35.45 g Cl CHAPTER 3 STOICHIOMETRY 581.73 = 2.01; Empirical formula = PNCl 0.862 2The empirical formula mass is ~31.0 + 14.0 + 2(35.5) = 116 molar mass 580  = 5; molecular formula = (PNCl ) =P N Cl empirical formula mass 116 2 5 5 5 1075. Assuming 100.00 g of compound (mass hydrogen = 100.00 g - 49.31 g C - 43.79 g O = 6.90 g H):1 mol C 1 mol H 49.31 g C × = 4.106 mol C; 6.90 g H × × = 6.85 mol H 12.01g C 1.008 g H1 mol O 43.79 g O × = 2.737 mol O 16.00 g O Dividing all mole values by 2.737 gives:4.106 6.85 2.737 = 1.500; = 2.50; = 1.000 2.737 2.737 2.737Since a whole number ratio is required, the empirical formula is C3H5O2.The empirical formula mass is: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol molar mass 146.1 = = 1.999; molecular formula = (C H O ) = C H O empirical formula mass 73.07 3 5 2 2 6 10 476. Assuming 100.00 g of compound (mass oxygen = 100.00 g  41.39 g C  3.47 g H = 55.14 g O): 1 mol C 1 mol H 41.39 g C × = 3.446 mol C; 3.47 g H × = 3.44 mol H 12.01g C 1.008 g H1 mol O 55.14 g O × = 3.446 mol O 16.00 g O All are the same mol values so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol.15.0 g molar mass = = 116 g/mol 0.129 mol molar mass 116  = 4.00; molecular formula = (CHO) = C H O empirical mass 29.02 4 4 4 4 59 CHAPTER 3 STOICHIOMETRY77. When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O. In the sample of propane combusted, the moles of C and H are:1 mol CO 2 1 mol C mol C = 2.641 g CO2 ×  = 0.06001 mol C 44.01g CO 2 mol CO 21 mol H 2O 2 mol H mol H = 1.442 g H2O ×  = 0.1600 mol H 18.02 g H 2O mol H 2O mol H 0.1600  = 2.666 mol C 0.06001Multiplying this ratio by three gives the empirical formula of C3H8.78. This compound contains nitrogen, and one way to determine the amount of nitrogen in the compound is to calculate composition by mass percent. We assume that all of the carbon in 33.5 mg CO2 came from the 35.0 mg of compound and all of the hydrogen in 41.1 mg H2O came from the 35.0 mg of compound.1 mol CO 1 mol C 12.01g C 2 2 3 3.35 × 10 g CO2 ×   = 9.14 × 10 g C 44.01g CO 2 mol CO 2 mol C9.14 103 g C %C = × 100 = 26.1% C 3.50 102 g compound1 mol H O 2 mol H 1.008 g H 2 2 3 4.11 × 10 g H2O ×   = 4.60 × 10 g H 18.02 g H 2O mol H 2O mol H4.60 103 g H %H = × 100 = 13.1% H 3.50 102 g compoundThe mass percent of nitrogen is obtained by difference:%N = 100.0 - (26.1 + 13.1) = 60.8% NNow perform the empirical formula determination by first assuming 100.0 g of compound. Out of 100.0 g of compound, there are:1 mol C 1 mol H 26.1 g C × = 2.17 mol C; 13.1 g H × = 13.0 mol H 12.01g C 1.008 g H1 mol N 60.8 g N × = 4.34 mol N 14.01g N CHAPTER 3 STOICHIOMETRY 602.17 13.0 4.34 Dividing all mol values by 2.17 gives: = 1.00; = 5.99; = 2.00 2.17 2.17 2.17The empirical formula is CH6N2. 79. The combustion data allow determination of the amount of hydrogen in cumene. One way to determine the amount of carbon in cumene is to determine the mass percent of hydrogen in the compound from the data in the problem; then determine the mass percent of carbon by difference (100.0 - mass %H = mass %C).1g 2.016 g H 1000 mg 42.8 mg H2O ×   = 4.79 mg H 1000 mg 18.02 g H 2O g4.79 mg H %H = × 100 = 10.1% H; %C = 100.0 - 10.1 = 89.9% C 47.6 mg cumeneNow solve this empirical formula problem. Out of 100.0 g cumene, we have:1 mol C 1 mol H 89.9 g C × = 7.49 mol C; 10.1 g H × = 10.0 mol H 12.01g C 1.008 g H10.0 4 = 1.34  , i.e., mol H to mol C are in a 4:3 ratio. Empirical formula = C3H4 7.49 3Empirical formula mass ≈ 3(12) + 4(1) = 40 g/molThe molecular formula is (C3H4)3 or C9H12 since the molar mass will be between 115 and 125 g/mol (molar mass ≈ 3 × 40 g/mol = 120 g/mol).80. First, we will determine composition by mass percent:1g 12.01g C 1000 mg 16.01 mg CO2 ×   = 4.369 mg C 1000 mg 44.01g CO 2 g 4.369 mg C %C = × 100 = 40.91% C 10.68 mg compound1g 2.016 g H 1000 mg 4.37 mg H2O ×   = 0.489 mg H 1000 mg 18.02 g H 2O g0.489 mg %H = × 100 = 4.58% H; %O = 100.00 - (40.91 + 4.58) = 54.51% O 10.68 mgSo, in 100.00 g of the compound, we have:1 mol C 1 mol H 40.91 g C × = 3.406 mol C; 4.58 g H × = 4.54 mol H 12.01g C 1.008 g H 61 CHAPTER 3 STOICHIOMETRY1 mol O 54.51 g O × = 3.407 mol O 16.00 g O 4.54 4 Dividing by the smallest number: = 1.33  ; the empirical formula is C3H4O3. 3.406 3The empirical formula mass of C3H4O3 is ≈ 3(12) + 4(1) + 3(16) = 88 g.176.1 Because = 2.0, the molecular formula is C6H8O6. 88Balancing Chemical Equations81. When balancing reactions, start with elements that appear in only one of the reactants and one of the products, then go on to balance the remaining elements. a. C6H12O6(s) + O2(g) → CO2(g) + H2O(g)Balance C atoms: C6H12O6 + O2 → 6 CO2 + H2OBalance H atoms: C6H12O6 + O2 → 6 CO2 + 6 H2OLastly, balance O atoms: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) b. Fe2S3(s) + HCl(g) → FeCl3(s) + H2S(g)Balance Fe atoms: Fe2S3 + HCl → 2 FeCl3 + H2SBalance S atoms: Fe2S3 + HCl → 2 FeCl3 + 3 H2SThere are 6 H and 6 Cl on right, so balance with 6 HCl on left:Fe2S3(s) + 6 HCl(g) → 2 FeCl3(s) + 3 H2S(g). c. CS2(l) + NH3(g) → H2S(g) + NH4SCN(s)C and S balanced; balance N:CS2 + 2 NH3 → H2S + NH4SCNH is also balanced. So: CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s)82. One of the most important parts to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. a. