Preview Student Solutions Manual for ZumdahlDeCostes Chemical Principles, 8th Edition by Steven S. Zumdahl, Donald J. DeCoste (2016)

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How to Use This Guide v

Chapter 2 Atoms, Molecules, and Ions 1

Chapter 3 Stoichiometry 19

Chapter 4 Types of Chemical Reactions and Solution Stoichiometry 59

Chapter 5 Gases 108

Chapter 6 Chemical Equilibrium 161

Chapter 7 Acids and Bases 207

Chapter 8 Applications of Aqueous Equilibria 273

Chapter 9 Energy, Enthalpy, and Thermochemisty 375

Chapter 10 Spontaneity, Entropy, and Free Energy 394

Chapter 11 Electrochemistry 442

Chapter 12 Quantum Mechanics and Atomic Theory 489

Chapter 13 Bonding: General Concepts 527

Chapter 14 Covalent Bonding: Orbitals 583

Chapter 15 Chemical Kinetics 624

Chapter 16 Liquids and Solids 677

Chapter 17 Properties of Solutions 723

Chapter 18 The Representative Elements 763

Chapter 19 Transition Metals and Coordination Chemistry 791

Chapter 20 The Nucleus: A Chemist’s View 823

Chapter 21 Organic Chemistry 843

iii

1

CHAPTER 2

ATOMS, MOLECULES, AND IONS

Development of the Atomic Theory

18 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 Avogadro’s 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 + 5 F2 → 2 X Two molecules of X contain 10 atoms of F and two atoms of Cl The formula of X is ClF5 for a balanced equation

20 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 Avogadro’s hypothesis (law) implies that volume ratios are proportional 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 Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at

constant temperature and pressure Here, 1 volume of N2 reacts with 3 volumes of H2 to produce 2 volumes of the gaseous product or in terms of molecule ratios:

1 N2 + 3 H2 → 2 product

In order for the equation to be balanced, the product must be NH3

22 For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with 1 g of

carbon From the formulas (two oxygen atoms per carbon atom in CO2 versus one oxygen atom per carbon atom in CO), CO2 will have twice the mass of oxygen that combines per gram of carbon as compared to CO For CO2 and C3O2, it is easiest to concentrate on the

mass of carbon that combines with 1 g of oxygen From the formulas (three carbon atoms per two oxygen atoms in C3O2 versus one carbon atom per two oxygen atoms in CO2), C3O2 will have three times the mass of carbon that combines per gram of oxygen as compared to CO2

As expected, the mass ratios are whole numbers as predicted by the law of multiple proportions

23 Hydrazine: 1.44 × 10−1 g H/g N; ammonia: 2.16 × 10−1g H/g N; hydrogen azide:

2.40 × 10−2g H/g N Let's try all of the ratios:

0240

0

144

0

= 6.00;

0240.0

216.0

= 9.00;

0240.0

0240.0

= 1.00;

144.0

216.0 = 1.50 =

23

All the masses of hydrogen in these three compounds can be expressed as simple whole- number ratios The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios

6 : 9 : 1

24 Compound 1: 21.8 g C and 58.2 g O (80.0 – 21.8 = mass O)

Compound 2: 34.3 g C and 45.7 g O (80.0 – 34.3 = mass O)

The mass of carbon that combines with 1.0 g of oxygen is:

Compound 1:

Og2.58

Cg8.21

= 0.375 g C/g O

Compound 2:

Og7.45

Cg3.34

= 0.751 g C/g O

The ratio of the masses of carbon that combine with 1 g of oxygen is

1

2375.0

751

supports the law of multiple proportions because this carbon ratio is a small whole number

25 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:

126

0

875

2

= 22.8; Mg:

126.0

500.1 = 11.9; O:

126.0

00.1 = 7.94

H O Na Mg Relative value 1.00 7.94 22.8 11.9

Accepted value 1.0079 15.999 22.99 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 Atom

26 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 β particles are electrons A cathode ray is a stream of electrons (β particles)

27 From section 2.6, the nucleus has “a diameter of about 10−13 cm” and the electrons “move

about the nucleus at an average distance of about 10−8 cm from it.” We will use these statements to help determine the densities Density of hydrogen nucleus (contains one proton only):

Vnucleus = 3 (3.14 (5 10 14cm)3 5 10 40cm3

3

4rπ3

24

g/cm103cm105

g1067.1

g/cm0.4cm

104

g109g1067

×

×+

×

28 From Section 2.6 of the text, the average diameter of the nucleus is approximately 10−13 cm,

and the electrons move about the nucleus at an average distance of approximately10−8cm From this, the diameter of an atom is about 2 × 10−8cm

cm101

cm102

in360,63grape1

ft5280grape

1040

6

1056

68.7 = 12.00;

640.0

84.3 = 6.00

Because all charges are whole-number multiples of 6.40 × 10−13 zirkombs, the charge on one electron could be 6.40 × 10−13 zirkombs However, 6.40 × 10−13 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 × 10−13 zirkombs or an integer fraction of 6.40 × 10−13

30 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

31 If the plum pudding model were correct (a diffuse positive charge with electrons scattered

throughout), then α 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 α particles were deflected at very large angles Rutherford reasoned that the large deflections of these α particles could be caused only by a center of concentrated positive charge that contains most

of the atom’s mass (the nuclear model of the atom)

