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1 CHAPTER TWO ATOMS, MOLECULES, AND IONS Development of the Atomic Theory 18. =1.000; = 1.999; = 2.999 The masses of fluorine are simple ratios of whole numbers to each other, 1:2:3. 19. From Avogadro’s hypothesis, volume ratios are equal to molecule ratios at constant temperature and 2 2 pressure. Therefore, we can write a balanced equation using the volume data, Cl + 3 F ÷ 3 2 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF for a balanced reaction. 20. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (i.e., depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro’s hypothesis implies that volume ratios are equal to molecule ratios at constant 2 2 temperature and pressure. H + Cl ÷ 2 HCl. From the balanced equation, the volume of HCl 2 2 produced will be twice the volume of H (or Cl ) reacted. 21. 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, i.e., = 1.00. To get Na, Mg, and O on the same scale, we do the same division. Na: = 22.8; Mg: = 11.9; O: = 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 the correct 2 2 formulas are H O, Na O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H.

Chapter Two Atoms, Molecules, And Ions

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Page 1: Chapter Two Atoms, Molecules, And Ions

1

CHAPTER TWO

ATOMS, MOLECULES, AND IONS

Development of the Atomic Theory

18. =1.000; = 1.999; = 2.999

The masses of fluorine are simple ratios of whole numbers to each other, 1:2:3.

19. From Avogadro’s hypothesis, volume ratios are equal to molecule ratios at constant temperature and

2 2pressure. Therefore, we can write a balanced equation using the volume data, Cl + 3 F ÷

32 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF fora balanced reaction.

20. a. The composition of a substance depends on the numbers of atoms of each element making upthe compound (i.e., depends on the formula of the compound) and not on the composition of themixture from which it was formed.

b. Avogadro’s hypothesis implies that volume ratios are equal to molecule ratios at constant

2 2temperature and pressure. H + Cl ÷ 2 HCl. From the balanced equation, the volume of HCl

2 2produced will be twice the volume of H (or Cl ) reacted.

21. 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, i.e., = 1.00. To get Na, Mg, and O on the same scale, we do the same division.

Na: = 22.8; Mg: = 11.9; O: = 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 the correct

2 2formulas are H O, Na O, and MgO. The smaller discrepancies result from the error in the assumedatomic mass of H.

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CHAPTER 2 ATOMS, MOLECULES, AND IONS2

The Nature of the Atom

22. Deflection of cathode rays by magnetic and electric fields led to the conclusion that they werenegatively charged. The cathode ray was produced at the negative electrode and repelled by thenegative pole of the applied electric field.

23. $ particles are electrons. A cathode ray is a stream of electrons ($ particles).

24. Density of hydrogen nucleus (contains one proton only):

nucleusV =

Density of H-atom (contains one proton and one electron):

atomV =

d =

Since electrons move about the nucleus at an average distance of about 1 × 10 cm, then the-8

diameter of an atom is about 2 × 10 cm. Let's set up a ratio:-8

, Solving:

diameter of model = 2 × 10 mm = 200 m5

25. First, divide all charges by the smallest quantity, 6.40 × 10 .-13

= 4.00; = 12.00; = 6.00

Since all charges are whole number multiples of 6.40 × 10 zirkombs then the charge on one-13

electron could be 6.40 × 10 zirkombs. However, 6.40 × 10 zirkombs could be the charge of two-13 -13

electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 ×10 zirkombs or an integer fraction of 6.40 × 10 . -13 -13

26. J. J. Thomson discovered the electrons. Henri Becquerel discovered radioactivity. Lord Rutherfordproposed the nuclear model of the atom. Dalton's original model proposed that atoms wereindivisible particles (that is, atoms had no internal structure). Thomson and Becquerel discoveredsubatomic particles and Rutherford's model attempted to describe the internal structure of the atomcomposed of these subatomic particles. In addition, the existence of isotopes, atoms of the sameelement but with different mass, had to be included in the model.

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CHAPTER 2 ATOMS, MOLECULES, AND IONS 3

27. The proton and neutron have similar mass with the mass of the neutron slightly larger than that ofthe proton. Each of these particles has a mass approximately 1800 times greater than that of anelectron. The combination of the protons and the neutrons in the nucleus makes up the bulk of themass of an atom, but the electrons make the greatest contribution to the chemical properties of theatom.

