Complete Solutions Manualto accompany

Chemical PrinciplesSixth EditionSteven S. Zumdahl

Thomas J. Hummel Steven S. Zumdahl University of Illinois at Urbana-Champaign

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TABLE OF CONTENTS

Page How to Use This Guide....................................................................................................................v Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Atoms, Molecules, and Ions.....................................................................................1 Stoichiometry.........................................................................................................16 Types of Chemical Reactions and Solution Stoichiometry ...................................51 Gases ......................................................................................................................96 Chemical Equilibrium..........................................................................................149 Acids and Bases ...................................................................................................192 Applications of Aqueous Equilibria.....................................................................254 Energy, Enthalpy, and Thermochemisty..............................................................341 Spontaneity, Entropy, and Free Energy ...............................................................369 Electrochemistry ..................................................................................................413 Quantum Mechanics and Atomic Theory ............................................................455 Bonding: General Concepts .................................................................................490 Covalent Bonding: Orbitals ................................................................................540 Chemical Kinetics................................................................................................576 Liquids and Solids................................................................................................624 Properties of Solutions.........................................................................................668 The Representative Elements...............................................................................707 Transition Metals and Coordination Chemistry...................................................730 The Nucleus: A Chemists View .........................................................................756 Organic Chemistry ...............................................................................................773 iii

HOW TO USE THIS GUIDESolutions to all of the end of chapter exercises are in this manual. This Solutions Guide can be very valuable if you use it properly. The way NOT to use it is to look at an exercise in the book and then immediately check the solution, often saying to yourself, Thats easy, I can do it. Developing problem solving skills takes practice. Dont look up a solution to a problem until you have tried to work it on your own. If you are completely stuck, see if you can find a similar problem in the Sample Exercises in the chapter. Only look up the solution as a last resort. If you do this for a problem, look for a similar problem in the end of chapter exercises and try working it. The more problems you do, the easier chemistry becomes. It is also in your self interest to try to work as many problems as possible. Most exams that you will take in chemistry will involve a lot of problem solving. If you have worked several problems similar to the ones on an exam, you will do much better than if the exam is the first time you try to solve a particular type of problem. No matter how much you read and study the text, or how well you think you understand the material, you dont really understand it until you have taken the information in the text and applied the principles to problem solving. You will make mistakes, but the good students learn from their mistakes. In this manual we have worked problems as in the textbook. We have shown intermediate answers to the correct number of significant figures and used the rounded answer in later calculations. Thus, some of your answers may differ slightly from ours. When we have not followed this convention, we have usually noted this in the solution. The most common exception is when working with the natural logarithm (ln) function, where we usually carried extra significant figures in order to reduce round-off error. In addition, we tried to use constants and conversion factors reported to at least one more significant figure as compared to numbers given in the problem. For some problems, this required the use of more precise atomic masses for H, C, N, and O as given in Chapter 3. This practice of carrying one extra significant figure in constants helps minimize round-off error. We are grateful to Claire Szoke for her outstanding effort in preparing the manuscript of this manual. We also thank Estelle Lebeau for her careful accuracy review and Jim Madru for his thorough copyediting of the Solutions Manual. We also are grateful to Don DeCoste for his assistance in creating solutions to some of the problems in this Solutions Guide. TJH SSZ

v

CHAPTER 2 ATOMS, MOLECULES, AND IONSDevelopment of the Atomic Theory18. Law of conservation of mass: mass is neither created nor destroyed. The total mass before a chemical reaction always equals the total mass after a chemical reaction. Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g hydrogen for every 8 g oxygen. Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers. For CO2 and CO discussed in section 2.2, the mass ratios of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio. 19. From Avogadros hypothesis (law), volume ratios are equal to molecule ratios at constant temperature and pressure. Therefore, we can write a balanced equation using the volume data, Cl2 + 3 F2 2 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF3 for a balanced reaction. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadros hypothesis (law) implies that volume ratios are equal to molecule ratios at constant temperature and pressure. H2 + Cl2 2 HCl. From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted. 21. Hydrazine: 1.44 10 1 g H/g N; Ammonia: 2.16 10 1 g H/g N; Hydrogen azide: 2.40 10 2 g H/g N; Let's try all of the ratios:0.144 0.216 0.216 3 = 6.00; = 9.00; = 1.50 = 0.0240 0.0240 0.144 2

20.

All the masses of hydrogen in these three compounds can be expressed as simple wholenumber ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6 : 9 : 1.

1

222.

CHAPTER 2 ATOMS, MOLECULES, AND IONSThe law of multiple proportions does not involve looking at the ratio of the mass of one element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound: Compound 1: Compound 2: 27.2 g C and 72.8 g O (100.0 - 27.2 = mass O) 42.9 g C and 57.1 g O (100.0 - 42.9 = mass O)

The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: Compound 2:27.2 g C = 0.374 g C/g O 72.8 g O

42.9 g C = 0.751 g C/g O 57.1 g O

0.751 2 = ; this supports the law of multiple proportions as this carbon ratio is a whole 0.374 1 number. 23. To get the atomic mass of H to be 1.00, we divide the mass that reacts with 1.00 g of oxygen by 0.126, that is, 0.126/0.126 = 1.00. To get Na, Mg, and O on the same scale, we do the same division. Na: 2.875 1.500 1.00 = 22.8; Mg: = 11.9; O: = 7.94 0.126 0.126 0.126 H Relative value Accepted value 1.00 1.0079 O 7.94 15.999 Na 22.8 22.99 Mg 11.9 24.31

The atomic masses of O and Mg are incorrect. The atomic masses of H and Na are close. Something must be wrong about the assumed formulas of the compounds. It turns out that the correct formulas are H2O, Na2O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H.

The Nature of the Atom24. Deflection of cathode rays by magnetic and electrical fields led to the conclusion that they were negatively charged. The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electrical field. particles are electrons. A cathode ray is a stream of electrons ( particles). Density of hydrogen nucleus (contains one proton only):

25.

CHAPTER 2

ATOMS, MOLECULES, AND IONS4 3 4 r = (3.14) (5 10 14 cm) 3 = 5 10 40 cm 3 3 3

3

Vnucleus =

d = density =

1.67 10 24 g = 3 1015 g/cm 3 40 3 5 10 cm

Density of H atom (contains one proton and one electron): Vatom = 4 (3.14) (1 10 8 cm) 3 = 4 10 24 cm 3 3

d=26.

1.67 10 24 g + 9 10 28 g = 0.4 g/cm 3 4 10 24 cm 3

From Fig. 2.13 of the text, the average diameter of the nucleus is 10 13 cm, and the average diameter of the volume where the electrons roam about is 10 8 cm.10 8 cm = 105; 13 10 cm1 mile 5280 ft 63,360 in = = 1 grape 1 grape 1 grape

Because the grape needs to be 105 times smaller than a mile, the diameter of the grape would need to be 63,360/(1 105) 0.6 in. This is a reasonable size for a grape. 27. First, divide all charges by the smallest quantity, 6.40 1013. 2.56 10 12 = 4.00; 6.40 10 13 7.68 = 12.00; 0.640 3.84 = 6.00 0.640

Because all charges are whole-number multiples of 6.40 1013 zirkombs, the charge on one electron could be 6.40 1013 zirkombs. However, 6.40 1013 zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 1013 zirkombs or an integer fraction of 6.40 1013. 28. The proton and neutron have similar mass, with the mass of the neutron slightly larger than that of the proton. Each of these particles has a mass approximately 1800 times greater than that of an electron. The combination of the protons and the neutrons in the nucleus makes up the bulk of the mass of an atom, but the electrons make the greatest contribution to the chemical properties of the atom. If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout), then particles should have traveled through the thin foil with very minor deflections in their path. This was not the case because a few of the particles were deflected at very large angles. Rutherford reasoned that the large deflections of these particles could be caused only by a center of concentrated positive charge that contains most of the atoms mass (the nuclear model of the atom).

29.

4

CHAPTER 2 ATOMS, MOLECULES, AND IONS

Elements and the Periodic Table30. a. A molecule has no overall charge (an equal number of electrons and protons are present). Ions, on the other hand, have electrons added to form anions (negatively charged ions) or electrons removed to form cations (positively charged ions). b. The sharing of electrons between atoms is a covalent bond. An ionic bond is the force of attraction between two oppositely charged ions. c. A molecule is a collection of atoms held together by covalent bonds. A compound is composed of two or more different elements having constant composition. Covalent and/or ionic bonds can hold the atoms together in a compound. Another difference is that molecules do not necessarily have to be compounds. H2 is two hydrogen atoms held together by a covalent bond. H2 is a molecule, but it is not a compound; H2 is a diatomic element. d. An anion is a negatively charged ion, for example, Cl, O2, and SO42 are all anions. A cation is a positively charged ion, for example, Na+, Fe3+, and NH4+ are all cations. 31. The atomic number of an element is equal to the number of protons in the nucleus of an atom of that element. The mass number is the sum of the number of protons plus neutrons in the nucleus. The atomic mass is the actual mass of a particular isotope (including electrons). As we will see in Chapter 3, the average mass of an atom is taken from a measurement made on a large number of atoms. The average atomic mass value is listed in the periodic table. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, and Am. Nonmetals: Si, B, At, Rn, and Br. b. Si, Ge, B, and At. The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At. Aluminum has mostly properties of metals, so it is generally not classified as a metalloid. 33. a. The noble gases are He, Ne, Ar, Kr, Xe, and Rn (helium, neon, argon, krypton, xenon, and radon). Radon has only radioactive isotopes. In the periodic table, the whole number enclosed in parentheses is the mass number of the longest-lived isotope of the element. b. promethium (Pm) and technetium (Tc) 34. Carbon is a nonmetal. Silicon and germanium are called metalloids as they exhibit both metallic and nonmetallic properties. Tin and lead are metals. Thus metallic character increases as one goes down a family in the periodic table. The metallic character decreases from left to right across the periodic table. a. Five: F, Cl, Br, I, and At b. Six: Li, Na, K, Rb, Cs, and Fr (H is not considered an alkali metal.) c. 14: Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, and Lu d. 40: All elements in the block defined by Sc, Zn, Uub, and Ac are transition metals.

32.

35.

CHAPTER 236.

ATOMS, MOLECULES, AND IONS

5

a. Cobalt is element 27. A = mass number = 27 + 31 = 58; 58 Co27

b. 37. a. b. d. f. h. 38. Symbol238 92 U

10 B 5 24 Mg: 12

c.

23 Mg 12

d.

132 I 53

e.

19 F 9

f.

65 Cu 29

12 protons, 12 neutrons, 12 electrons 12 p, 12 n, 10 e 27 p, 32 n, 24 e c. e. g. i.59 Co2+: 27 59 Co: 27

24 Mg2+: 12 59 Co3+: 27 79 Se: 34 63 Ni: 28

27 p, 32 n, 25 e

27 p, 32 n, 27 e 34 p, 45 n, 36 e 28 p, 31 n, 26 e

34 p, 45 n, 34 e 28 p, 35 n, 28 e

79 2 Se : 34 59 2+ Ni : 28

Number of Protons in Nucleus92

Number of Neutrons in Nucleus146

Number of Electrons92

Net Charge0

40 2+ Ca 20 51 3+ V 23 89 Y 39 79 Br 35 31 3 P 15

20

20

18

2+

23

28

20

3+

39

50

39

0

35

44

36

1

15

16

18

3

39.

