Chapter Two Atoms, Molecules, And Ions

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<p>CHAPTER TWO ATOMS, MOLECULES, AND IONS</p> <p>Development of the Atomic Theory18. =1.000; = 1.999; = 2.999</p> <p>The masses of fluorine are simple ratios of whole numbers to each other, 1:2:3. 19. From Avogadros hypothesis, 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 (i.e., depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadros hypothesis 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. 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., Na: = 1.00. To get Na, Mg, and O on the same scale, we do the same division. = 22.8; Mg: H Relative Value Accepted Value 1.00 1.0079 = 11.9; O: O 7.94 15.999 = 7.94 Na 22.8 22.99 Mg 11.9 24.31</p> <p>20.</p> <p>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 formulas are H2 O, Na2 O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H.</p> <p>1</p> <p>2</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>The Nature of the Atom22. Deflection of cathode rays by magnetic and electric 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 electric field. $ particles are electrons. A cathode ray is a stream of electrons ($ particles). Density of hydrogen nucleus (contains one proton only): Vnucleus =</p> <p>23. 24.</p> <p>Density of H-atom (contains one proton and one electron): Vatom =</p> <p>d= Since electrons move about the nucleus at an average distance of about 1 10-8 cm, then the diameter of an atom is about 2 10-8 cm. Let's set up a ratio: , Solving:</p> <p>diameter of model = 2 105 mm = 200 m 25. First, divide all charges by the smallest quantity, 6.40 10-13. = 4.00; = 12.00; = 6.00</p> <p>Since all charges are whole number multiples of 6.40 10-13 zirkombs then 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 . 26. J. J. Thomson discovered the electrons. 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.</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>3</p> <p>27.</p> <p>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 alpha particles should have traveled through the thin foil with very minor deflections in their path. This was not the case as a few of the alpha particles were deflected at very large angles. Rutherford reasoned that the large deflections of these alpha 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).</p> <p>28.</p> <p>Elements and the Periodic Table29. 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 three, 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. promethium (Pm) and technetium (Tc) a. Eight; Li to Ne c. Eighteen; K to Kr e. Five; O, S, Se, Te, Po b. Eight; Na to Ar d. Four; Fe, Ru, Os, Hs f. Four; Ni, Pd, Pt, Uun (#110)</p> <p>30. 31.</p> <p>32.</p> <p>a. Element #5 is boron.</p> <p>B Cl</p> <p>b. d. f.</p> <p>Z = 7; A = 7 + 8 = 15; A = 92 + 143 = 235; Z = 15; A = 31; P</p> <p>N U</p> <p>c. Z = 17; A = 17 + 18 = 35; e. Z = 6; A = 14; C</p> <p>33.</p> <p>a.</p> <p>Mg: 12 protons, 12 neutrons, 12 electrons</p> <p>b.</p> <p>: 12 p, 12 n, 10 e</p> <p>c.</p> <p>: 27 p, 32 n, 25 e</p> <p>d.</p> <p>: 27 p, 32 n, 24 e</p> <p>e.</p> <p>Co: 27 p, 32 n, 27 e</p> <p>4</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>f.</p> <p>Se: 34 p, 45 n, 34 e</p> <p>g.</p> <p>: 34 p, 45 n, 36 e</p> <p>h. 34. Symbol</p> <p>Ni: 28 p, 35 n, 28 e</p> <p>i.</p> <p>: 28 p, 31 n, 26 e</p> <p>Number of protons in nucleus92</p> <p>Number of neutrons in nucleus146</p> <p>Number of electrons92</p> <p>Net charge0</p> <p>U</p> <p>Ca 2+</p> <p>20</p> <p>20</p> <p>18</p> <p>2+</p> <p>V 3+</p> <p>23</p> <p>28</p> <p>20</p> <p>3+</p> <p>Y</p> <p>39</p> <p>50</p> <p>39</p> <p>0</p> <p>Br -</p> <p>35</p> <p>44</p> <p>36</p> <p>1-</p> <p>P 3-</p> <p>15</p> <p>16</p> <p>18</p> <p>3-</p> <p>35.