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) b. 3 Pb(NO3)2(aq) + 2 Na3PO4(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq) c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) CHAPTER 3 STOICHIOMETRY 62 d. Sr(OH)2(aq) + 2 HBr(aq) → 2H2O(l) + SrBr2(aq)83. a. 3 Ca(OH)2(aq) + 2 H3PO4(aq) → 6 H2O(l) + Ca3(PO4)2(s) b. Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l) c. 2 AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2 HNO3(aq)84. a. 2 KO2(s) + 2 H2O(l) → 2 KOH(aq) + O2(g) + H2O2(aq) or4 KO2(s) + 6 H2O(l) → 4 KOH(aq) + O2(g) + 4 H2O2(aq) b. Fe2O3(s) + 6 HNO3(aq) → 2 Fe(NO3)3(aq) + 3 H2O(l) c. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) d. PCl5(l) + 4 H2O(l) → H3PO4(aq) + 5 HCl(g) e. 2 CaO(s) + 5 C(s) → 2 CaC2(s) + CO2(g) f. 2 MoS2(s) + 7 O2(g) → 2 MoO3(s) + 4 SO2(g) g. FeCO3(s) + H2CO3(aq) → Fe(HCO3)2(aq)85. a. The formulas of the reactants and products are C6H6(l) + O2(g) → CO2(g) + H2O(g). To balance this combustion reaction, notice that all of the carbon in C 6H6 has to end up as carbon in CO2 and all of the hydrogen in C6H6 has to end up as hydrogen in H2O. To balance C and H, we need 6 CO2 molecules and 3 H2O molecules for every 1 molecule of C6H6. We do oxygen last. Because we have 15 oxygen atoms in 6 CO2 molecules and 3 H2O molecules, we need 15/2 O2 molecules in order to have 15 oxygen atoms on the reactant side.15 C6H6(l) + O2(g) → 6 CO2(g) + 3 H2O(g); Multiply by two to give whole numbers. 22 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) b. The formulas of the reactants and products are C4H10(g) + O2(g) → CO2(g) + H2O(g). 13 C4H10(g) + O2(g) → 4 CO2(g) + 5 H2O(g); Multiply by two to give whole numbers. 22 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) c. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(g) 3 d. 2 Fe(s) + O2(g) → Fe2O3(s); For whole numbers: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) 2 1 e. 2 FeO(s) + O2(g) → Fe2O3(s); For whole numbers, multiply by two. 2 63 CHAPTER 3 STOICHIOMETRY4 FeO(s) + O2(g) → 2 Fe2O3(s)86. a. 16 Cr(s) + 3 S8(s) → 8 Cr2S3(s) b. 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) c. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) d. 2 Eu(s) + 6 HF(g) → 2 EuF3(s) + 3 H2(g)87. a. SiO2(s) + C(s) → Si(s) + CO(g)Balance oxygen atoms: SiO2 + C → Si + 2 COBalance carbon atoms: SiO2(s) + 2 C(s) → Si(s) + 2 CO(g) b. SiCl4(l) + Mg(s) → Si(s) + MgCl2(s)Balance Cl atoms: SiCl4 + Mg → Si + 2 MgCl2Balance Mg atoms: SiCl4(l) + 2 Mg(s) → Si(s) + 2 MgCl2(s) c. Na2SiF6(s) + Na(s) → Si(s) + NaF(s)Balance F atoms: Na2SiF6 + Na → Si + 6 NaFBalance Na atoms: Na2SiF6(s) + 4 Na(s) → Si(s) + 6 NaF(s)88. CaSiO3(s) + 6 HF(aq) → CaF2(aq) + SiF4(g) + 3 H2O(l)Reaction Stoichiometry89. The stepwise method to solve stoichiometry problems is outlined in the text. Instead of calculating intermediate answers for each step, we will combine conversion factors into one calculation. This practice reduces round-off error and saves time.Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s)1 mol Fe 2 mol Al 26.98 g Al 15.0 g Fe × = 0.269 mol Fe; 0.269 mol Fe ×  = 7.26 g Al 55.85 g Fe 2 mol Fe mol Al1 mol Fe2O3 159.70 g Fe2O3 0.269 mol Fe ×  = 21.5 g Fe2O3 2 mol Fe mol Fe2O31 mol Al2O3 101.96 g Al2O3 0.269 mol Fe ×  = 13.7 g Al2O3 2 mol Fe mol Al2O3 CHAPTER 3 STOICHIOMETRY 6490. 10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s)1 mol KClO3 3 mol P4O10 283.88 g P4O10 52.9 g KClO3 ×   = 36.8 g P4O10 122.55 g KClO3 10 mol KClO3 mol P4O101000 g Al 1mol Al 3 mol NH ClO 117.49 g NH ClO 91. 1.000 kg Al ×   4 4  4 4 = 4355 g kg Al 26.98 g Al 3 mol Al mol NH4ClO492. a. Ba(OH)2  8H2O(s) + 2 NH4SCN(s) → Ba(SCN)2(s) + 10 H2O(l) + 2 NH3(g)1 mol Ba(OH)  8H O b. 6.5 g Ba(OH)  8H O × 2 2 = 0.0206 mol = 0.021 mol 2 2 315.4 g2 mol NH4SCN 76.13 g NH4SCN 0.021 mol Ba(OH)2  8H2O ×  1 mol Ba(OH)2  8H2O mol NH4SCN= 3.2 g NH4SCN 3.0 g NH  1000 g 1 mol NH  1mol C H O N 4 4 4 5 7 2 93. 1.0 × 10 kg waste ×      × 100 kg waste kg 18.04 g NH4 55 mol NH4113.12 g C H O N 5 7 2 4 + = 3.4 × 10 g tissue if all NH4 converted mol C5H7O2N+ Since only 95% of the NH4 ions react: mass of tissue = (0.95) (3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue75 g Ca (PO ) 1mol Ca (PO ) 94. 