(13.034) = amu

CHAPTER STOICHIOMETRY Questions. Isotope Mass Abundance 1 C 1.000 0 amu 98.89% 1 C 1.0 amu 1.11% Average mass = 0.9889 ( 1.0000) + 0.0111(1.0) = 1.01 amu From the relative abundances, there would be 9889 atoms of 1 C and 111 atoms of 1 C in the 10,000 atom sample. The average mass of carbon is independent of the sample size; it will always be 1.01 amu. Total mass = 10,000 atoms 1.01 amu atom = 1.01 10 5 amu For e of carbon (6.01 10 atoms C), the average mass would still be 1.01 amu. The number of 1 C atoms would be 0.9889(6.01 10 ) = 5.955 10 atoms 1 C, and the number of 1 C atoms would be 0.0111(6.01 10 ) = 6.68 10 1 atoms 1 C. Total mass = 6.01 10 atoms 1.01 amu atom = 7. 10 amu Total mass in g = 6.01 10 atoms 1.01 amu atom 1g 6.01 10 amu = 1.01 g/mol By using the carbon-1 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.. Consider a sample of glucose, C 6 H 1 O 6. 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 mole of glucose contains 6 mol C, 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 (6.0 10 ) allows one to convert between moles of compound and number of molecules.

CHAPTER STOICHIOMETRY 5 5. Avogadro s number of dollars = 6.0 10 dollars mol dollars 6.0 10 dollars dollars mol dollars = 1 10 1 dollars/person 9 6 10 people 1 trillion = 1,000,000,000,000 = 1 10 1 ; each person would have 100 trillion dollars. 6. The molar mass is the mass of e of the compound. The empirical mass is the mass of 1 mole 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 must be multiplied by the same whole number used to convert the empirical formula to the molecular formula. For example, C 6 H 1 O 6 is the molecular formula for glucose, and CH O is the empirical formula. The whole-number multiplier is 6. This same factor of 6 is the multiplier used to equate the empirical mass (0 g/mol) of glucose to the molar mass (180 g/mol). 7. The mass percent of a compound is a constant no matter what amount of substance is present. Compounds always have constant composition. 8. A balanced equation starts with the correct formulas of the reactants and products. The coefficients necessary to balance the equation 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. 9. The theoretical yield is the stoichiometric amount of product that should form if the limiting reactant is completely consumed and the reaction has 100% yield. 0. A reactant is present in excess if there is more of that reactant present than is needed to combine with the limiting reactant for the process. By definition, the limiting reactant cannot be present in excess. An excess of any reactant does not affect the theoretical yield for a process; the theoretical yield is determined by the limiting reactant. 1. 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. a. Mass of CB produced = 1.00 10 molecules A B 6.0 10 A B molecules A B mol CB A B molar mass of mol CB CB b. Atoms of A produced = 1.00 10 molecules A B atoms A ecule A B

6 CHAPTER STOICHIOMETRY c. Mol of C reacted = 1.00 10 molecules A B 6.0 10 A B molecules A mol C A B actual mass d. Percent yield = 100; the theoretical mass of CB produced was theoretical mass calculated in part a. If the actual mass of CB produced is given, then the percent yield can be determined for the reaction using the percent yield equation.. One method is to determine the actual mole ratio of XY to Y present and compare this ratio to the required :e 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. Exercises When each reactant is assumed limiting and the amount of product is calculated, there are two possible answers (assuming two reactants). The correct answer (the amount of product that could be produced) is always the smaller number. Even though there is enough of the other reactant to form more product, once the small quantity is reached, the limiting reactant runs out, and the reaction cannot continue. Atomic Masses and the Mass Spectrometer. A = 0.010(0.97) + 0.10(05.975) + 0.10(06.9759) + 0.50(07.9766) A =.86 + 9.6 + 5.7 + 109.0 = 07. amu; from the periodic table, the element is Pb. B. A = 0.0800(5.9569) + 0.070(6.95176) + 0.780(7.9797) + 0.0550(8.9781) + 0.050(9.979) = 7.88 amu This is element Ti (titanium). 5. Let A = mass of 185 Re: 186.07 = 0.660(186.956) + 0.70(A), 186.07! 117.0 = 0.70(A) A = 69. 0.70 = 185 amu (A = 18.95 amu without rounding to proper significant figures.) 6. Abundance 8 Si = 100.00 (.70 +.09) = 9.1%; from the periodic table, the average atomic mass of Si is 8.09 amu.

CHAPTER STOICHIOMETRY 7 8.09 = 0.91(7.98) + 0.070(atomic mass 9 Si) + 0.009(9.97) Atomic mass 9 Si = 9.01 amu The mass of 9 Si is actually a little less than 9 amu. There are other isotopes of silicon that are considered when determining the 8.09 amu average atomic mass of Si listed in the atomic table. 7. Let x = % of 151 Eu and y = % of 15 Eu, then x + y = 100 and y = 100 x. 151.96 = x( 150.9196) + (100 x)(15.909) 100 15196 = (150.9196)x + 159.09 (15.909)x, 96 = (.001)x x = 8%; 8% 151 Eu and 100 8 = 5% 15 Eu 8. If silver is 51.8% 107 Ag, then the remainder is 109 Ag (8.18%). The average atomic mass is then: 51.8(106.905) + 8.18(A) 107.868 = 100 10786.8 = 550. + (8.18)A, A = 108.9 amu = atomic mass of 109 Ag 9. There are three peaks in the mass spectrum, each mass units apart. This is consistent with two isotopes differing in mass by two mass units. The peak at 157.8 corresponds to a Br molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 157.8/ or 78.9. This corresponds to 79 Br. The second isotope is 81 Br with mass equal to 161.8/ = 80.9. The peaks in the mass spectrum correspond to 79 Br, 79 Br 81 Br, and 81 Br in order of increasing mass. The intensities of the highest and lowest masses tell us the two isotopes are present in about equal abundance. The actual abundance is 50.68% 79 Br and 9.% 81 Br. 0. GaAs can be either 69 GaAs or 71 GaAs. The mass spectrum for GaAs will have peaks at 1 (= 69 + 75) and 16 (= 71 + 75) with intensities in the ratio of 60 : 0 or :. 1 16 Ga As can be 69 Ga As, 69 Ga 71 GaAs, or 71 Ga As. The mass spectrum will have peaks at 88, 90, and 9 with intensities in the ratio of 6 : 8 : 16 or 9 : 1 :. We get this ratio from the following probability table:

8 CHAPTER STOICHIOMETRY 69 Ga (0.60) 71 Ga (0.0) 69 Ga (0.60) 0.6 0. 71 Ga (0.0) 0. 0.16 Moles and Molar Masses 88 90 9 1. 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. 500. atoms Fe Fe 55.85 g Fe =.6 6.0 10 atoms Fe mol Fe 0 10 g Fe. 500.0 g Fe Fe 55.85 g Fe = 8.95 mol Fe 8.95 mol Fe 6.0 10 atoms Fe mol Fe = 5.91 10 atoms Fe. 1.00 carat 0.00 g C carat C 6.0 10 atoms C = 1.00 10 atoms C 1.01g C mol C. 5.0 10 1 atoms C C 6.0 10 atoms C = 8. 10 mol C 8. 10 mol C 1.01g C mol C = 0.10 g C 5. Al O : (6.98) + (16.00) = 101.96 g/mol Na AlF 6 : (.99) + 1(6.98) + 6(19.00) = 09.95 g/mol 6. HFC! 1a, CH FCF : (1.01) + (1.008) + (19.00) = 10.0 g/mol HCFC!1, CHClFCF : (1.01) + 1(1.008) + 1(5.5) + (19.00) = 16.8 g/mol 7. a. The formula is NH. 1.01 g/mol + (1.008 g/mol) = 17.0 g/mol