Atomic number = 63 (Eu); net charge = +63 60 = 3+; mass number = 63 + 88 = 151; symbol: 151 Eu3+ 63 Atomic number = 50 (Sn); mass number = 50 + 68 = 118; net charge = +50 48 = 2+; symbol: 118 Sn2+. 50

640.

CHAPTER 2 ATOMS, MOLECULES, AND IONSAtomic number = 16 (S); net charge = +16 18 = 2; mass number = 16 + 18 = 34; symbol: 34 S216

Atomic number = 16 (S); net charge = +16 18 = 2; Mass number = 16 + 16 = 32; symbol: 32 S216

41.

In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A and 3A metals form stable 1+, 2+ and 3+ charged cations, respectively. Group 5A, 6A, and 7A nonmetals form 3!, 2! and 1! charged anions, respectively. a. Lose 2 e to form Ra2+. d. Gain 2 e to form Te 2 . b. Lose 3 e to form In3+. e. Gain 1 e to form Br!. c. Gain 3 e to form P 3 . f. Lose 1 e to form Rb+.

42.

See Exercise 41 for a discussion of charges various elements form when in ionic compounds. a. Element 13 is Al. Al forms 3+ charged ions in ionic compounds. Al3+ b. Se2 c. Ba2+ d. N3 e. Fr+ f. Br

Nomenclature43. AlCl3, aluminum chloride; CrCl3, chromium(III) chloride; ICl3, iodine trichloride; AlCl3 and CrCl3 are ionic compounds following the rules for naming ionic compounds. The major difference is that CrCl3 contains a transition metal (Cr) that generally exhibits two or more stable charges when in ionic compounds. We need to indicate which charged ion we have in the compound. This is generally true whenever the metal in the ionic compound is a transition metal. ICl3 is made from only nonmetals and is a covalent compound. Predicting formulas for covalent compounds is extremely difficult. Because of this, we need to indicate the number of each nonmetal in the binary covalent compound. The exception is when there is only one of the first species present in the formula; when this is the case, mono- is not used (it is assumed). a. Dinitrogen monoxide is correct. N and O are both nonmetals resulting in a covalent compound. We need to use the covalent rules of nomenclature. The other two names are for ionic compounds. b. Copper(I) oxide is correct. With a metal in a compound, we have an ionic compound. Because copper, like most transition metals, forms at least a couple of different stable charged ions, we must indicate the charge on copper in the name. Copper oxide could be CuO or Cu2O, hence why we must give the charge of most transition metal compounds. Dicopper monoxide is the name if this were a covalent compound, which it is not. c. Lithium oxide is correct. Lithium forms 1+ charged ions in stable ionic compounds. Because lithium is assumed to form 1+ ions in compounds, we do not need to indicate the charge of the metal ion in the compound. Dilithium monoxide would be the name if Li2O were a covalent compound (a compound composed of only nonmetals).

44.

CHAPTER 245.

ATOMS, MOLECULES, AND IONSb. dinitrogen tetroxide d. tetraphosphorus hexoxide b. magnesium phosphate d. sulfur difluoride f. j. l. sodium hydrogen phosphate magnesium hydroxide silver chromate c. cobalt(II) iodide h. sodium hypochlorite h. lithium nitride

7

a. sulfur difluoride c. iodine trichloride

46.

a. sodium perchlorate c. aluminum sulfate e. sulfur hexafluoride g. sodium dihydrogen phosphate i. sodium hydroxide k. aluminum hydroxide

47.

a. copper(I) iodide d. sodium carbonate f. i. tetrasulfur tetranitride barium chromate

b. copper(II) iodide g. sulfur hexafluoride j. ammonium nitrate

e. sodium hydrogen carbonate or sodium bicarbonate

48.

a. acetic acid d. iodine monochloride g. sulfuric acid j. tin(IV) oxide b. SO3 f.

b. ammonium nitrite e. lead(II) phosphate h. strontium nitride k. sodium chromate c. Na2SO3 g. Cr(C2H3O2)2 l. NaH

c. colbalt(III) sulfide f. i. l. potassium iodate aluminum sulfite hypochlorous acid

49.

a. SO2 e. Li3N i. j. (NH4)2HPO4

d. KHSO3 h. SnF4

Cr2(CO3)3

NH4HSO4: composed of NH4+ and HSO4 ions k. KClO4 n. HBr b. Na2O2 f. PbO k. ZnS n. P2O5 b. CuSO4; copper(II) sulfate d. MgSO4; magnesium sulfate f. CaSO4; calcium sulfate c. KCN g. PbO23+

m. HBrO 50. a. Na2O e. SiCl4 i. j. CdSe

d. Cu(NO3)2 h. CuCl

GaAs: We would predict the stable ions to be Ga and As3. l. Hg2Cl2: Mercury(I) exists as Hg22+.

m. HNO2 51.

a. Pb(C2H3O2)2; lead(II) acetate c. CaO; calcium oxide e. Mg(OH)2; magnesium hydroxide

g. N2O; dinitrogen monoxide or nitrous oxide (common name)

852.

CHAPTER 2 ATOMS, MOLECULES, AND IONSa. Iron forms 2+ and 3+ charged ions; we need to include a Roman numeral for iron. Iron(III) chloride is correct. b. This is a covalent compound so use the covalent rules. Nitrogen dioxide is correct. c. This is an ionic compound, so use the ionic rules. Calcium oxide is correct. Calcium only forms stable 2+ ions when in ionic compounds, so no Roman numeral is needed. d. This is an ionic compound, so use the ionic rules. Aluminum sulfide is correct. e. This is an ionic compound, so use the ionic rules. Mg is magnesium. Magnesium acetate is correct. f. Because phosphate has a 3 charge, the charge on iron is 3+. Iron(III) phosphate is correct.

g. This is a covalent compound, so use the covalent rules. Diphosphorus pentasulfide is correct. h. Because each sodium is 1+ charged, we have the O22 (peroxide) ion present. Sodium peroxide is correct. Note that sodium oxide would be Na2O. i. j. HNO3 is nitric acid, not nitrate acid. Nitrate acid does not exist. H2S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name). H2SO4 is sulfuric acid. b. perchloric acid, HClO4 e. phosphoric acid, H3PO4 c. acetic acid, HC2H3O2

53.

a. nitric acid, HNO3 d. sulfuric acid, H2SO4

Additional Exercises54. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons and neutrons, which can be broken down into quarks. For our purpose, electrons, neutrons, and protons are the key smaller parts of an atom. b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have 0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions were included, then different ions/atoms of H could have different numbers of electrons. c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2 protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium.

CHAPTER 2

ATOMS, MOLECULES, AND IONS

9

d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1 g hydrogen for every 16 g of O present. These are distinctly different compounds, each with its own unique relative number and types of atoms present. e. A chemical equation involves a reorganization of the atoms. Bonds are broken between atoms in the reactants, and new bonds are formed in the products. The number and types of atoms between reactants and products do not change. Because atoms are conserved in a chemical reaction, mass is also conserved. 55. A compound will always have a constant composition by mass. From the initial data given, the mass ratio of H : S : O in sulfuric acid (H2SO4) is:2.02 32.07 64.00 : : = 1 : 15.9 : 31.7 2.02 2.02 2.02

If we have 7.27 g H, then we will have 7.27 15.9 = 116 g S and 7.27 31.7 = 230. g O in the second sample of H2SO4. 56. Mass is conserved in a chemical reaction. Mass: Chromium(III) oxide + aluminum chromium + aluminum oxide 34.0 g 12.1 g 23.3 ?

Mass aluminum oxide produced = (34.0 + 12.1) ! 23.3 = 22.8 g 57. From the Na2X formula, X has a 2! charge. Because 36 electrons are present, X has 34 protons, 79 ! 34 = 45 neutrons, and is selenium. a. True. Nonmetals bond together using covalent bonds and are called covalent compounds. b. False. The isotope has 34 protons. c. False. The isotope has 45 neutrons. d. False. The identity is selenium, Se. 58. a. Fe2+: 26 protons (Fe is element 26.); protons electrons = charge, 26 2 = 24 electrons; FeO is the formula because the oxide ion has a 2 charge. b. Fe3+: 26 protons; 23 electrons; Fe2O3 d. Cs+: 55 protons; 54 electrons; Cs2O f. P3: 15 protons; 18 electrons; AlP c. Ba2+: 56 protons; 54 electrons; BaO e. S2: 16 protons; 18 electrons; Al2S3 g. Br: 35 protons; 36 electrons; AlBr3

h. N3: 7 protons; 10 electrons; AlN 59. From the XBr2 formula, the charge on element X is 2+. Therefore, the element has 88 protons, which identifies it as radium, Ra. 230 88 = 142 neutrons.

1060. 61.

CHAPTER 2 ATOMS, MOLECULES, AND IONSThe solid residue must have come from the flask. In the case of sulfur, SO42 is sulfate, and SO32 is sulfite. By analogy: SeO42: selenate; SeO32: selenite; TeO42: tellurate; TeO32: tellurite

62.

The polyatomic ions and acids in this problem are not named in the text. However, they are all related to other ions and acids named in the text that contain a same group element. Because HClO4 is perchloric acid, HBrO4 would be perbromic acid. Because ClO3 is the chlorate ion, KIO3 would be potassium iodate. Since ClO2 is the chlorite ion, NaBrO2 would be sodium bromite. And finally, because HClO is hypochlorous acid, HIO would be hypoiodous acid. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or: 4.784 g In A = , A = atomic mass of In = 76.54 16.00 1.000 g O If the formula is In2O3, then two times the atomic mass of In will combine with three times the atomic mass of O, or: 2A 4.784 g In = , A = atomic mass of In = 114.8 (3)16.00 1.000 g O The latter number is the atomic mass of In used in the modern periodic table.

63.

64.

a. Ca2+ and N3: Ca3N2, calcium nitride c. Rb+ and F: RbF, rubidium fluoride e. Ba2+ and I: BaI2, barium iodide g. Cs+ and P3: Cs3P, cesium phosphide

b. K+ and O2: K2O, potassium oxide d. Mg2+ and S2: MgS, magnesium sulfide f. Al3+ and Se2: Al2Se3, aluminum selenide

h. In3+ and Br: InBr3, indium(III) bromide; In also forms In+ ions, but you would predict In3+ ions from its position in the periodic table. 65. The cation has 51 protons and 48 electrons. The number of protons corresponds to the atomic number. Thus this is element 51, antimony. There are 3 fewer electrons than protons. Therefore, the charge on the cation is 3+. The anion has one-third the number of protons of the cation which corresponds to 17 protons; this is element 17, chlorine. The number of electrons in this anion of chlorine is 17 + 1 = 18 electrons. The anion must have a charge of 1. The formula of the compound formed between Sb3+ and Cl is SbCl3. The name of the compound is antimony(III) chloride. The Roman numeral is used to indicate the charge of Sb because the predicted charge is not obvious from the periodic table. 66. a. This is element 52, tellurium. Te forms stable 2 charged ions in ionic compounds (like other oxygen family members).

CHAPTER 2b. c. d. 67.

ATOMS, MOLECULES, AND IONS

11

Rubidium. Rb, element 37, forms stable 1+ charged ions. Argon. Ar is element 18. Astatine. At is element 85.