Elements, Ions, and the Periodic Table

32 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 SO4 − are all anions A cation is a positively charged ion, for example, Na+, Fe3+, and NH4+ are all cations

33 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

is discussed 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

34 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

35 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)

36 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

37 Use the periodic table to identify the elements

a Cl; halogen b Be; alkaline earth metal

c Eu; lanthanide metal d Hf; transition metal

e He; noble gas f U; actinide metal

g Cs; alkali metal

38 The number and arrangement of electrons in an atom determine how the atom will react with

other atoms, i.e., the electrons determine the chemical properties of an atom The number of

neutrons present determines the isotope identity and the mass number

39 For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number

of neutrons When the number of protons and neutrons is equal to each other, the mass

number (protons + neutrons) will be twice the atomic number (protons) Therefore, for

lighter isotopes, the ratio of the mass number to the atomic number is close to 2 For

example, consider 28Si, which has 14 protons and (28 – 14 =) 14 neutrons Here, the mass

number to atomic number ratio is 28/14 = 2.0 For heavier isotopes, there are more neutrons

than protons in the nucleus Therefore, the ratio of the mass number to the atomic number

increases steadily upward from 2 as the isotopes get heavier and heavier For example, 238U

has 92 protons and (238 – 92 =) 146 neutrons The ratio of the mass number to the atomic

number for 238U is 238/92 = 2.6

40 a transition metals b alkaline earth metals c alkali metals

44

Symbol Number of Protons in

Nucleus

Number of Neutrons in Nucleus

Number of Electrons

Net Charge

47 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 2e−to form Ra2+ b Lose 3e− to form In3+ c Gain 3e− to form P3−

d Gain 2e− to form 2−

Te e Gain 1e− to form Br− f Lose 1e− to form Rb+

48 See Exercise 47 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−

Nomenclature

49 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)

50 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)

51 a mercury(I) oxide b iron(III) bromide

c cobalt(II) sulfide d titanium(IV) chloride

52 a barium sulfite b sodium nitrite

c potassium permanganate d potassium dichromate

g Pb(CO3)2 h NH4C2H3O2

53 a sulfur difluoride b dinitrogen tetroxide

c iodine trichloride d tetraphosphorus hexoxide

e sulfur hexafluoride f sodium hydrogen phosphate

g sodium dihydrogen phosphate h lithium nitride

55 a copper(I) iodide b copper(II) iodide c cobalt(II) iodide

d sodium carbonate e sodium hydrogen carbonate or sodium bicarbonate

f tetrasulfur tetranitride g selenium tetrabromide h sodium hypochlorite

i barium chromate j ammonium nitrate

56 a acetic acid b ammonium nitrite c colbalt(III) sulfide

d iodine monochloride e lead(II) phosphate f potassium chlorate

g sulfuric acid h strontium nitride i aluminum sulfite

j tin(IV) oxide k sodium chromate l hypochlorous acid

e Li3N f Cr2(CO3)3 g Cr(C2H3O2)2 h SnF4

i NH4HSO4: composed of NH4+ and HSO4 − ions

j (NH4)2HPO4 k KClO4 l NaH

i GaAs: We would predict the stable ions to be Ga3+ and As3−

j CdSe k ZnS l Hg2Cl2: Mercury(I) exists as Hg22+

m HNO2 n P2O5

59 a Pb(C2H3O2)2; lead(II) acetate b CuSO4; copper(II) sulfate

c CaO; calcium oxide d MgSO4; magnesium sulfate

e Mg(OH)2; magnesium hydroxide f CaSO4; calcium sulfate

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

60 a 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

i HNO3 is nitric acid, not nitrate acid Nitrate acid does not exist

j H2S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name)

H2SO4 is sulfuric acid

61 a nitric acid, HNO3 b perchloric acid, HClO4 c acetic acid, HC2H3O2

d sulfuric acid, H2SO4 e phosphoric acid, H3PO4

Additional Exercises

62 Yes, 1.0 g H would react with 37.0 g 37Cl, and 1.0 g H would react with 35.0 g 35Cl

No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37Cl and 1 g H/35 g Cl for 35Cl

As long as we had pure 37Cl or pure 35Cl, the ratios will always hold If we have a mixture (such as the natural abundance of chlorine), the ratio will also be constant as long as the composition of the mixture of the two isotopes does not change

63 a This represents ionic bonding Ionic bonding is the electrostatic attraction between

anions and cations

b This represents covalent bonding where electrons are shared between two atoms This could be the space-filling model for H2O or SF2 or NO2, etc

64 J J Thomson discovered electrons He postulated that all atoms must contain electrons, but

Thomson also postulated that atoms must contain positive charge in order for the atom to be electrically neutral Henri Becquerel discovered radioactivity Lord Rutherford proposed the nuclear model of the atom Dalton's original model proposed that atoms were indivisible particles (that is, atoms had no internal structure) Thomson and Becquerel discovered subatomic particles, and Rutherford's model attempted to describe the internal structure of the atom composed of these subatomic particles In addition, the existence of isotopes, atoms of the same element but with different mass, had to be included in the model

