1. Atoms, Molecules and Ions Chapter 2

2. Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements. 2. Compounds are composed of atoms of more than one element. The relative number of atoms of each element in a given compound is always the same. 3. Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed in chemical reactions. 2.1

3. 2

4. 8 X 2 Y 16 X8 Y + 2.1

5. J.J. Thomson, measured mass/charge of e - (1906 Nobel Prize in Physics) 2.2

6. Cathode Ray Tube 2.2

7. Laws nConservation of Mass nLaw of Definite Proportion- compounds have a constant composition by mass. nThey react in specific ratios by mass. nMultiple Proportions- When two elements form more than one compound, The different masses of one element that combine with the same mass of the other element are in the ration of small whole numbers.

8. What?! nWater has 8 g of oxygen per 1g of hydrogen. nHydrogen peroxide has 16 g of oxygen per 1g of hydrogen. nComparing the grams of O per 1 g of H in each compound: n16/8 = 2/1 nSmall whole number ratios

9. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 9.10 1 Analyze List the knowns and the unknown. Apply the law of multiple proportions to the two compounds. For each compound, find the grams of carbon that combine with 1.00 g of oxygen. Then find the ratio of the masses of carbon in the two compounds. Confirm that the ratio is the lowest whole-number ratio. KNOWNS Compound A = 2.41 g C and 3.22 g O Compound B = 6.71 g C and 17.9 g O UNKNOWN Mass ratio of C per g O in two compounds = ?

10. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 9.10 Solve Solve for the unknown. First, calculate grams of carbon per gram of oxygen in compound A. 2 2.41 g C 3.22 g O 0.748 g C 1.00 g O =

11. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 9.10 Solve Solve for the unknown. Then, calculate grams of carbon per gram of oxygen in compound B. 2 6.71 g C 17.9 g O 0.375 g C 1.00 g O =

12. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 9.10 Solve Solve for the unknown. Calculate the mass ratio to compare the two compounds. 2 To calculate the mass ratio, compare the masses of one element per gram of the other element in each compound. 0.748 g C 0.375 g O 1.99 1 =≈ 2 1

13. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 9.10 Evaluate Does this result make sense? The ratio is a low whole-number ratio, as expected. For a given mass of oxygen, compound A contains twice the mass of carbon as compound B. 3

14. Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% by mass carbon and 57.1% by mass oxygen. The second compound contains 27.3% by mass carbon and 72.7% by mass oxygen. Show that the data are consistent with the Law of Multiple Proportions. The Law of Multiple Proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers.

15. Therefore, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio. In 100 g of the first compound (100 is chosen to make calculations easier) there are 57.1 g O and 42.9 g C. The mass of O per gram C is:57.1 g O / 42.9 g 1.33 g O per g C In the 100 g of the second compound, there are 72.7 g O and 27.3 g C. The mass of oxygen per gram of carbon is:72.7 g O / 27.3 g C = 2.66 g O per g C Dividing the mass O per g C of the second (larger value) compound:2.66 / 1.33 = 2

16. nGay-Lussac- under the same conditions of temperature and pressure, compounds always react in whole number ratios by volume. nAvagadro- interpreted that to mean nat the same temperature and pressure, equal volumes of gas contain the same number of particles n(called Avagadro’s hypothesis) A Helpful Observation

17. e - charge = -1.60 x 10 -19 C Thomson’s charge/mass of e - = -1.76 x 10 8 C/g e - mass = 9.10 x 10 -28 g Measured mass of e - (1923 Nobel Prize in Physics) 2.2

18. (Uranium compound) 2.2 0 -1 e 4 2 He

19. 2.2

20. The modern view of the atom was developed by Ernest Rutherford (1871-1937).

21. 1.atoms positive charge is concentrated in the nucleus 2.proton (p) has opposite (+) charge of electron (-) 3.mass of p is 1840 x mass of e - (1.67 x 10 -24 g)  particle velocity ~ 1.4 x 10 7 m/s (~5% speed of light) (1908 Nobel Prize in Chemistry) 2.2

22. atomic radius ~ 100 pm = 1 x 10 -10 m nuclear radius ~ 5 x 10 -3 pm = 5 x 10 -15 m Rutherford’s Model of the Atom 2.2 “If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.”

