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Element #8 : Oxygen, Isotopes

Element #8 : Oxygen, Isotopes. 16 8 O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 17 8 O 8 Protons 9 Neutrons 0.037% 16.9997341 amu 18 8 O 8 Protons 10 Neutrons

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Element #8 : Oxygen, Isotopes

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  1. Element #8 : Oxygen, Isotopes • 168O 8 Protons 8 Neutrons 99.759% 15.99491462 amu • 178O 8 Protons 9 Neutrons 0.037% 16.9997341 amu • 188O8 Protons 10 Neutrons 0.204 % 17.999160 amu

  2. Fig.2.A

  3. Fig.2.B

  4. Calculating the “Average” Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%). 24Mg (78.7%) 23.98504 amu 25Mg (10.2%) 24.98584 amu 26Mg (11.1%) 25.98636 amu Total = With Significant Digits = amu

  5. Calculating the “Average” Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%). 24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amux 0.102 = 2.548556 amu 26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu 24.309268 amu With Significant Digits = 24.3 amu

  6. Calculate the Average Atomic Mass of Zirconium, Element #40 Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr. Isotope (% abd.) Mass (amu) (%) Fractional Mass 90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu 91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu 92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu 94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu 96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu 91.2215 amu With Significant Digits = 91.22 amu

  7. Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is 79.904 amu. Since exact masses of isotopes not give, estimate from: mass in amu = #p+ + #n: 79Br = 79 g/mol and 81Br = 81 g/mol (approximately). Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0 Solution:

  8. Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is 79.904 amu. Since exact masses of isotopes not give, estimate from: mass in amu = #p+ + #n: 79Br = 79 g/mol and 81Br = 81 g/mol (approximately). Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0 Solution: X(79) + Y(81) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(79) + Y(81) = 79.904 79 - 79 Y + 81Y = 79.904 2 Y = 0.904 = 1 w/ sig. figs. so Y = 0.5 X = 1.00 - Y = 1.00 - 0.5 = 0.5 %X = % 79Br = 0.5 x 100% = 50% (Actual: 50.67% = 79Br) %Y = % 81Br = 0.5 x 100% = 50% (Actual: 49.33% = 81Br)

  9. Modern Reassessment of the Atomic Theory 1. All matter is composed of atoms. Although atoms are composed of smaller particles (electrons, protons, and neutrons), the atom is the smallest body thatretains the unique identity of the element. 2. Atoms of one element cannot be converted into atoms of another element ina chemical reaction. Elements can only be converted into other elements in Nuclear reactions in which protons are changed. 3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number, but not in chemical behavior (much). A sample of the element is treated as though its atoms have an average mass. 4. Compounds are formed by the chemical combination of two or more elements in specific ratios, as originally stated by Dalton.

  10. Definitions • ELEMENT - A substance that cannot be separated into simpler substances by chemical means • COMPOUND - A substance composed of atoms of two or more elements chemically united in fixed proportions • PERIODIC TABLE - “MENDELEEV TABLE” - A tabular arrangement of the elements, vertical groups or families of elements based upon their chemical properties - actually combining ratios with oxygen

  11. Fig2.16

  12. Fig.2.17

  13. Groups in the Periodic Table Main Group Elements (Vertical Groups) Group IA - Group IIA - Group IIIA - Group IVA - Group VA - Group VIA - Group VIIA - Group VIIIA - Other Groups ( Vertical and Horizontal Groups) Group IB - 8B - Period 6 Group - Period 7 Group -

  14. Groups in the Periodic Table Main Group Elements (Vertical Groups) Group IA - Alkali Metals Group IIA - Alkaline Earth Metals Group IIIA - Boron Family Group IVA - Carbon Family Group VA - Nitrogen Family Group VIA - Oxygen Family (Calcogens) Group VIIA - Halogens Group VIIIA - Noble Gases Other Groups ( Vertical and Horizontal Groups) Group IB - 8B - Transition Metals Period 6 Group -Lanthanides (Rare Earth Elements) Period 7 Group - Actinides

  15. Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr The Periodic Table of the Elements H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Du Sg Bo Ha Me The Halogens The Alkali Metals The Alkaline Earth Metals The Noble Gases

  16. The Periodic Table of the Elements H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Du Sg Bo Ha Me Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Boron family Nitrogen family Carbon Family Oxygen Family

  17. The Periodic Table of the Elements H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Du Sg Bo Ha Me The Transition Metals Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Lanthanides: The Rare Earth Elements The Actinides

  18. Fig.2.18

  19. Fig.2.19

  20. Fig.2.20

  21. Predicting the Ion an Element will form in Chemical Reactions Problem: What monoatomic ions will each of the elements form? (a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9) Plan: We use the “z” value to find the element in the periodic table and which is the nearest noble gas. Elements that lie after a noble gas will loose electrons, and those before a noble gas will gain electrons. Solution: (a) Ba (b) S (c) Ti (d) F

  22. Predicting the Ion an Element will form in Chemical Reactions Problem: What monoatomic ions will each of the elements form? (a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9) Plan: We use the “z” value to find the element in the periodic table and which is the nearest noble gas. Elements that lie after a noble gas will loose electrons, and those before a noble gas will gain electrons. Solution: (a) Ba+2, Barium is an alkaline earth element, Group 2A, and is expected to loose two electrons to attain the same number of electrons as the noble gas Xenon! (b) S -2, Sulfur is in the Oxygen family, Group 6A, and is expected to gain two electrons to attain the same number of electrons as the noble gas Argon! (c) Ti+4, Titanium is in Group 4B, and is expected to loose 4 electrons to attain the same number of electrons as the noble gas Argon! (d) F -, Fluorine is in a halogen, Group 7A, and is expected to gain one electron, to attain the same number of electrons as the noble gas Neon!