65 The equation for the reaction between the elements of sodium and chlorine is 2 Na(s) + Cl2(g)

→ 2 NaCl(s) The sodium reactant exists as singular sodium atoms packed together very tightly and in a very organized fashion This type of packing of atoms represents the solid phase The chlorine reactant exists as Cl2 molecules In the picture of chlorine, there is a lot

of empty space present This only occurs in the gaseous phase When sodium and chlorine react, the ionic compound NaCl is the product NaCl exists as separate Na+ and Cl− ions Because the ions are packed very closely together and are packed in a very organized fashion, NaCl is depicted in the solid phase

66 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

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

67 From the law of definite proportions, a given compound always contains exactly the same

proportion of elements by mass The first sample of chloroform has a total mass of 12.0 g C + 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures) The mass percent of carbon in this sample of chloroform is:

totalg41.119

Cg0.12

× 100 = 10.05% C by mass

From the law of definite proportions, the second sample of chloroform must also contain

10.05% C by mass Let x = mass of chloroform in the second sample:

x

Cg0

30

× 100 = 10.05, x = 299 g chloroform

68 Mass is conserved in a chemical reaction

chromium(III) oxide + aluminum → chromium + aluminum oxideMass: 34.0 g 12.1 g 23.3 ?

Mass aluminum oxide produced = (34.0 + 12.1) − 23.3 = 22.8 g

69 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

70 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 c Ba2+: 56 protons; 54 electrons; BaO

d Cs+: 55 protons; 54 electrons; Cs2O e S2−: 16 protons; 18 electrons; Al2S3

f P3−: 15 protons; 18 electrons; AlP g Br−: 35 protons; 36 electrons; AlBr3

h N3−: 7 protons; 10 electrons; AlN

71 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

72 Number of electrons in the unknown ion:

2.55 × 26

10− g ×

kg1011.9

electron1

g1000

kg1

proton1g1000

kg1

The number of electrons in the unknown atom:

3.92 × 10−26 g ×

kg011.9

electron1

g1000

kg1

31

−

×

In a neutral atom, the number of protons and electrons is the same Therefore, this is element

proton1g1000

kg1

73 The halogens have a high affinity for electrons, and one important way they react is to form

anions of the type X− The alkali metals tend to give up electrons easily and in most of their compounds exist as M+ cations Note: These two very reactive groups are only one electron

away (in the periodic table) from the least reactive family of elements, the noble gases

74 The solid residue must have come from the flask

75 In the case of sulfur, SO4 − is sulfate, and SO3 − is sulfite By analogy:

SeO4 −: selenate; SeO3 −: selenite; TeO4 −: tellurate; TeO3 −: tellurite

76 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 iodous acid

hypo-77 If the formula is InO, then one atomic mass of In would combine with one atomic mass of O,

or:

Og000.1

Ing784.400

Ing784.400.16)

3

(

A

The latter number is the atomic mass of In used in the modern periodic table

78 a Ca2+ and N3−: Ca3N2, calcium nitride b K+ and O2−: K2O, potassium oxide

c Rb+ and F−: RbF, rubidium fluoride d Mg2+ and S2−: MgS, magnesium sulfide

e Ba2+ and I−: BaI2, barium iodide

f Al3+ and Se2−: Al2Se3, aluminum selenide

g Cs+ and P3−: Cs3P, cesium phosphide

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

79 The 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: 27.2 g C and 72.8 g O (100.0 - 27.2 = mass O) Compound 2: 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:

Og8.72

Cg2.27

= 0.374 g C/g O

Compound 2:

Og1.57

Cg9.42

= 0.751 g C/g O

1

2374

80 a This is element 52, tellurium Te forms stable 2− charged ions in ionic compounds (like

other oxygen family members)

b Rubidium Rb, element 37, forms stable 1+ charged ions

c Argon Ar is element 18 d Astatine At is element 85

81 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

Number of electrons

b False; this was J J Thomson

c False; a proton is about 1800 times more massive than an electron

d The nucleus contains the protons and the neutrons

84 carbon tetrabromide, CBr4; cobalt(II) phosphate, Co3(PO4)2;

magnesium chloride, MgCl2; nickel(II) acetate, Ni(C2H3O2)2;

calcium nitrate, Ca(NO3)2

85 Co(NO2)2, cobalt(II) nitrite; AsF5, arsenic pentafluoride; LiCN, lithium cyanide;

K2SO3, potassium sulfite; Li3N, lithium nitride; PbCrO4, lead(II) chromate

86 K will lose 1 e− to form K+ Cs will lose 1 e− to form Cs+

Br will gain 1 e− to form Br− Sulfur will gain 2 e− to formS2 −

Se will gain 2 e− to form Se2 −

c Ga is a metal and is expected to lose electrons when forming ions

d True

e Titanium(IV) oxide is correct for this transition metal ionic compound

Challenge Problems

88 Because the gases are at the same temperature and pressure, the volumes are directly

proportional to the number of molecules present Let’s 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:

89 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

d 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)

90 For each experiment, divide the larger number by the smaller In doing so, we get:

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 ×

XYg)4.02.4(

Yg2.4+ = 19.2 g Y (and 1.8 g X)

If the compound in experiment 3 has the XY formula, then:

21 g XY ×

XYg)0.20.7(

Yg0.7+ = 16.3 g Y (and 4.7 g X)

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

91 Compound I:

Qg00.3

Rg0.14

=

Qg00.1

Rg67.4

; Compound II:

Qg50.4

Rg00.7

=

Qg00.1

Rg56.1

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

56.1

67

4 = 2.99 ≈ 3

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

92 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 × 10−24 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

1

g1031

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]

93 Avogadro proposed that equal volumes of gases (at constant temperature and pressure)

contain equal numbers of molecules In terms of balanced equations, Avogadro’s hypothesis (law) implies that volume ratios will be identical to molecule ratios Assuming one molecule

of octane reacts, 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

94 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 + Y b → 2 XeYf

From the balanced equations, a = 2c = e and b = d = 2f

Substituting into the balanced equations:

00.1+ = 0.3043, y = 1.14

Compound II = X2Y: If X has relative mass of 1.00,

y

+00.2

00.2

602.0 = 2.04; A = 2.04 when B = 1.00

172.0

401.0 = 2.33; C = 2.33 when B = 1.00

320.0

374.0 = 1.17; C = 1.17 when A = 1.00

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:

602.0B

mass

)Amass(6

4

2 = ; mass A2 = 0.340(mass B4)

172.0

401.0B

mass

)Cmass(4

4

3 = ; mass C3 = 0.583(mass B4)

320.0

374.0)Amass(3

)Cmass(2

Note that any set of balanced reactions that confirms the initial mass data is correct This

is just one possibility

19

CHAPTER 3

STOICHIOMETRY

Atomic Masses and the Mass Spectrometer

23 Average atomic mass = A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947)

+ 0.0550(48.947841) + 0.0540(49.944792) = 47.88 u

This is element Ti (titanium) Note: u is an abbreviation for amu (atomic mass units)

24 Because we are not given the relative masses of the isotopes, we need to estimate the masses

of the isotopes A good estimate is to assume that only the protons and neutrons contribute to the overall mass of the atom and that the atomic mass of a proton and neutron are each 1.00 u

So the masses are about: 54Fe, 54.00 u; 56Fe, 56.00 u; 57Fe, 57.00 u; 58Fe, 58.00 u Using these masses and the abundances given in the mass spectrum, the calculated average atomic mass would be:

0.0585(54.00) + 0.9175(56.00) + 0.0212(57.00) + 0.0028(58.00) = 55.91 u

The average atomic mass listed in the periodic table is 55.85 u

25 If silver is 51.82% 107Ag, then the remainder is 109Ag (48.18%) Determining the atomic

mass (A) of 109Ag:

107.868 =

100

)A(18.48)905.106(82

10786.8 = 5540 + (48.18)A, A = 108.9 u = atomic mass of 109Ag

26 Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100 − x

151.96 =

100

)9209.152)(

100()9196.150

15196 = (150.9196)x + 15292.09 − (152.9209)x, −96 = −(2.0013)x

x = 48%; 48% 151Eu and 100 − 48 = 52% 153Eu

= 185 u (A = 184.95 u without rounding to proper significant figures)

28 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 u; from the periodic table, the element is Pb

29 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 [= 0.2534(100)/(0.2534 + 0.2466)] and 49.32% 81Br [= 0.2466(100)/0.5000]

31 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

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:

69Ga (0.60) 71Ga (0.40)

288 290 292

Moles and Molar Masses

32 Molar mass of C6H8O6 = 6(12.011) + 8(1.0079) + 6(15.999) = 176.123 g/mol

10 tablets ×

g176.12

mol1mg1000

g1tablet

mg0

10− mol C6H8O6

8 tablets ×

g176.12

mol1tablet

g5000

mol

molecules10

022

mol1mg1000

g1

022

6 × 23

= 1.67 × 1021 molecules

34 4.24 g C6H6 ×

g11.78

mol1 = 5.43 × 2

022

= 3.92 × 1023 atoms total

0.224 mol H2O ×

mol

g02.18

= 4.04 g H2O

0.224 mol H2O ×

mol

molecules10

022

= 4.05 × 1023 atoms total

2.71 × 1022 molecules CO2 ×

molecules10

022.6

mol1

= 1.98 g CO2

2.71 × 1022 molecules CO2 ×

2

COmolecule

totalatoms3

= 8.13 × 1022 atoms total

3.35 × 1022 atoms total ×

totalatoms6

molecule1

= 5.58 × 1021 molecules CH3OH

5.58 × 1021 molecules CH3OH ×

molecules10

022.6

mol1

= 0.297 g CH3OH

35 a 20.0 mg C8H10N4O2 ×

g20.194

mol1mg1000

g1

× = 1.03 × 10−4mol C8H10N4O2

b 2.72 × 1021 molecules C2H5OH ×

molecules10

022.6

mol1

mol1

= 3.41 × 10−2mol CO2

36 a A chemical formula gives atom ratios as well as mole ratios We will use both ratios to