28. If the plum pudding model were correct (a diffuse positive charge with electrons scatteredthroughout), then alpha particles should have traveled through the thin foil with very minordeflections in their path. This was not the case as a few of the alpha particles were deflected at verylarge angles. Rutherford reasoned that the large deflections of these alpha particles could be causedonly by a center of concentrated positive charge that contains most of the atom’s mass (the nuclearmodel of the atom).

Elements and the Periodic Table

29. The atomic number of an element is equal to the number of protons in the nucleus of an atom of thatelement. The mass number is the sum of the number of protons plus neutrons in the nucleus. Theatomic mass is the actual mass of a particular isotope (including electrons). As we will see inchapter three, the average mass of an atom is taken from a measurement made on a large number ofatoms. The average atomic mass value is listed in the periodic table.

30. promethium (Pm) and technetium (Tc)

31. a. Eight; Li to Ne b. Eight; Na to Ar

c. Eighteen; K to Kr d. Four; Fe, Ru, Os, Hs

e. Five; O, S, Se, Te, Po f. Four; Ni, Pd, Pt, Uun (#110)

32. a. Element #5 is boron. B b. Z = 7; A = 7 + 8 = 15; N

c. Z = 17; A = 17 + 18 = 35; Cl d. A = 92 + 143 = 235; U

e. Z = 6; A = 14; C f. Z = 15; A = 31; P

33. a. Mg: 12 protons, 12 neutrons, 12 electrons

b. : 12 p, 12 n, 10 e c. : 27 p, 32 n, 25 e

d. : 27 p, 32 n, 24 e e. Co: 27 p, 32 n, 27 e

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CHAPTER 2 ATOMS, MOLECULES, AND IONS4

f. Se: 34 p, 45 n, 34 e g. : 34 p, 45 n, 36 e

h. Ni: 28 p, 35 n, 28 e i. : 28 p, 31 n, 26 e

34.

SymbolNumber of protons in

nucleusNumber of neutrons in

nucleusNumber ofelectrons

Net charge

U92 146 92 0

Ca 20 20 18 2+2+

V 23 28 20 3+3+

Y 39 50 39 0

Br 35 44 36 1--

P 15 16 18 3-3-

35. Metals lose electrons to form cations in ionic compounds and nonmetals gain electrons to formanions. Group IA, IIA and IIIA metals form stable +1, +2 and +3 charged cations, respectively.Group VA, VIA and VIIA nonmetals form -3, -2 and -1 charged anions, respectively.

a. Lose one e to form Na . b. Lose two e to form .- + -

c. Lose two e to form . d. Gain one e to form I .- - -

e. Lose three e to form . f. Gain two e to form .- -

g. Gain three e to form . h. Lose one e to form Cs . - - +

i. Gain two e to form .-

36. Carbon is a nonmetal. Silicon and germanium are called metalloids as they exhibit both metallic andnonmetallic properties. Tin and lead are metals. Thus, metallic character increases as one goesdown a family in the periodic table. The metallic character decreases from left to right across theperiodic table.

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CHAPTER 2 ATOMS, MOLECULES, AND IONS 5

Nomenclature

37. a. sulfur difluoride b. dinitrogen tetroxide

c. iodine trichloride d. tetraphosphorus hexoxide

38. a. sodium perchlorate b. magnesium phosphate

c. aluminum sulfate d. sulfur difluoride

e. sulfur hexafluoride f. sodium hydrogen phosphate

g. sodium dihydrogen phosphate h. lithium nitride

i. sodium hydroxide j. magnesium hydroxide

k. aluminum hydroxide l. silver chromate

39. 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. sulfur hexafluoride h. sodium hypochlorite

i. barium chromate j. ammonium nitrate

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

d. iodine monochloride e. lead(II) phosphate f. potassium iodate

g. sulfuric acid h. strontium nitride i. aluminum sulfite

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

2 3 2 3 341. a. SO b. SO c. Na SO d. KHSO

3 2 3 3 2 3 2 2 4e. Li N f. Cr (CO ) Cr(C H O ) h. SnFg.

4 4 4 4i. NH HSO : Composed of NH and HSO ions.+ -

4 2 4 4j. (NH ) HPO k. KClO l. NaH

m. HBrO n. HBr

2 2 2 3 242. a. Na O b. Na O c. KCN d. Cu(NO )

4 2e. SiCl f. PbO g. PbO h. CuCl

i. GaAs: We would predict the stable ions to be Ga and As .3+ 3-

2 2 2j. CdSe k. ZnS l. Hg Cl : Mercury(I) exists as Hg .2+

2 2 5m. HNO P On.