  23. The Periodic Table of the Elements Most Probable Oxidation State (same concept as Fig. 2.20) +1 0 +2 +3 +_4 - 3 - 2 - 1 H He Li Be B C N O F Ne +3 +4 +5 +1 + 2 Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cu Zn Ga Ge As Se Br Cr Mn Fe Co Ni Kr Rb Sr Zr Nb Ag I Mo Tc Ru Rh Pd Xe Y Cd In Sn Sb Te Ba Au Hg W Re Os Ir Pt Rn Cs La Hf Ta Tl Pb Bi Po At Fr Ra Ac Rf Du Sg Bo Ha Me +3 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr +3

  24. Chemical Compounds and Bonds Chemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound. Covalent Compounds - Electrons are shared between atoms of different elements to form Covalent Cpds. Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds. “Cation” - An atom that has lost electron(s) to form “ + ” ions. May be 1 or more e-s. Common with metal elements. “Anion” - An atom which has gained electron(s), to form “ - ” ions. Common w/ nonmetal elements. Later we’ll learn that group of atoms can also be anion or cation Mono-atomic (monatomic) ions form binary ionic compounds.

  25. Fig.2.20

  26. Fig 2.22 (P 65) The polyatomic ion

  27. Chemical Formulas Empirical Formula- Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.

  28. Empirical and Molecular Formulas NameMolecularEmpiricalwater H2O hydrogen H2O2 peroxideethane C2H6 sulfur S8acetic acid CH3COOH

  29. Empirical and Molecular Formulas NameMolecularEmpiricalwater H2O H2Ohydrogen H2O2 HO peroxideethane C2H6 CH3sulfur S8 Sacetic acid CH3COOH COH2

  30. Fig.2.23

  31. Start trying to learn those in bold. Best done by looking at name w/ use.

  32. Give the Name and Chemical Formulas of the Compounds formed from the following pairs of Elements a) Sodium and Oxygen Na2O Sodium Oxide b) Zinc and Chlorine c) Calcium and Fluorine d) Strontium and Nitrogen e) Hydrogen and Iodine f) Scandium and Sulfur

  33. Give the Name and Chemical Formulas of the Compounds formed from the following pairs of Elements a) Sodium and Oxygen Na2O Sodium Oxide b) Zinc and Chlorine ZnCl2 Zinc Chloride c) Calcium and Fluorine CaF2 Calcium Fluoride d) Strontium and Nitrogen Sr3N2 Strontium Nitride e) Hydrogen and Iodine HI Hydrogen Iodide f) Scandium and Sulfur Sc2S3 Scandium Sulfide

  34. Start learning these boldface ones.

  35. Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion. Give the systematic names for the formulas or the formulas for the names of the following compounds. a) Iron III Sulfide - Fe is +3, and S is -2 therefore the compound is: Fe2S3 b) CoF2 - c) Stannic Oxide - d) NiCl3 -

  36. Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion. Give the systematic names for the formulas or the formulas for the names of the following compounds. a) Iron III Sulfide - Fe is +3, and S is -2 therefore the compound is: Fe2S3 b) CoF2 - the anion is Fluoride (F -1) and there are two F -1, the cation is Cobalt and it must be Co+2 therefore the compound is: Cobalt (II) Fluoride c) Stannic Oxide - Stannic is the common name for Tin (IV), Sn+4, the Oxide ion is O-2, therefore the formula of the compound is: SnO2 d) NiCl3 - The anion is chloride (Cl-1), there are three anions, so the Nickel cation is Ni+3, therefore the name of the compound is: Nickel (III) Chloride

  37. Rules for Families of Oxoanions Families with Two Oxoanions The ion with more O atoms takes the nonmetal root and the suffix “-ate”. The ion with fewer O atoms takes the nonmetal root and the suffix “-ite”. Families with Four Oxoanions (usually a Halogen) The ion with most O atoms has the prefix “per-”, the nonmetal root and the suffix “-ate”. The ion with one less O atom has just the suffix “-ate”. The ion with two less O atoms has the just the suffix “-ite”. The ion with three less O atoms has the prefix “hypo-” and the suffix “-ite”.

  38. Fig.2.24

  39. NAMING OXOANIONS - EXAMPLES Prefixes Root Suffixes Chlorine Bromine Iodine per “ ” ate perchlorate perbromate periodate [ ClO4-] [ BrO4-] [ IO4-] “ ” ate chlorate bromate iodate [ ClO3-] [BrO3-] [ IO3-] “ ” ite chlorite bromite iodite [ ClO2-] [ BrO2-] [ IO2-] hypo “ ” ite hypochlorite hypobromite hypoiodite [ ClO -] [ BrO -] [ IO -] No. of O atoms

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