Because this is a relatively small number of neutrons, the number of protons will be very close to the number of neutrons present. The heavier elements have significantly more neutrons than protons in their nuclei. Because this element forms anions, it is a nonmetal and will be a halogen because halogens form stable 1! charged ions in ionic compounds. From the halogens listed, chlorine, with an average atomic mass of 35.45, fits the data. The two isotopes are 35Cl and 37Cl, and the number of electrons in the 1! ion is 18. Note that because the atomic mass of chlorine listed in the periodic table is closer to 35 than 37, we can assume that 35Cl is the more abundant isotope. This is discussed in Chapter 3.

Challenge Problems68. Because the gases are at the same temperature and pressure, the volumes are directly proportional to the number of molecules present. Lets consider hydrogen and oxygen to be monatomic gases and that water has the simplest possible formula (HO). We have the equation: H + O HO But the volume ratios are also equal to the molecule ratios, which correspond to the coefficients in the equation: 2 H + O 2 HO Because atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correct this, we can make oxygen a diatomic molecule: 2 H + O2 2 HO This does not require hydrogen to be diatomic. Of course, if we know water has the formula H2O, we get: 2 H + O2 2 H2O The only way to balance this is to make hydrogen diatomic: 2 H2 + O2 2 H2O 69. a. Both compounds have C2H6O as the formula. Because they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties because the atoms are bonded together differently. These compounds are called isomers of each other. b. When wood burns, most of the solid material in wood is converted to gases, which escape. The gases produced are most likely CO2 and H2O. c. The atom is not an indivisible particle but is instead composed of other smaller particles, for example, electrons, neutrons, and protons.

12

CHAPTER 2 ATOMS, MOLECULES, AND IONSd. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Although the compounds are composed of different isotopes, their properties are similar because different isotopes of the same element have similar properties (except, of course, their mass).

70.

For each experiment, divide the larger number by the smaller. In doing so, we get: experiment 1 experiment 2 experiment 3 X = 1.0 Y = 1.4 X = 1.0 Y = 10.5 Z = 1.0 Y = 3.5

Our assumption about formulas dictates the rest of the solution. For example, if we assume that the formula of the compound in experiment 1 is XY and that of experiment 2 is YZ, we get relative masses of: X = 2.0; Y = 21; Z = 15 (= 21/1.4) and a formula of X3Y for experiment 3 [three times as much X must be present in experiment 3 as compared to experiment 1 (10.5/3.5 = 3)]. However, if we assume the formula for experiment 2 is YZ and that of experiment 3 is XZ, then we get: X = 2.0; Y = 7.0; Z = 5.0 (= 7.0/1.4) and a formula of XY3 for experiment 1. Any answer that is consistent with your initial assumptions is correct. The answer to part d depends on which (if any) of experiments 1 and 3 have a formula of XY in the compound. If the compound in expt. 1 has formula XY, then: 21 g XY 4.2 g Y = 19.2 g Y (and 1.8 g X) (4.2 + 0.4) g XY

If the compound in experiment 3 has the XY formula, then: 21 g XY H 7 .0 g Y = 16.3 g Y (and 4.7 g X) (7.0 + 2.0) g XY

Note that it could be that neither experiment 1 nor experiment 3 has XY as the formula. Therefore, there is no way of knowing an absolute answer here. 71. Compound I: 14.0 g R 4.67 g R 7.00 g R 1.56 g R = ; Compound II: = 3.00 g Q 1.00 g Q 4.50 g Q 1.00 g Q 4.67 = 2.99 3. 1.56

The ratio of the masses of R that combines with 1.00 g Q is

CHAPTER 2

ATOMS, MOLECULES, AND IONS

13

As expected from the law of multiple proportions, this ratio is a small whole number. Because compound I contains three times the mass of R per gram of Q as compared with compound II (RQ), the formula of compound I should be R3Q. 72. Most of the mass of the atom is due to the protons and the neutrons in the nucleus, and protons and neutrons have about the same mass (1.67 1024 g). The ratio of the mass of the molecule to the mass of a nuclear particle will give a good approximation of the number of nuclear particles (protons and neutrons) present. 7.31 10 23 g = 43.8 44 nuclear particles 1.67 10 24 g Thus there are 44 protons and neutrons present. If the number of protons equals the number of neutrons, we have 22 protons in the molecule. One possibility would be the molecule CO2 [6 + 2(8) = 22 protons]. 73. Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equal numbers of molecules. In terms of balanced equations, Avogadros hypothesis (law) implies that volume ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule of CxHy produces 8 molecules of CO2 and 9 molecules of H2O. CxHy + n O2 8 CO2 + 9 H2O. Because all the carbon in octane ends up as carbon in CO2, octane must contain 8 atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in H2O, so one molecule of octane must contain 9 2 = 18 atoms of H. Octane formula = C8H18 and the ratio of C:H = 8:18 or 4:9. Let Xa be the formula for the atom/molecule X, Yb be the formula for the atom/molecule Y, XcYd be the formula of compound I between X and Y, and XeYf be the formula of compound II between X and Y. Using the volume data, the following would be the balanced equations for the production of the two compounds. Xa + 2 Yb 2 XcYd; 2 Xa + Yb 2 XeYf From the balanced equations, a = 2c = e and b = d = 2f. Substituting into the balanced equations: X2c + 2 Y2f 2 XcY2f 2 X2c + Y2f 2 X2cYf For simplest formulas, assume that c = f = 1. Thus: X2 + 2 Y2 2 XY2 and 2 X2 + Y2 2 X2Y Compound I = XY2: If X has relative mass of 1.00, 1.00 = 0.3043, y = 1.14. 1.00 + 2 y

74.

14

CHAPTER 2 ATOMS, MOLECULES, AND IONSCompound II = X2Y: If X has relative mass of 1.00, 2.00 = 0.6364, y = 1.14. 2.00 + y

The relative mass of Y is 1.14 times that of X. Thus if X has an atomic mass of 100, then Y will have an atomic mass of 114.

Marathon Problem75. a. For each set of data, divide the larger number by the smaller number to determine relative masses. 0.602 = 2.04; A = 2.04 when B = 1.00 0.2950.401 = 2.33; C = 2.33 when B = 1.00 0.172

0.374 = 1.17; C = 1.17 when A = 1.00 0.320 To have whole numbers, multiply the results by 3. Data set 1: A = 6.1 and B = 3.0 Data set 2: C = 7.0 and B = 3.0 Data set 3: C = 3.5 and A = 3.0 or C = 7.0 and A = 6.0 Assuming 6.0 for the relative mass of A, the relative masses would be A = 6.0, B = 3.0, and C = 7.0 (if simplest formulas are assumed). b. Gas volumes are proportional to the number of molecules present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is: 6 A2 + B4 4 A3B B4 + 4 C3 4 BC3 3 A2 + 2 C3 6 AC In any correct set of reactions, the calculated mass data must match the mass data given initially in the problem. Here, the new table of relative masses would be: 6 ( mass A 2 ) 0.602 = ; mass A2 = 0.340(mass B4) mass B 4 0.2954 ( mass C 3 ) 0.401 = ; mass C3 = 0.583(mass B4) mass B 4 0.172

2 (mass C 3 ) 0.374 = ; mass A2 = 0.570(mass C3) 3 (mass A 2 ) 0.320

CHAPTER 2

ATOMS, MOLECULES, AND IONS

15

Assume some relative mass number for any of the masses. We will assume that mass B = 3.0, so mass B4 = 4(3.0) = 12. Mass C3 = 0.583(12) = 7.0, mass C = 7.0/3 Mass A2 = 0.570(7.0) = 4.0, mass A = 4.0/2 = 2.0 When we assume a relative mass for B = 3.0, then A = 2.0 and C = 7.0/3. The relative masses having all whole numbers would be A = 6.0, B = 9.0, and C = 7.0. Note that any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility.

CHAPTER 3 STOICHIOMETRYAtomic Masses and the Mass Spectrometer23. A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu This is element Ti (titanium). 24. Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100 x. 151.96 =

x(150.9196) + (100 x)(152.9209) 100

15196 = (150.9196)x + 15292.09 (152.9209)x, 96 = (2.0013)x x = 48%; 48% 151Eu and 100 48 = 52% 153Eu 25. 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 - 117.0 = 0.3740(A) A= 26.69.2 = 185 amu (A = 184.95 amu without rounding to proper significant figures) 0.3740

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.

27.

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 Br2 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 161.84/2 = 80.92. The peaks in the mass spectrum correspond to 79Br2, 79Br81Br, and 81Br2 in order of increasing mass. The intensities of the highest and lowest masses tell us the two isotopes are present at about equal abundance. The actual abundance is 50.68% 79Br and 49.32% 81Br.

16

CHAPTER 328.

STOICHIOMETRY

17

Scaled Intensity Compound Mass Intensity Largest Peak = 100 __________________________________________________________________ H2120Te H2122Te H2123Te H2124Te H2125Te H2126Te H2128Te H2130Te100

121.92 123.92 124.92 125.92 126.92 127.92 129.92 131.93

0.09 2.46 0.87 4.61 6.99 18.71 31.79 34.48

0.3 7.1 2.5 13.4 20.3 54.3 92.2 100.0

50

0 122 124 126 128 130 132 134

29.

GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have two peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60 : 40 or 3 : 2.

144

146

Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have three 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:

1869 69 71

CHAPTER 3Ga (0.60) 0.36 0.2471

STOICHIOMETRY

Ga (0.40) 0.24 0.16

Ga (0.60) Ga (0.40)

288

290

292

Moles and Molar Masses30. 4.24 g C6H6 1 mol = 5.43 10 2 mol C6H6 78.11 g

5.43 10 2 mol C6H6

6.022 10 23 molecules = 3.27 1022 molecules C6H6 mol

Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 total atoms. 3.27 1022 molecules C6H6 0.224 mol H2O 0.224 mol H2O 12 atoms total = 3.92 1023 atoms total molecule

18.02 g = 4.04 g H2O mol6.022 10 23 molecules = 1.35 1023 molecules H2O mol3 atoms total = 4.05 1023 atoms total molecule 1 mol = 4.50 10 2 mol CO2 6.022 10 23 molecules

1.35 1023 molecules H2O 2.71 1022 molecules CO2 4.50 10 2 mol CO2

44.01 g = 1.98 g CO2 mol3 atoms total = 8.13 1022 atoms total molecule CO 2

2.71 1022 molecules CO2 3.35 1022 atoms total

1 molecule = 5.58 1021 molecules CH3OH 6 atoms total

CHAPTER 3

STOICHIOMETRY1 mol = 9.27 10 3 mol CH3OH 6.022 10 23 molecules

19

5.58 1021 molecules CH3OH

9.27 10 3 mol CH3OH

32.04 g = 0.297 g CH3OH mol

31.

a. 20.0 mg C8H10N4O2

1g 1 mol = 1.03 10 4 mol C8H10N4O2 1000 mg 194.20 g 1 mol = 4.52 10 3 mol C2H5OH 23 6.022 10 molecules

b. 2.72 1021 molecules C2H5OH

c. 1.50 g CO2

1 mol = 3.41 10 2 mol CO2 44.01 g

32.