CHAPTER STOICHIOMETRY 9 b. The formula is N H. (1.01) + (1.008) =.05 g/mol c. (NH ) Cr O 7 : (1.01) + 8(1.008) + (5.00) + 7(16.00) = 5.08 g/mol 8. a. The formula is P O 6. (0.97 g/mol) + 6(16.00 g/mol) = 19.88 g/mol b. Ca (PO ) : (0.08) + (0.97) + 8(16.00) = 10.18 g/mol c. Na HPO : (.99) + 1(1.008) + 1(0.97) + (16.00) = 11.96 g/mol NH 9. a. 1.00 g NH 17.0 g NH b. 1.00 g N H N.05 g N H H = 0.0587 mol NH = 0.0 N H c. 1.00 g (NH ) Cr O 7 (NH) CrO 7 5.08 g (NH ) Cr O 7 =.97 10 mol (NH ) Cr O 7 50. a. 1.00 g P O 6 P O 6 19.88 g =.55 10 mol P O 6 b. 1.00 g Ca (PO ) 1 mol Ca (PO) =. 10 mol Ca (PO ) 10.18 g c. 1.00 g Na HPO Na HPO 11.96 g = 7.0 10 mol Na HPO 51. a. 5.00 mol NH 17.0 g NH mol NH = 85. g NH b. 5.00 mol N H.05 g N mol N H H = 160. g N H c. 5.00 mol (NH ) Cr O 7 5.08 g (NH ) Cr O (NH ) Cr O 7 7 = 160 g (NH ) Cr O 7 5. a. 5.00 mol P O 6 19.88 g P O 6 = 1.10 10 g P O 6 b. 5.00 mol Ca (PO ) 10.18 g mol Ca (PO) = 1.55 10 g Ca (PO ) c. 5.00 mol Na HPO 11.96 g mol Na HPO = 7.10 10 g Na HPO

50 CHAPTER STOICHIOMETRY 5. Chemical formulas give atom ratios as well as mole ratios. a. 5.00 mol NH N mol NH 1.01 g N = 70.1 g N mol N b. 5.00 mol N H mol N mol N H 1.01 g N = 10. g N mol N c. 5.00 mol (NH ) Cr O 7 mol N mol (NH ) Cr O 7 1.01 g N = 10. g N mol N 5. a. 5.00 mol P O 6 mol P mol P O 6 0.97 g P = 619 g P mol P b. 5.00 mol Ca (PO ) mol P mol Ca (PO ) 0.97 g P = 10. g P mol P c. 5.00 mol Na HPO P mol Na HPO 0.97 g P = 155 g P mol P 55. a. 1.00 g NH NH 6.0 10 molecules NH 17.0 g NH mol NH =.5 10 molecules NH b. 1.00 g N H c. 1.00 g (NH ) Cr O 7 NH 6.0 10 molecules NH.05 g NH mol NH = 1.88 10 molecules N H (NH) CrO 7 5.08 g (NH ) Cr O 7 6.0 10 formula units (NH) CrO7 =.9 10 1 formula units (NH ) Cr O 7 mol (NH ) Cr O 7 56. a. 1.00 g P O 6 PO 6 6.0 10 molecules =.7 10 ecules P O 6 19.88 g mol P O 6 b. 1.00 g Ca (PO ) Ca(PO) 6.0 10 10.18 g mol Ca formula units (PO ) = 1.9 10 1 formula units Ca (PO ) c. 1.00 g Na HPO Na HPO 6.0 10 formula units 11.96 g mol Na HPO =. 10 1 formula units Na HPO

CHAPTER STOICHIOMETRY 51 57. Using answers from Exercise 55: a..5 10 molecules NH 1atom N molecule NH =.5 10 atoms N b. 1.88 10 molecules N H atoms N molecule N H =.76 10 atoms N c..9 10 1 formula units (NH ) Cr O 7 58. Using answers from Exercise 56: atoms N formula unit (NH ) CrO 7 =.78 10 1 atoms N a..7 10 ecules P O 6 atoms P molecule P O 6 = 1.10 10 atoms P b. 1.9 10 1 formula units Ca (PO ) atoms P formula unit Ca (PO ) =.88 10 1 atoms P c.. 10 1 formula units Na HPO 1atom P formula unit Na HPO =. 10 1 atoms P 59. Molar mass of CCl F = 1.01 + (5.5) + (19.00) = 10.91 g/mol 5.56 mg CCl F 1g 1000 mg 10.91 g 6.0 10 molecules mol =.77 10 19 molecules CCl F 5.56 10 g CCl F CCl F 10.91 g mol Cl CCl F 5.5 g Cl mol Cl =.6 10 g =.6 mg Cl 60. The H O is part of the formula of bauxite (they are called waters of hydration). Combining elements together, the chemical formula for bauxite would be Al O 5 H. a. Molar mass = (6.98) + 5(16.00) + (1.008) = 17.99 g/mol b. 0.58 mol Al O H O mol Al mol Al O H O 6.98 g Al mol Al = 1 g Al c. 0.58 mol Al O H O mol Al mol Al O H O 6.0 10 atoms mol Al = 7.0 10 atoms Al

5 CHAPTER STOICHIOMETRY d..1 10 formula units Al O H O 1mol Al 6.0 10 O HO 17.99 g formula units mol = 80 g Al O H O 61. a. 150.0 g Fe O 159.70 g = 0.99 mol Fe O b. 10.0 mg NO 1g 1000 mg =.17 10 mol NO 6.01 g c. 1.5 10 16 molecules BF 6.0 10 molecules =.5 8 10 mol BF 6. a. 0.0 mg C 8 H 10 N O 1g 1000 mg = 1.0 10 mol C 8 H 10 N O 19.0 g b..7 10 ecules C H 5 OH c. 1.50 g CO.01 g =.1 6.0 10 10 mol CO molecules =.5 10 mol C H 5 OH 6. 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 C H 5 O N = (1.01) + 5(1.008) + (16.00) + 1.0l = 75.07 g/mol 5.00 g C H 5 O N C H5ON 75.07 g C H O N 6.0 5 10 molecules CH mol C H O N 5 1atom N =.01 10 atoms N molecule C H O N 5 5 O N b. Molar mass of Mg N = (.1) + (1.01) = 100.95 g/mol MgN 6.0 10 formula units MgN 5.00 g Mg N 100.95 g MgN mol MgN atoms N = 5.97 10 atoms N mol Mg N c. Molar mass of Ca(NO ) = 0.08 + (1.01) + 6(16.00) = 16.10 g/mol

CHAPTER STOICHIOMETRY 5 Ca(NO ) mol N 6.0 10 atoms N 5.00 g Ca(NO ) 16.10 g Ca(NO ) mol Ca(NO ) mol N =.67 10 atoms N d. Molar mass of N O = (1.01) + (16.00) = 9.0 g/mol 5.00 g N O N O mol N 6.0 10 atoms N 9.0 g NO mol NO mol N = 6.5 10 atoms N 6.. g C 6 H 6 5. 78.11 g 10 mol C 6 H 6 = 5. 10 mol C 6 H 6 6.0 10 molecules mol =.7 10 molecules C 6 H 6 Each molecule of C 6 H 6 contains 6 atoms C + 6 atoms H = 1 total atoms..7 10 molecules C 6 H 6 0. mol H O 0. mol H O 18.0 g mol 1.5 10 molecules H O 1 atoms total molecule =.0 g H O =.9 10 atoms total 6.0 10 molecules = 1.5 10 molecules H O mol atoms total molecule =.05 10 atoms total.71 10 molecules CO 6.0 10 molecules =.50 10 mol CO.50 10 mol CO.01 g mol = 1.98 g CO.71 10 molecules CO atoms total molecule CO = 8.1 10 atoms total.5 10 atoms total ecule 6 atoms total = 5.58 10 ecules CH OH 5.58 10 ecules CH OH 6.0 10 molecules = 9.7 10 mol CH OH 9.7 10 mol CH OH.0 g mol = 0.97 g CH OH

5 CHAPTER STOICHIOMETRY 65. a. (1.01) + (1.008) + (5.5) + (16.00) = 165.9 g/mol b. 500.0 g 165.9 g =.0 mol C H Cl O c..0 10 - mol 165.9 g mol =. g C H Cl O d. 5.0 g C H Cl O 165.9 g 6.0 10 molecules mol atoms Cl molecule = 5.5 10 atoms of chlorine e. 1.0 g Cl Cl 5.5 g C H Cl O 165.9 g C H Cl O = 1.6 g chloral hydrate mol Cl mol CHClO f. 500 molecules 6.0 10 molecules 165.9 g mol = 1.7 10 19 g 66. As we shall see in later chapters, the formula written as (CH ) N O tries to tell us something about how the atoms are attached to each other. For our purposes in this problem, we can write the formula as C H 6 N O. a. (1.01) + 6(1.008) + (1.01) + 1(16.00) = 7.09 g/mol 1g b. 50 mg =. 10 mol c. 0.050 mol 1000 mg 7.09 g 7.09 g mol =.7 g d. 1.0 mol C H 6 N O 6.0 10 molecules CH mol C H N O 6 6 N O 6 atoms of H molecule C H N 6 O =.6 10 atoms of hydrogen e. 1.0 10 6 7.09 g molecules = 1. 10 16 g 6.0 10 molecules mol 7.09 g f. ecule = 1.0 10 g 6.0 10 molecules mol Percent Composition 67. a. C H O : Molar mass = (1.01) + (1.008) + (16.00) = 6.0 +.0 +.00 = 7.06 g/mol