23. Chadwick’s Experiment (1932) H atoms - 1 p; He atoms - 2 p mass He/mass H should = 2 measured mass He/mass H = 4  + 9 Be 1 n + 12 C + energy neutron (n) is neutral (charge = 0) n mass ~ p mass = 1.67 x 10 -24 g 2.2 There must be something else in the nucleus

24. mass p = mass n = 1840 x mass e - 2.2

25. Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei X A Z H 1 1 H (D) 2 1 H (T) 3 1 U 235 92 U 238 92 Mass Number Atomic Number Element Symbol 2.3

27. 6 protons, 8 (14 - 6) neutrons, 6 electrons 6 protons, 5 (11 - 6) neutrons, 6 electrons Do You Understand Isotopes? 2.3 How many protons, neutrons, and electrons are in C 14 6 ? How many protons, neutrons, and electrons are in C 11 6 ?

28. Average atomic mass from isotopes

29. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Atomic Mass The atomic mass of an element is a weighted average mass of the atoms in a naturally occurring sample of the element. A weighted average mass reflects both the mass and the relative abundance of the isotopes as they occur in nature.

30. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Atomic Mass To calculate the atomic mass of an element, multiply the mass of each isotope by its natural abundance, expressed as a decimal, and then add the products.

31. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Atomic Mass Carbon has two stable isotopes: carbon-12, which has a natural abundance of 98.89 percent, and carbon-13, which has a natural abundance of 1.11 percent. The mass of carbon-12 is 12.000 amu; the mass of carbon-13 is 13.003 amu. The atomic mass of carbon is calculated as follows: Atomic mass of carbon = (12.000 amu x 0.9889) + 13.003 amu x 0.0111) = (11.867 amu) + (0.144 amu) = 12.011 amu

32. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculating Atomic Mass Element X has two naturally occurring isotopes. The isotope with a mass of 10.012 amu ( 10 X) has a relative abundance of 19.91 percent. The isotope with a mass of 11.009 amu ( 11 X) has a relative abundance of 80.09 percent. Calculate the atomic mass of element X. Sample Problem 4.5

33. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 4.5 The mass each isotope contributes to the element’s atomic mass can be calculated by multiplying the isotope’s mass by its relative abundance. The atomic mass of the element is the sum of these products. Analyze List the knowns and the unknown. 1 atomic mass of X = ? KNOWNSUNKNOWN Isotope 10 X: mass = 10.012 amu relative abundance = 19.91% = 0.1991 Isotope 11 X: mass = 11.009 amu relative abundance = 80.09% = 0.8009

34. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Use the atomic mass and the decimal form of the percent abundance to find the mass contributed by each isotope. for 10 X: 10.012 amu x 0.1991 = 1.993 amu for 11 X: 11.009 amu x 0.8009 = 8.817 amu Sample Problem 4.5 Calculate Solve for the unknowns. 2

35. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Add the atomic mass contributions for all the isotopes. Sample Problem 4.5 Calculate Solve for the unknowns. 2 For element X, atomic mass = 1.953 amu + 8.817 amu = 10.810 amu

36. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 4.5 Evaluate Does the result make sense? 3 The calculated value is closer to the mass of the more abundant isotope, as would be expected.

37. Period Group Alkali Metal Noble Gas Halogen Alkali Earth Metal 2.4

38. A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds H2H2 H2OH2ONH 3 CH 4 A diatomic molecule contains only two atoms H 2, N 2, O 2, Br 2, HCl, CO A polyatomic molecule contains more than two atoms O 3, H 2 O, NH 3, CH 4 2.5

39. ELEMENTS THAT EXIST AS DIATOMIC MOLECULES Remember: BrINClHOF These elements only exist as PAIRS. Note that when they combine to make compounds, they are no longer elements so they are no longer in pairs! P: 1 or 4 S: 1 or 8

40. An ion is an atom, or group of atoms, that has a net positive or negative charge. cation – ion with a positive charge If a neutral atom loses one or more electrons it becomes a cation. anion – ion with a negative charge If a neutral atom gains one or more electrons it becomes an anion. Na 11 protons 11 electrons Na + 11 protons 10 electrons Cl 17 protons 17 electrons Cl - 17 protons 18 electrons 2.5

41. Forming Cations & Anions A CATION forms when an atom loses one or more electrons. An ANION forms when an atom gains one or more electrons Mg --> Mg 2+ + 2 e- F + e- --> F -