</p> <p>Metals lose electrons to form cations in ionic compounds and nonmetals gain electrons to form anions. 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+. c. Lose two e- to form e. Lose three e- to form g. Gain three e- to form i. Gain two e- to form . . . . b. Lose two e- to form d. Gain one e- to form I-. f. Gain two e- to form . .</p> <p>h. Lose one e- to form Cs+.</p> <p>36.</p> <p>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.</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>5</p> <p>Nomenclature37. a. sulfur difluoride c. iodine trichloride 38. a. sodium perchlorate c. aluminum sulfate e. sulfur hexafluoride g. sodium dihydrogen phosphate i. sodium hydroxide b. dinitrogen tetroxide d. tetraphosphorus hexoxide b. magnesium phosphate d. sulfur difluoride f. sodium hydrogen phosphate</p> <p>h. lithium nitride j. l. magnesium hydroxide silver chromate c. cobalt(II) iodide</p> <p>k. aluminum hydroxide 39. a. copper(I) iodide d. sodium carbonate f. i. 40. tetrasulfur tetranitride barium chromate</p> <p>b. copper(II) iodide</p> <p>e. sodium hydrogen carbonate or sodium bicarbonate g. sulfur hexafluoride j. ammonium nitrate c. colbalt(III) sulfide f. i. l. potassium iodate aluminum sulfite hypochlorous acid h. sodium hypochlorite</p> <p>a. acetic acid d. iodine monochloride g. sulfuric acid j. tin(IV) oxide b. SO3 f. Cr2 (CO3 )3</p> <p>b. ammonium nitrite e. lead(II) phosphate h. strontium nitride k. sodium chromate c. Na2 SO3 g. Cr(C2 H3 O2 )2 d. KHSO3 h. SnF4</p> <p>41.</p> <p>a. SO2 e. Li3 N i. j.</p> <p>NH4 HSO4 : Composed of NH4 + and HSO4 - ions. (NH4 )2 HPO4 k. KClO4 n. HBr b. Na2 O2 f. PbO c. KCN g. PbO2 d. Cu(NO3 )2 h. CuCl l. NaH</p> <p>m. HBrO 42. a. Na2 O e. SiCl4 i. j.</p> <p>GaAs: We would predict the stable ions to be Ga3+ and As3-. CdSe k. ZnS n. P2 O5 b. CuSO4 : copper(II) sulfate d. MgSO4 : magnesium sulfate f. CaSO4 : calcium sulfate l. Hg2 Cl2 : Mercury(I) exists as Hg2 2+ .</p> <p>m. HNO2 43.</p> <p>a. Pb(C2 H3 O2 )2 : lead(II) acetate c. CaO: calcium oxide e. Mg(OH)2 : magnesium hydroxide</p> <p>6</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>g. N2 O: dinitrogen monoxide or nitrous oxide (common name)</p> <p>Additional Exercises44. There should be no difference. The chemical composition of insulin from both sources will be the same and therefore, it will have the same activity regardless of the source. As a practical note, trace contaminants in the two types of insulin may be different. These trace components may be important. a. nitric acid, HNO3 d. sulfuric acid, H2 SO4 46. b. perchloric acid, HClO4 e. phosphoric acid, H3 PO4 c. acetic acid, HC2 H3 O2</p> <p>45.</p> <p>a. Fe2+ : 26 protons (Fe is element 26.); protons - electrons = charge, 26 -2 = 24 electrons; FeO is the formula since the oxide ion has a -2 charge. b. Fe3+: 26 protons; 23 electrons; Fe2 O3 d. Cs+: 55 protons; 54 electrons; Cs2 O f. P3-: 15 protons; 18 electrons; AlP c. Ba2+: 56 protons; 54 electrons; BaO e. S2-: 16 protons; 18 electrons; Al2 S3 g. Br-: 35 protons; 36 electrons; AlBr3</p> <p>h. N3-: 7 protons; 10 electrons; AlN 47. 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 The solid residue must have come from the flask. In the case of sulfur, SO42- is sulfate and SO32- is sulfite. By analogy: SeO4 2-: selenate; SeO3 2-: selenite; TeO4 2- tellurate; TeO3 2-: tellurite 50. The PO4 3- ion is phosphate and PO3 3- is phosphite. By analogy: Mg3 (AsO4 )2 : magnesium arsenate; H3 AsO4 : arsenic acid; Na3 SbO4 : sodium antimonate; Na3 AsO3 : sodium asenite; Na2 HAsO4 : 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 If the formula is In2 O3 , then two times the atomic mass of In will combine with three times the atomic mass of O, or:</p> <p>48. 