CHAPTER STOICHIOMETRY 55 Mass % C = Mass % H = 6.0 g C 7.06 g compound.0 g H 7.06 g compound 100 = 50.00% C 100 = 5.595% H Mass % O = 100.00 (50.00 + 5.595) =.1% O or: % O =.00 g 7.06 g 100 =.1% O b. C H 6 O : Molar mass = (1.01) + 6(1.008) + (16.00) = 8.0 + 6.08 +.00 = 86.09 g/mol Mass % C = 8.0 g 86.09 g 100 = 55.80% C; mass % H = Mass % O = 100.00 (55.80 + 7.05) = 7.18% O 6.08 g 86.09 g 100 = 7.05% H c. C H N: Molar mass = (1.01) + (1.008) + 1(1.01) = 6.0 +.0 + 1.01 = 5.06 g/mol Mass % C = 6.0 g 5.06 g 100 = 67.90% C; mass % H =.0 g 5.06 g 100 = 5.699% H Mass % N = 1.01 g 5.06 g 100 = 6.0% N or % N = 100.00! (67.90 + 5.699) = 6.0% N 68. In e of YBa Cu O 7, there are e of Y, moles of Ba, moles of Cu, and 7 moles of O. 88.91g Y 17. g Ba Molar mass = Y + mol Ba mol Y mol Ba Molar mass = 88.91 + 7.6 + 190.65 + 11.00 = 666. g/mol 6.55 g Cu + mol Cu 16.00 g O + 7 mol O mol Cu mol O Mass % Y = 88.91 g 666. g 100 = 1.5% Y; mass % Ba = 7.6 g 666. g 100 = 1.% Ba Mass % Cu = 190.65 g 666. g 100 = 8.6% Cu; mass % O = 11.0 g 666. g 100 = 16.81% O

56 CHAPTER STOICHIOMETRY 69. a. NO: Mass % N = b. NO : Mass % N = 1.01 g N 0.01 g NO 1.01 g N 6.01 g NO 100 = 6.68% N 100 = 0.5% N c. N O : Mass % N = (1.01) g N 9.0 g N O 100 = 0.5% N d. N O: Mass % N = (1.01) g N.0 g N O 100 = 6.65% N The order from lowest to highest mass percentage of nitrogen is: NO = N O < NO < N O. 70. a. C 8 H 10 N O : Molar mass = 8(1.01) + 10(1.008) + (1.0l) + (16.00) = 19.0 g/mol Mass % C = 8(1.01) g C 19.0 g C 8H10NO 100 = 96.08 19.0 100 = 9.7% C b. C 1 H O 11 : Molar mass = 1(1.01) + (1.008) + 11(16.00) =.0 g/mol Mass % C = 1(1.01) g C.0 g C 1HO11 100 =.10% C c. C H 5 OH: Molar mass = (1.01) + 6(1.008) + 1(16.00) = 6.07 g/mol Mass % C = (1.01) g C 6.07 g C H OH 5 100 = 5.1% C The order from lowest to highest mass percentage of carbon is: sucrose (C 1 H O 11 ) < caffeine (C 8 H 10 N O ) < ethanol (C H 5 OH) 71. There are 0.90 g Cu for every 100.000 g of fungal laccase. Assuming 100.00 g fungal laccase: Cu fungal laccase Mol fungal laccase = 0.90 g Cu = 1.5 10 mol 6.55 g Cu mol Cu x g fungallaccase mol fungal laccase 100.000 g =, x = molar mass = 6.5 10 g/mol 1.5 10 mol 7. There are 0.7 g Fe for every 100.000 g hemoglobin (Hb). Assuming 100.000 g hemoglobin: Fe Hb Mol Hb = 0.7 g Fe = 1.55 10 mol Hb 55.85 g Fe mol Fe x g Hb mol Hb = 100.000 g Hb 1.55 10 mol Hb, x = molar mass = 6.5 10 g/mol

CHAPTER STOICHIOMETRY 57 Empirical and Molecular Formulas 1.01g C 7. a. Molar mass of CH O = C mol C 1.008 g H + mol H mol H 16.00 g O + O mol O = 0.0 g/mol % C = 1.01g C 0.0 g CH O 100 = 9.99% C; % H =.016 g H 0.0 g CH O 100 = 6.71% H % O = 16.00 g O 0.0 g CH O 100 = 5.8% O or % O = 100.00 (9.99 + 6.71) = 5.0% b. Molar mass of C 6 H 1 O 6 = 6(1.01) + 1(1.008) + 6(16.00) = 180.16 g/mol % C = 76.06 g C 180.16 g C 6H1O6 100 = 0.00%; % H = 1.(1.008) g 180.16 g 100 = 6.71% % O = 100.00 (0.00 + 6.71) = 5.9% c. Molar mass of HC H O = (1.01) + (1.008) + (16.00) = 60.05 g/mol % C =.0 g 60.05 g 100 = 0.00%; % H =.0 g 60.05 g 100 = 6.71% % O = 100.00 (0.00 + 6.71) = 5.9% 7. All three compounds have the same empirical formula, CH O, 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. 75. a. The molecular formula is N O. The smallest whole number ratio of the atoms (the empirical formula) is NO. b. Molecular formula: C H 6 ; empirical formula: CH c. Molecular formula: P O 10 ; empirical formula: P O 5 d. Molecular formula: C 6 H 1 O 6 ; empirical formula: CH O 76. a. SNH: Empirical formula mass =.07 + 1.01 + 1.008 = 7.09 g 188.5 g 7.09 g =.000; so the molecular formula is (SNH) or S N H.

58 CHAPTER STOICHIOMETRY b. NPCl : Empirical formula mass = 1.01 + 0.97 + (5.5) = 115.88 g/mol 7.6 g 115.88 g =.0000; molecular formula is (NPCl ) or N P Cl 6. c. CoC O : 58.9 + (1.01) + (16.00) = 170.97 g/mol 1.9 g 170.97 g =.0000; molecular formula: Co C 8 O 8 d. SN:.07 + 1.01 = 6.08 g/mol; 77. Out of 100.00 g of compound, there are: 18. g 6.08 g =.000; molecular formula: S N 8.6 g C C 1.01g C =.050 mol C; 8.16 g H H 1.008 g H = 8.10 mol H % O = 100.00 8.6 8.16 =.0%;.0 g O Dividing each mole value by the smallest number: O 16.00 g O =.700 mol O.050.700 = 1.500; 8.10.700 =.00;.700.700 = 1.000 Because a whole number ratio is required, the C : H : O ratio is 1.5 : : 1 or : 6 :. So the empirical formula is C H 6 O. 78. Assuming 100.00 g of nylon-6: 6.68 g C C 1.01g C = 5.0 mol C; 1.8 g N N 1.01 g N = 0.887 mol N 9.80 g H H 1.008 g H = 9.7 mol H; 1.1 g O O 16.00 g O = 0.888 mol O Dividing each mole value by the smallest number: 5.0 0.887 = 6.000; 9.7 0.887 = 11.0 The empirical formula for nylon-6 is C 6 H 11 NO 79. Compound I: Mass O = 0.698 g Hg x O y 0.6018 g Hg = 0.080 g O

CHAPTER STOICHIOMETRY 59 0.6018 g Hg Hg 00.6 g Hg =.000 10 mol Hg 0.080 g O O 16.00 g O =.00 10 mol O The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO. Compound II: Mass Hg = 0.17 g Hg x O y 0.016 g O = 0.01 g Hg 0.01 g Hg Hg 00.6 g Hg =.00 10 mol Hg; 0.016 g O O 16.00 g O = 1.0 10 mol O The mole ratio between Hg and O is : 1, so the empirical formula is Hg O. 80. 1.11 g N N 1.01 g N = 8.001 10 H mol N; 0.161 g H 1.008 g H = 1.60 1 10 mol H 0.80 g C C 1.01g C =.00 10 O mol C; 0.60 g O 16.00 g O =.00 10 mol O Dividing all mole values by the smallest number: 8.001 10.00 10 =.00; 1.60 10.00 10 1 =.00;.00 10.00 10 = 1.00 The empirical formula is N H CO. 81. Out of 100.0 g, there are: 69.6 g S S.07 g S =.17 mol S; 0. g N N 1.01 g N =.17 mol N The empirical formula is SN because the mole values are in a 1 : e ratio. The empirical formula mass of SN is ~ 6 g. Because 18/6 =.0, the molecular formula is S N. 8. Assuming 100.0 g of compound: 6.7 g P P 0.97 g P = 0.86 mol P; 1.1 g N N 1.01 g N = 0.86 mol N 61. g Cl Cl 5.5 g Cl = 1.7 mol Cl 1.7 0.86 =.01; the empirical formula is PNCl.