1.0 × 103 g phosphorite × 3 4 2  3 4 2  100 g phosphorite 310.18 g Ca3 (PO4 )21 mol P4 123.88 g P4 × = 150 g P4 2 mol Ca3 (PO4 )2 mol P41 mol C H O 1 mol C H O 102.09 g C H O 2 7 6 3 4 6 3 4 6 3 95. a. 1.00 × 10 g C7H6O3 ×   138.12 g C7 H 6O3 1 mol C7 H 6O3 1 mol C 4 H 6O3= 73.9 g C4H6O31 mol C H O 1 mol C H O 180.15 g C H O 2 7 6 3 9 8 4 9 8 4 b. 1.00 × 10 g C7H6O3 ×   138.12 g C7 H 6O3 1 mol C7 H 6O3 mol C9 H8O 4= 1.30 × 102 g aspirin96. 2 LiOH(s) + CO2(g) → Li2CO3(aq) + H2O(l) 65 CHAPTER 3 STOICHIOMETRYThe total volume of air exhaled each minute for the 7 astronauts is 7 × 20. = 140 L/min.1 mol LiOH 1 mol CO 44.01g CO 100 g air 25,000 g LiOH ×  2  2   23.95 g LiOH 2 mol LiOH mol CO2 4.0 g CO21 mL air 1 L 1min 1 hr    = 68 hr = 2.8 days 0.0010 g air 1000 mL 140 L air 60 minLimiting Reactants and Percent Yield97. The product formed in the reaction is NO2; the other species present in the product represent- tation is excess O2. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules react with 3 O2 molecules to form 6 NO2 molecules.6 NO(g) + 3 O2(g) → 6 NO2(g)For smallest whole numbers, the balanced reaction is:2 NO(g) + O2(g) → 2 NO2(g)98. In the following table, we have listed three rows of information. The Initial row is the number of molecules present initially, the Change row is the number of molecules that react to reach completion, and the Final row is the number of molecules present at completion. To determine the limiting reactant, let’s calculate how much of one reactant is necessary to react with the other.4 molecules NH 3 10 molecules O2 × = 8 molecules NH3 to react with all of the O2 5 molecules O 2Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all of the O2, O2 is limiting.4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)Initial 10 molecules 10 molecules 0 0 Change 8 molecules 10 molecules +8 molecules +12 molecules Final 2 molecules 0 8 molecules 12 moleculesThe total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 8 molecules NO + 12 molecules H2O = 22 molecules total.1 mol BaO 2 3 99. 1.50 g BaO2 × = 8.86 × 10 mol BaO2 169.3 g BaO 20.0272 g HCl 1 mol HCl 25.0 mL ×  = 1.87 ×102 mol HCl mL 36.46 g HCl CHAPTER 3 STOICHIOMETRY 66The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual ratio is: 1.87 102 mol HCl 3 = 2.11 8.86 10 mol BaO2Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2) is the limiting reagent. 1 mol H O 34.02 g H O 3 2 2 2 2 8.86 ×10 mol BaO2 ×  = 0.301 g H2O2 mol BaO2 mol H2O2The amount of HCl reacted is: 2 mol HCl 3 2 8.86 ×10 mol BaO2 × = 1.77 × 10 mol HCl mol BaO2 excess mol HCl = 1.87 × 102 mol  1.77 × 102 mol = 1.0 × 103 mol HCl 36.46 g HCl mass of excess HCl = 1.0 × 3 mol HCl × = 3.6 × 2 g HCl 10 mol HCl 10100. Ca3(PO4)2 + 3 H2SO4 → 3 CaSO4 + 2 H3PO41 mol Ca (PO ) 3 3 4 2 1.0 × 10 g Ca3(PO4)2 × = 3.2 mol Ca3(PO4)2 310.18 g Ca 3 (PO 4 ) 298 g H SO 1 mol H SO 3 2 4 2 4 1.0 × 10 g conc. H2SO4 ×  = 10. mol H2SO4 100 g conc. H2SO4 98.09 g H2SO4The required mole ratio from the balanced equation is 3 mol H2SO4 to 1 mol Ca3(PO4)2. The 10. mol H SO actual ratio is: 2 4 = 3.1 3.2 mol Ca 3 (PO 4 ) 2This is larger than the required mole ratio, so Ca3(PO4)2 (the denominator), is the limiting reagent.3 mol CaSO4 136.15 CaSO4 3.2 mol Ca3(PO4)2 ×  = 1300 g CaSO4 produced mol Ca 3 (PO4 )2 mol CaSO42 mol H 3PO 4 97.99 g H 3PO 4 3.2 mol Ca3(PO4)2 ×  = 630 g H3PO4 produced mol Ca 3 (PO 4 ) 2 mol H 3PO 4101. An alternative method to solve limiting reagent problems is to assume each reactant is limiting and calculate how much product could be produced from each reactant. The reactant that produces the smallest amount of product will run out first and is the limiting reagent. 67 CHAPTER 3 STOICHIOMETRY1 mol NH 2 mol HCN 6 3 5 5.00 × 10 g NH3 ×  = 2.94 × 10 mol HCN 17.03 g NH3 2 mol NH31 mol O 2 mol HCN 6 2 5 5.00 × 10 g O2 ×  = 1.04 × 10 mol HCN 32.00 g O 2 3 mol O 21 mol CH 2 mol HCN 6 4 5 5.00 × 10 g CH4 ×  = 3.12 × 10 mol HCN 16.04 g CH 4 2 mol CH 4O2 is limiting because it produces the smallest amount of HCN. Although more product could 5 be produced from NH3 and CH4, only enough O2 is present to produce 1.04 × 10 mol HCN. The mass of HCN produced is:27.03 g HCN 1.04 × 105 mol HCN × = 2.81 × 106 g HCN mol HCN1 mol O 6 mol H O 18.02 g H O 6 2 2 2 6 5.00 × 10 g O2 ×   = 5.63 × 10 g H2O 32.00 g O2 3 mol O2 1 mol H2O102. We will use the strategy utilized in the previous problem to solve this limiting reactant problem.If C3H6 is limiting:1 mol C3H6 2 mol C3H3N 53.06 g C3H3N 15.0 g C3H6 ×   = 18.9 g C3H3N 42.08 g C3H6 2 mol C3H6 mol C3H3NIf NH3 is limiting:1 mol NH3 2 mol C3H3N 53.06 g C3H3N 5.00 g NH3 ×   = 15.6 g C3H3N 17.03 g NH3 2 mol NH3 mol C3H3NIf O2 is limiting:1 mol O2 2 mol C3H3N 53.06 g C3H3N 10.0 g O2 ×   = 11.1 g C3H3N 32.00 g O2 3 mol O2 mol C3H3NO2 produces the smallest amount of product, thus O2 is limiting and 11.1 g C3H3N can be produced.103. C7H6O3 + C4H6O3 → C9H8O4 + HC2H3O2 1mol C H O 7 6 3 2 1.50 g C7H6O3 × = 1.09 × 10 mol C7H6O3 138.12 g C7 H 6O31mol C