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/mol 5.00 g C2H5O2N

1 mol C 2 H 5 O 2 N 6.022 10 23 molecules C 2 H 5 O 2 N 75.07 g C 2 H 5 O 2 N mol C 2 H 5 O 2 NH1 atom N = 4.01 1022 atoms N molecule C 2 H 5 O 2 N

b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol 5.00 g Mg3N2 1 mol Mg 3 N 2 6.022 10 23 formula units Mg 3 N 2 100.95 g Mg 3 N 2 mol Mg 3 N 2 2 atoms = 5.97 1022 atoms N mol Mg 3 N 2

c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol 5.00 g Ca(NO3)2 1 mol Ca ( NO 3 ) 2 164.10 g Ca ( NO 3 ) 2 2 mol N mol Ca ( NO 3 ) 2 6.022 10 23 atoms N = 3.67 1022 atoms N mol N

20

CHAPTER 3d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4

STOICHIOMETRY

6.022 10 23 atoms N 1 mol N 2 O 4 2 mol N mol N 92.02 g N 2 O 4 mol N 2 O 4 = 6.54 1022 atoms N

33.

4.0 g H2

1 mol H 2 2 mol H 6.022 10 23 atoms H = 2.4 1024 atoms 2.016 g H 2 1 mol H 2 1 mol H 1 mol He 6.022 10 23 atoms He = 6.0 1023 atoms 4.003 g He 1 mol He 2 mol F 6.022 10 23 atoms F = 1.2 1024 atoms 1 mol F2 1 mol F 1 mol CO 2 3 mol atoms (1 C + 2 O) 6.022 10 23 atoms 44.01 g CO 2 1 mol CO 2 1 mol atoms = 1.81 1024 atoms

4.0 g He

1.0 mol F2 H

44.0 g CO2

146 g SF6

1 mol SF6 7 mol atoms (1 S + 6 F) 6.022 10 23 atoms = 4.21 1024 atoms 146.07 g SF6 1 mol SF6 1 mol atoms

The order is: 4.0 g He < 1.0 mol F2 < 44.0 g CO2 < 4.0 g H2 < 146 g SF6 34. a. 14 mol C H 12.011 g 1.0079 g 14.007 g + 18 mol H + 2 mol N mol C mol H mol N + 5 mol O b. 10.0 g C14H18N2O5 15.999 g = 294.305 g/mol mol O

1 mol C14 H18 N 2 O 5 = 3.40 102 mol C14H18N2O5 294.3 g C14 H18 N 2 O 5

c.

1.56 mol

294.3 g = 459 g C14H18N2O5 mol

d. 5.0 mg

1g 1 mol 6.02 10 23 molecles = 1.0 1019 molecules C14H18N2O5 1000 mg 294.3 g mol 1 mol C14 H18 N 2 O 5 2 mol N 6.02 10 23 atoms N 294.3 g C14 H18 N 2 O 5 mol C14 H18 N 2 O 5 mol N = 4.9 1021 atoms N

e.

1.2 g C14H18N2O5

CHAPTER 3f.

STOICHIOMETRY1 mol 294.3 g = 4.9 1013 g 23 mol 6.02 10 atoms

21

1.0 109 molecules

g. 1 molecule

1 mol 294.305 g = 4.887 1022 g C14H18N2O5 23 mol 6.022 10 atoms

35.

a.

2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol 1 mol = 3.023 mol C2H3Cl3O2 165.39 g165.39 g = 3.3 g C2H3Cl3O2 mol

b. 500.0 g

c.

2.0 10-2 mol

d. 5.0 g C2H3Cl3O2

1 mol 6.02 10 23 molecules 3 atoms Cl 165.39 g mol molecule = 5.5 1022 atoms of chlorine

e.

1.0 g Cl

1 mol Cl 1 mol C 2 H 3Cl 3O 2 165.39 g C 2 H 3Cl 3O 2 = 1.6 g chloral hydrate 35.45 g 3 mol Cl mol C 2 H 3Cl 3O 2

f.

500 molecules

1 mol 165.39 g = 1.373 1019 g 23 mol 6.022 10 molecules

36.

1.0 lb flour

454 g flour 30.0 10 9 g C 2 H 4 Br2 1 mol C 2 H 4 Br2 lb flour g flour 187.9 g C 2 H 4 Br2 6.02 10 23 molecules = 4.4 1016 molecules C2H4Br2 mol C 2 H 4 Br2

Percent Composition37. Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol Mass % C = 20(12.01) g C 100 = 71.40% C 336.43 g compound29(1.008) g H 100 = 8.689% H 336.43 g compound

Mass % H =

22Mass % F = 19.00 g F 100 = 5.648% F 336.43 g compound

CHAPTER 3

STOICHIOMETRY

Mass % O = 100.00 ! (71.40 + 8.689 + 5.648) = 14.26% O or: Mass % O = 38. a. 3(16.00) g O 100 = 14.27% O 336.43 g compound

C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol Mass % C =8(12.01) g C 96.08 100 = 100 = 49.47% C 194.20 g C8 H10 N 4 O 2 194.20

b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C =c. 12(12.01) g C 100 = 42.10% C 342.30 g C12 H 22 O11

C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol Mass % C = 2(12.01) g C 100 = 52.14% C 46.07 g C 2 H 5 OH

The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH) 39. For each compound, determine the mass of 1 mole of compound, and then calculate the mass percentage of N in 1 mole of that compound. a. NO: % N = b. NO2: % N = c. N2O4: % N = 14.007 g N 100 = 46.681% N 30.006 g NO 14.007 g N 100 = 30.447% N 46.005 g NO 228.014 g N 100 = 30.447% N 92.010 g N 2 O 4

d. N2O: % N = 40.

28.014 g N 100 = 63.649% N 44.013 g N 2 O

Assuming 100.00 g cyanocobalamin: mol cyanocobalamin = 4.34 g Co 1 mol Co 1 mol cyanocobalamin 58.93 g Co mol Co = 7.36 102 mol cyanocobalamin

CHAPTER 3

STOICHIOMETRY

23

x g cyanocobalamin 100.00 g , x = molar mass = 1360 g/mol = 1 mol cyanocobalamin 7.36 10 2 mol

41.

There are 0.390 g Cu for every 100.000 g of fungal laccase. Lets assume 100.000 g fungal laccase. Mol fungal laccase = 0.390 g Cu 1 mol Cu 1 mol fungal laccase = 1.53 103 mol 63.55 g Cu 4 mol Cu

x g fungal laccase 100.000 g = , x = molar mass = 6.54 104 g/mol 1 mol fungal laccase 1.53 10 3 mol

42.

If we have 100.0 g of Portland cement, we have 50. g Ca3SiO5, 25 g Ca2SiO4, 12 g Ca3Al2O6, 8.0 g Ca2AlFeO5, and 3.5 g CaSO42H2O. Mass percent Ca: 50. g Ca3SiO5 1 mol Ca 3SiO 5 3 mol Ca 40.08 g Ca = 26 g Ca 228.33 g Ca 3SiO 5 1 mol Ca 3SiO 5 1 mol Ca

25 g Ca2SiO4

80.16 g Ca = 12 g Ca 172.25 g Ca 2SiO 4 120.24 g Ca = 5.3 g Ca 270.20 g Ca 3 Al2 O 680.16 g Ca = 2.6 g Ca 242.99 g Ca 2 AlFeO5

12 g Ca3Al2O6

8.0 g Ca2AlFeO5

3.5 g CaSO42H2O

40.08 g Ca = 0.81 g Ca 172.18 g CaSO 4 2 H 2 O

Mass of Ca = 26 + 12 + 5.3 + 2.6 + 0.81 = 47 g Ca % Ca =47 g Ca 100 = 47% Ca 100.0 g cement

Mass percent Al: 12 g Ca3 Al2O6 53.96 g Al = 2.4 g Al 270.20 g Ca 3 Al 2 O 6

8.0 g Ca2AlFeO5

26.98 g Al = 0.89 g Al 242.99 g Ca 2 AlFeO5

24% Al = 2.4 g + 0.89 g 100 = 3.3% Al 100.0 g

CHAPTER 3

STOICHIOMETRY

Mass percent Fe: 8.0 g Ca2AlFeO5 1.8 g 55.85 g Fe = 1.8 g Fe; % Fe = 100 = 1.8% Fe 242.99 g Ca 2 AlFeO5 100.0 g

Empirical and Molecular Formulas43. 12.011 g 1.0079 g a. Molar mass of CH2O = 1 mol C mol C + 2 mol H mol H

15.999 g + 1 mol O mol O = 30.026 g/mol %C= 12.011 g C 2.0158 g H 100 = 40.002% C; % H = 100 = 6.7135% H 30.026 g CH 2 O 30.026 g CH 2 O

%O=

15.999 g O 100 = 53.284% O or % O = 100.000 (40.002 + 6.7135) 30.026 g CH 2 O = 53.285%

b. Molar mass of C6H12O6 = 6(12.011) + 12(1.0079) + 6(15.999) = 180.155 g/mol %C= 72.066 g C 100 = 40.002%; 180.155 g C 6 H12 O 6 %H= 12(1.0079) g 100 = 6.7136% 180.155 g

% O = 100.00 (40.002 + 6.7136) = 53.284% c. Molar Mass of HC2H3O2 = 2(12.011) + 4(1.0079) + 2(15.999) = 60.052 g/mol %C= 24.022 g 100 = 40.002%; 60.052 g %H= 4.0316 g 100 = 6.7135% 60.052 g

% O = 100.000 (40.002 + 6.7135) = 53.285% 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. 44. a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical formula) is NO2.

CHAPTER 3

STOICHIOMETRY

25

b. Molecular formula: C3H6; empirical formula: CH2 c. Molecular formula: P4O10; empirical formula: P2O5 d. Molecular formula: C6H12O6; empirical formula: CH2O 45. Compound I: mass O = 0.6498 g HgxOy 0.6018 g Hg = 0.0480 g O 0.6018 g Hg 1 mol Hg = 3.000 103 mol Hg 200.6 g Hg

0.0480 g O

1 mol O = 3.00 103 mol O 16.00 g O

The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO. Compound II: mass Hg = 0.4172 g HgxOy 0.016 g O = 0.401 g Hg 0.401 g Hg 1 mol Hg 1 mol O = 2.00 103 mol Hg; 0.016 g O = 1.0 103 mol O 200.6 g Hg 16.00 g O

The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O. 46. Out of 100.00 g of adrenaline, there are: 56.79 g C 1 mol C 1 mol H = 4.728 mol C; 6.56 g H = 6.51 mol H 12.011 g C 1.008 g H 1 mol O 1 mol N = 1.773 mol O; 8.28 g N H = 0.591 mol N 15.999 g O 14.01 g N

28.37 g O H

Dividing each mole value by the smallest number: 4.728 6.51 1.773 0.591 = 8.00; = 11.0; = 3.00; = 1.00 0.591 0.591 0.591 0.591 This gives adrenaline an empirical formula of C8H11O3N. 47. First, we will determine composition in mass percent. We assume that all the carbon in the 0.213 g CO2 came from the 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. 0.213 g CO2 12.01 g C 0.0581 g C = 0.0581 g C; % C = 100 = 37.0% C 44.01 g CO 2 0.157 g compound

260.0310 g H2O

CHAPTER 3

STOICHIOMETRY

2.016 g H 3.47 10 3 g = 3.47 103 g H; % H = 100 = 2.21% H 18.02 g H 2 O 0.157 g

We get the mass percent of N from the second experiment: 0.0230 g NH3 H 14.01 g N = 1.89 102 g N 17.03 g NH 3

%N=

1.89 10 2 g 100 = 18.3% N 0.103 g

The mass percent of oxygen is obtained by difference: % O = 100.00 (37.0 + 2.21 + 18.3) = 42.5% O So, out of 100.00 g of compound, there are: 37.0 g C 1 mol C 1 mol H = 3.08 mol C; 2.21 g H = 2.19 mol H 12.01 g C 1.008 g H1 mol N 1 mol O = 1.31 mol N; 42.5 g O = 2.66 mol O 14.01 g N 16.00 g O

