1. Chapter 4 – Atomic Structure

2. Greek Atomic Theory (400 BC) “There is no smallest among the small and no largest among the large, but always something still smaller and something still larger.” Opposed Democritus’ idea of the atom Anassagoras “These particles are absolutely small, so small that their size cannot be diminished; absolutely full and incompressible, differing only in shape, arrangement, position, and magnitude. Democritus called these particles atoms (“indivisible”). Democritus

3. Evidence Supporting Atomic Theory (1700s) Law of conservation of mass –Total mass of the products is the same as the sum of the masses of the reactants. –Matter cannot be neither created nor destroyed, just rearranged! + Sulfur atom 32 mass units Oxygen molecule 32 mass units Sulfur dioxide molecule 64 mass units

4. Evidence Supporting Atomic Theory (1700s) Law of definite composition –A compound contains the same elements in exactly the same proportions by mass regardless of the size of the sample or source of the compound.

5. Evidence Supporting Atomic Theory (1700s) Law of multiple proportions –Applies to different compounds made from the same elements. –Mass ratio for one of the elements that combines with a fixed mass of the other element can be expressed in small whole numbers.

6. Evidence Supporting Atomic Theory (1700s) Law of multiple proportions Ratio of mass of oxygen in H 2 O 2 to mass of oxygen in H 2 O is 2 mass units 1 mass units

7. Modern Atomic Theory (1800s) John Dalton Dalton argued that these three experimental laws could not be explained without assuming that all compounds are made from tiny particles such as atoms. This reasoning led to the development of the modern theory of the atom.

8. Dalton’s Atomic Theory (1800s) All matter is made of indivisible and indestructible atoms. Atoms of a given element are identical in their physical and chemical properties. Atoms of different elements have different physical and chemical properties. Atoms of different elements combine in simple, whole-number ratios to form chemical compounds. Atoms cannot be subdivided, created, or destroyed when they are combined, separated, or rearranged in chemical reactions.

9. Atomic Model Modifications (1910) e-e- e-e- e-e- e-e- e-e- e-e- e-e- Rutherford Electron Cloud Positive nucleus (proton) e-e- e-e- e-e- e-e- e-e- e-e- e-e- Thompson Embedded electrons Positive sphere

10. 10 -13 cm Rutherford’s Atomic Model (1910) 10 -8 cm e-e- e-e- e-e- e-e- e-e- An atom is mostly empty space!!

11. Discovered the neutron by comparing electric charges and masses of different atoms. The neutron is found in the nucleus with the proton. The neutron has the same mass as a proton. The neutron has no charge. James Chadwick (1932) James Chadwick Nobel Laureate 1935

12. Chadwick Neutral neutron Positive Proton Atomic Model Modifications (1932) Rutherford Electron Cloud Positive nucleus (proton) e-e- e-e- e-e- e-e- e-e- e-e- e-e-

13. Subatomic Particles

14. Anassagoras (400 BC) Democritus (400 BC) Law of Definite Proportions Law of Conservation of Mass Law of Multiple Proportions Dalton (1800s) Thompson (1897)Rutherford (1911) 1700s e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- Chadwick (1932) Discovery of electron, plum pudding model Discovery of nucleus (proton) Discovery of neutron Modern atomic theory

15. Check for understanding Which Chemist from 500 BC developed the theory that atoms were indivisible, the smallest particle? Which Chemist discovered the electron and created the Plum pudding model? Which Chemist had a large forehead and discovered the neutron?

16. How do the structures of atoms differ? 6 C Carbon 12.011 Atomic Number Chemical Symbol Element Name Atomic mass

17. How do the structures of atoms differ? 6 C Carbon 12.011 Atomic Number Number of protons in an atom. In carbon, there are 6 protons in each carbon atom. Since atoms are electrically neutral, the number of protons must equal the number of electrons in an atom!

18. How do the structures of atoms differ? 6 C Carbon 12.011 Atomic mass Sum of protons and neutrons in an atom. Standard unit is atomic mass unit (amu). 1 amu = 1.661 x 10 -24 grams 1 amu = 1/12 th of the mass of a carbon-12 atom

19. How do the structures of atoms differ? 6 C Carbon 12.011 8 O Oxygen 15.9994 Determine the atomic number, the number of protons, electrons, and neutrons for each element.

