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Chapter 4 Atomic Structure Theories about matter were based on the ideas of Greek philosophers: Democritus (400 B.C. ) – coins the term “atom” saying.

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Presentation on theme: "Chapter 4 Atomic Structure Theories about matter were based on the ideas of Greek philosophers: Democritus (400 B.C. ) – coins the term “atom” saying."— Presentation transcript:

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2 Chapter 4 Atomic Structure

3 Theories about matter were based on the ideas of Greek philosophers: Democritus (400 B.C. ) – coins the term “atom” saying matter can be subdivided only as small as an elemental particle. from Greek “a” means not and “tomos” means cutting atom means “indivisible”

4 Greeks continued… Aristotle thought that all matter consisted of 4 basic elements: ◦ earth, air, fire, water *Their views were not backed by any evidence.

5 Dalton’s Atomic Theory (experiment based!) 3)Atoms of different elements combine in simple whole-number ratios to form chemical compounds 4)In chemical reactions, atoms are combined, separated, or rearranged – but never changed into atoms of another element. 1)All elements are composed of tiny indivisible particles called atoms 2)Atoms of the same element are identical. Atoms of any one element are different from those of any other element. John Dalton (1766 – 1844)

6 In 1981 we see atoms! ◦ Using the scanning electron microscope ◦ OR (STM – scanning tunneling microscope) ◦ See page. 100 (Ni atoms) ◦ *Note: only seeing electron cloud

7 Section 2: The Structure of the Atom consists of:ProtonsNeutronsElectrons symbols:p + n o e - location: in nucleus in nucleus outside nucleus gain/loss: not gained can be a can be gained or lost different number or lost (isotopes) (forms ions)

8 consists of:ProtonsNeutronsElectrons mass number: 1 amu 1 amu 0 amu *important # p + *holds nucleus *involved in info: determines togetherchemical type of elementreactions (atomic number)

9 Nuclear Force: Very strong, short range force that holds the particles of the nucleus together Question: What is it about the composition of the nucleus that requires the concept of nuclear forces? A: Like charges do repel each other, so protons would not be expected to be close to other protons. Nuclear forces prevent the protons from repelling each other.

10 Question: What are the attraction relationships in the nucleus (application of nuclear forces)? A: proton-proton, neutron-proton, neutron-neutron

11 Section 4.2 Structure of the Nuclear Atom One change to Dalton’s atomic theory is that atoms are divisible into subatomic particles: ◦ Electrons, protons, and neutrons are examples of these fundamental particles

12 Discovery of the Electron In 1897, J.J. Thomson used a cathode ray tube to deduce the presence of a negatively charged particle: the electron p. 106

13 Modern Cathode Ray Tubes  Cathode ray tubes pass electricity through a gas that is contained at a very low pressure. TelevisionComputer Monitor

14 Mass of the Electron 1916 – Robert Millikan determines the mass of the electron: 1/1840 the mass of a hydrogen atom; has one unit of negative charge The oil drop apparatus Mass of the electron is 9.11 x 10 -28 g

15 Conclusions from the Study of the Electron: a) Cathode rays have identical properties regardless of the element used to produce them. All elements must contain identically charged electrons. b) Atoms are neutral, so there must be positive particles in the atom to balance the negative charge of the electrons c) Electrons have so little mass that atoms must contain other particles that account for most of the mass

16 Conclusions from the Study of the Electron:  Eugen Goldstein in 1886 observed what is now called the “proton” - particles with a positive charge, and a relative mass of 1 (or 1840 times that of an electron)  1932 – James Chadwick confirmed the existence of the “neutron” – a particle with no charge, but a mass nearly equal to a proton

17 Subatomic Particles ParticleCharge Mass (g) Location Electron (e - ) (e - ) 9.11 x 10 -28 9.11 x 10 -28 Electron cloud Proton (p + ) +1 1.67 x 10 -24 1.67 x 10 -24Nucleus Neutron (n o ) (n o )0 1.67 x 10 -24 1.67 x 10 -24Nucleus

18 Thomson’s Atomic Model Thomson believed that the electrons were like plums embedded in a positively charged “pudding,” thus it was called the “plum pudding” model. J. J. Thomson

