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Chapter 4 Structure of the Atom

Chapter 4 Structure of the Atom. 4.3 How Atoms Differ. Section 4.3 How Atoms Differ. The number of protons and the mass number define the type of atom. Objectives. Explain the role of atomic number in determining the identity of an atom .

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Chapter 4 Structure of the Atom

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  1. Chapter 4Structure of the Atom • 4.3 How Atoms Differ

  2. Section 4.3 How Atoms Differ The number of protons and the mass number define the type of atom. Objectives • Explain the role of atomic number in determining the identity of an atom. • Define an isotope; be able to identify and give an example of one • Write an isotope in any one of the 3 standard notations for them. • Explain how the atomic mass unit (amu) is defined • Describe in semi-quantitative terms the relative masses of the neutron, the proton and the electron.

  3. Section 4.3 How Atoms Differ The number of protons and the mass number define the type of atom. Objectives (cont) • Explain why atomic masses are not whole numbers. • Calculate the number of electrons, protons, and neutrons in an atom given its mass number and atomic number. • Calculate the atomic mass of an element given the isotope masses and abundances (and variations of this problem). • Explainwhat a mass spectrum is, identify the instrument used to obtain itand describethe basic ideas behind its operation.

  4. Section 4.3 How Atoms Differ Key Concepts • The atomic number of an atom is given by its number of protons. The mass number of an atom is the sum of its neutrons and protons. • atomic number = number of protons = number of electrons • mass number = atomic number + number of neutrons • Atoms of the same element with different numbers of neutrons are called isotopes. • The atomic mass unit (amu) is defined as 1/12 the mass of a carbon-12 atom • The atomic mass of an element is a weighted average of the masses of all of its naturally occurring isotopes.

  5. Atomic Number (AN) • AN = # of protons • For any neutral element • # of protons = # of electrons • Different ANs  different elements • Lithium, Li AN = 3 • Carbon, C AN = 6

  6. Practice • Atomic Number • Practice Problems, page 116 • 12 – 15 • Chapter Assessment, page 128-9 • 58, 59, 66, 75 • Appendix Suppl. problems, pp 977-8 • 1(a-f), 2&3(a-d)

  7. Isotopes / Mass Number • Isotopes have same number of protons but a differing number of neutrons • Same # protons Same element • Mass number = sum of number of protons and neutrons in the nucleus • Mass number = atomic number + number of neutrons • Mass number not the same as atomic mass

  8. Isotopes / Mass Number • To distinguish a particular isotope from another, use element name followed by dash followed by mass number • Potassium-39 • Hydrogen-3 (aka tritium) • Can also use chemical symbol • K-39 • H-3

  9. K-39 K-40 K-41 19e- 19e- 19e- 19p+ 21n0 19p+ 20n0 19p+ 20n0 19p+ 22n0 Potassium, K (Figure 4.17) • K AN = 19 (19 protons, 19 electrons)

  10. Mass Number Atomic Number Alternate Notation for Isotopes Rather than using silver-107, silver-109 for isotopes, often preferable to use Note that numbers are to left of symbol

  11. 22 Ne symbol 10 Example Problem 4.2, page 118 • One of neon’s isotopes found to have atomic number 10 & mass number 22 • Find # protons, electrons, neutrons • Name isotope and give its symbol • AN = # protons = 10 • Neutral atom, # electrons = protons=10 • Mass number = # protons + # neutrons • # neutrons = 22 – 10 = 12 • Isotope = neon-22

  12. Practice • Atomic & mass number and isotope symbols • Practice Problems, page 118 • 16, 17 • Chapter Assessment, page 128-9 • 60 – 62, 64, 67-69, 72-74 • Appendix Suppl. Problems, page 978 • 4, 5, 6(a-f)

  13. Mass of Individual Atoms • So far, only discussed mass number • Atomic mass has definition in terms of a chosen atomic standard • Carbon-12 atom assigned a mass of exactly 12 atomic mass units (amu) • => One amu = 1/12 mass of 126C • All masses of atoms or atomic particles expressed in terms of amu

  14. Masses of Subatomic Particles Protons and neutrons do not have mass of exactly 1 amu Proton (p+1) and neutron (n0) masses slightly different Electron mass ~1/1840 (p+1 or n0)

  15. Mass Spectrum & Mass Spectrometer • Q. How do you “weigh” these atoms to get their masses? • Mass spectroscopy • Charge (ionize) atom or molecule • Accelerate in electric field • Laws of physics predict path of ion in a known magnetic field • Specific path and place where it strikes a detector depends on ion’s mass • See page 125

  16. Mass Spectrometer Detector Electric field accelerates ions Least massive ions + ions Most massive ions Magnetic Field Slits Heating coil to vaporize sample

  17. Mass Spectrometer Launch video from misc Mass Spectroscopy (Royal Soc Chem) (7m 58s) Essentials: from start to 1:54 & from 3:25 to 4:48 Mass Spectroscopy (Royal Soc Chem)

  18. Mass Spectrometer Acceleration Ionization Electromagnet To vacuum pump Vaporized Sample Deflection Detection

  19. Mass Spectrum – Mercury Isotopes

  20. Relative Abundance Mass Number Mass Spectrum - Mercury Natural abundance Hg-196, 0.146% Hg-198, 10.02% Hg-199, 16.84% Hg-200, 23.13% Hg-201, 13.22% Hg-202, 29.80% Hg-204, 6.85%

  21. Atomic Mass - Elements • Atomic mass of element is weighted average of the isotopes of that element • AM(element) = • AVGwt = Mass(1) x Abundance(1) + • Mass(2) x Abundance(2) + • … • where Mass(i) = atomic mass of isotope(i)

  22. Atomic Mass - Elements Atomic masses don’t have integer values because: a) Protons and neutrons have masses close to but not exactly 1 amu, so mass of a given isotope not integer b) Even if isotope masses had integer values, process of doing weighted average over isotopes generally gives result which is not an integer

  23. 3517Cl Atomic mass = 34.969 amu % abundance = 75.770% Contribution to weighted avg = 26.496 amu 3717Cl Atomic mass = 36.966 amu % abundance = 24.230% Contribution to weighted avg = 8.9569 amu AVGwt = 26.496 + 8.9569 = 35.453amu This is value listed in period table for Cl • Chlorine Example – Fig 4.18, page 119

  24. Example Problem 4.3, page 121 • Unknown element X • 6X 6.015 amu 7.59% abundance • 7X 7.016 amu 92.41% abundance • Calculate contributions to weighted avg • 6.015 amu 0.0759 = 0.457 amu • 7.016 amu 0.9241 = 6.483 amu • Sum to find mass; Atomic mass = • 0.457 + 6.483 = 6.940 amu • Matches atomic mass of lithium (Li)

  25. Practice • At mass of from isotope abundance • Practice Problems, page 121 • 18 - 19 • Section Assessment, page 121 • 23 - 24 • Chapter Assessment, page 129 • 71 (data source?), 76 – 78 • Appendix Suppl. Problems, page 978 • 7, 8

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