H O 4 6 3 2 2.00 g C4H6O3 × = 1.96 × 10 mol C4H6O3 102.09 g C 4 H 6O3 CHAPTER 3 STOICHIOMETRY 68C7H6O3 is the limiting reagent because the actual moles of C7H6O3 present are below the required 1:1 mol ratio. The theoretical yield of aspirin is:1 mol C H O 180.15 g C H O -2 9 8 4 9 8 4 1.09 × 10 mol C7H6O3 ×  = 1.96 g C9H8O4 mol C7 H 6O3 mol C9 H8O 4 1.50 g % yield = × 100 = 76.5% 1.96 g1 mol C6H5Cl 104. a. 1142 g C6H5Cl × = 10.1 mol C6H5Cl 112.55 g C6H5Cl1mol C2HOCl3 485 g C2HOCl3 × = 3.29 mol C2HOCl3 147.38 g C2HOCl32 mol C H Cl From the balanced equation, the required mole ratio is 6 5 = 2. The actual 1 mol C2HOCl310.1 mol C H Cl mole ratio present is 6 5 = 3.07. The actual mole ratio is greater than 3.29 mol C2HOCl3 the required mole ratio, so the denominator of actual mole ratio (C2HOCl3) is limiting.1 mol C14H9Cl5 354.46 g C14H9Cl5 3.29 mol C2HOCl3 ×  = 1170 g C14H9Cl5 (DDT) mol C2HOCl3 mol C14H9Cl5 b. C2HOCl3 is limiting and C6H5Cl is in excess.2 mol C6H5Cl 112.55 g C6H5Cl c. 3.29 mol C2HOCl3 ×  = 741 g C6H5Cl reacted mol C2HOCl3 mol C6H5Cl1142 g  741 g = 401 g C6H5Cl in excess200.0 g DDT d. % yield = × 100 = 17.1% 1170 g DDT1000 kg 1000 g 1 mol Cu 3FeS3 3 mol Cu 63.55 g 105. 2.50 metric tons Cu3FeS3 ×     metric ton kg 342.71g 1 mol Cu 3FeS3 mol Cu= 1.39 × 106 g Cu (theoretical) 86.3 g Cu (actual) 1.39 × 106 g Cu (theoretical) × = 1.20 × 106 g Cu = 1.20 × 103 kg Cu 100. g Cu (theoretical) = 1.20 metric tons Cu (actual)106. P4(s) + 6 F2(g) → 4 PF3(g); The theoretical yield of PF3 is: 69 CHAPTER 3 STOICHIOMETRY100.0 g PF3 (theoretical) 120. g PF3 (actual) × = 154 g PF3 (theoretical) 78.1g PF3 (actual)1 molPF3 6 mol F2 38.00 g F2 154 g PF3 ×   = 99.8 g F2 87.97 g PF3 4 mol PF3 mol F299.8 g F2 are needed to produce an actual PF3 yield of 78.1%.Additional Exercises0.368 g XeFn 1 mol XeF 107. molar mass XeFn = 9.031020 molecules XeF  n = 245 g/mol n 6.022 1023 molecules245 g = 131.3 g + n(19.00 g), n = 5.98; formula = XeF6108. In one hour, the 1000. kg of wet cereal produced contains 580 kg H2O and 420 kg of cereal. We want the final product to contain 20.% H2O. Let x = mass of H2O in final product. x = 0.20, x = 84 + 0.20 x, x = 105 ≈ 110 kg H O 420  x 2The amount of water to be removed is 580 - 110 = 470 kg/hr.109. 2 H2(g) + O2(g) → 2 H2O(g)1molecule O 2 a. 50 molecules H2 × = 25 molecules O2 2 molecules H 2 Stoichiometric mixture. Neither is limiting.1molecule O 2 b. 100 molecules H2 × = 50 molecules O2; 2 molecules H 2O2 is limiting since only 40 molecules O2 are present. c. From b, 50 molecules of O2 will react completely with 100 molecules of H2. We have 100 molecules (an excess) of O2. So, H2 is limiting.1 mol O 2 d. 0.50 mol H2 × = 0.25 mol O2; H2 is limiting because 0.75 mol O2 are present. 2 mol H 21 mol O 2 e. 0.80 mol H2 × = 0.40 mol O2; H2 is limiting because 0.75 mol O2 are present. 2 mol H 2 CHAPTER 3 STOICHIOMETRY 701 mol H 2 1mol O 2 f. 1.0 g H2 ×  = 0.25 mol O2 2.016 g H 2 2 mol H 2 Stoichiometric mixture, neither is limiting.1 mol H 2 1mol O 2 32.00 g O 2 g. 5.00 g H2 ×   = 39.7 g O2 2.016 g H 2 2 mol H 2 mol O 2H2 is limiting because 56.00 g O2 are present.0.262 g C7H5BiO4 1mol C7H5BiO4 1 mol Bi 209.0 g Bi 110. 2 tablets ×    tablet 362.11g C7H5BiO4 1 mol C7H5BiO4 mol Bi = 0.302 g Bi consumed 111. Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; Because 104.14/13.02 = 7.998 ≈ 8, the molecular formula for styrene is (CH)8 = C8H8.23 1mol C8H8 8 mol H 6.002 10 atoms H 22 2.00 g C8H8 ×   = 9.25 × 10 atoms H 104.14 g C8H8 mol C8H8 mol H12.01mg C 11.46 mg 112. 41.98 mg CO2 × = 11.46 mg C; %C = × 100 = 57.85% C 44.01 mg CO 2 19.81 mg 2.016 mg H 0.772 mg 6.45 mg H2O × = 0.722 mg H; %H = × 100 = 3.64% H 18.02 mg H 2O 19.81mg%O = 100.00 - (57.85 + 3.64) = 38.51% OOut of 100.00 g terephthalic acid, there are:1 mol C 1 mol H 57.85 g C × = 4.817 mol C; 3.64 g H × = 3.61 mol H 12.01g C 1.008 g H1 mol O 38.51 g O × = 2.407 mol O 16.00 g O4.817 3.61 2.407 = 2.001; = 1.50; = 1.000 2.407 2.407 2.407The C:H:O mole ratio is 2:1.5:1 or 4:3:2. Empirical formula: C4H3O2Mass of C4H3O2 ≈ 4(12) + 3(1) + 2(16) = 8341.5 g 166 Molar mass = = 166 g/mol; = 2; Molecular formula: C8H6O4 0.250 mol 831 mol H 1 mol C 113. 17.3 g H × = 17.2 mol H; 82.7 g C × = 6.89 mol C 1.008 g H 12.01g C 71 CHAPTER 3 STOICHIOMETRY17.2 = 2.50; The empirical formula is C2H5. 