42. A monatomic ion contains only one atom A polyatomic ion contains more than one atom 2.5 Na +, Cl -, Ca 2+, O 2-, Al 3+, N 3- OH -, CN -, NH 4 +, NO 3 -

43. 13 protons, 10 (13 – 3) electrons 34 protons, 36 (34 + 2) electrons Do You Understand Ions? 2.5 How many protons and electrons are in ?Al 27 13 3+ How many protons and electrons are in ?Se 78 34 2-

44. 2.5

45. 2.6

46. A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in a substance H2OH2O H2OH2O molecularempirical C 6 H 12 O 6 CH 2 O O3O3 O N2H4N2H4 NH 2 2.6

47. ionic compounds consist of a combination of cation(s) and an anion(s) the formula is always the same as the empirical formula the sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero The ionic compound NaCl 2.6

48. Formula of Ionic Compounds Al 2 O 3 2.6 2 x +3 = +63 x -2 = -6 Al 3+ O 2- CaBr 2 1 x +2 = +22 x -1 = -2 Ca 2+ Br - Na 2 CO 3 1 x +2 = +21 x -2 = -2 Na + CO 3 2-

49. 2.6

50. 2.7

51. Examples of Older Names of Cations formed from Transition Metals From Zumdahl

52. Chemical Nomenclature Ionic Compounds –often a metal + nonmetal –anion (nonmetal), add “ide” to element name BaCl 2 barium chloride K2OK2O potassium oxide Mg(OH) 2 magnesium hydroxide KNO 3 potassium nitrate 2.7

53. Transition metal ionic compounds –indicate charge on metal with Roman numerals FeCl 2 2 Cl - -2 so Fe is +2 iron(II) chloride FeCl 3 3 Cl - -3 so Fe is +3 iron(III) chloride Cr 2 S 3 3 S -2 -6 so Cr is +3 (6/2)chromium(III) sulfide 2.7

54. Molecular compounds nonmetals or nonmetals + metalloids common names H 2 O, NH 3, CH 4, C 60 element further left in periodic table is 1 st element closest to bottom of group is 1 st if more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom last element ends in ide 2.7

55. HIhydrogen iodide NF 3 nitrogen trifluoride SO 2 sulfur dioxide N 2 Cl 4 dinitrogen tetrachloride NO 2 nitrogen dioxide N2ON2Odinitrogen monoxide Molecular Compounds 2.7 TOXIC ! Laughing Gas

56. 2.7

57. An acid can be defined as a substance that yields hydrogen ions (H + ) when dissolved in water. HCl Pure substance, hydrogen chloride Dissolved in water (H + Cl - ), hydrochloric acid An oxoacid is an acid that contains hydrogen, oxygen, and another element. HNO 3 nitric acid H 2 CO 3 carbonic acid H 2 SO 4 sulfuric acid 2.7 HNO 3

58. 2.7

61. A base can be defined as a substance that yields hydroxide ions (OH - ) when dissolved in water. NaOH sodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide 2.7

63. Mixed Practice 1.Dinitrogen monoxide 2.Potassium sulfide 3.Copper (II) nitrate 4.Dichlorine heptoxide 5.Chromium (III) sulfate 6.Ferric sulfite 7.Calcium oxide 8.Barium carbonate 9.Iodine monochloride 1.N 2 O 2.K 2 S 3.Cu(NO 3 ) 2 4.Cl 2 O 7 5.Cr 2 (SO 4 ) 3 6.Fe 2 (SO 3 ) 3 7.CaO 8.BaCO 3 9.ICl

64. Mixed Practice 1.BaI 2 2.P 4 S 3 3.Ca(OH) 2 4.FeCO 3 5.Na 2 Cr 2 O 7 6.I 2 O 5 7.Cu(ClO 4 ) 2 8.CS 2 9.B 2 Cl 4 1.Barium iodide 2.Tetraphosphorus trisulfide 3.Calcium hydroxide 4.Iron (II) carbonate 5.Sodium dichromate 6.Diiodine pentoxide 7.Cupric perchlorate 8.Carbon disulfide 9.Diboron tetrachloride

65. 65 Organic chemistry is the branch of chemistry that deals with carbon compounds C H H H OH C H H H NH 2 C H H H COH O methanol methylamineacetic acid Functional Groups

66. 66