49.</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>7</p> <p>, A = atomic mass of In = 114.8 The latter number is the atomic mass of In used in the modern periodic table. 52. a. Ca2+ and N3-: Ca3 N2 , calcium nitride c. Rb+ and F-: RbF, rubidium fluoride e. Ba2+ and I-: BaI2 , barium iodide g. Cs+ and P3-: Cs3 P, 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. 53. 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: = 1.50 = = 6.00; = 9.00 b. d. f. K+ and O2-: K2 O, potassium oxide Mg2+ and S2-: MgS, magnesium sulfide Al3+ and Se2-: Al2Se3, aluminum selenide</p> <p>All the masses of hydrogen in these three compounds can be expressed as simple whole number ratios 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 other oxygen family members). b. c. d. 55. Rubidium. Rb, element 37, forms stable +1 charged ions. Argon. Ar is element 18. Astatine. At is element 85.</p> <p>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 37 Cl and 1 g H/35 g Cl for 35 Cl. As long as we had pure 35 Cl or pure 37 Cl, the above 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.</p> <p>8</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>Challenge Problems56. Because the gases are at the same temperature and pressure, the volumes are directly related to the number of moles. 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 the mole ratios, which correspond to coefficients in the equation: 2H + O 2HO Since atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correct this, we can make oxygen a diatomic molecule: 2H + O2 2HO This does not require hydrogen to be diatomic. Of course, if we know water has the formula H2 O, we get: 2H + O2 2H2 O. The only was to balance this is to make hydrogen diatomic: 2H2 + O2 2H2 O a. Both compounds have C2 H6 O as the formula. Since they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties since 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 H2 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. 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). 58. For each experiment, divide the larger number by the smaller. In doing so, we get: expt. 1 expt. 2 expt. 3 x = 1.0 y = 1.4 x = 1.0 y = 10.5 z = 1.0 y = 3.5</p> <p>57.</p> <p>Our assumption about formulas dictate the rest of the solution. For example, if we assume the formula 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)</p> <p>and a formula of X3 Y for expt. 3 (three times as much X must be present in expt. 3 as compared to expt. 1 (10.5/3.5 = 3).</p> <p>CHAPTER 2</p> <p>ATOMS, MOLECULES, AND IONS</p> <p>9</p> <p>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)</p> <p>and a formula of XY3 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 the compound. If the compound in expt. 1 has formula XY, then: 21g XY = 19.2 g Y (and 1.8 g X)</p> <p>If the compound in expt. 3 has the XY formula, then: 21 g XY = 16.3 g Y (and 4.7 g X)</p> <p>Note, it could be that neither experiment 1 or 3 have XY as the formula. Therefore, there is no way of knowing an absolute answer here.</p> <p>59.</p> <p>Compound I:</p> <p>; Compound II:</p> <p>The ratio of the masses of R that combines with 1.00 g Q is:</p> <p>= 2.99 . 3</p> <p>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 (RQ), then the formula of compound I should be R3 Q. 60. 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. We...</p>