6.89 The empirical formula mass is ~29 g, so two times the empirical formula would put the compound in the correct range of the molar mass. Molecular formula = (C2H5)2 = C4H101 molecule C H 1 mol C H 23 4 10 4 10 2 2.59 × 10 atoms H ×  = 4.30 × 10 mol C4H10 10 atoms H 6.022 1023 molecules58.12 g 2 4.30 × 10 mol C4H10 × = 2.50 g C4H10 mol C 4 H10114. Assuming 100.00 g E3H8: 1 mol H 3 mol E mol E = 8.73 g H ×  = 3.25 mol E 1.008 g H 8 mol H x g E 91.27 g E  , x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 amu 1 mol E 3.25 mol E115. Mass of H2O = 0.755 g CuSO4  xH2O - 0.483 g CuSO4 = 0.272 g H2O1 mol CuSO 4 0.483 g CuSO4 × = 0.00303 mol CuSO4 159.62 g CuSO 41 mol H2O 0.272 g H2O × = 0.0151 mol H2O 18.02 g H2O0.0151mol H2O 4.98 mol H 2O = ; Compound formula = CuSO4  5 H2O, x = 5 0.00303 g CuSO4 1 mol CuSO 4116. a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer:1 mol C3H 3 N 53.06 g C3H 3 N 8.80 g N ×  = 33.3 g C3H3N 14.01g N 1 mol C3H 3 N33.3 g C3H 3 N % C H N = = 33.3% C H N 3 3 100.00 g polymer 3 3Only butadiene in the polymer reacts with Br2:1 mol Br2 1mol C4H6 54.09 g C4H6 0.605 g Br2 ×   = 0.205 g C4H6 159.80 g Br2 mol Br2 mol C4H6 0.205 g % C H = × 100 = 17.1% C H 4 6 1.20 g 4 6 CHAPTER 3 STOICHIOMETRY 72 b. If we have 100.0 g of polymer:1 mol C H N 33.3 g C H N × 3 3 = 0.628 mol C H N 3 3 53.06 g 3 31 mol C 4 H 6 17.1 g C4H6 × = 0.316 mol C4H6 54.09 g C 4 H 61mol C8H8 49.6 g C8H8 × = 0.476 mol C8H8 104.14 g C8H8 0.628 0.316 0.476 Dividing by 0.316: = 1.99; = 1.00; = 1.51 0.316 0.316 0.316 This is close to a mol ratio of 4:2:3. Thus, there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in the polymer or (A4B2S3)n.1 mol CO 2 1 mol C 1 mol C 24 H 30 N 3O 376.51g 117. 1.20 g CO2 ×    44.01g mol CO 2 24 mol C mol C 24 H 30 N 3O= 0.428 g C24H30N3O0.428 g C 24 H 30 N 3O × 100 = 42.8% C H N O 1.00 g sample 24 30 3118. a. CH4(g) + 4 S(s) → CS2(l) + 2 H2S(g) or 2 CH4(g) + S8(s) → 2 CS2(l) + 4 H2S(g)1 mol CH 4 1 mol S b. 120. g CH4 × = 7.48 mol CH4; 120. g S × = 3.74 mol S 16.04 g CH 4 32.07 g SThe required S to CH4 mole ratio is 4:1. The actual S to CH4 mol ratio is: 3.74 mol S = 0.500 7.48 mol CH 4 This is well below the required ratio so sulfur is the limiting reagent.1 mol CS2 76.15 g CS2 The theoretical yield of CS2 is: 3.74 mol S ×  = 71.2 g CS2 4 mol S mol CS2The same amount of CS2 would be produced using the balanced equation with S8.1 mol 1 mol 119. 126 g B H × = 2.00 mol B H ; 192 g O × = 6.00 mol O 5 9 63.12 g 5 9 2 32.00 g 2 mol O 6.00 2 (actual) = = 3.00 mol B5H9 2.00The required mol O2 to mol B5H9 ratio is 12/2 = 6. The actual mole ratio is less than the required mole ratio, thus the numerator (O2) is limiting. 73 CHAPTER 3 STOICHIOMETRY9 mol H2O 18.02 g H2 O 6.00 mol O2 ×  = 81.1 g H2O 12 mol O2 mol H2O1 mol 120. 25.0 g Ag O × = 0.108 mol Ag O 2 231.8 g 21 mol 50.0 g C H N SO × = 0.200 mol C H N SO 10 10 4 2 250.29 g 10 10 4 2 mol C H N SO 0.200 10 10 4 2 (actual) = = 1.85 mol Ag2O 0.108The actual mole ratio is less than the required mole ratio (2), so C10H10N4SO2 is limiting.2 mol AgC10H9N4SO2 357.18 g 0.200 mol C10H10N4SO2 ×  2 mol C10H10 N4SO2 mol AgC10H9N4SO2= 71.4 g AgC10H9N4SO2 produced1 mol Fe 1 mol Fe2O3 159.70 g Fe2O3 121. 453 g Fe ×   = 648 g Fe2O3 55.85 g Fe 2 mol Fe mol Fe2O3648g Fe2O3 mass %Fe O = × 100 = 86.2% 2 3 752 g ore 65.38g Zn 122. a. Mass of Zn in alloy = 0.0985 g ZnCl2 × = 0.0473 g Zn 136.28g ZnCl2 0.0473g Zn %Zn = × 100 = 9.34% Zn; %Cu = 100.00  9.34 = 90.66% Cu 0.5065g brass b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted copper could be measured.123. Assuming one mol of vitamin A (286.4 g vitamin A):0.8386 g C 1 mol C mol C = 286.4 g vitamin A ×  = 20.00 mol C g vitamin A 12.01g C0.1056 g H 1 mol H mol H = 286.4 g vitamin A ×  = 30.00 mol H g vitamin A 1.008 g HSince one mol of vitamin A contains 20 mol C and 30 mol H, the molecular formula of vitamin A is C20H30E. To determine E, let’s calculate the molar mass of E.286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/molFrom the periodic table, E = oxygen and the molecular formula of vitamin A is C20H30O. CHAPTER 3 STOICHIOMETRY 74Challenge Problems atoms 85Rb 124. = 2.591; Assuming 100 atoms, let x = number of 85Rb atoms atoms 87Rb and 100  x = number of 87Rb atoms. x 259.1 = 2.591, x = 259.1  2.591 x, x = = 72.15% 85Rb 100  x 3.591 85.4678  61.26 0.7215 (84.9117) + 0.2785 (A) = 85.4678, A = = 86.92 amu 0.2785 = atomic mass of 87Rb125. First, we will determine composition in mass percent. We assume all the carbon in the 0.213 g CO2 came from 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2O came from the 0.157 g of the compound.12.01 g C 0.0581 g C 0.213 g CO2 × = 0.0581 g C; %C = × 100 = 37.0% C 44.01g CO2 0.1571g compound 2.016 g H 3 -3 3.4710 g 0.0310 g H2O × = 3.47 × 10 g H; %H = = 2.21% H 18.02 g H 2O 0.157 gWe get %N from the second experiment:14.01 g N 2 0.0230 g NH3 × = 1.89 × 10 g N 17.03 g NH31.89102 g %N = × 100 = 18.3% N 0.103 gThe mass percent of oxygen is obtained by difference:%O = 100.00 - (37.0 + 2.21 + 18.3) = 42.5%So out of 100.00 g of compound, there are:1 mol C 1 mol H 37.0 g C × = 3.08 mol C; 2.21 g H × = 2.19 mol H 12.01g C 1.008 g H 1mol N 1 mol O 18.3 g N × = 1.31 mol N; 42.5 g O × = 2.66 mol O 14.01g N 16.00 g OThe last, and often the hardest part, is to find simple whole number ratios. Divide all mole values by the smallest number: 75 CHAPTER 3 STOICHIOMETRY3.08 2.19 1.31 2.66 = 2.35; = 1.67; = 1.00; = 2.03 1.31 1.31 1.31 1.31Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6.1000g HNO 1mol HNO 6 3 3 7 126. 