60 CHAPTER STOICHIOMETRY The empirical formula mass is 1.0 + 1.0 + (5.5) = 116. Molar mass Empirical formula mass 580 = = 5.0; the molecular formula is (PNCl ) 5 = P 5 N 5 Cl 10. 116 8. Assuming 100.00 g of compound: 7.08 g C 6. g Cl C 1.01g C Cl 5.5 g Cl =.90 mol C; 6.59 g H = 1.07 mol Cl Dividing all mole values by 1.07 gives: H 1.008 g H = 6.5 mol H.90 1.07 =.999; 6.5 1.07 = 5.00; 1.07 1.07 = 1.000 The empirical formula is C H 5 Cl. The empirical formula mass is (1.01) + 5(1.008) + 1(5.5) = 76.5 g/mol. Molar mass Empirical formula mass = 15 =.00 ; the molecular formula is (C H 5 Cl) = C 6 H 10 Cl. 76.5 8. Assuming 100.00 g of compound (mass oxygen = 100.00 g! 1.9 g C!.7 g H = 55.1 g O): 1.9 g C C 1.01g C =.6 mol C;.7 g H H 1.008 g H =. mol H 55.1 g O O 16.00 g O =.6 mol O All are the same mole values, so the empirical formula is CHO. The empirical formula mass is 1.01 + 1.008 + 16.00 = 9.0 g/mol. Molar mass = 15.0 g 0.19 mol = 116 g/mol Molar mass Empirical mass 116 = =.00; molecular formula = (CHO) = C H O 9.0 85. When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon in CO and all the hydrogen in the compound ends up as hydrogen in H O. In the sample of propane combusted, the moles of C and H are:

CHAPTER STOICHIOMETRY 61 mol C =.61 g CO CO.01g CO C = 0.0600 C mol CO H O mol H mol H = 1. g H O = 0.1600 mol H 18.0 g HO mol HO mol H 0.1600 = =.666 mol C 0.06001 Multiplying this ratio by gives the empirical formula of C H 8. 86. 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 the carbon in.5 mg CO came from the 5.0 mg of compound and all the hydrogen in 1.1 mg H O came from the 5.0 mg of compound..5 10 g CO CO.01g CO C 1.01g C = 9.1 mol CO mol C 10 g C Mass % C = 9.1 10.50 10 g C g compound 100 = 6.1% C.11 10 g H O H O 18.0 g H O mol H 1.008 g H =.60 10 g H mol H O mol H Mass % H =.60 10.50 10 g H g compound 100 = 1.1% H The mass percent of nitrogen is obtained by difference: Mass % N = 100.0 (6.1 + 1.1) = 60.8% N Now perform the empirical formula determination by first assuming 100.0 g of compound. Out of 100.0 g of compound, there are: 6.1 g C C 1.01g C =.17 mol C; 1.1 g H H 1.008 g H = 1.0 mol H N 60.8 g N =. mol N 1.01 g N.17 Dividing all mole values by.17 gives:.17 = 1.00; 1.0.17 = 5.99;. =.00.17 The empirical formula is CH 6 N. 87. 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

6 CHAPTER STOICHIOMETRY the compound from the data in the problem; then determine the mass percent of carbon by difference (100.0 mass % H = mass % C)..8 mg H O 1g 1000 mg.016 g H 1000 mg =.79 mg H 18.0 g H O g Mass % H =.79 mg H 7.6 mg cumene 100 = 10.1% H; mass % C = 100.0 10.1 = 89.9% C Now solve the empirical formula problem. Out of 100.0 g cumene, we have: 89.9 g C C 1.01g C = 7.9 mol C; 10.1 g H H 1.008 g H = 10.0 mol H 10.0 7.9 = 1.. ; the mol H to mol C ratio is :. The empirical formula is C H. Empirical formula mass (1) + (1) = 0 g/mol. The molecular formula must be (C H ) or C 9 H 1 because the molar mass of this formula will be between 115 and 15 g/mol (molar mass 0 g/mol = 10 g/mol). 88. There are several ways to do this problem. We will determine composition by mass percent: 1g 1.01g C 1000 mg 16.01 mg CO =.69 mg C 1000 mg.01g CO g % C =.69 mg C 10.68 mg compound 100 = 0.91% C.7 mg H O 1g 1000 mg.016 g H 1000 mg = 0.89 mg H 18.0 g H O g % H = 0.89 mg 100 =.58% H; % O = 100.00 (0.91 +.58) = 5.51% O 10.68 mg So, in 100.00 g of the compound, we have: 0.91 g C C 1.01g C =.06 mol C;.58 g H H 1.008 g H =.5 mol H 5.51 g O O 16.00 g O Dividing by the smallest number: =.07 mol O.5.06 = 1.. ; the empirical formula is C H O.

CHAPTER STOICHIOMETRY 6 The empirical formula mass of C H O is (1) + (1) + (16) = 88 g. Because 176.1 =.0, the molecular formula is C 6 H 8 O 6. 88 Balancing Chemical Equations 89. When balancing reactions, start with elements that appear in only one of the reactants and one of the products, and then go on to balance the remaining elements. a. C 6 H 1 O 6 (s) + O (g) CO (g) + H O(g) Balance C atoms: C 6 H 1 O 6 + O 6 CO + H O Balance H atoms: C 6 H 1 O 6 + O 6 CO + 6 H O Lastly, balance O atoms: C 6 H 1 O 6 (s) + 6 O (g) 6 CO (g) + 6 H O(g) b. Fe S (s) + HCl(g) FeCl (s) + H S(g) Balance Fe atoms: Fe S + HCl FeCl + H S Balance S atoms: Fe S + HCl FeCl + H S There are 6 H and 6 Cl on right, so balance with 6 HCl on left: Fe S (s) + 6 HCl(g) FeCl (s) + H S(g). c. CS (l) + NH (g) H S(g) + NH SCN(s) C and S balanced; balance N: CS + NH H S + NH SCN H is also balanced. CS (l) + NH (g) H S(g) + NH SCN(s) 90. An important part to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. a. C H 5 OH(l) + O (g) CO (g) + H O(g) b. Pb(NO ) (aq) + Na PO (aq) Pb (PO ) (s) + 6 NaNO (aq) c. Zn(s) + HCl(aq) ZnCl (aq) + H (g) d. Sr(OH) (aq) + HBr(aq) H O(l) + SrBr (aq) 91. H O (aq) MnO catalyst H O(l) + O (g)