19 Ernest Rutherford’s Gold Foil Experiment - 1911  Alpha particles are helium nuclei - The alpha particles were fired at a thin sheet of gold foil  Particles that hit on the detecting screen (film) are recorded

20 Rutherford’s Findings a) The nucleus is small b) The nucleus is dense c) The nucleus is positively charged  Most of the particles passed right through  A few particles were deflected  VERY FEW were greatly deflected “Like howitzer shells bouncing off of tissue paper!” Conclusions:

21 The Rutherford Atomic Model Based on his experimental evidence: The atom is mostly empty space All the positive charge, and almost all the mass is concentrated in a small area in the center. He called this a “nucleus” The nucleus is composed of protons and neutrons (they make the nucleus!) The electrons distributed around the nucleus, and occupy most of the volume His model was called a “nuclear model”

22 Rutherford’s atom – nuclear model

23 Isotopes *Dalton was wrong about all elements of the same type being identical Isotopes are atoms of the same element that have different numbers of neutrons in the nucleus.  Same chemical reactions  Differ in mass  Examples: carbon-12 and carbon-13

24 Isotope hyphen notation: Element name-dash mass number Example: carbon-14 where14 is the mass number. (NOTE: mass number = #p + and #n 0 ) ________ protons ________electrons ________ neutrons

25 How many protons, neutrons, and electrons are there in cobalt-60? _______ protons _______ electrons _______ neutrons

26 isotope symbols isotope symbols Contain the symbol of the element, the mass number and the atomic number. X Mass number Atomic number Subscript → Superscript →

27 Symbols n Find each of these: a) number of protons b) number of neutrons c) number of electrons d) Atomic number e) Mass Number Br 80 35

28 Symbols n n If an element has 78 electrons and 117 neutrons what is the a) Atomic number b) Mass number c) number of protons d) complete symbol

29 Symbols If an element has 91 protons and 140 neutrons what is the a) Atomic number b) Mass number c) number of electrons d) complete symbol

30 *Another word for isotope is nuclide. Mass # = p + + n 0 Nuclide p+p+p+p+ n0n0n0n0 e-e-e-e- Mass # Oxygen - 10 -3342 - 31 - 3115 8 8 18 Arsenic 7533 75 Phosphorus 15 31 16

31 Hydrogen isotopes have names. All other isotopes are named with the mass number. IsotopeProtonsElectronsNeutronsNucleus Hydrogen–1 (protium) (protium)110 Hydrogen-2(deuterium)111 Hydrogen-3(tritium)112

32 Isotopes of hydrogen Visual Examples + - + - + - Hydrogen (Protium)Hydrogen (Deuterium)Hydrogen (Tritium)

33 Measuring Atomic Mass Instead of grams, the unit we use is the Atomic Mass Unit (amu) It is defined as one-twelfth the mass of a carbon- 12 atom. ◦ Carbon-12 chosen because of its isotope purity. Each isotope has its own atomic mass, thus we determine the average from percent abundance.

34 Atomic Masses pg. 116 IsotopeSymbolComposition of the nucleus % in nature Carbon-12 12 C6 protons 6 neutrons 98.89% Carbon-13 13 C6 protons 7 neutrons 1.11% Carbon-14 14 C6 protons 8 neutrons <0.01% Atomic mass is the average of all the naturally occurring isotopes of that element. Carbon = 12.011

35 How many isotopes does carbon have? Carbon (C) has 15 known isotopes, from 8 C to 22 C

36 To calculate the average: p. 118 Multiply the atomic mass of each isotope by it’s % abundance (expressed as a decimal), then add the results. *If not told otherwise, the mass of the isotope is expressed in atomic mass units (amu)

37 Example problem: Element X has two natural isotopes. The isotope with a mass of 10.012 amu (X-10) has a relative abundance of 19.91%. The isotope with a mass of 11.009 amu (X-11) has a relative abundance of 80.09%. Calculate the atomic mass of this element. See page 119

38 Example problem: The relative abundances and atomic masses are 0.337% (mass = 35.978 amu), 0.063% (mass = 37.963 amu), and 99.600% (mass = 39.962 amu), respectively. Calculate the average atomic mass of argon.

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