1.0 × 10 kg HNO3 ×  = 1.6 × 10 mol HNO3 kg HNO3 63.02 g HNO3We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all 3 equations.2 mol HNO 2 mol NO 4 mol NO 16 mol HNO 3  2   3 3 mol NO2 2 mol NO 4 mol NH3 24 mol NH3Thus, we can produce 16 mol HNO3 for every 24 mol NH3 we begin with: 24 mol NH 17.03 g NH 7 3 3 8 5 1.6 × 10 mol HNO3 ×  = 4.1 × 10 g or 4.1 × 10 kg 16 mol HNO3 mol NH3 This is an oversimplified answer. In practice, the NO produced in the third step is recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1:1, i.e., 1 mol of NH3 produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer 5 of 2.7 × 10 kg NH3 reacted.127. The two relevant equations are:4 FeO(s) + O2(g) → 2 Fe2O3(s) and 4 Fe3O4(s) + O2(g) → 6 Fe2O3(s)Let x = mass FeO, so 5.430 - x = mass Fe3O4. The mol of each are: x 5.430  x moles FeO = and moles Fe3O4 = 71.85 231.55Thus, moles Fe2O3 is: x 2 moles Fe2O3   5.430  x 6 moles Fe2O3          71.85 4 moles FeO   231.35 4 moles Fe3O4  and mass Fe2O3 is: x 2   5.430  x 6   159.70 g/mol         = 5.779 g  71.85 4   231.35 4   2.10 g Solving: x = 2.10 g; Thus, the mixture is × 100 = 38.7% FeO by mass. 5.430 g128. 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) CHAPTER 3 STOICHIOMETRY 7630.07 g/mol 44.09 g/molLet x = mass C2H6, so 9.780  x = mass C3H8. Use the balanced reaction to set up an equation for the mol of O2 required. x 7 9.780  x 5    = 1.120 mol O2 30.07 2 44.09 1 3.7 g Solving: x = 3.7 g C H ; × 100 = 38% C H by mass 2 6 9.780 g 2 6129. The two relevant equations are:Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) and Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)Let x = mass Mg, so 10.00  x = mass Zn. From the balanced equations, moles H2 = moles Zn + moles Mg.1 mol H2 mol H2 = 0.5171 g H2 × = 0.2565 mol H2 2.016 g H2 x 10.00  x Thus, 0.2565 = + ; Solving, x = 4.008 g Mg. 24.31 65.384.008 g × 100 = 40.08% Mg 10.00 g 130. Let M = unknown element mass M 2.077 mass %M =  100  × 100 = 56.01% M total mass compound 3.708100.00 – 56.01 = 43.99% OAssuming 100.00 g compound:1 mol O 43.99 g O × = 2.749 mol O 16.00 g O 56.01g M If MO is the formula of the oxide, then M has a molar mass = 2.749 mol M = 20.37 g/mol. This is too low for the molar mass. We must have fewer moles of M than mol O present in the formula. Some possibilities are MO2, M2O3, MO3, etc. It is a guessing game as to which to try. Let’s assume an MO2 formula. Then the molar mass of M is:56.01g M 1 mol M = 40.75 g/mol 2.749 mol O  2 mol O 77 CHAPTER 3 STOICHIOMETRYThis is close to calcium but calcium forms an oxide having the CaO formula, not CaO2.If MO3 is assumed to be the formula then the molar mass of M calculates to be 61.12 g/mol which is too large. Therefore, the mol O to mol M ratio must be between 2 and 3. Some reasonable possibilities are 2.25, 2.33, 2.5, 2.67, 2.75 (these are reasonable since they will lead to whole number formulas). Trying a mol O to mol M ratio of 2.5 to 1 gives a molar mass of: 56.01g M 1mol M = 50.94 g/mol. 2.749 mol O  2.5 mol OThis is the molar mass of vanadium and V2O5 is a reasonable formula for an oxide of vanadium. The other choices for the O:M mol ratios between 2 and 3 do not give as reasonable results. Therefore, M is vanadium and the formula is V2O5.131. We know water is a product, so one of the elements in the compound is hydrogen.XaHb + O2 → H2O + ? 2 To balance the H atoms, the mole ratio between XaHb : H2O = . b 1.39 g 1.21g mol compound = = 0.0224 mol; mol H O = = 0.0671 mol 62.09 g / mol 2 18.02 g / mol2 0.0224  and b = 6; XaH6 has a molar mass of 62.09 g/mol. b 0.067162.09 = a × molar mass of X + 6 × 1.008, a × molar mass of X = 56.04Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), Li (a = 8). N fits the data best so N4H6 is the formula.132. The balanced equation is: 2 Sc(s) + 2x HCl(aq) → 2 ScClx(aq)+ x H2(g) 2 The mol ratio of Sc : H2 = . x 1 mol Sc moles Sc = 2.25 g Sc × = 0.0500 mol Sc 44.96 g Sc1 mol H2 mol H2 = 0.1502 g H2 × = 0.07450 mol H2 2.016 g H2 2 0.0500 = , x = 3; The formula is ScCl3. x 0.07450133. Total mass of copper used: CHAPTER 3 STOICHIOMETRY 78(8.0 cm 16.0 cm  0.060 cm) 8.96 g 10,000 boards ×  = 6.9 × 105 g Cu board cm3Amount of Cu removed = 0.80 × 6.9 × 105 g = 5.5 × 105 g Cu1mol Cu 1mol Cu(NH3 ) 4 Cl 2 202.59g Cu(NH3 ) 4 Cl2 5.5 × 105 g Cu ×   63.55 g Cu mol Cu mol Cu(NH3 ) 4 Cl 26 = 1.8 × 10 g Cu(NH3)4Cl21mol Cu 4 mol NH 17.03g NH 5 3 3 5 5.5 × 10 g Cu ×   = 5.9 × 10 g NH3 63.55 g Cu mol Cu mol NH3134. a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every 4 mol of C6H5O3N reacted. The actual yield is 3 moles of acetaminophen compared to a theoretical yield of 4 moles of acetaminophen. Solving for percent yield by mass (where M = molar mass acetaminophen): 3 mol  M % yield = × 100 = 75% 4 mol  M b. The product of the percent yields of the individual steps must equal the overall yield, 75%.