18.3 g N

Lastly, and often the hardest part, we need to find simple whole number ratios. Divide all mole values by the smallest number:3.08 2.19 1.31 2.66 = 2.35; = 1.67; = 1.00; = 2.03 1.31 1.31 1.31 1.31

Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6. 48. Assuming 100.00 g of compound (mass oxygen = 100.00 g 41.39 g C 3.47 g H = 55.14 g O): 41.39 g C 55.14 g O 1 mol C 1 mol H = 3.446 mol C; 3.47 g H = 3.44 mol H 12.011 g C 1.008 g H 1 mol O = 3.446 mol O 15.999 g O

All are the same mole values, so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol. Molar mass = 15.0 g = 116 g/mol 0.129 mol

CHAPTER 3

STOICHIOMETRY

27

Molar mass 116 = = 4.00; molecular formula = (CHO)4 = C4H4O4 29.02 Empirical mass 49. Assuming 100.00 g of compound (mass hydrogen = 100.00 g 49.31 g C 43.79 g O = 6.90 g H): 49.31 g C 1 mol C 1 mol H = 4.105 mol C; 6.90 g H = 6.85 mol H 12.011 g C 1.008 g H 1 mol O = 2.737 mol O 15.999 g O

43.79 g O

Dividing all mole values by 2.737 gives: 4.105 6.85 2.737 = 1.500; = 2.50; = 1.000 2.737 2.737 2.737 Because a whole number ratio is required, the empirical formula is C3H5O2. Empirical formula mass 3(12.0) + 5(1.0) +2(16.0) = 73.0 g/molMolar mass 146.1 = = 2.00; molecular formula = (C3H5O2)2 = C6H10O4 Empirical formula mass 73.0

50.

41.98 mg CO2

12.011 mg C 11.46 mg = 11.46 mg C; % C = 100 = 57.85% C 44.009 mg CO 2 19.81 mg

6.45 mg H2O

2.016 mg H 0.722 mg = 0.722 mg H; % H = 100 = 3.64% H 18.02 mg H 2 O 19.81 mg

% O = 100.00 (57.85 + 3.64) = 38.51% O Out of 100.00 g terephthalic acid, there are: 57.85 g C 1 mol C 1 mol H = 4.816 mol C; 3.64 g H = 3.61 mol H 12.011 g C 1.008 g H 1 mol O = 2.407 mol O 15.999 g O3.61 2.407 = 1.50; = 1.000 2.407 2.407

38.51 g O 4.816 = 2.001; 2.407

The C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2. The empirical formula is C4H3O2.

28Mass of C4H3O2 4(12) + 3(1) + 2(16) = 83 Molar mass = 51.

CHAPTER 3

STOICHIOMETRY

41.5 g 166 = 166 g/mol; = 2.0; the molecular formula is C8H6O4. 0.250 mol 83

First, we will determine composition by mass percent: 16.01 mg CO2 1g 1000 mg 12.011 g C 1000 mg = 4.369 mg C 44.009 g CO 2 g

%C=

4.369 mg C 100 = 40.91% C 10.68 mg compound 1g 1000 mg 2.016 g H 1000 mg = 0.489 mg H 18.02 g H 2 O g

4.37 mg H2O

%H=

0.489 mg 100 = 4.58% H; % O = 100.00 - (40.91 + 4.58) = 54.51% O 10.68 mg

So, in 100.00 g of the compound, we have: 40.91 g C 54.51 g O 1 mol H 1 mol C = 3.406 mol C; 4.58 g H = 4.54 mol H 12.011 g C 1.008 g H

1 mol O = 3.407 mol O 15.999 g O 4.54 4 Dividing by the smallest number: = 1.33 . ; the empirical formula is C3H4O3. 3.406 3

The empirical formula mass of C3H4O3 is 3(12) + 4(1) + 3(16) = 88 g. Because 52. 176.1 = 2.0, the molecular formula is C6H8O6. 88

a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N 1 mol C 3 H 3 N 53.06 g C 3 H 3 N = 33.3 g C3H3N 14.01 g N 1 mol C 3 H 3 N33.3 g C 3 H 3 N = 33.3% C3H3N 100.00 g polymer

% C3H3N =

Only butadiene in the polymer reacts with Br2: 0.605 g Br2 1 mol C 4 H 6 54.09 g C 4 H 6 1 mol Br2 = 0.205 g C4H6 159.8 g Br2 mol Br2 mol C 4 H 6

CHAPTER 3

STOICHIOMETRY% C4H6 = 0.205 g 100 = 17.1% C4H6 1.20 g

29

b. If we have 100.0 g of polymer: 33.3 g C3H3N 1 mol C 3 H 3 N = 0.628 mol C3H3N 53.06 g

17.1 g C4H6

1 mol C 4 H 6 = 0.316 mol C4H6 54.09 g C 4 H 6

49.6 g C8H8

1 mol C8 H 8 = 0.476 mol C8H8 104.14 g C8 H 8 0.628 0.316 0.476 = 1.99; = 1.00; = 1.51 0.316 0.316 0.316

Dividing by 0.316:

This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in the polymer, or (A4B2S3)n.

Balancing Chemical Equations53. 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. C6H12O6(s) + O2(g) CO2(g) + H2O(g) Balance C atoms: C6H12O6 + O2 6 CO2 + H2O Balance H atoms: C6H12O6 + O2 6 CO2 + 6 H2O Lastly, 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 + H2S Balance S atoms: Fe2S3 + HCl 2 FeCl3 + 3 H2S There 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 are balanced; balance N:

30CS2 + 2 NH3 H2S + NH4SCN

CHAPTER 3

STOICHIOMETRY

H is also balanced. CS2(l) + 2 NH3(g) H2S(g) + NH4SCN(s). 54. An important part 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) 55. 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) e. 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g) 56. a. 2 KO2(s) + 2 H2O(l) 2 KOH(aq) + O2(g) + H2O2(aq) or 4 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)

Reaction Stoichiometry57. 1.000 kg Al 3 mol NH 4 ClO 4 117.49 g NH 4 ClO 4 1000 g Al 1 mol Al kg Al 26.98 g Al 3 mol Al mol NH 4 ClO 4 = 4355 g NH4ClO4 58. 1.0 106 kg HNO3 1000 g HNO 3 1 mol HNO 3 = 1.6 107 mol HNO3 kg HNO 3 63.0 g HNO 3

We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all three equations:

CHAPTER 3

STOICHIOMETRY

31

2 mol HNO 3 16 mol HNO 3 2 mol NO 2 4 mol NO = 3 mol NO 2 2 mol NO 4 mol NH 3 24 mol NH 3 Thus we can produce 16 mol HNO3 for every 24 mol NH3 that we begin with: 1.6 107 mol HNO3 24 mol NH 3 17.0 g NH 3 = 4.1 108 g or 4.1 105 kg NH3 16 mol HNO 3 mol NH 3

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; that is, 1 mol of NH3 produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer of 2.7 105 kg NH3 reacted. 59. Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s) 15.0 g Fe 1 mol Fe 2 mol Al 26.98 g Al = 0.269 mol Fe; 0.269 mol Fe = 7.26 g Al 55.85 g Fe 2 mol Fe mol Al1 mol Fe 2 O 3 159.70 g Fe 2 O 3 = 21.5 g Fe2O3 2 mol Fe mol Fe 2 O 3

0.269 mol Fe

0.269 mol Fe 60.

1 mol Al 2 O 3 101.96 g Al2 O 3 = 13.7 g Al2O3 2 mol Fe mol Al2 O 3

10 KClO3(s) + 3 P4(s) 3 P4O10(s) + 10 KCl(s) 52.9 g KClO3 1 mol KClO3 3 mol P4 O10 283.88 g P4 O10 = 36.8 g P4O10 122.55 g KClO3 10 mol KClO3 mol P4 O10

61.

2 LiOH(s) + CO2(g) Li2CO3(aq) + H2O(l) The total volume of air exhaled each minute for the 7 astronauts is 7 20. = 140 L/min. 25,000 g LiOH 1 mol LiOH 1 mol CO 2 44.01 g CO 2 100 g air 23.95 g LiOH 2 mol LiOH mol CO 2 4.0 g CO 2

1 mL air 1L 1 min 1h = 68 h = 2.8 days 0.0010 g air 1000 mL 140 L air 60 min 62. 1.0 104 kg waste + + 1 mol C 5 H 7 O 2 N 3.0 kg NH 4 1 mol NH 4 1000 g + + 100 kg waste kg 18.04 g NH 4 55 mol NH 4

113.1 g C 5 H 7 O 2 N = 3.4 104 g tissue if all NH4+ converted mol C 5 H 7 O 2 N

32Because only 95% of the NH4+ ions react:

CHAPTER 3

STOICHIOMETRY

mass of tissue = (0.95)(3.4 104 g) = 3.2 104 g or 32 kg bacterial tissue 63. 1.0 103 g phosphorite 75 g Ca 3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) 2 100 g phosphorite 310.18 g Ca 3 (PO 4 ) 2

1 mol P4 2 mol Ca 3 (PO 4 ) 2 64. Total mass of copper used: 10,000 boards

123.88 g P4 = 150 g P4 mol P4

(8.0 cm 16.0 cm 0.060 cm) 8.96 g = 6.9 105 g Cu 3 board cm

Amount of Cu to be recovered = 0.80 (6.9 105 g) = 5.5 105 g Cu 5.5 105 g Cu 1 mol Cu ( NH 3 ) 4 Cl 2 202.6 g Cu ( NH 3 ) 4 Cl 2 1 mol Cu 63.55 g Cu mol Cu mol Cu ( NH 3 ) 4 Cl 2 = 1.8 106 g Cu(NH3)4Cl2 5.5 105 g Cu 4 mol NH 3 17.03 g NH 3 1 mol Cu = 5.9 105 g NH3 63.55 g Cu mol Cu mol NH 3

Limiting Reactants and Percent Yield65. The product formed in the reaction is NO2; the other species present in the product representtation 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) 66. 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, lets calculate how much of one reactant is necessary to react with the other. 10 molecules O2 4 molecules NH 3 = 8 molecules NH3 to react with all the O2 5 molecules O 2

CHAPTER 3

STOICHIOMETRY

33

Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all the O2, O2 is limiting. Initial Change Final 4 NH3(g) 10 molecules !8 molecules 2 molecules + 5 O2(g) 4 NO(g) 10 molecules 0 !10 molecules +8 molecules 0 8 molecules + 6 H2O(g) 0 +12 molecules 12 molecules

The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 8 molecules NO + 12 molecules H2O = 22 molecules. 67. 1.50 g BaO2 1 mol BaO 2 = 8.86 103 mol BaO2 169.3 g BaO 2

25.0 mL

0.0272 g HCl 1 mol HCl = 1.87 102 mol HCl mL 36.46 g HCl

The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual mole ratio is: 1.87 10 2 mol HCl = 2.11 8.86 10 3 mol BaO 2 Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2) is the limiting reagent. 8.86 103 mol BaO2 H 1 mol H 2 O 2 34.02 g H 2 O 2 = 0.301 g H2O2 mol BaO 2 mol H 2 O 2

The amount of HCl reacted is: 8.86 103 mol BaO2 2 mol HCl = 1.77 102 mol HCl mol BaO 2