illustrate 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 ×

NOHCmol

NOHCmolecules10

022.6NOHCg07.75

NOHCmol1

2 5 2

2 5 2 23

2 5 2

2 5

×

NOHCmolecule

Natom1

2 5 2

= 4.01 × 1022 atoms N

b Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol

5.00 g Mg3N2 ×

2 3

2 3 23

2 3

2 3

NMgmol

NMgunitsformula10

022.6NMgg95.100

NMgmol

2

3NMgmol

atoms2

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 ×

2 3

2 3

)NO(Cag10.164

)NO(Camol1

2

3)NO(Camol

Nmol2

×

Nmol

Natoms10

022

2

4 2

ONmol

Nmol2ONg02.92

ONmol

Nmol

Natoms10

022

Hatoms10

022.6Hmol1

Hmol2Hg016.2

Hmol

2 2

= 2.4 × 1024 atoms

4.0 g He ×

Hemol1

Heatoms10

022.6Heg003.4

Hemol

1 × × 23

= 6.0 × 1023 atoms

1.0 mol F2 ×

Fmol1

Fatoms10

022.6Fmol1

Fmol

atoms10

022.6CO

mol1

)O2C1(atomsmol3COg01.44

COmol

2 2

atoms10

022.6SF

mol1

)F6S1(atomsmol7SFg07.146

SFmol

6 6

= 4.21 × 1024 atoms

The order is: 4.0 g He < 1.0 mol F < 44.0 g CO < 4.0 g H < 146 g SF

38 a 14 mol C ×

Cmol

g011.12

+ 18 mol H ×

Hmol

g0079.1

+ 2 mol N ×

Nmol

g007.14

+ 5 mol O ×

Omol

g999.15

= 294.305 g/mol

b 10.0 g C14H18N2O5 ×

5 2 18 14

5 2 18 14

ONHCg3.294

ONHCmol1

= 3.40 × 10−2 mol C14H18N2O5

c 1.56 mol ×

mol

g.294 = 459 g C14H18N2O5

d 5.0 mg ×

mol

molecles10

02.6g3.294

mol1mg1000

Natoms10

02.6ONHCmol

Nmol2O

NHCg3.294

ONHCmol

5 2 18 14 5

2 18 14

5 2 18

02.6

mol1

022.6

mol1

mol1 = 3.023 mol C2H3Cl3O2

c 2.0 × 10-2 mol ×

mol

g39.165

= 3.3 g C2H3Cl3O2

d 5.0 g C2H3Cl3O2 ×

molecule

Clatoms3mol

molecules10

02.6g39.165

2 3 3 2 2

3 3 2

OClHCmol

OClHCg39.165Cl

mol3

OClHCmol1g45.35

Clmol

f 500 molecules ×

mol

g39.165molecules10

022.6

mol1

2 4 2 2

4 2 9

BrHCg9.187

BrHCmol1flour

g

BrHCg100.30flour

lb

flourg454

molecules10

02

6 ×

× = 4.4 × 1016 molecules C2H4Br2

Percent Composition

41 Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol

Mass % C =

compoundg

43.336

Cg)01.12(20

× 100 = 71.40% C

Mass % H =

compoundg

43.336

Hg)008.1(29

× 100 = 8.689% H

Mass % F =

compoundg

43.336

Fg00.19

× 100 = 5.648% F Mass % O = 100.00 − (71.40 + 8.689 + 5.648) = 14.26% O or:

Mass % O =

compoundg

43.336

Og)00.16(3

× 100 = 14.27% O

42 a C3H4O2: Molar mass = 3(12.011) + 4(1.0079) + 2(15.999) = 36.033 + 4.0316 + 31.998

= 72.063 g/mol Mass % C =

compoundg

72.063

Cg36.033

× 100 = 50.002% C

Mass % H =

compoundg

72.063

Hg4.0316

Mass % O = 100.000 − (50.002 + 5.5945) = 44.404% O or:

g72.063

g31.998

g48.044

× 100 = 55.807% C; mass % H =

g86.089

g6.0474

g36.033

× 100 = 67.905% C; mass % H =

g53.064

g3.0237

× 100 = 5.6982% H

Mass % N =

g53.064

g14.007 × 100 = 26.396% N or % N = 100.000 − (67.905 + 5.6982)

= 26.397% N

43 In 1 mole of YBa2Cu3O7, there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu, and 7 moles

of O

Molar mass = 1 mol Y  +2molBa137mol.3gBaBa

Ymol

Yg91.88

Cug55.63

Og00.16

Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol

Mass % Y =

g2.666

g91.88

× 100 = 13.35% Y; mass % Ba =

g2.666

g6.274

× 100 = 41.22% Ba

Mass % Cu =

g2.666

g65.190

× 100 = 28.62% Cu; mass % O =

g2.666

g0.112

× 100 = 16.81% O

44 a C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol

Mass % C =

2 4 10

Cg20.194

Cg)01.12(8

× 100 =

20.194

08.96

× 100 = 49.47% C

b C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol

Mass % C =

11 22

Cg30.342

Cg)01.12(12

× 100 = 42.10% C

c C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol

Mass % C =

OHHCg07.46

Cg)01.12(2

5 2

× 100 = 52.14% C The order from lowest to highest mass percentage of carbon is:

sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH)

45 NO: Mass % N =

NOg01.30

Ng01.14

× 100 = 46.68% N

NO2: Mass % N =

2

NOg01.46

Ng01.14

× 100 = 30.45% N

N2O: Mass % N =

ONg02.44

Ng)01.14(2

46 Assuming 100.00 g cyanocobalamin:

mol cyanocobalamin = 4.34 g Co

Comol

amincyanocobalmol

1Cog58.93

Comol

×

= 7.36 × 10−2 mol cyanocobalamin

mol1036.7

g00.100amin

cyanocobalmol

1

amincyanocobalg

, x = molar mass = 1360 g/mol

47 There are 0.390 g Cu for every 100.000 g of fungal laccase Let’s assume 100.000 g fungal

laccase

Mol fungal laccase = 0.390 g Cu

Cumol4

laccasefungalmol1Cug63.55

Cumol

laccasefungal

mol

1

laccasefungal

g

x

=

mol1053.1

g000.100

3

−

× , x = molar mass = 6.54 × 104 g/mol

48 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 CaSO4 •2H2O

Mass percent Ca:

50 g Ca3SiO5 ×

Camol1

Cag08.40SiOCamol1

Camol3SiO

Cag33.228

SiOCamol1

5 3 5

Cag16.80

= 12 g Ca

12 g Ca3Al2O6 ×

6 2

3Al OCag20.270

Cag24.120

= 5.3 g Ca

8.0 g Ca2AlFeO5 ×

5

2AlFeOCa

g99.242

Cag16.80

= 2.6 g Ca

3.5 g CaSO4 •2H2O ×

OH2CaSOg18.172

Cag08.40

2

4• = 0.81 g Ca Mass of Ca = 26 + 12 + 5.3 + 2.6 + 0.81 = 47 g Ca

% Ca =

cementg0.100

Cag47

× 100 = 47% Ca Mass percent Al:

12 g Ca3 Al2O6 ×

6 2

3Al OCag20.270

Alg96.53

= 2.4 g Al

8.0 g Ca2AlFeO5 ×

5

2AlFeOCa

g99.242

Alg98.26

= 0.89 g Al

% Al =

g0.100

g89.0g4

× 100 = 3.3% Al Mass percent Fe:

8.0 g Ca2AlFeO5 ×

5

2AlFeOCa

g99.242

Feg85.55

= 1.8 g Fe; % Fe =

g0.100

g8.1

× 100 = 1.8% Fe

Empirical and Molecular Formulas

49 a Molar mass of CH2O = 1 mol C12mol.011Cg+ 2 mol H1mol.0079Hg

g999.15

= 30.026 g/mol

% C =

OCHg026.30

Cg011.12

2

× 100 = 40.002% C; % H =

OCHg026.30

Hg0158.2

2

× 100 = 6.7135% H

% O =

OCHg026.30

Og999.15

Cg155.180

Cg066.72

× 100 = 40.002%; % H =

g155.180

g)0079.1(12

g022.24

× 100 = 40.002%; % H =

g052.60

g0316.4

× 100 = 6.7135%

% 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

50 a The molecular formula is N2O4 The smallest whole number ratio of the atoms (the

empirical formula) is NO2

b Molecular formula: C3H6; empirical formula: CH2

c Molecular formula: P4O10; empirical formula: P2O5

d Molecular formula: C6H12O6; empirical formula: CH2O

51 a SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g/mol

g09.47

g35.188

= 4.000; so the molecular formula is (SNH)4 or S4N4H4

b NPCl2: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol

g88.115

g64.347

= 3.0000; molecular formula is (NPCl2)3 or N3P3Cl6

c CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol

g97.170

g94.341

= 2.0000; molecular formula: Co2C8O8

d SN: 32.07 + 14.01 = 46.08 g/mol;

g08.46

g32.184

= 4.000; molecular formula: S4N4

52 1.121 g N ×

Ng14.007

Nmol1

= 8.003 × 10−2mol N; 0.161 g H ×

Hg008.1

Hmol1

= 1.60 × 10−1 mol H

0.480 g C ×

Cg01.12

Cmol1

= 4.00 × 10−2mol C; 0.640 g O ×

Og00.16

Omol1

= 4.00 × 10−2mol O

Dividing all mole values by the smallest number:

2 2

1000

4

10003

1060.1

1000.4

Hgmol1

= 3.000 × 10−3 mol Hg

0.0480 g O ×

Og00.16

Omol1

= 3.00 × 10−3 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 = 0.4172 g HgxOy − 0.016 g O = 0.401 g Hg

0.401 g Hg ×

Hgg6.200

Hgmol1

= 2.00 × 10−3 mol Hg; 0.016 g O ×

Og00.16

Omol1

= 1.0 × 10−3 mol O

The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O

54 Out of 100.00 g of adrenaline, there are:

56.79 g C ×

Cg011.12

Cmol1

= 4.728 mol C; 6.56 g H ×

Hg008.1

Hmol1

= 6.51 mol H

28.37 g O ×

Og999.15

Omol1

= 1.773 mol O; 8.28 g N ×

Ng01.14

Nmol1

= 0.591 mol N Dividing each mole value by the smallest number:

51.6 = 11.0;

591.0

773.1 = 3.00;

591.0

591.0 = 1.00 This gives adrenaline an empirical formula of C8H11O3N

55 Assuming 100.00 g of nylon-6:

63.68 g C ×

Cg01.12

Cmol1

= 5.302 mol C; 12.38 g N ×

Ng01.14

Nmol1

= 0.8837 mol N

9.80 g H ×

Hg008.1

Hmol1

= 9.72 mol H; 14.14 g O ×

Og00.16

Omol1

= 0.8838 mol O Dividing each mole value by the smallest number:

72.9 = 11.0;