(0.87) (0.98) (x) = 0.75, x = 0.88; Step III has a % yield = 88%.135. 10.00 g XCl2 + excess Cl2 → 12.55 g XCl4; 2.55 g Cl reacted with XCl2 to form XCl4. XCl4 contains 2.55 g Cl and 10.00 g XCl2. From mol ratios, 10.00 g XCl2 must also contain 2.55 gCl; mass X in XCl2 = 10.00  2.55 = 7.45 g X.1 mol Cl 1 mol XCl2 1 mol X 2.55 g Cl ×   = 3.60 × 102 mol X 35.45 g Cl 2 mol Cl mol XCl2So, 3.60 × 10 2 mol X has a mass equal to 7.45 g X. The molar mass of X is:7.45 g X = 207 g/mol X; Atomic mass = 207 amu so X is Pb. 3.60 102 mol X136. 4.000 g M2S3 → 3.723 g MO2There must be twice as many mol of MO2 as mol of M2S3 in order to balance M in the reaction. Setting up an equation for 2 mol MO2 = mol M2S3 where A = molar mass M: 4.000 g  3.723 g 8.000 3.723 2   ,   2 A  3(32.07)  A  2(16.00) 2 A  96.21 A  32.00 79 CHAPTER 3 STOICHIOMETRY8.000 A + 256.0 = 7.446 A + 358.2, 0.554 A = 102.2, A = 184 g/mol; atomic mass = 184 amu 137. Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2 anions. The simplest compound of the two elements is Al2O3. Similarly, we would expect the formula of any group 6A element with Al to be Al2X3. Assuming this, out of 100.00 g of compound there are 18.56 g Al and 81.44 g of the unknown element, X. Let’s use this information to determine the molar mass of X which will allow us to identify X from the periodic table.1 mol Al 3 mol X 18.56 g Al ×  = 1.032 mol X 26.98 g Al 2 mol Al81.44 g of X must contain 1.032 mol of X.81.44 g X The molar mass of X = = 78.91 g/mol X. 1.032 mol XFrom the periodic table, the unknown element is selenium and the formula is Al2Se3.138. NaCl(aq) + Ag+(aq) → AgCl(s); KCl(aq) + Ag+(aq) → AgCl(s) 1 mol AgCl 1 mol Cl  8.5904 g AgCl ×  = 5.991 × 102 mol Cl 143.4 g AgCl 1 mol AgClThe molar masses of NaCl and KCl are 58.44 and 74.55 g/mol, respectively. Let x = mass NaCl and y = mass KCl: x y 2  x + y = 4.000 g; 58.44 + 74.55 = 5.991 × 10 total mol Cl or 1.276 x + y = 4.466 Solving using simultaneous equations:1.276 x + y = 4.466  x  y = -4.000 0.276 x = 0.466, x = 1.69 g NaCl and y = 2.31 g KCl1.69 g %NaCl = × 100 = 42.3% NaCl; %KCl = 57.7% 4.00 g139. The balanced equations are:4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) and 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g)Let 4x = number of mol of NO formed, and let 4y = number of mol of NO2 formed. Then:4x NH3 + 5x O2 → 4x NO + 6x H2O and 4y NH3 + 7y O2 → 4y NO2 + 6y H2OAll the NH3 reacted, so 4x + 4y = 2.00. 10.00  6.75 = 3.25 mol O2 reacted, so 5x + 7y CHAPTER 3 STOICHIOMETRY 80= 3.25.Solving by the method of simultaneous equations:20 x + 28 y = 13.0  20 x  20 y = 10.0 8 y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12 mol NO = 4x = 4 × 0.12 = 0.48 mol NO formed140. CxHyOz + oxygen → x CO2 + y/2 H2O 1 mol C 1 mol CO2 12.01g C 2.20 g CO2    mass %C in aspirin = 44.01g CO2 mol CO2 mol C = 60.0% C 1.00 g aspirin2 mol H 1 mol H2O 1.008 g H 0.400 g H2O    mass %H in aspirin = 18.02 g H2O mol H2O mol H = 4.48% H 1.00 g aspirin mass %O = 100.00  (60.0 + 4.48) = 35.5% OAssuming 100.00 g aspirin:1 mol C 1 mol H 60.0 g C × = 5.00 mol C; 4.48 g H × = 4.44 mol H 12.01g C 1.008 g H1 mol O 35.5 g O × = 2.22 mol O 16.00 g O 5.00 4.44 Dividing by the smallest number: = 2.25; = 2.00 2.22 2.22Empirical formula = (C2.25 H2.00O)4 = C9H8O4Empirical mass 9(12) + 8(1) + 4(16) = 180 g/mol; This is in the 170 – 190 g/mol range so the molecular formula is also C9H8O4.Balance the aspirin synthesis reaction to determine the formula for salicylic acid.CaHbOc + C4H6O3 → C9H8O4 + C2H4O2, CaHbOc = salicylic acid = C7H6O3Integrative Problems1 mol Fe 6.022 1023 atoms Fe 141. a. 1.05 × 10 20 g Fe ×  = 113 atoms Fe 55.85 g Fe mol Fe 81 CHAPTER 3 STOICHIOMETRY b. The total number of platinum atoms is 14 × 20 = 280 atoms (exact number). The mass of these atoms is:1 mol Pt 195.1g Pt 280 atoms Pt ×  = 9.071 × 10 20 g Pt 6.022 1023 atoms Pt mol Pt1 mol Ru 6.022 1023 atoms Ru c. 9.071 × 10 20 g Ru ×  = 540.3 = 540 atoms Ru 101.1g Ru mol Ru142. Assuming 100.00 g of tetrodotoxin:1 mol C 1mol N 41.38 g C × = 3.445 mol C; 13.16 g N × = 0.9393 mol N 12.01g C 14.01g N1 mol H 1 mol O 5.37 g H × = 5.33 mol H; 40.09 g O × = 2.506 mol O 1.008 g H 16.00 g ODivide by the smallest number:3.445 5.33 2.506 = 3.668; = 5.67; = 2.668 0.9393 0.9393 0.9393To get whole numbers for each element, multiply through by 3. empirical formula = (C3.668H5.67NO2.668)3 = C11H17N3O8The mass of the empirical formula is 319.3 g/mol.1.59 1021 g molar mass tetrodotoxin = 1 mol = 319 g/mol 3 molecules  6.022 1023 molecules Because the empirical mass and molar mass are the