Excess mol HCl = 1.87 102 mol 1.77 102 mol = 1.0 103 mol HCl Mass of excess HCl = 1.0 103 mol HCl 68. 25.0 g Ag2O 1 mol = 0.108 mol Ag2O 231.8 g1 mol = 0.200 mol C10H10N4SO2 250.29 g 36.46 g HCl = 3.6 102 g HCl mol HCl

50.0 g C10H10N4SO2

34Mol C10 H10 N 4SO 2 0.200 = 1.85 (actual) = Mol Ag 2 O 0.108

CHAPTER 3

STOICHIOMETRY

The actual mole ratio is less than the required mole ratio (2), so C10H10N4SO2 is limiting. 2 mol AgC10 H 9 N 4SO 2 357.18 g 2 mol C10 H10 N 4SO 2 mol AgC10 H 9 N 4SO 2 = 71.4 g AgC10H9N4SO2 produced 69. 2.50 metric tons Cu3FeS3 1 mol Cu 3 FeS3 1000 kg 1000 g 3 mol Cu metric ton kg 342.71 g 1 mol Cu 3 FeS3 63.55 g = 1.39 106 g Cu (theoretical) mol Cu 86.3 g Cu (actual) = 1.20 106 g Cu = 1.20 103 kg Cu 1.39 106 g Cu (theoretical) 100. g Cu ( theoretical) = 1.20 metric tons Cu (actual) 1 mol C 6 H 5 Cl = 10.15 mol C6H5Cl 112.55 g C 6 H 5 Cl

0.200 mol C10H10N4SO2

70.

a. 1142 g C6H5Cl

485 g C2HOCl3

1 mol C 2 HOCl3 = 3.29 mol C2HOCl3 147.38 g C 2 HOCl3 2 mol C 6 H 5 Cl From the balanced equation, the required mole ratio is = 2. The actual 1 mol C 2 HOCl 3 mole ratio present is10.15 mol C 6 H 5 Cl = 3.09. The actual mole ratio is greater than 3.29 mol C 2 HOCl3

the required mole ratio, so the denominator of the actual mole ratio (C2HOCl3) is limiting. 3.29 mol C2HOCl3 1 mol C14 H 9 Cl 5 354.46 g C14 H 9 Cl 5 = 1170 g C14H9Cl5 (DDT) mol C 2 HOCl3 mol C14 H 9 Cl 5

b. C2HOCl3 is limiting, and C6H5Cl is in excess. c. 3.29 mol C2HOCl3 2 mol C 6 H 5 Cl 112.55 g C 6 H 5 Cl = 741 g C6H5Cl reacted mol C 2 HOCl3 mol C 6 H 5 Cl

1142 g ! 741 g = 401 g C6H5Cl in excess d. Percent yield =200.0 g DDT 100 = 17.1% 1170 g DDT

CHAPTER 371.

STOICHIOMETRY

35

An alternative method to solve limiting-reagent problems is to assume that 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. 5.00 106 g NH3 1 mol NH 3 2 mol HCN = 2.94 105 mol HCN 17.03 g NH 3 2 mol NH 3

5.00 106 g O2

1 mol O 2 2 mol HCN = 1.04 105 mol HCN 32.00 g O 2 3 mol O 2

5.00 106 g CH4

1 mol CH 4 2 mol HCN = 3.12 105 mol HCN 16.04 g CH 4 2 mol CH 4

O2 is limiting because it produces the smallest amount of HCN. Although more product could be produced from NH3 and CH4, only enough O2 is present to produce 1.04 105 mol HCN. The mass of HCN that can be produced is: 1.04 105 mol HCN 27.03 g HCN = 2.81 106 g HCN mol HCN

5.00 106 g O2 72.

1 mol O 2 6 mol H 2 O 18.02 g H 2 O = 5.63 106 g H2O 32.00 g O 2 3 mol O 2 1 mol H 2 O

2 C3H6(g) + 2 NH3(g) + 3 O2(g) 2 C3H3N(g) + 6 H2O(g) a. We will solve this limiting reagent problem using the same method as described in Exercise 3.71. 1.00 103 g C3H6 1 mol C 3 H 6 42.08 g C 3 H 6 2 mol C 3 H 3 N = 23.8 mol C3H3N 2 mol C 3 H 6

1.50 103 g NH3

1 mol NH 3 2 mol C 3 H 3 N = 88.1 mol C3H3N 17.03 g NH 3 2 mol NH 3

2.00 103 g O2

2 mol C 3 H 3 N 1 mol O 2 = 41.7 mol C3H3N 32.00 g O 2 3 mol O 2

C3H6 is limiting because it produces the smallest amount of product, and the mass of acrylonitrile that can be produced is: 23.8 mol 53.06 g C 3 H 3 N = 1.26 103 g C3H3N mol

36b. 23.8 mol C3H3N

CHAPTER 3

STOICHIOMETRY

6 mol H 2 O 18.02 g H 2 O = 1.29 103 g H2O 2 mol C 3 H 3 N mol H 2 O

Amount NH3 needed: 23.8 mol C3H3N 2 mol NH 3 17.03 g NH 3 = 405 g NH3 2 mol C 3 H 3 N mol NH 3

Amount NH3 in excess = 1.50 103 g 405 g = 1.10 103 g NH3 Amount O2 needed: 23.8 mol C3H3N 3 mol O 2 32.00 g O 2 = 1.14 103 g O2 2 mol C 3 H 3 N mol O 2

Amount O2 in excess = 2.00 103 g 1.14 103 g = 860 g O2 1.10 103 g NH3 and 860 g O2 are in excess. 73. P4(s) + 6 F2(g) 4 PF3(g); the theoretical yield of PF3 is: 120. g PF3 (actual) 100.0 g PF3 ( theoretical) = 154 g PF3 (theoretical) 78.1 g PF3 (actual)

154 g PF3

1 mol PF3 6 mol F2 38.00 g F2 = 99.8 g F2 87.97 g PF3 4 mol PF3 mol F2

99.8 g F2 are needed to actually produce 120. g of PF3 if the percent yield is 78.1%. 74. 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 mol of acetaminophen compared with a theoretical yield of 4 mol of acetaminophen. Solving for percent yield by mass (where M = molar mass acetaminophen): percent yield = 3 mol M 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 percent yield of 88%.

CHAPTER 3

STOICHIOMETRY

37

Additional Exercises75. 17.3 g H H 1 mol H 1 mol C = 17.2 mol H; 82.7 g C = 6.89 mol C 1.008 g H 12.01 g C

17.2 = 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 = C4H10 2.59 1023 atoms H 1 molecule C 4 H10 1 mol C 4 H10 10 atoms H 6.022 10 23 molecules

= 4.30 102 mol C4H10 4.30 102 mol C4H10 76. Assuming 100.00 g E3H8: Mol E = 8.73 g H 1 mol H 3 mol E = 3.25 mol E 1.008 g H 8 mol H58.12 g = 2.50 g C4H10 mol C 4 H10

77.

xgE 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 E 0.105 g = 70.9 g/mol Molar mass X2 = 1 mol 20 8.92 10 molecules 6.022 10 23 molecules

The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine. Assuming 100.00 g of MX3 compound: 54.47 g Cl 1 mol = 1.537 mol Cl 35.45 g 1 mol M = 0.5123 mol M 3 mol Cl 45.53 g M = 88.87 g/mol M 0.5123 mol M

1.537 mol Cl

Molar mass of M =

M is the element yttrium (Y), and the name of YCl3 is yttrium(III) chloride. The balanced equation is 2 Y + 3 Cl2 2 YCl3. Assuming Cl2 is limiting:

381.00 g Cl2

CHAPTER 3

STOICHIOMETRY

2 mol YCl3 195.26 g YCl3 1 mol Cl 2 = 1.84 g YCl3 70.90 g Cl 2 3 mol Cl 2 1 mol YCl3 2 mol YCl3 195.26 g YCl3 1 mol Y = 2.20 g YCl3 88.91 g Y 2 mol Y 1 mol YCl3

Assuming Y is limiting: 1.00 g Y

Because Cl2, when it all reacts, produces the smaller amount of product, Cl2 is the limiting reagent, and the theoretical yield is 1.84 g YCl3. 78. 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. 2.00 g C8H8 79. 1 mol C8 H 8 8 mol H 6.022 10 23 atoms = 9.25 1022 atoms H 104.14 g C8 H 8 mol C8 H 8 mol H

Mass of H2O = 0.755 g CuSO4xH2O 0.483 g CuSO4 = 0.272 g H2O 0.483 g CuSO4 1 mol CuSO 4 = 0.00303 mol CuSO4 159.62 g CuSO 4

0.272 g H2O

1 mol H 2 O = 0.0151 mol H2O 18.02 g H 2 O

0.0151 mol H 2 O 4.98 mol H 2 O = ; compound formula = CuSO45H2O, x = 5 0.00303 mol CuSO 4 1 mol CuSO 4

80.

In 1 hour, the 1000. kg of wet cereal 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 H2O 420 + x

The amount of water to be removed is 580 110 = 470 kg/h. 81. 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 a 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. Lets use this information to determine the molar mass of X, which will allow us to identify X from the periodic table. 18.56 g Al 1 mol Al 3 mol X = 1.032 mol X 26.98 g Al 2 mol Al

81.44 g of X must contain 1.032 mol of X.

CHAPTER 3

STOICHIOMETRY81.44 g X = 78.91 g/mol 1.032 mol X

39

Molar mass of X =

From the periodic table, the unknown element is selenium, and the formula is Al2Se3. 82. The reaction is BaX2(aq) + H2SO4(aq) BaSO4(s) + 2 HX(aq). 0.124 g BaSO4 137.3 g Ba 0.0729 g Ba = 0.0729 g Ba; % Ba = 100 = 46.1% 233.4 g BaSO 4 0.158 g BaX 2

The formula is BaX2 (from positions of the elements in the periodic table), and 100.0 g of compound contains 46.1 g Ba and 53.9 g of the unknown halogen. There must also be: 1 mol Ba 2 mol X = 0.672 mol of the halogen in 53.9 g of halogen 137.3 g Ba mol Ba 53.9 g Therefore, the molar mass of the halogen is = 80.2 g/mol. 0.672 mol 46.1 g Ba This molar mass is close to that of bromine. Thus the formula of the compound is BaBr2. 83. 1.20 g CO2 1 mol C 24 H 30 N 3O 1 mol CO 2 1 mol C 376.51 g 44.01 g mol CO 2 24 mol C mol C 24 H 30 N 3O

= 0.428 g C24H30N3O 0.428 g C 24 H 30 N 3O 100 = 42.8% C24H30N3O (LSD) 1.00 g sample 84. Ca3(PO4)2(s) + 3 H2SO4(aq) 3 CaSO4(s) + 2 H3PO4(aq) 1.0 103 g Ca3(PO4)2 1 mol Ca 3 (PO 4 ) 2 = 3.2 mol Ca3(PO4)2 310.2 g Ca 3 (PO 4 ) 2

1.0 103 g conc. H2SO4

98 g H 2SO 4 1 mol H 2SO 4 = 10. mol H2SO4 100 g conc. H 2SO 4 98.1 g H 2SO 4

The required mole ratio from the balanced equation is 3 mol H2SO4 to 1 mol Ca3(PO4)2. The 10. mol H 2SO 4 actual ratio is = 3.1. 3.2 mol Ca 3 (PO 4 ) 2 This is larger than the required mole ratio, so Ca3(PO4)2 is the limiting reagent. 3.2 mol Ca3(PO4)2 3 mol CaSO 4 136.2 g CaSO 4 = 1300 g CaSO4 produced mol Ca 3 (PO 4 ) 2 mol CaSO 4

403.2 mol Ca3(PO4)2

CHAPTER 3

STOICHIOMETRY

2 mol H 3 PO 4 98.0 g H 3 PO 4 = 630 g H3PO4 produced mol Ca 3 (PO 4 ) 2 mol H 3 PO 4

85.