6 CHAPTER STOICHIOMETRY 9. Fe O (s) + H (g) Fe(s) + H O(g) Fe O (s) + CO(g) Fe(s) + CO (g) 9. a. Ca(OH) (aq) + H PO (aq) 6 H O(l) + Ca (PO ) (s) b. Al(OH) (s) + HCl(aq) AlCl (aq) + H O(l) c. AgNO (aq) + H SO (aq) Ag SO (s) + HNO (aq) 9. a. KO (s) + H O(l) KOH(aq) + O (g) + H O (aq) or KO (s) + 6 H O(l) KOH(aq) + O (g) + H O (aq) b. Fe O (s) + 6 HNO (aq) Fe(NO ) (aq) + H O(l) c. NH (g) + 5 O (g) NO(g) + 6 H O(g) d. PCl 5 (l) + H O(l) H PO (aq) + 5 HCl(g) e. CaO(s) + 5 C(s) CaC (s) + CO (g) f. MoS (s) + 7 O (g) MoO (s) + SO (g) g. FeCO (s) + H CO (aq) Fe(HCO ) (aq) 95. a. The formulas of the reactants and products are C 6 H 6 (l) + O (g) CO (g) + H O(g). To balance this combustion reaction, notice that all of the carbon in C 6 H 6 has to end up as carbon in CO and all of the hydrogen in C 6 H 6 has to end up as hydrogen in H O. To balance C and H, we need 6 CO molecules and H O molecules for every ecule of C 6 H 6. We do oxygen last. Because we have 15 oxygen atoms in 6 CO molecules and H O molecules, we need 15/ O molecules in order to have 15 oxygen atoms on the reactant side. 15 C 6 H 6 (l) + O (g) 6 CO (g) + H O(g); multiply by two to give whole numbers. C 6 H 6 (l) + 15 O (g) 1 CO (g) + 6 H O(g) b. The formulas of the reactants and products are C H 10 (g) + O (g) CO (g) + H O(g). 1 C H 10 (g) + O (g) CO (g) + 5 H O(g); multiply by two to give whole numbers. C H 10 (g) + 1 O (g) 8 CO (g) + 10 H O(g) c. C 1 H O 11 (s) + 1 O (g) 1 CO (g) + 11 H O(g) d. Fe(s) + O (g) Fe O (s); for whole numbers: Fe(s) + O (g) Fe O (s) e. FeO(s) + 1 O (g) Fe O (s); for whole numbers, multiply by two.

CHAPTER STOICHIOMETRY 65 FeO(s) + O (g) Fe O (s) 96. a. 16 Cr(s) + S 8 (s) 8 Cr S (s) b. NaHCO (s) Na CO (s) + CO (g) + H O(g) c. KClO (s) KCl(s) + O (g) d. Eu(s) + 6 HF(g) EuF (s) + H (g) 97. a. SiO (s) + C(s) Si(s) + CO(g) Balance oxygen atoms: SiO + C Si + CO Balance carbon atoms: SiO (s) + C(s) Si(s) + CO(g) b. SiCl (l) + Mg(s) Si(s) + MgCl (s) Balance Cl atoms: SiCl + Mg Si + MgCl Balance Mg atoms: SiCl (l) + Mg(s) Si(s) + MgCl (s) c. Na SiF 6 (s) + Na(s) Si(s) + NaF(s) Balance F atoms: Na SiF 6 + Na Si + 6 NaF Balance Na atoms: Na SiF 6 (s) + Na(s) Si(s) + 6 NaF(s) 98. CaSiO (s) + 6 HF(aq) CaF (aq) + SiF (g) + H O(l) Reaction Stoichiometry 99. 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. Fe O (s) + Al(s) Fe(l) + Al O (s) 15.0 g Fe Fe 55.85 g Fe = 0.69 mol Fe; 0.69 mol Fe mol Al mol Fe 6.98 g Al = 7.6 g Al mol Al 0.69 mol Fe 0.69 mol Fe FeO mol Fe AlO mol Fe 159.70 g FeO = 1.5 g Fe O mol Fe O 101.96 g AlO = 1.7 g Al O mol Al O 100. 10 KClO (s) + P (s) P O 10 (s) + 10 KCl(s)

66 CHAPTER STOICHIOMETRY 5.9 g KClO KClO 1.55 g KClO mol P O 8.88 g P O 10 10 = 6.8 g P O 10 10 mol KClO mol PO10 101. 1.000 kg Al 1000 g Al kg Al Al 6.98 g Al mol NHClO mol Al 117.9 g NHClO mol NH ClO = 55 g =.55 kg NH ClO 10. a. Ba(OH) 8H O(s) + NH SCN(s) Ba(SCN) (s) + 10 H O(l) + NH (g) b. 6.5 g Ba(OH) 8H O Ba(OH) 8H O 15. g = 0.006 mol = 0.0 0.0 Ba(OH) 8H O 10. a. 1.0 10 mg NaHCO 1g 1000 mg mol NH SCN Ba(OH) 8H O NaHCO 8.01 g NaHCO 6 8 7 76.1 g NHSCN mol NHSCN =. g NH SCN C6H8O mol NaHCO 19.1 g C6H8O7 = 0.076 g or 76 mg C 6 H 8 O 7 mol C H O 7 b. 0.10 g NaHCO NaHCO 8.01 g NaHCO mol CO mol NaHCO.01g CO mol CO = 0.05 g or 5 mg CO 10. 1.0 10 g phosphorite 75 g Ca(PO) 100 g phosphorite Ca(PO) 10.18 g Ca (PO ) P mol Ca (PO ) 1.88 g P mol P = 150 g P 105. 1.0 ton CuO 907 kg ton 1000 g kg CuO 79.55 g CuO C mol CuO 1.01g C mol C 100. g coke 95 g C = 7. 10 g or 7 kg coke 106. LiOH(s) + CO (g) Li CO (aq) + H O(l) The total volume of air exhaled each minute for the 7 astronauts is 7 0. = 10 L/min. 5,000 g LiOH LiOH.95 g LiOH CO mol LiOH.01g CO mol CO 100 g air.0 g CO 1 ml air 1 L 1 min 1 h = 68 h =.8 days 0.0010 g air 1000 ml 10 L air 60 min

CHAPTER STOICHIOMETRY 67 Limiting Reactants and Percent Yield 107. The product formed in the reaction is NO ; the other species present in the product representtation is excess O. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules react with O molecules to form 6 NO molecules. 6 NO(g) + O (g) 6 NO (g) For smallest whole numbers, the balanced reaction is: NO(g) + O (g) NO (g) 108. 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. 10 molecules O molecules NH 5 molecules O = 8 molecules NH to react with all of the O Because we have 10 molecules of NH and only 8 molecules of NH are necessary to react with all of the O, O is limiting. NH (g) + 5 O (g) NO(g) + 6 H O(g) Initial 10 molecules 10 molecules 0 0 Change!8 molecules!10 molecules +8 molecules +ecules Final molecules 0 8 molecules ecules The total number of molecules present after completion = molecules NH + 0 molecules O + 8 molecules NO + ecules H O = molecules. 109. a. 1.00 10 g N 5.00 10 g H N = 5.7 mol N 8.0 g N H = 8 mol H.016 g H The required mole ratio from the balanced equation is mol H to N. The actual mole ratio is: 8 mol H 5.7 mol N = 6.95 This is well above the required mole ratio, so N in the denominator is the limiting reagent. mol NH 17.0 g NH 5.7 mol N = 1. 10 g NH produced mol N mol NH

68 CHAPTER STOICHIOMETRY mol H.016 g H b. 5.7 mol N = 16 g H reacted mol N mol H Excess H = 500. g H initially 16 g H reacted = 8 g H in excess (unreacted) 110. Ca (PO ) + H SO CaSO + H PO 1.0 10 g Ca (PO ) Ca 10.18 g Ca (PO (PO ) ) =. mol Ca (PO ) 1.0 10 98 g HSO g conc. H SO 100 g conc. H SO HSO = 10. mol H SO 98.09 g H SO The required mole ratio from the balanced equation is mol H SO to Ca (PO ). The 10. mol H SO actual mole ratio is: =.1. mol Ca (PO ) This is larger than the required mole ratio, so Ca (PO ) (in the denominator) is the limiting reagent.. mol Ca (PO ). mol Ca (PO ) mol CaSO mol Ca (PO ) mol HPO mol Ca (PO ) 16.15 CaSO = 100 g CaSO produced mol CaSO 97.99 g HPO = 60 g H PO produced mol H PO 111. An alternative method to solve limiting-reagent problems is to assume each reactant is limiting and then 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. 5.00 10 6 g NH NH 17.0 g NH mol HCN =.9 10 5 mol HCN mol NH 5.00 10 6 g O O.00 g O mol HCN = 1.0 10 5 mol HCN mol O 5.00 10 6 g CH CH 16.0 g CH mol HCN =.1 10 5 mol HCN mol CH O is limiting because it produces the smallest amount of HCN. Although more product could be produced from NH and CH, only enough O is present to produce 1.0 10 5 mol HCN. The mass of HCN produced is: 1.0 10 5 7.0 g HCN mol HCN =.81 10 6 g HCN mol HCN