same, the molecular formula is the same as the empirical formula, C11H17N3O8.1kg 10. μg 1106 g 1 mol 6.022 1023 molecules 165 lb ×     2.2046 lb kg μg 319.3 g 1 mol18 = 1.4 × 10 molecules tetrodotoxin is the LD50 dosage0.105 g 1 mol 143. molar mass X2 = 8.92 1020 molecules  = 70.9 g/mol 6.022 1023 molecules The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine. CHAPTER 3 STOICHIOMETRY 82Assuming 100.00 g of MX3 compound:1 mol 54.47 g Cl × = 1.537 mol Cl 35.45 g1mol M 1.537 mol Cl × = 0.5123 mol M 3 mol Cl45.53 g M molar mass M = = 88.87 g/mol M 0.5123 mol MM is the element yttrium (Y) and the name of YCl3 is yttrium(III) chloride.The balanced equation is: 2 Y + 3 Cl2 → 2 YCl3Assuming Cl2 is limiting:1 mol Cl2 2 mol YCl3 195.26 g YCl3 1.00 g Cl2 ×   = 1.84 g YCl3 70.90 g Cl2 3 mol Cl2 1 mol YCl3 Assuming Y is limiting:1 mol Y 2 mol YCl3 195.26 g YCl3 1.00 g Y ×   = 2.20 g YCl3 88.91g Y 2 mol Y 1mol YCl3Cl2 is the limiting reagent and the theoretical yield is 1.84 g YCl3.144. 2 As + 4 AsI3 → 3 As2I4Volume of As cube = (3.00 cm)3 = 27.0 cm35.72 g As 1 mol As 27.0 cm3 ×  = 2.06 mol As cm3 74.92 g As1 mol AsI 24 3 1.01 × 10 molecules AsI3 × 23 = 1.68 mol AsI3 6.022 10 molecules AsI3From the balanced equation, we need twice the number of moles of AsI3 as As to react. Because the mole of AsI3 present are less than the mole of As present, AsI3 is limiting.3 mol As2I4 657.44 g As2I4 1.68 mol AsI3 ×  = 828 g As2I4 4 mol AsI3 2 mol As2I4 actual yield 0.756 = , actual yield = 0.756 × 828 g = 626 g As I 828 g 2 4Marathon Problems 83 CHAPTER 3 STOICHIOMETRY145. To solve the limiting reagent problem, we must determine the formulas of all the compounds so we can get a balanced reaction. a. 40 million trillion = 40 × 106 × 1012 = 4.000 × 1019 (assuming 4 S.F.)1mol A 4.000 × 1019 molecules A × = 6.642 × 10-5 mol A 6.02211023 molecules A4.26 103 g A Molar mass of A = = 64.1 g/mol 6.642 105 mol AMass of carbon in one mol of A is:37.5 g C 64.1 g A × = 24.0 g carbon = 2 mol carbon in substance A 100.0 g AThe remainder of the molar mass (64.1 g - 24.0 g = 40.1 g) is due to the alkaline earth metal. From the periodic table, calcium has a molar mass of 40.08 g/mol. The formula of substance A is CaC2. b. 5.36 g H + 42.5 g O = 47.9 g; Substance B only contains H and O. Determining the empirical formula of B:1 mol H 5.32 5.36 g H × = 5.32 mol H; = 2.00 1.008 g H 2.661 mol O 2.66 42.5 g O × = 2.66 mol O; = 1.00 16.00 g O 2.66Empirical formula = H2O; The molecular formula of substance B could be H2O, H4O2, H6O3, etc. The most reasonable choice is water (H2O) for substance B. c. Substance C + O2 → CO2 + H2O; Substance C must contain carbon and hydrogen, and may contain oxygen. Determining the mass of carbon and hydrogen in substance C:1 mol CO2 1 mol C 12.01g C 33.8 g CO2 ×   = 9.22 g carbon 44.01g CO2 mol CO2 mol C1 mol H2O 2 mol H 1.008 g H 6.92 g H2O ×   = 0.774 g hydrogen 18.02 g H2O mol H2O mol H9.22 g carbon + 0.774 g hydrogen = 9.99 g; Substance C initially weighed 10.0 g, so there is no oxygen present in substance C. Determining the empirical formula for substance C: CHAPTER 3 STOICHIOMETRY 841 mol C 9.22 g × = 0.768 mol carbon 12.01g C1 mol H 0.774 g H × = 0.768 mol hydrogen 1.008 g H mol C/mol H = 1.00; The empirical formula is CH which has an empirical formula mass ≈ 13. The mass spectrum data indicates a molar mass of 26 g/mol, thus the molecular formula for substance C is C2H2. d. Substance D is Ca(OH)2. Now we can answer the question. The balanced equation is:CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq)1 mol CaC2 45.0 g CaC2 × = 0.702 mol CaC2 64.10 g CaC21 mol H2O 23.0 g H2O × = 1.28 mol H2O 18.02 g H2O mol H O 1.28 2  = 1.82 mol CaC2 0.702Because the actual mole ratio present is smaller than the required 2:1 mole ratio from the balanced equation, H2O is limiting.1 mol C2H2 26.04 g C2H2 1.28 mol H2O ×  = 16.7 g C2H2 = mass of product C 2mol H2O mol C2H2146. a. i. If the molar mass of A is greater than the molar mass of B, then we cannot determine the limiting reactant, because, while we have a smaller number of moles of A, we also need fewer moles of A (from the balanced reaction). ii. If the molar mass of B is greater than the molar mass of A, then B is the limiting reactant because we have a smaller number of moles of B and we need more B (from the balanced reaction). b. A + 5 B → 3 CO2 + 4 H2OTo conserve mass : 44.01 + 5(B) = 3(44.01) + 4(18.02); solving: B = 32.0 g/molBecause it is diatomic, the best choice for B is O2. c. We can solve this without mass percent data simply by balancing the equation: 85 CHAPTER 3 STOICHIOMETRYA + 5 O2 → 3 CO2 + 4 H2OA must be C3H8. This is also the empirical formula.3(12.01) Note: × 100 = 81.71%. So this checks. 3(12.01)  8(1.008)

Chapter 3 Stoichiometry 83

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