2 3 2 2 443. a. Pb(C H O ) : lead(II) acetate b. CuSO : copper(II) sulfate

4c. CaO: calcium oxide d. MgSO : magnesium sulfate

2 4e. Mg(OH) : magnesium hydroxide f. CaSO : calcium sulfate

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CHAPTER 2 ATOMS, MOLECULES, AND IONS6

2g. N O: dinitrogen monoxide or nitrous oxide (common name)

Additional Exercises

44. There should be no difference. The chemical composition of insulin from both sources will be thesame and therefore, it will have the same activity regardless of the source. As a practical note, tracecontaminants in the two types of insulin may be different. These trace components may beimportant.

3 4 2 3 245. a. nitric acid, HNO b. perchloric acid, HClO c. acetic acid, HC H O

2 4 3 4d. sulfuric acid, H SO e. phosphoric acid, H PO

46. a. Fe : 26 protons (Fe is element 26.); protons - electrons = charge, 26 -2 = 24 electrons; FeO is2+

the formula since the oxide ion has a -2 charge.

2 3b. Fe : 26 protons; 23 electrons; Fe O c. Ba : 56 protons; 54 electrons; BaO3+ 2+

2 2 3d. Cs : 55 protons; 54 electrons; Cs O e. S : 16 protons; 18 electrons; Al S+ 2-

3f. P : 15 protons; 18 electrons; AlP g. Br : 35 protons; 36 electrons; AlBr3- -

h. N : 7 protons; 10 electrons; AlN3-

247. From the XBr formula, the charge on element X is +2. Therefore, the element has 88 protons,which identifies it as radium, Ra. 230 - 88 = 142 neutrons

48. The solid residue must have come from the flask.

4 349. In the case of sulfur, SO is sulfate and SO is sulfite. By analogy:2- 2-

4 3 4 3SeO : selenate; SeO : selenite; TeO tellurate; TeO : tellurite2- 2- 2- 2-

4 350. The PO ion is phosphate and PO is phosphite. By analogy:3- 3-

3 4 2 3 4 3 4Mg (AsO ) : magnesium arsenate; H AsO : arsenic acid; Na SbO : sodium antimonate;

3 3 2 4Na AsO : sodium asenite; Na HAsO : sodium hydrogen arsenate

51. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or:

, A = atomic mass of In = 76.54

2 3If the formula is In O , then two times the atomic mass of In will combine with three times theatomic mass of O, or:

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CHAPTER 2 ATOMS, MOLECULES, AND IONS 7

, A = atomic mass of In = 114.8

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

3 2 252. a. Ca and N : Ca N , calcium nitride b. K and O : K O, potassium oxide2+ 3- + 2-

c. Rb and F : RbF, rubidium fluoride d. Mg and S : MgS, magnesium sulfide+ - 2+ 2-

2 2 3e. Ba and I : BaI , barium iodide f. Al and Se : Al Se , aluminum selenide2+ - 3+ 2-

3g. Cs and P : Cs P, cesium phosphide+ 3-

3h. In and Br : InBr , indium(III) bromide; In also forms In ions, but you would predict In3+ - + 3+

ions from its position in the periodic table.

53. Hydrazine: 1.44 × 10 g H/g N; Ammonia: 2.16 × 10 g H/g N -1 -1

Hydrogen azide: 2.40 × 10 g H/g N-2

Let's try all of the ratios:

= 1.50 = = 6.00; = 9.00

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

54. a. This is element 52, tellurium. Te forms stable -2 charged ions in ionic compounds (like otheroxygen family members).

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

c. Argon. Ar is element 18.

d. Astatine. At is element 85.

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

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

As long as we had pure Cl or pure Cl, the above ratios will always hold. If we have a mixture35 37

(such as the natural abundance of chlorine), the ratio will also be constant as long as the compositionof the mixture of the two isotopes does not change.