453 g Fe

1 mol Fe 2 O 3 159.70 g Fe 2 O 3 1 mol Fe = 648 g Fe2O3 55.85 g Fe 2 mol Fe mol Fe 2 O 3 648 g Fe 2 O 3 100 = 86.2% 752 g ore

Mass % Fe2O3 =

86.

12

C21H6: 2(12.000000) + 6(1.007825) = 30.046950 amu C1H216O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 amu N16O: 1(14.003074) + 1(15.994915) = 29.997989 amu

12 14

The peak results from 12C1H216O.85

87.

87

Rb atoms = 2.591; Rb atoms

If we had exactly 100 atoms, x = number of 85Rb atoms and 100 x = number of 87Rb atoms.

x 259.1 = 2.591, x = 259.1 (2.591)x, x = = 72.15; 72.15% 85Rb 3.591 100 x

0.7215(84.9117) + 0.2785(A) = 85.4678, A = 88. Assuming 100.00 g of tetrodotoxin: 41.38 g C

85.4678 61.26 = 86.92 amu 0.2785

1 mol C 1 mol N = 3.445 mol C; 13.16 g N = 0.9395 mol N 12.011 g C 14.007 g N

5.37 g H

1 mol H 1 mol O = 5.33 mol H; 40.09 g O = 2.506 mol O 1.008 g H 15.999 g O

Divide by the smallest number:3.445 5.33 2.506 = 3.667; = 5.67; = 2.667 0.9395 0.9395 0.9395

To get whole numbers for each element, multiply through by 3. Empirical formula: (C3.667H5.67NO2.667)3 = C11H17N3O8; the mass of the empirical formula is 319.3 g/mol.

CHAPTER 3

STOICHIOMETRY1.59 10 21 g = 319 g/mol 1 mol 3 molecules 6.022 10 23 molecules

41

Molar mass tetrodotoxin =

Because the empirical mass and molar mass are the same, the molecular formula is the same as the empirical formula, C11H17N3O8. 165 lb 1 kg 10. g 1 10 6 g 1 mol 6.022 10 23 molecules 2.2046 lb kg g 319.3 g 1 mol = 1.4 1018 molecules tetrodotoxin is the LD50 dosage 89. The volume of a gas is proportional to the number of molecules of gas. Thus the formulas are: I: NH3 The mass ratios are: I: 4.634 g N gH II: 6.949 g N gH III: 41.7 g N gH II: N2H4 III: HN3

If we set the atomic mass of H equal to 1.008, then the atomic mass for nitrogen is: I: 14.01 II: 14.01 III. 14.0A 4.634 = , A = 14.01 3(1.008) 1

For example, for compound I: 90.

PaO2 + O2 PaxOy (unbalanced) 0.200 g PaO2 231 g Pa = 0.1757 g Pa (We will carry an extra significant figure.) 263 g PaO 2 1 mol O = 2.025 103 mol O 16.00 g O

0.2081 g PaxOy 0.1757 g Pa = 0.0324 g O; 0.0324 g O H 0.1757 g Pa H1 mol Pa = 7.61 104 mol Pa 231 g Pa

2.025 10 3 mol O Mol O 2 8 mol O = 2.66 . 2 = ; empirical formula: Pa3O8 = 4 Mol Pa 3 3 mol Pa 7.61 10 mol Pa 91. 1.375 g AgI 1.375 g AgI 1 mol AgI = 5.856 103 mol AgI = 5.856 103 mol I 234.8 g AgI 126.9 g I = 0.7431 g I; XI2 contains 0.7431 g I and 0.257 g X. 234.8 g AgI

425.856 103 mol I 1 mol X = 2.928 103 mol X 2 mol I

CHAPTER 3

STOICHIOMETRY

Molar mass =

0.257 g X 87.8 g = ; atomic mass = 87.8 amu (X is Sr.) 3 mol 2.928 10 mol X

92.

40.0/A x (40.0)A z mol X =2= = or Az = 3Ax X2Z: 40.0% X and 60.0% Z by mass; mol Z 60.0/A z (60.0)A x where A = molar mass For XZ2, molar mass = Ax + 2Az = Ax + 2(3Ax) = 7Ax. Mass % X =Ax 100 = 14.3% X; % Z = 100.0 14.3 = 85.7% Z 7A x

93.

Assuming 1 mole of vitamin A (286.4 g vitamin A): mol C = 286.4 g vitamin A 0.8396 g C 1 mol C = 20.00 mol C g vitamin A 12.011 g C

mol H = 286.4 g vitamin A

0.1056 g H 1 mol H = 30.01 mol H g vitamin A 1.0079 g H

Because 1 mole of vitamin A contains 20 mol C and 30 mol H, the molecular formula of vitamin A is C20H30E. To determine E, lets calculate the molar mass of E: 286.4 g = 20(12.01) + 30(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 C20H30O. 94. We would see the peaks corresponding to:10 10

B35Cl3 [mass 10 + 3(35) = 115 amu], 10B35Cl237Cl (117), 10B35Cl37Cl2 (119), B37Cl3 (121), 11B35Cl3 (116), 11B35Cl237Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122)

We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122.

Challenge Problems95. When the discharge voltage is low, the ions present are in the form of molecules. When the discharge voltage is increased, the bonds in the molecules are broken, and the ions present are in the form of individual atoms. Therefore, the high discharge data indicate that the ions 16 + 18 + O , O , and 40Ar+ are present. The only combination of these individual ions that can explain the mass data at low discharge is 16O16O+ (mass = 32), 16O18O+ (mass = 34), and 40Ar+ (mass = 40). Therefore, the gas mixture contains 16O16O, 16O18O, and 40Ar. To determine the

CHAPTER 3

STOICHIOMETRY

43

percent composition of each isotope, we use the relative intensity data from the high discharge data to determine the percentage that each isotope contributes to the total relative intensity. For 40Ar:1.0000 1.0000 100 = 100 = 57.094% 40Ar 1.7515 0.7500 + 0.0015 + 1.0000

For 16O:

0.7500 0.0015 100 = 42.82% 16O; for 18O: 100 = 8.6 102% 18O 1.7515 1.7515

Note: 18F instead of 18O could also explain the data. However, OF(g) is not a stable compound. This is why 18O is the best choice because O2(g) does form.

96.

Fe(s) +

1 2

O 2 (g) FeO(s) ; 2 Fe(s) +1 mol Fe = 0.3581 mol 55.85 g

3 2

O 2 (g ) Fe 2 O 3 (s)

20.00 g Fe

(11.20 3.24) g O 2

1 mol O 2 = 0.2488 mol O2 consumed (1 extra sig. fig.) 32.00 g

Assuming x mol of FeO is produced from x mol of Fe, so that 0.3581 x mol of Fe reacts to form Fe2O3:x Fe +

1 x O 2 x FeO 2 3 0.3581 x 0.3581 x (0.3581 x) mol Fe + mol O 2 mol Fe 2 O 3 2 2 2

Setting up an equation for total moles of O2 consumed:1 2

x +

3 4

(0.3581 x) = 0.2488 mol O 2 ,

x = 0.0791 = 0.079 mol FeO

0.079 mol FeO

71.85 g FeO = 5.7 g FeO produced mol

Mol Fe2O3 produced =

0.3581 0.079 = 0.140 mol Fe2O3 2

0.140 mol Fe2O3 97.

159.70 g Fe 2 O 3 = 22.4 g Fe2O3 produced mol

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 mole ratios, 10.00 g XCl2 must also contain 2.55 g Cl; mass X in XCl2 = 10.00 ! 2.55 = 7.45 g X.

442.55 g Cl

CHAPTER 3

STOICHIOMETRY

1 mol XCl 2 1 mol Cl 1 mol X = 3.60 10 2 mol X 35.45 g Cl 2 mol Cl mol XCl 2

So, 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 X 98. 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. Zn + moles Mg. Mol H2 = 0.5171 g H2 0.2565 = From the balanced equations, moles H2 = moles

1 mol H 2 = 0.2565 mol H2 2.0158 g H 2

x 10.00 x + ; solving, x = 4.008 g Mg. 24.31 65.38

4.008 g 100 = 40.08% Mg 10.00 g 99. For a gas, density and molar mass are directly proportional to each other. Molar mass XHn = 2.393(32.00) = 8.91 10 2 mol H 4 mol H = 2 mol XH n 2.23 10 mol XH n Molar mass X = 76.58 4(1.008 g) = 72.55 g/mol; the element is Ge. 100. The balanced equation is 2 Sc(s) + 2x HCl(aq) 2 ScClx(aq)+ x H2(g). The mol ratio of Sc to H2 = Mol Sc = 2.25 g Sc 2 . x 2 mol H 76.58 g ; 0.803 g H2O mol 18.02 g H 2 O = 8.91 102 mol H

1 mol Sc = 0.0500 mol Sc 44.96 g Sc1 mol H 2 = 0.07451 mol H2 2.0158 g H 2

Mol H2 = 0.1502 g H2

2 0.0500 = , x = 3; the formula is ScCl3. x 0.07451

CHAPTER 3101.

STOICHIOMETRY

45

4.000 g M2S3 3.723 g MO2 There must be twice as many moles of MO2 as moles of M2S3 in order to balance M in the reaction. Setting up an equation for 2(mol M2S3) = mol MO2 where A = molar mass M: 4.000 g 3.723 g 8.000 3.723 2 2A + 3(32.07) = A + 2(16.00) , 2A + 96.21 = A + 32.00 (8.000)A + 256.0 = (7.446)A + 358.2, (0.554)A = 102.2, A = 184 g/mol; atomic mass = 184 amu

102.

We know that water is a product, so one of the elements in the compound is hydrogen. XaHb + O2 H2O + ? To balance the H atoms, the mole ratio between XaHb and H2O = Mol compound = 2 . b

1.39 g 1.21 g = 0.0224 mol; mol H2O = = 0.0671 mol 62.09 g / mol 18.02 g / mol

2 0.0224 = , b = 6; XaH6 has a molar mass of 62.09 g/mol. b 0.0671 62.09 = a(molar mass of X) + 6(1.008), a(molar mass of X ) = 56.04 Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), and Li (a = 8). N fits the data best so N4H6 is the most likely formula. 103. The balanced equations are: C(s) + 1/2 O2(g) CO(g) and C(s) + O2(g) CO2(g) If we have 100.0 mol of products, then we have 72.0 mol CO2, 16.0 mol CO, and 12.0 mol O2. The initial moles of C equals 72.0 (from CO2) + 16.00 (from CO) = 88.0 mol C and the initial moles of O2 equals 72.0 (from CO2) + 16.0/2 (from CO) + 12.0 (unreacted O2) = 92.0 mol O2. The initial reaction mixture contained:92.0 mol O 2 = 1.05 mol O2/mol C 88.0 mol C

104.