CHAPTER STOICHIOMETRY 69 5.00 10 6 g O O.00 g O 6 mol HO 18.0 g HO = 5.6 10 6 g H O mol O H O 11. We will use the strategy utilized in the previous problem to solve this limiting reactant problem. If C H 6 is limiting: 15.0 g C H 6 C H6.08 g C H 6 mol CHN 5.06 g CHN = 18.9 g C H N mol C H mol C H N 6 If NH is limiting: 5.00 g NH NH 17.0 g NH mol CHN 5.06 g CHN = 15.6 g C H N mol NH mol C H N If O is limiting: 10.0 g O O.00 g O mol CHN 5.06 g CHN = 11.1 g C H N mol O mol C H N O produces the smallest amount of product; thus O is limiting, and 11.1 g C H N can be produced. 11. C H 6 (g) + Cl (g) C H 5 Cl(g) + HCl(g) 00. g C H 6 C H 6 0.07 g CH = 9.98 mol C H 6 ; 650. g Cl 6 70.90 g Cl Cl = 9.17 mol Cl The balanced equation requires a 1 : e ratio between reactants. To react with all the Cl present, 9.17 mol of C H 6 is needed. Because 9.98 mol C H 6 is actually present, C H 6 is in excess, and Cl is the limiting reagent. The theoretical yield of C H 5 Cl is: Percent yield = C H5Cl 6.51g CH5Cl 9.17 mol Cl = 59 g C H 5 Cl mol Cl mol C H Cl actual theoretical 100 = 90. g 59 g 5 100 = 8.8% 11. a. 11 g C 6 H 5 Cl C6 H5Cl = 10.15 mol C 6 H 5 Cl 11.55 g C H Cl 6 5 85 g C HOCl CHOCl 17.8 g C HOCl =.9 mol C HOCl

70 CHAPTER STOICHIOMETRY From the balanced equation, the required mole ratio is mol C6H5Cl C HOCl =. The actual 10.15 mol C6H5Cl mole ratio present is =.09. The actual mole ratio is greater than.9 mol CHOCl the required mole ratio, so the denominator of the actual mole ratio (C HOCl ) is limiting. C1H9Cl5 5.6 g C H Cl.9 mol C HOCl mol C HOCl b. C HOCl is limiting, and C 6 H 5 Cl is in excess. 1 9 5 = 1170 g C 1 H 9 Cl 5 (DDT) mol C1H9Cl5 mol C6 H5Cl 11.55 g C6H5Cl c..9 mol C HOCl = 71 g C 6 H 5 Cl reacted mol C HOCl mol C H Cl 11 g! 71 g = 01 g C 6 H 5 Cl in excess 6 5 d. Percent yield = 00.0 g DDT 1170 g DDT 100 = 17.1% 115..50 metric tons Cu FeS 1000 kg metric ton 1000 g kg CuFeS.71 g mol Cu Cu FeS 6.55 g = 1.9 10 6 g Cu (theoretical) mol Cu 1.9 10 6 g Cu (theoretical) 86. g Cu (actual) 100. g Cu (theoretical) = 1.0 10 6 g Cu = 1.0 10 kg Cu = 1.0 metric tons Cu (actual) 116. P (s) + 6 F (g) PF (g); the theoretical yield of PF is: 100.0 g PF (theoretical) 10. g PF (actual) = 15 g PF (theoretical) 78.1g PF (actual) PF 6 mol F 8.00 g F 15 g PF = 99.8 g F 87.97 g PF mol PF mol F 99.8 g F is needed to actually produce 10. g of PF if the percent yield is 78.1%. Connecting to Biochemistry 117. Molar mass of C 6 H 8 O 6 = 6(1.01) + 8(1.008) + 6(16.00) = 176.1 g/mol 500.0 mg 1g 1000 mg =.89 10 mol 176.1 g

CHAPTER STOICHIOMETRY 71.89 10 mol 6.0 10 molecules = 1.710 10 ecules mol 118. a. 9(1.01) + 8(1.008) + (16.00) = 180.15 g/mol b. 500. mg 1g 1000 mg =.78 10 mol 180.15 g.78 10 mol 6.0 10 molecules mol = 1.67 10 ecules 119. a. C H 1.01 g + 18 mol H mol C 1.008 g mol H + mol N 1.01 g mol N b. 10.0 g C 1 H 18 N O 5 c. 1.56 mol 9.g mol 1mol C H 9.0 g C N O 1 18 5 1H18NO5 = 59 g C 1 H 18 N O 5 + 5 mol O 16.00 g mol O = 9.0 g =.0 10 mol C 1 H 18 N O 5 d. 5.0 mg 1g 1000 mg 9.0 g 6.0 10 molecules mol = 1.0 10 19 molecules C 1 H 18 N O 5 e. The chemical formula tells us that ecule of C 1 H 18 N O 5 contains atoms of N. If we have e of C 1 H 18 N O 5 molecules, then moles of N atoms are present. 1. g C 1 H 18 N O 5 C1H18NO5 9.0 g C H N O 1 18 5 mol N mol C H N 1 18 O 5 6.0 10 atoms N =.9 10 1 atoms N mol N f. 1.0 10 9 molecules 6.0 10 atoms 9.0 g mol =.9 10 1 g 9.0 g g. ecule =.887 10 g C 1 H 18 N O 5 6.0 10 atoms mol 10. Molar mass = 0(1.01) + 9(1.008) + 19.00 + (16.00) = 6. g/mol

7 CHAPTER STOICHIOMETRY Mass % C = Mass % H = Mass % F = 0(1.01) g C 6. g compound 9(1.008) g H 6. g compound 19.00 g F 6. g compound 100 = 71.0% C 100 = 8.689% H 100 = 5.68% F Mass % O = 100.00! (71.0 + 8.689 + 5.68) = 1.6% O or: % O = (16.00) g O 6. g compound 100 = 1.7% O 11. There are many valid methods to solve this problem. We will assume 100.00 g of compound, and then determine from the information in the problem how many moles of compound equals 100.00 g of compound. From this information, we can determine the mass of one mole of compound (the molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin: mol cyanocobalamin =. g Co x g cyanocobalamin cyanocobalamin 1. Out of 100.00 g of adrenaline, there are: = Co 58.9 g Co cyanocobalamin mol Co = 7.6 10 mol cyanocobalamin 100.00 g, x = molar mass = 1.6 10 g/mol 7.6 10 mol 56.79 g C C 1.01g C =.79 mol C; 6.56 g H H 1.008 g H = 6.5 H 8.7 g O O 16.00 g O = 1.77 mol O; 8.8 g N Dividing each mole value by the smallest number: N 1.01 g N = 0.59 N.79 0.591 = 8.00; 6.51 0.591 = 11.0; 1.77 0.591 =.00; 0.591 0.591 = 1.00 This gives adrenaline an empirical formula of C 8 H 11 O N. 1. Assuming 100.00 g of compound (mass hydrogen = 100.00 g - 9.1 g C -.79 g O = 6.90 g H): C H 9.1 g C =.106 mol C; 6.90 g H = 6.85 mol H 1.01g C 1.008 g H.79 g O O 16.00 g O =.77 mol O