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CHAPTER 2 ATOMS, MOLECULES, AND IONS8

Challenge Problems

56. Because the gases are at the same temperature and pressure, the volumes are directly related to thenumber of moles. Let’s consider hydrogen and oxygen to be monatomic gases, and that water hasthe simplest possible formula (HO). We have the equation:

H + O ÷ HOBut, the volume ratios are also the mole ratios, which correspond to coefficients in the equation:

2H + O ÷ 2HOSince atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correctthis, we can make oxygen a diatomic molecule:

22H + O ÷ 2HO

2This does not require hydrogen to be diatomic. Of course, if we know water has the formula H O,we get:

2 22H + O ÷ 2H O.The only was to balance this is to make hydrogen diatomic:

2 2 22H + O ÷ 2H O

2 657. a. Both compounds have C H O as the formula. Since they have the same formula, their masspercent composition will be identical. However, these are different compounds with differentproperties since the atoms are bonded together differently. These compounds are called isomersof each other.

b. When wood burns, most of the solid material in wood is converted to gases, which escape. The

2 2gases produced are most likely CO and H O.

c. The atom is not an indivisible particle, but is instead composed of other smaller particles, e.g.,electrons, neutrons, protons.

d. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Althoughthe compounds are composed of different isotopes, their properties are similar because differentisotopes of the same element have similar properties (except, of course, their mass).

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

expt. 1 x = 1.0 y = 10.5expt. 2 y = 1.4 z = 1.0expt. 3 x = 1.0 y = 3.5

Our assumption about formulas dictate the rest of the solution. For example, if we assume theformula of the compound in expt. 1 is XY and that of expt. 2 is YZ, we get relative masses of:

X = 2.0 Y = 21 Z = 15 (= 21/1.4)

3and a formula of X Y for expt. 3 (three times as much X must be present in expt. 3 as compared toexpt. 1 (10.5/3.5 = 3).

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CHAPTER 2 ATOMS, MOLECULES, AND IONS 9

However, if we assume the formula for expt. 2 is YZ and that of expt. 3 is XZ, then we get:

X = 2.0 Y = 7.0 Z = 5.0 (= 7.0/1.4)

3and a formula of XY for expt. 1.

Any answer which 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 thecompound. If the compound in expt. 1 has formula XY, then:

21g XY × = 19.2 g Y (and 1.8 g X)

If the compound in expt. 3 has the XY formula, then:

21 g XY × = 16.3 g Y (and 4.7 g X)

Note, it could be that neither experiment 1 or 3 have XY as the formula. Therefore, there is no wayof knowing an absolute answer here.

59. Compound I: ; Compound II:

The ratio of the masses of R that combines with 1.00 g Q is: = 2.99 . 3

As expected from the law of multiple proportions, this ratio is a small whole number.

Since compound I contains three times the mass of R per gram of Q as compared to compound II

3(RQ), then the formula of compound I should be R Q.

60. The law of multiple proportions does not involve looking at the ratio of the mass of one element withthe total mass of the compounds. We can show this supports the law of multiple proportions bycomparing 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)Compound 2: 42.9 g C and 57.1 g O (100.0 - 42.9)

Reduce this ratio so that each is compared to 1.0 g oxygen

Compound 1: = 0.374 g C/g O

Compound 2: = 0.751 g C/g O

, thus supporting the law of multiple proportions

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CHAPTER 2 ATOMS, MOLECULES, AND IONS10

61. Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equalnumbers of molecules. In terms of balanced equations, Avogadro’s hypothesis implies that volume

x yratios will be identical to molecule ratios. Assuming one molecule of octane (C H ) reacting, then

x y 2 2 x y 21 molecule of C H produces 8 molecules of CO and 9 molecules of H O. C H + n O ÷

2 2 28 CO + 9 H O. Since all the carbon in octane ends up as carbon in CO , then octane contains 8atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in , so one molecule of

8 18octane contains 9 × 2 = 18 atoms of H. Octane formula = C H and the ratio of C:H = 8:18 or 4:9.

Marathon Problem

62. a. = 2.04 A = 2.04 B = 1.00

= 2.33 B = 1.00 C = 2.33

= 1.17 A = 1.00 C = 1.17

To determine whole numbers, multiply through by 3 first.

A = 6.1 B = 3.0B = 3.0 C = 7.0A = 3.0 C = 3.5 or A = 6.0 and C = 7.0

From these numbers, the relative masses would be: A = 6.0, B = 3.0, C = 7.0

b. There are many consistent solutions. One such solution is:

2 4 36 A + B ÷ 4 A B

4 3 3 B + 4 C ÷ 4 BC

2 33 A + 2 C ÷ 6 AC

In any set of reactions, the calculated mass data must match the mass data given initially in theproblem. Here, the new table of relative masses would be:

A = 6.0 B = 9.0 C = 7.0

Any set of balanced reactions that confirms the initial mass data is correct.