20. How do the structures of atoms differ? 8 O Oxygen 15.9994 Atomic number = # protons Oxygen has 8 protons and 8 electrons (electrically neutral). Atomic mass = # protons + # neutrons 15.9994 amu = 8 protons + 8 neutrons. Oxygen has 8 protons, 8 electrons, and 8 neutrons

21. How do the structures of atoms differ? 14 Si Silicon 28.086 92 U Uranium 238.029 Determine the atomic number, the number of protons, electrons, and neutrons for each element.

22. Average Masses and Isotopes Q: If carbon has 6 protons and 6 neutrons, why is its atomic mass 12.011 and not 12.000? A: The mass listed for any element on the Periodic Table is the weighted average of the masses of all its naturally occurring isotopes.

23. Isotopes are atoms of the same element that have the same number of protons and electrons but different numbers of neutrons. Carbon has two naturally occurring stable isotopes, carbon-12 and carbon-13. Where 12 and 13 represent mass numbers. Isotopes

24. Isotope Notation X A Z Chemical symbol for the element Mass Number Atomic Number Ex. Write carbon-12 and carbon-13 in isotope notation. Determine the number of protons, electrons and neutrons.

25. Isotope Notation C 12 6 Carbon - 12Carbon - 13 C 13 6 Protons = 6 Electrons = 6 Neutrons = 12-6 = 6 Protons = 6 Electrons = 6 Neutrons = 13-6 = 7

26. Average Masses and Isotopes Q: If carbon has 6 protons and 6 neutrons, why is its atomic mass 12.011 and not 12.000? A: The mass listed for any element on the Periodic Table is the weighted average of the masses of all its naturally occurring isotopes.

27. Weighted Average A 100 x MAMA + () B x () MBMB Abundance of isotope A Mass of isotope A Abundance of isotope B Mass of isotope B Atomic Mass follows the general form of… + (……

28. Weighted Average Carbon 12 98.89% 12 13 1.11% 13 IsotopeAbundance Mass A 100 x MAMA + () B x () MBMB Abundance of Carbon-12 (A) Mass of Carbon-12 (A) Abundance of Carbon-13 (B) Mass of Carbon-13 (B)

29. Weighted Average Carbon 12 98.89% 12 13 1.11% 13 IsotopeAbundance Mass 98.89 100 C = x 12+ () 1.11 100 x () 13 = 12.011 Abundance of Carbon-12 Mass of Carbon-12 Abundance of Carbon-13 Mass of Carbon-13

30. Isotopes Diamond, C Carbon-12 (98.89%) Carbon-13 (1.11%)

31. Carbon-12 (98.89%) Carbon-13 (1.11%) + + + + + + + + + + + +

32. You Try It Determine the atomic mass for silicon if 92.21% of its atoms have a mass of 27.98 amu, 4.70% have a mass of 28.98 amu, and 3.09% have a mass of 29.97 amu. 92.21 100 Si = x 27.98 + () 4.70 100 x 28.98 ( ) + 3.09 100 x 29.97 ( ) Si = 28.1amu

33. A Bit More Difficult Boron has two naturally occurring isotopes, boron-10 and boron-11. The abundance of boron-10 is 19.97% and the mass of boron-10 is 10.0129 amu. The abundance of boron-11 is 80.17%. Determine the mass of boron-11 if the atomic mass of boron is 10.811 amu.

34. Atomic Mass Can atoms be counted or measured? C 12 6 Carbon - 12 Protons = 6 Electrons = 6 Neutrons = 12-6 = 6 ParticlesMass Protons = 6 x (1.673 x 10 -24 g) Electrons = 6 x (9.11 x 10 -27 g) Neutrons = 6 x (1.675 x 10 -24 g) Mass Carbon-12 = 1.99 x 10 -23 g

35. Atomic Mass Can atoms be counted or measured? C 12 6 Carbon - 12 Mass 1 Carbon-12 atom = 1.99 x 10 -23 g Instead of using actual atomic masses, chemists work with relative masses. In establishing this relative scale, scientists chose the carbon-12 atom as the standard. Based on this, 1 atomic mass unit (amu) = 1/12 th the mass of a carbon-12 atom. Mass 1 Carbon-12 atom = 1.99 x 10 -23 g Mass 1 Carbon-12 atom = 12 amu