CHAPTER STOICHIOMETRY 7 Dividing all mole values by.77 gives:.106.77 = 1.500; 6.85.77 =.50;.77.77 = 1.000 Because a whole number ratio is required, the empirical formula is C H 5 O. Empirical formula mass: (1.01) + 5(1.008) +(16.00) = 7.07 g/mol Molar mass Empirical formula mass = 16.1 7.07 = 1.999; molecular formula = (C H 5 O ) = C 6 H 10 O 1. 1.0 10 kg waste +.0 kg NH 100 kg waste 1000 g kg NH 18.0 g NH + + C5H7O N 55 mol NH 11.1 g C5 H7ON =. 10 g tissue if all NH + converted mol C H O N 5 7 + Because only 95% of the NH + ions react: mass of tissue = (0.95)(. 10 g) =. 10 g or kg bacterial tissue 15. a. 1.00 10 g C 7 H 6 O C7H6O 18.1 g C H O 7 6 CH6O C H O 7 6 10.09 g CH6O C H O 6 = 7.9 g C H 6 O b. 1.00 10 g C 7 H 6 O C7H6O 18.1 g C H O 7 6 C9H8O C H O 7 6 180.15 g C9H8O mol C H O 9 8 = 1.0 10 g aspirin 16. C 7 H 6 O + C H 6 O C 9 H 8 O + HC H O C7H 6O 1.50 g C 7 H 6 O = 1.09 10 mol C 7 H 6 O 18.1 g C H O 7 6.00 g C H 6 O C 10.09 g C H 6 H O 6 O = 1.96 10 mol C H 6 O C 7 H 6 O is the limiting reagent because the actual moles of C 7 H 6 O present is below the required 1 : e ratio with C H 6 O. The theoretical yield of aspirin is: 1.09 10 mol C 7 H 6 O C9H8O mol C H O 7 6 180.15 g C9H8O = 1.96 g C 9 H 8 O mol C H O 9 8 Percent yield = 1.50 g 1.96 g 100 = 76.5%

7 CHAPTER STOICHIOMETRY 17. 1.50 g BaO BaO 169. g BaO = 8.86 10 mol BaO 5.0 ml 0.07 g HCl ml HCl = 1.87 10 mol HCl 6.6 g HCl The required mole ratio from the balanced reaction is mol HCl to BaO. The actual mole ratio is: 1.87 10 8.86 10 mol HCl mol BaO =.11 Because the actual mole ratio is larger than the required mole ratio, BaO in the denominator is the limiting reagent. 8.86 10 mol BaO HO mol BaO.0 g HO = 0.01 g H O mol H O The amount of HCl reacted is: 8.86 10 mol BaO mol HCl mol BaO = 1.77 10 mol HCl Excess mol HCl = 1.87 10 mol! 1.77 10 mol = 1.0 10 mol HCl Mass of excess HCl = 1.0 10 mol HCl 6.6 g HCl mol HCl =.6 10 g HCl unreacted 18. 5.0 g Ag O 1.8 g = 0.108 mol Ag O 50.0 g C 10 H 10 N SO 50.9 g = 0.00 mol C 10 H 10 N SO Mol C10 H10NSO (actual) = Mol Ag O 0.00 0.108 = 1.85 The actual mole ratio is less than the required mole ratio (), so C 10 H 10 N SO is limiting. 0.00 mol C 10 H 10 N SO mol AgC mol C 10 10 H H N SO 10 9 N SO mol AgC 57.18 g 10 H N SO 9 = 71. g AgC 10 H 9 N SO produced

CHAPTER STOICHIOMETRY 75 Additional Exercises 19. 1 C 1 H 6 : (1.000000) + 6(1.00785) = 0.06950 amu 1 C 1 H 16 O: 1(1.000000) + (1.00785) + 1(15.99915) = 0.010565 amu 1 N 16 O: 1(1.0007) + 1(15.99915) = 9.997989 amu The peak results from 1 C 1 H 16 O. 10. We would see the peaks corresponding to: 10 B 5 Cl [mass 10 + (5) = 115 amu], 10 B 5 Cl 7 Cl (117), 10 B 5 Cl 7 Cl (119), 10 B 7 Cl (11), 11 B 5 Cl (116), 11 B 5 Cl 7 Cl (118), 11 B 5 Cl 7 Cl (10), 11 B 7 Cl (1) We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 10, 11, and 1. 0.68 g XeFn 11. Molar mass XeF n = = 5 g/mol 0 XeFn 9.0 10 molecules XeFn 6.0 10 molecules 5 g = 11. g + n(19.00 g), n = 5.98; formula = XeF 6 1. In 1 hour, the 1000. kg of wet cereal contains 580 kg H O and 0 kg of cereal. We want the final product to contain 0.% H O. Let x = mass of H O in final product. x = 0.0, x = 8 + (0.0)x, x = 105 110 kg H O 0 + x The amount of water to be removed is 580-110 = 70 kg/h. 1. H (g) + O (g) H O(g) a. 50 molecules H ecule O molecules H Stoichiometric mixture. Neither is limiting. b. 100 molecules H ecule O molecules H = 5 molecules O = 50 molecules O ; O is limiting because only 0 molecules O are present. c. From b, 50 molecules of O will react completely with 100 molecules of H. We have 100 molecules (an excess) of O. So H is limiting. d. 0.50 mol H O mol H = 0.5 mol O ; H is limiting because 0.75 mol O is present.

76 CHAPTER STOICHIOMETRY e. 0.80 mol H O mol H = 0.0 mol O ; H is limiting because 0.75 mol O is present. f. 1.0 g H H.016 g H O = 0.5 mol O mol H Stoichiometric mixture. Neither is limiting. g. 5.00 g H H.016 g H O.00 g O = 9.7 g O mol H mol O H is limiting because 56.00 g O is present. 1. tablets 0.6 g C H5BiO tablet C7H5BiO 6.11g C H BiO Bi 09.0 g Bi C7H5BiO mol Bi = 0.0 g Bi consumed 7 7 5 15. Empirical formula mass = 1.01 + 1.008 = 1.0 g/mol; because 10.1/1.0 = 7.998 8, the molecular formula for styrene is (CH) 8 = C 8 H 8..00 g C 8 H 8 C H8 10.1 g C H 8 8 8 mol H 6.00 10 atoms H = 9.5 10 atoms H mol C H mol H 8 8 8 16. 1.98 mg CO 1.01 mg C.01 mg CO = 11.6 mg C; % C = 11.6 mg 19.81 mg 100 = 57.85% C 6.5 mg H O.016 mg H = 0.7 mg H; % H = 18.0 mg H O 0.77 mg 19.81 mg 100 =.6% H % O = 100.00 (57.85 +.6) = 8.51% O Out of 100.00 g terephthalic acid, there are: 57.85 g C C 1.01g C =.817 mol C;.6 g H H 1.008 g H =.6 H 8.51 g O O 16.00 g O =.07 mol O.817.07 =.001;.61.07 = 1.50;.07.07 = 1.000 The C : H : O mole ratio is : 1.5 : 1 or : :. The empirical formula is C H O. Mass of C H O (1) + (1) + (16) = 8.

CHAPTER STOICHIOMETRY 77 Molar mass = 1.5 g 0.50 mol = 166 g/mol; 166 =.0; the molecular formula is C8 H 6 O. 8 17. 17. g H H 1.008 g H = 17. mol H; 8.7 g C C 1.01g C = 6.89 mol C 17. =.50; the empirical formula is C H 5. 6.89 The empirical formula mass is ~9 g, so two times the empirical formula would put the compound in the correct range of the molar mass. Molecular formula = (C H 5 ) = C H 10..59 10 atoms H ecule CH 10 atoms H 10 CH10 =.0 10 mol C H 10 6.0 10 molecules.0 10 mol C H 10 58.1 g mol C H 10 =.50 g C H 10 18. Assuming 100.00 g E H 8 : mol E = 8.7 g H H 1.008 g H mol E =.5 mol E 8 mol H x g E E = 91.7 g E.5 mol E, x = molar mass of E = 8.1 g/mol; atomic mass of E = 8.1 amu 19. Mass of H O = 0.755 g CuSO C xh O 0.8 g CuSO = 0.7 g H O 0.8 g CuSO CuSO 159.6 g CuSO = 0.000 mol CuSO H O 0.7 g H O = 0.015 H O 18.0 g H O 0.015 HO 0.000 g CuSO.98 mol H O = ; compound formula = CuSO C 5H O, x = 5 CuSO 10. a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N % C H N = C HN 5.06 g CHN =. g C H N 1.01 g N C H N. g C HN =.% C H N 100.00 g polymer Only butadiene in the polymer reacts with Br :

78 CHAPTER STOICHIOMETRY 0.605 g Br Br 159.8 g Br C H 5.09 g C 6 6 = 0.05 g C H 6 mol Br mol CH6 H % C H 6 = 0.05 g 1.0 g b. If we have 100.0 g of polymer: 100 = 17.1% C H 6. g C H N 1 mol C H N = 0.68 mol C H N 5.06 g 17.1 g C H 6 9.6 g C 8 H 8 CH6 5.09 g C H 8 6 C8H8 10.1 g C H 8 = 0.16 mol C H 6 = 0.76 mol C 8 H 8 Dividing by 0.16: 0.68 0.16 = 1.99; 0.16 0.16 = 1.00; 0.76 0.16 = 1.51 11. 1.0 g CO This is close to a mole ratio of : :. Thus there are acrylonitrile to butadiene to styrene molecules in the polymer, or (A B S ) n. CO.01 g C mol CO CH0NO mol C 0.8 g C H 0N O 100 =.8% C H 0 N O (LSD) 1.00 g sample 76.51 g mol C H N 0 O = 0.8 g C H 0 N O 1. a. CH (g) + S(s) CS (l) + H S(g) or CH (g) + S 8 (s) CS (l) + H S(g) CH b. 10. g CH 16.0 g CH = 7.8 mol CH ; 10. g S S.07 g S =.7 mol S The required S to CH mole ratio is : 1. The actual S to CH mole ratio is:.7 mol S = 0.500 7.8 mol CH This is well below the required mole ratio, so sulfur is the limiting reagent. CS The theoretical yield of CS is:.7 mol S mol S 76.15 g CS = 71. g CS mol CS The same amount of CS would be produced using the balanced equation with S 8.