8837.0

8838.0 = 1.000

The empirical formula for nylon-6 is C6H11NO

56 When combustion data are given, it is assumed that all the carbon in the compound ends up

as carbon in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O In the sample of fructose combusted, the masses of C and H are:

mass C = 2.20 g CO2 ×

Cmol

Cg12.01 COmol

Cmol1CO

g44.01

COmol1

2 2

mass H = 0.900 g H2O ×

Hmol

Hg1.008 OHmol

Hmol2OHg18.02

OHmol1

2 2

Cmol1

= 0.0500 mol C; 0.101 g H ×

Hg008.1

Hmol1

= 0.100 mol H

0.799 g O ×

Og00.16

Omol1

= 0.0499 mol O

Dividing by the smallest number:

0.0499

0.100 = 2.00; the empirical formula is CH2O

57 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 ×

2

COg01.44

Cg01.12

= 0.0581 g C; % C =

compoundg

157.0

Cg0581.0

× 100 = 37.0% C

0.0310 g H2O ×

OHg02.18

Hg016.2

2

= 3.47 × 10−3 g H; % H =

g157.0

g1047

Ng01.14

= 1.89 × 10−2 g N

% N =

g103.0

g1089

1 × −2

× 100 = 18.3% N 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 ×

Cg01.12

Cmol1

= 3.08 mol C; 2.21 g H ×

Hg008.1

Hmol1

= 2.19 mol H

18.3 g N ×

Ng01.14

Nmol1

= 1.31 mol N; 42.5 g O ×

Og00.16

Omol1

19.2 = 1.67;

31.1

31.1 = 1.00;

31.1

66.2 = 2.03 Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6

58 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 ×

Cg011.12

Cmol1

= 3.446 mol C; 3.47 g H ×

Hg008.1

Hmol1

= 3.44 mol H

55.14 g O ×

Og999.15

Omol1

g0.15

= 116 g/mol

massEmpirical

massMolar

= 02.29116

= 4.00; molecular formula = (CHO)4 = C4H4O4

59 Assuming 100.0 g of compound:

26.7 g P ×

Pg97.30

Pmol1

= 0.862 mol P; 12.1 g N ×

Ng01.14

Nmol1

= 0.864 mol N

61.2 g Cl ×

Clg45.35

Clmol1

formulaEmpirical

massMolar = = 5.0; the molecular formula is (PNCl2)5 = P5N5Cl10

60 41.98 mg CO2 ×

2

COmg009.44

Cmg011.12

= 11.46 mg C; % C =

mg81.19

mg46.11

× 100 = 57.85% C

6.45 mg H2O ×

OHmg02.18

Hmg016.2

2

= 0.722 mg H; % H =

mg81.19

mg722.0

Cmol1

= 4.816 mol C; 3.64 g H ×

Hg008.1

Hmol1

= 3.61 mol H

38.51 g O ×

Og999.15

Omol1

61.3 = 1.50;

407.2

407.2 = 1.000 The C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2 The empirical formula is C4H3O2 Mass of

C4H3O2 ≈ 4(12) + 3(1) + 2(16) = 83

Molar mass =

mol250.0

g5.41

= 166 g/mol;

83

166 = 2.0; the molecular formula is C8H6O4

61 First, we will determine composition by mass percent:

16.01 mg CO2 ×

g

mg1000CO

g009.44

Cg011.12mg

1000

g1

68.10

Cmg369.4

× 100 = 40.91% C

4.37 mg H2O ×

g

mg1000O

Hg02.18

Hg016.2mg

1000

g1

2

×

% H =

mg68.10

mg489.0

× 100 = 4.58% H; % O = 100.00 - (40.91 + 4.58) = 54.51% O

So, in 100.00 g of the compound, we have:

40.91 g C ×

Cg011.12

Cmol1

= 3.406 mol C; 4.58 g H ×

Hg008.1

Hmol1

= 4.54 mol H

54.51 g O ×

Og999.15

Omol1

= 3.407 mol O

Dividing by the smallest number:

406.3

54.4 = 1.33 =

62 a Only acrylonitrile contains nitrogen If we have 100.00 g of polymer:

8.80 g N ×

NHCmol1

NHCg06.53Ng01.14

NHCmol1

3 3

3 3 3

% C3H3N =

polymerg

00.100

NHCg3

= 33.3% C3H3N Only butadiene in the polymer reacts with Br2:

0.605 g Br2 ×

6 4

6 4

2

6 4

2

2

HCmol

HCg09.54Br

mol

HCmol1Brg8.159

Brmol

% C4H6 =

g20.1

g205.0

× 100 = 17.1% C4H6

b If we have 100.0 g of polymer:

33.3 g C3H3N ×

g06.53

NHCmol

= 0.628 mol C3H3N

17.1 g C4H6 ×

6 4

6 4

HCg09.54

HCmol1

= 0.316 mol C4H6

49.6 g C8H8 ×

8 8

8 8

HCg14.104

HCmol1

= 0.476 mol C8H8

Dividing by 0.316:

316.0

628.0 = 1.99;

316.0

316.0 = 1.00;

316.0

476.0 = 1.51 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 Equations

63 Only one product is formed in this representation This product has two Ys bonded to an X

The other substance present in the product mixture is just the excess of one of the reactants (Y) The best equation has smallest whole numbers Here, answer c would be this smallest whole number equation (X + 2 Y → XY2) Answers a and b have incorrect products listed, and for answer d, an equation only includes the reactants that go to produce the product; excess reactants are not shown in an equation