CxHyOz + oxygen x CO2 + y/2 H2O2.20 g CO 2 1 mol C 1 mol CO 2 12.01 g C 44.01 g CO 2 mol CO 2 mol C = 60.0% C 1.00 g aspirin

Mass % C in aspirin =

460.400 g H 2 O

CHAPTER 3

STOICHIOMETRY

Mass % H in aspirin =

1 mol H 2 O 2 mol H 1.008 g H 18.02 g H 2 O mol H 2 O mol H = 4.48% H 1.00 g aspirin

Mass % O = 100.00 ! (60.0 + 4.48) = 35.5% O Assuming 100.00 g aspirin: 60.0 g C 1 mol C 1 mol H = 5.00 mol C; 4.48 g H = 4.44 mol H 12.01 g C 1.008 g H

1 mol O = 2.22 mol O 16.00 g O 5.00 4.44 Dividing by the smallest number: = 2.25; = 2.00 2.22 2.22

35.5 g O

Empirical formula: (C2.25 H2.00O)4 = C9H8O4. Empirical mass 9(12) + 8(1) + 4(16) = 180 g/mol; this is in the 170190 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 = C7H6O3 105. LaH2.90 is the formula. If only La3+ is present, LaH3 would be the formula. If only La2+ is present, LaH2 would be the formula. Let x = mol La2+ and y = mol La3+: (La2+)x(La3+)yH(2x + 3y) where x + y = 1.00 and 2x + 3y = 2.90 Solving by simultaneous equations: 2x + 3y = 2.90 2x 2y = 2.00 y = 0.90 and x = 0.10 LaH2.90 contains 106. 1 9 La2+, or 10.% La2+, and La3+, or 90.% La3+. 10 10

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) 30.07 g/mol 44.09 g/mol Let x = mass C2H6, so 9.780 ! x = mass C3H8. Use the balanced reaction to set up an equation for the moles of O2 required.x 7 9.780 x 5 + = 1.120 mol O2 30.07 2 44.09 1

CHAPTER 3

STOICHIOMETRY3.7 g 100 = 38% C2H6 by mass 9.780 g

47

Solving: x = 3.7 g C2H6; 107 .

Let x = mass KCl and y = mass KNO3. Assuming 100.0 g of mixture, x + y = 100.0 g. Molar mass KCl = 74.55 g/mol; molar mass KNO3 = 101.11 g/mol Mol KCl =x y ; mol KNO3 = 74.55 101.11

Knowing that the mixture is 43.2% K, then in the 100.0 g mixture:y x 39.10 + = 43.2 74.55 101.11

We have two equations and two unknowns: (0.5245)x + (0.3867)y = 43.2x + y = 100.0

Solving, x = 32.9 g KCl; 108.

32.9 g 100 = 32.9% KCl 100.0 g

Let M = unknown element Mass % M = mass M 2.077 100 = 100 = 56.01% M total mass compound 3.708

100.00 56.01 = 43.99% O Assuming 100.00 g compound: 43.99 g O 1 mol O = 2.750 mol O 15.999 g O

If MO is the formula of the oxide, then M has a molar mass of

56.01 g M = 20.37 g/mol. 2.750 mol M This is too low for the molar mass. We must have fewer moles of M than moles O present in the formula. Some possibilities are MO2, M2O3, MO3, etc. It is a guessing game as to which

to try. Lets assume an MO2 formula. Then the molar mass of M is: 56.01 g M = 40.73 g/mol 1 mol M 2.750 mol O 2 mol O This is close to calcium, but calcium forms an oxide having the CaO formula, not CaO2.

48

CHAPTER 3

STOICHIOMETRY

If MO3 is assumed to be the formula, then the molar mass of M calculates to be 61.10 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, and 2.75 (these are reasonable because 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: 56.01 g M = 50.92 g/mol 1 mol M 2.750 mol O 2.5 mol O This is the molar mass of vanadium and V2O5 is a reasonable formula for an oxide of vanadium. The other choices for the O : M mole ratios between 2 and 3 do not give as reasonable results. Therefore, M is vanadium, and the formula is V2O5. 109. 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 moles of NO formed, and let 4y = number of moles of NO2 formed. Then: 4x NH3 + 5x O2 4x NO + 6x H2O and 4y NH3 + 7y O2 4y NO2 + 6y H2O All the NH3 reacted, so 4x + 4y = 2.00. 10.00 6.75 = 3.25 mol O2 reacted, so 5x + 7y = 3.25. Solving by the method of simultaneous equations: 20x + 28y = 13.0 20x 20y = 10.0 8y = 3.0, y = 0.38; 4x + 4 0.38 = 2.00, x = 0.12 Mol NO = 4x = 4 0.12 = 0.48 mol NO formed 110.a N2H4 + b NH3 + (10.00 4.062) O2 c NO2 + d H2O

Setting up four equations to solve for the four unknowns: 2a + b = c 2c + d = 2(10.00 4.062) 4 a + 3b = 2da(32.05) + b(17.03) = 61.00

(N mol balance) (O mol balance) (H mol balance) (mass balance)

Solving the simultaneous equations gives a = 1.12 = 1.1 mol N2H4.1.1 mol N 2 H 4 32.05 g / mol N 2 H 4 100 = 58% N2H4 61.00 g

CHAPTER 3

STOICHIOMETRY

49

Marathon Problems111. To solve the limiting-reagent problem, we must determine the formulas of all the compounds so that we can get a balanced reaction. a. 40 million trillion = (40 106) 1012 = 4.000 1019 (assuming 4 sig. figs.) 4.000 1019 molecules A 1 mol A = 6.642 10-5 mol A 23 6.022 10 molecules A

Molar mass of A =

4.26 10 3 g A = 64.1 g/mol 6.642 10 5 mol A

Mass of carbon in 1 mol of A is: 64.1 g A 37.5 g C = 24.0 g carbon = 2 mol carbon in substance A 100.0 g A

The 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: 5.36 g H 1 mol H 5.32 = 5.32 mol H; = 2.00 1.008 g H 2.66

42.5 g O

1 mol O 2.66 = 2.66 mol O; = 1.00 2.66 16.00 g O

Empirical 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: 33.8 g CO2 1 mol CO 2 1 mol C 12.01 g C = 9.22 g carbon 44.01 g CO 2 mol CO 2 mol C

6.92 g H2O

1 mol H 2 O 2 mol H 1.008 g H = 0.774 g hydrogen 18.02 g H 2 O mol H 2 O mol H

9.22 g carbon + 0.774 g hydrogen = 9.99 g; because substance C initially weighed 10.0 g, there is no oxygen present in substance C. Determining the empirical formula for substance C:

509.22 g 1 mol C = 0.768 mol carbon 12.01 g C

CHAPTER 3

STOICHIOMETRY

0.774 g H

1 mol 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. Because the mass spectrum data indicate a molar mass of 26 g/mol, 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) 45.0 g CaC2 1 mol CaC 2 = 0.702 mol CaC2 64.10 g CaC 21 mol H 2 O mol H 2 O 1.28 = = 1.28 mol H2O; = 1.82 18.02 g H 2 O mol CaC 2 0.702

23.0 g H2O

Because the actual mole ratio present is smaller than the required 2 : 1 mole ratio from the balanced equation, H2O is limiting. 1.28 mol H2O 112. a. i.1 mol C 2 H 2 26.04 g C 2 H 2 = 16.7 g C2H2 = mass of product C 2 mol H 2 O mol C 2 H 2

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 fewer 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 fewer number of moles of B and we need more B (from the balanced reaction). b. A + 5 B 3 CO2 + 4 H2O To conserve mass: 44.01 + 5(B) = 3(44.01) + 4(18.02); solving: B = 32.0 g/mol Because B is diatomic, the best choice for B is O2. c. We can solve this without mass percent data simply by balancing the equation: A + 5 O2 3 CO2 + 4 H2O A must be C3H8 (which has a similar molar mass to CO2). This is also the empirical formula.Note:

3(12.01) 100 = 81.71% C. So this checks. 3(12.01) + 8(1.008)

CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRYAqueous Solutions: Strong and Weak Electrolytes10. Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, for example, concentrated sugar water. Acids are either strong or weak electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in water and do not dissolve. b. KF is a soluble ionic compound, so it is a strong electrolyte. KF(aq) actually exists as separate hydrated K+ ions and hydrated F ions in solution. C6H12O6 is a polar covalent molecule that is a nonelectrolyte. C6H12O6 is hydrated as described in part a. c. RbCl is a soluble ionic compound, so it exists as separate hydrated Rb+ ions and hydrated Cl ions in solution. AgCl is an insoluble ionic compound so the ions stay together in solution and fall to the bottom of the container as a precipitate. d. HNO3 is a strong acid and exists as separate hydrated H+ ions and hydrated NO3 ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a. 12. 13. MgSO4(s) Mg2+(aq) + SO42(aq); NH4NO3(s) NH4+(aq) + NO3(aq) a. Ba(NO3)2(aq) Ba2+(aq) + 2 NO3(aq); picture iv represents the Ba2+ and NO3 ions present in Ba(NO3)2(aq). b. NaCl(aq) Na+(aq) + Cl(aq); picture ii represents NaCl(aq).

11.

51

52

CHAPTER 4

SOLUTION STOICHIOMETRY

c. K2CO3(aq) 2 K+(aq) + CO32(aq); picture iii represents K2CO3(aq). d. MgSO4(aq) Mg2+(aq) + SO42(aq); picture i represents MgSO4(aq).

Solution Concentration: Molarity14. 75.0 mL

0.79 g 1 mol 1.3 mol = 5.2 M C2H5OH = 1.3 mol C2H5OH; molarity = mL 46.1 g 0.250 L0.250 mol NaOH 40.00 g NaOH = 20.0 g NaOH L mol

15.

a. 2.00 L

Place 20.0 g NaOH in a 2-L volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way. b. 2.00 L 0.250 mol NaOH 1 L stock = 0.500 L L 1.00 mol NaOH

Add 500. mL of 1.00 M NaOH stock solution to a 2-L volumetric flask; fill to the mark with water, mixing several times along the way. c. 2.00 L 0.100 mol K 2 CrO 4 194.20 g K 2 CrO 4 = 38.8 g K2CrO4 L mol K 2 CrO 4

Similar to the solution made in part a, instead using 38.8 g K2CrO4. d. 2.00 L 0.100 mol K 2 CrO 4 1 L stock = 0.114 L L 1.75 mol K 2 CrO 4

Similar to the solution made in part b, instead using 114 mL of the 1.75 M K2CrO4 stock solution. 16. a.

M Ca ( NO3 ) 2 =

0.100 mol Ca ( NO3 ) 2 = 1.00 M 0.100 L3

Ca(NO3)2(s) Ca2+(aq) + 2 NO3(aq); M Ca 2 + = 1.00 M; M NO = 2(1.00) = 2.00 M b.M Na 2SO 4 =

2.5 mol Na 2SO 4 = 2.0 M 1.25 L4 2

Na2SO4(s) 2 Na+(aq) + SO42(aq); M Na + = 2(2.0) = 4.0 M; M SO c. 5.00 g NH4Cl 1 mol NH 4 Cl = 0.0935 mol NH4Cl 53.49 g NH 4 Cl

= 2.0 M

CHAPTER 4

SOLUTION STOICHIOMETRY0.0935 mol NH 4 Cl = 0.187 M 0.5000 L4 +

53

M NH 4Cl =

NH4Cl(s) NH4+(aq) + Cl(aq); M NH d. 1.00 g K3PO4

= M Cl = 0.187 M

1 mol K 3 PO 4 = 4.71 103 mol K3PO4 212.2