CHAPTER STOICHIOMETRY 79 1. 16 g B 5 H 9 6.1 g =.00 mol B 5 H 9 ; 19 g O.00 g = 6.00 mol O Mol O Mol B H 5 9 (actual) = 6.00.00 =.00 The required mol O to mol B 5 H 9 ratio is 1/ = 6. The actual mole ratio is less than the required mole ratio; thus the numerator (O ) is limiting. 9 mol H O 18.0 g HO 6.00 mol O = 81.1 g H O O mol H O 1. NaNO (s) NaNO (s) + O (g); the amount of NaNO in the impure sample is: 0.86 g NaNO NaNO 69.00 g NaNO mol NaNO mol NaNO 85.00 g NaNO mol NaNO Mass percent NaNO = 0.58 g NaNO 100 = 8.0% 0.0 g sample = 0.58 g NaNO 15. 5 g Fe Fe 55.85 g Fe FeO 159.70 g FeO = 68 g Fe O mol Fe mol Fe O 68g Fe O Mass percent Fe O = 100 = 86.% 75 g ore 65.8g Zn 16. a. Mass of Zn in alloy = 0.0985 g ZnCl = 0.07 g Zn 16.8g ZnCl 0.07g Zn % Zn = 100 = 9.% Zn; % Cu = 100.00! 9. = 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. 17. Assuming e of vitamin A (86. g vitamin A): mol C = 86. g vitamin A 0.886 g C C = 0.00 mol C g vitamin A 1.01g C mol H = 86. g vitamin A 0.1056 g H g vitamin A H = 0.00 mol H 1.008 g H Because e of vitamin A contains 0 mol C and 0 mol H, the molecular formula of vitamin A is C 0 H 0 E. To determine E, let s calculate the molar mass of E: 86. g = 0(1.01) + 0(1.008) + molar mass E, molar mass E = 16.0 g/mol From the periodic table, E = oxygen, and the molecular formula of vitamin A is C 0 H 0 O.

80 CHAPTER STOICHIOMETRY 18. X Z: 0.0% X and 60.0% Z by mass; where A = molar mass. mol X mol Z For XZ, molar mass = A x + A z = A x + (A x ) = 7A x. 0.0/A (0.0)A x z = = = or A z = A x 60.0/A z (60.0)A x Mass percent X = Challenge Problems A x 7A x 100 = 1.% X; % Z = 100.0 1. = 85.7% Z 19. The volume of a gas is proportional to the number of molecules of gas. Thus the formulas are: I: NH II: N H III: HN The mass ratios are: I: 8.5 g N 17.75 g H =.6 g N g H II: 6.99 g N g H III: 1.7 g N g H If we set the atomic mass of H equal to 1.008, then the atomic mass for nitrogen is: I: 1.01 II: 1.01 III. 1.0 For example, for compound I: A (1.008) =.6, A = 1.01 1 150. 85 87 Rb atoms Rb atoms =.591; if we had exactly 100 atoms, x = number of 85 Rb atoms, and 100 x = number of 87 Rb atoms. x 100 x =.591, x = 59.1 (.591)x, x = 59.1.591 = 7.15; 7.15% 85 Rb 85.678 61.6 0.715(8.9117) + 0.785(A) = 85.678, A = 0.785 = 86.9 amu 151. First, we will determine composition in mass percent. We assume that all the carbon in the 0.1 g CO came from the 0.157 g of the compound and that all the hydrogen in the 0.010 g H O came from the 0.157 g of the compound. 0.1 g CO 1.01g C.01 g CO = 0.0581 g C; % C = 0.0581g C 0.157 g compound 100 = 7.0% C 0.010 g H O.016 g H 18.0 g H O =.7 10 g H; % H =.7 10 0.157 g g 100 =.1% H We get the mass percent of N from the second experiment:

CHAPTER STOICHIOMETRY 81 0.00 g NH H 1.01 g N 17.0g NH = 1.89 10 g N % N = 1.89 10 g 0.10 g 100 = 18.% N The mass percent of oxygen is obtained by difference: % O = 100.00 (7.0 +.1 + 18.) =.5% O So, out of 100.00 g of compound, there are: 7.0 g C C 1.01 g C =.08 mol C;.1 g H H 1.008 g H =.19 mol H 18. g N N 1.01 g N = 1. N;.5 g O O 16.00 g O =.66 mol O Lastly, and often the hardest part, we need to find simple whole-number ratios. Divide all mole values by the smallest number:.08 1.1 =.5;.19 1.1 = 1.67; 1.1 1.1 = 1.00;.66 1.1 =.0 Multiplying all these ratios by gives an empirical formula of C 7 H 5 N O 6. 15. 1.0 10 6 kg HNO 1000g HNO kg HNO 1mol HNO = 1.6 10 7 mol HNO 6.0 g HNO We need to get the relationship between moles of HNO and moles of NH. We have to use all three equations. mol HNO mol NO mol NO mol NO mol NO mol NH = 16 mol HNO mol NH Thus we can produce 16 mol HNO for every mol NH, we begin with: 1.6 10 7 mol HNO mol NH 16 mol HNO 17.0 g NH =.1 10 8 g or.1 10 5 kg NH mol NH 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 NH and mol HNO turns out to be 1 : 1; that is, e of NH produces e of HNO. Taking into consideration that NO is recycled back gives an answer of.7 10 5 kg NH reacted.

8 CHAPTER STOICHIOMETRY 1 15. Fe(s) + O (g) FeO(s) ; Fe(s) + (g) Fe O (s) 0.00 g Fe Fe 55.85 g = 0.58 O O 11.0.) g O.00 g = 0.88 mol O consumed (1 extra sig. fig.) ( Assuming x mol of FeO is produced from x mol of Fe so that 0.581 x mol of Fe reacts to form Fe O : x Fe + 1 x O x FeO 0.581 x 0.581 x ( 0.581 x ) mol Fe + mol O Setting up an equation for total moles of O consumed: 1 x + ( 0.581 x) = 0.88 mol O, x = 0.0791 = 71.85 g FeO 0.079 mol FeO = 5.7 g FeO produced mol mol FeO 0.079 mol FeO Mol Fe O produced = 0.581 0.079 = 0.10 mol Fe O 0.10 mol Fe O 159.70 g Fe O =. g Fe O produced mol 15. C H 6 (g) + 7 O (g) CO (g) + 6 H O(l); C H 8 (g) + 5 O (g) CO (g) + H O(l) 0.07 g/mol.09 g/mol Let x = mass C H 6, so 9.780! x = mass C H 8. Use the balanced equation to set up a mathematical expression for the moles of O required. x 0.07 7 9.780 x +.09 Solving: x =.7 g C H 6 ; 5 1.7 g 9.780 g = 1.10 mol O 100 = 8% C H 6 by mass 155. The two relevant equations are: Zn(s) + HCl(aq) ZnCl (aq) + H (g) and Mg(s) + HCl(aq) MgCl (aq) + H (g) Let x = mass Mg, so 10.00! x = mass Zn. From the balanced equations, moles H = moles Zn + moles Mg.