64 a SiO2(s) + C(s) → Si(s) + CO(g); Si is balanced

Balance oxygen atoms: SiO2 + C → Si + 2 CO

Balance carbon atoms: SiO2(s) + 2 C(s) → Si(s) + 2 CO(g)

b SiCl4(l) + Mg(s) → Si(s) + MgCl2(s); Si is balanced

Balance Cl atoms: SiCl4 + Mg → Si + 2 MgCl2

Balance Mg atoms: SiCl4(l) + 2 Mg(s) → Si(s) + 2 MgCl2(s)

c Na2SiF6(s) + Na(s) → Si(s) + NaF(s); Si is balanced

Balance F atoms: Na2SiF6 + Na → Si + 6 NaF

Balance Na atoms: Na2SiF6(s) + 4 Na(s) → Si(s) + 6 NaF(s)

65 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

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:

CS2 + 2 NH3 → H2S + NH4SCN

H is also balanced CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s)

66 Fe3O4(s) + 4 H2(g) → 3 Fe(s) + 4 H2O(g); Fe3O4(s) + 4 CO(g) → 3 Fe(s) + 4 CO2(g)

67 a The formulas of the reactants and products are C6H6(l) + O2(g) → CO2(g) + H2O(g) To

balance this combustion reaction, notice that all of the carbon in C6H6 has to end up as carbon in CO2 and all of the hydrogen in C6H6 has to end up as hydrogen in H2O To balance C and H, we need 6 CO2 molecules and 3 H2O molecules for every 1 molecule of

C6H6 We do oxygen last Because we have 15 oxygen atoms in 6 CO2 molecules and 3

H2O molecules, we need 15/2 O2 molecules in order to have 15 oxygen atoms on the reactant side

68 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)

69 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)

70 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)

4 4 4

4

ClONHmol

ClONHg49.117Al

mol3

ClONHmol3Alg98.26

Almol1Al

kg

Alg1000

O8HBa(OH)mol

= 0.0206 mol = 0.021 mol

0.021 mol Ba(OH)2 •8H2O ×

SCNNHmol

SCNNHg76.13O

8HBa(OH)mol

1

SCNNHmol2

4 4

2 2

3

3

NaHCOmol

3

OHCmol1NaHCOg

01.84

NaHCOmol

1mg1000

g

×

7 8 6

7 8 6

OHCmol

OHCg12.192

COg01.44NaHCOmol

3

COmol3NaHCO

g01.84

NaHCOmol

×

= 0.052 g or 52 mg CO2

74 1.0 × 106 kg HNO3

3 3

3

3

HNOg0.63

HNOmol1HNO

kg

HNOg

2

3

NHmol4

NOmol4NOmol2

NOmol2NOmol3

HNOmol

3

3

NHmol24

HNOmol16

Thus we can produce 16 mol HNO3 for every 24 mol NH3 that we begin with:

1.6 × 107 mol HNO3 ×

3 3

3

3

NHmol

NHg0.17HNOmol16

NHmol24

× = 4.1 × 108 g or 4.1 × 105 kg NH3

This is an oversimplified answer In practice, the NO produced in the third step is recycled back continuously into the process in the second step If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1 : 1; 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

75 Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s)

15.0 g Fe ×

Feg85.55

Femol1

= 0.269 mol Fe; 0.269 mol Fe ×

Almol

Alg98.26Femol2

Almol

0.269 mol Fe ×

3 2

3 2 3

2

OFemol

OFeg70.159Fe

mol2

OFemol

0.269 mol Fe ×

3 2

3 2 3

2

OAlmol

OAlg96.101Fe

mol2

OAlmol1

76 10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s)

52.9g KClO3 ×

10 4

10 4

3

10 4

3

3

OPmol

OPg88.283KClO

mol10

OPmol3KClOg55.122

KClOmol

77 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

2 2

2 2

COg0.4

airg100CO

mol

COg01.44LiOHmol2

COmol1LiOHg95.23

LiOHmol1

min60

h1airL140

min1mL1000

L1air

g0010.0

airmL1

4

4 4

NHmol55

NOHCmol1NHg04.18

NHmol1kg

g1000waste

kg100

NHkg0

NOHCmol

NOHCg1

= 3.4 × 104 g tissue if all NH4+ converted

Because only 95% of the NH4+ ions react:

mass of tissue = (0.95)(3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue

2 4 3

2 4 3 2

4 3

)(POCag310.18

)(POCamol1e

phosphoritg

100

)(POCag75

2 4 3

4

)PO(Camol2

Pmol1

×

4

4

Pmol

Pg88.123

)cm060.0cm0.16cm0.8

Amount of Cu to be recovered = 0.80 × (6.9 × 105 g) = 5.5 × 105 g Cu

5.5 × 105 g Cu ×

2 4 3

2 4 3 2

4 3

Cl)NH(Cumol

Cl)NH(Cug6.202Cu

mol

Cl)NH(Cumol1Cug55.63

Cumol1

NHmol

NHg03.17Cu

mol

NHmol4Cug55.63

Cumol

Limiting Reactants and Percent Yield

81 The product formed in the reaction is NO2; the other species present in the product picture 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)

82 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 O2 ×

2

3

Omolecules5

NHmolecules4

= 8 molecules NH3 to react with all the O2

Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all the O2, O2 is limiting