36. Bridge between the “invisible” world of atoms and the macroscopic world of materials. 1 Equality that indicates quantity not mass 2 1 dozen _____________ = 12 _____________ XX eggs 3 Conversion factor 1 dozen X 12 X The Mole (mol,  )

37. Bridge between the “invisible” world of atoms and the macroscopic world of materials. Conversion factor that indicates quantity not mass 2 1 1 mol _____________ = 6.02 × 10 23 _____________ XX atoms 3 Conversion factor 1 mol X 6.02 × 10 23 X “Avogadro’s number” 4

38. 6.02 X 10 23 “Avogadro’s number” 1 602,000,000,000,000,000,000,000 “602 Billion Trillion” 2 20,000,000,000,000 $100 bills6,342 years to distribute 200,000,000,000 pounds2,500,000 semi-trucks 1.33 times around the Earth

39. 6.02 X 10 23 There is 1 mol of atoms in the atomic mass of an element when that mass is expressed in grams. 3 C 12 6 Carbon - 12 12 amu 1 mol carbon-12 atoms = ___________ atoms of carbon-12 = 6.02 x 10 23 ___________ grams of carbon-12. 12

40. 6.02 X 10 23 There is 1 mol of atoms in the atomic mass of an element when that mass is expressed in grams. 3 F 19 9 Fluorine - 19 19 amu 1 mol fluorine-19 atoms = __________ atoms of fluorine-19 = 6.02 x 10 23 __________ grams of fluorine-19. 19

41. 6.02 X 10 23 There is 1 mol of atoms in the atomic mass of an element when that mass is expressed in grams. 3 Carbon - 12Hydrogen - 1Fluorine - 19 1 mol 6.02 x 10 23 atoms 6.02 x 10 23 atoms 6.02 x 10 23 atoms 12 grams1 gram19 grams The mole represents quantity not mass!

42. 6 C Carbon 12.011 Atomic Mass C 14 6 Carbon - 14 14 amu vs. A specific isotope of CThe element C 1 mol of carbon-14 = 14 grams 1 mol of carbon = 12.0 grams 6.02 x 10 23 atoms

43. 9 F Fluorine 18.9940 Atomic Mass F 22 9 Fluorine - 22 22 amu vs. A specific isotope of FThe element F 1 mol of fluorine-22 = 22 grams 1 mol of fluorine = 19.0 grams 6.02 x 10 23 atoms

44. Determine how many golf balls are in 34 dozen. Conversion Factor ? 1. Given2. Unknown 34 dozen# of golf balls 3. Transformation4. Set-up and solve 12 golf balls per 1 dozen 12 golf balls 1 dozen given X unknown given 34 dozen 12 golf balls 1 dozen

45. Determine how many atoms are in 3 moles of Al. Conversion Factor ? 1. Given2. Unknown 3 moles of Alatoms 3. Transformation4. Set-up and solve 6.02 x 10 23 atoms per 1 mol 6.02 x 10 23 atoms 1 mol given X unknown given 3 moles Al 6.02 x 10 23 atoms 1 mole Al

46. Determine how many moles are in 4 x 10 34 atoms of C. Conversion Factor ? 1. Given2. Unknown 4 x 10 34 atoms of Cmoles 3. Transformation4. Set-up and solve 6.02 x 10 23 atoms per 1 mol 6.02 x 10 23 atoms 1 mol given X unknown given 4 x 10 34 atoms 1 mol 6.02 x 10 23 atoms

47. Atomic Mass Can atoms be counted or measured? Instead of using actual atomic masses, chemists work with relative masses. In establishing this relative scale, scientists chose the carbon-12 atom as the standard. Based on this, 1 atomic mass unit (amu) = 1/12 th the mass of a carbon-12 atom. F 19 9 Fluorine - 19 Mass 1 Fluorine-19 atom = 3.18 x 10 -23 g Mass 1 Fluorine-19 atom = 19 amu