PreAP Chemistry Chapter 4 Chapter 4 Annotation Questions Due to the box NOW!

1. PreAP Chemistry Chapter 4 Chapter 4 Annotation Questions Due to the box NOW! Late is only worth 60%, but is better than 0%!

2. Democritus was the early (around 400BC) Greek philosopher who is credited with the concept of the atom (atomos) –which means invisible

3. Dalton(around 1800AD) is an English school teacher who proposed the law of conservation of mass, the law of definite proportions, and the law of multiple proportions. His many experiments with gases proved these laws are true, if atoms exist.

4. Dalton is also known as the Father of the (Modern) Atomic Theory

5. Dalton’s atomic theory: 1.All matter is composed of very small particles called atoms

6. Dalton’s atomic theory: • All matter is composed of very small particles called atoms • Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in these properties.

7. 3.Atoms cannot be subdivided, created, or destroyed

8. Atoms cannot be subdivided, created, or destroyed • 4.Atoms of different elements combine in simple whole-number ratios to form chemical compounds.

9. Atoms cannot be subdivided, created, or destroyed • 4.Atoms of different elements combine in simple whole-number ratios to form chemical compounds. • 5.In chemical reactions, atoms are combined, separated, or rearranged.

10. Two aspects of Dalton’s atomic theory proven to be incorrect: • We now know atoms are divisible.

11. Two aspects of Dalton’s atomic theory proven to be incorrect: • We now know atoms are divisible. • b. Atoms of the same element can have different masses.

12. Atom --smallest particle of an element that retains the properties of that element.

13. J. J. Thomson is the man credited with the discovery of the electrons in the late 1800’s, using cathode ray tubes

14. Millikan calculated the mass of the electron (very, very small)

15. Knowledge of electrons led to two inferences about atomic structure: 1.Because atoms are electrically neutral, they must contain positive charge to balance the negative electrons.

16. Knowledge of electrons led to two inferences about atomic structure: • 1.Because atoms are electrically neutral, they must contain positive charge to balance the negative electrons. • 2. Because electrons have so little mass, atoms must contain other particles to account for most of their mass

17. What would this look like?

18. What would this look like?

19. Nucleus of the atom—discovered by Lord Ernest Rutherford

20. Nucleus of the atom—discovered by Lord Ernest Rutherford Gold foil experiment—actually done by Hans Geiger and Ernest Marsden

21. Observations: • Majority of the alpha (α)particles penetrated foil undeflected. • b.About 1 in 20,000 were slightly deflected • c.About 1 in 20,000 were deflected back to emitter

23. Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus

24. Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus 2. Most of the atom is empty

25. Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus • Most of the atom is empty • 3.Magnitude of charge on the nucleus is different for different atoms

26. 4. Number of electrons outside the nucleus = number of units of nuclear charge (to account for the fact that the atom is electrically neutral)

27. 4. Number of electrons outside the nucleus = number of units of nuclear charge (to account for the fact that the atom is electrically neutral) Atoms are electrically neutral because they contain equal numbers of protons and electrons

28. A couple years later Rutherford presented evidence for a neutral particle which was also in the nucleus and contained a similar mass to that of a proton – called a neutron

29. Mass of one proton = mass of neutron = mass of 1837 electrons

30. Mass of one proton = mass of neutron = mass of 1837 electrons Thus the total mass of an atom is basically the sum of the protons and neutrons, called the atomic mass or mass number, abbreviated A

31. Atomic number—the number of protons in the nucleus of the atom.

32. Atomic number—the number of protons in the nucleus of the atom. --number of protonsidentifies the element and is equal to the number of electrons (of a neutral atom) --symbol is Z

33. Isotopes are atoms of the same element that have different masses because they have different numbers of neutrons but they still have similar chemical properties

34. Isotopes of Carbon

35. Isotopes of Carbon

36. Nuclide is the general term for any isotope of any element

37. Nuclide is the general term for any isotope of any element Atomic Mass Unit (amu) is exactly 1/12 the mass of a carbon-12 atom

38. Average atomic mass is the weight average of the atomic masses of the naturally occurring isotopes of an element. Ave. Atomic mass = %abundance(mass of isotope 1) + %abundance(mass of isotope 2) +…..

39. Example 1: Element Sciencium has two isotopes. Sciencium-301 has an abundance of 59.5%, and Sciencium-304 is the other. What is the average atomic mass? 301amu x .595 = 179 amu 304amu x .405 = 123 amu 302 amu

40. Example 2: Element Pepsium has an average atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope? 327amu x 0.305 = 99.7 amu ?amu x 0.695 = ? amu 335 amu 1 – 0.305 = 0.695

41. Example 2: Element Pepsium has an average atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope? 327amu x 0.305 = 99.7 amu ?amu x 0.695 = ? amu 335 amu Step 1: Find the missing weighted mass

42. Example 2: Element Pepsium has an average atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope? 327amu x 0.305 = 99.7 amu ?amu x 0.695 = 235amu 335 amu 335 – 99.7 = 235.3

43. Example 2: Element Pepsium has an average atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope? 327amu x 0.305 = 99.7 amu ?amu x 0.695 = 235amu 335 amu Step 2: Find the missing mass

44. Example 2: Element Pepsium has an average atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope? 327amu x 0.305 = 99.7 amu 338amu x 0.695 = 235amu 335 amu 235 ÷ 0.695 = 338

45. Mass Spectrometry How we know isotopes.

46. A mass spectrometer has three parts: 1. ionizer 2. magnetic field 3. detector And might look like this

47. If a sample of a pure element is placed in the spectrometer, then all the ions formed will have the same charge. For example Cl → Cl- Cl Cl- Ionizer

48. The ions then pass through a magnetic field that will change their paths. Which will change direction more, something heavy or something light? - Cl Cl- Cl- + Magnetic Field Detector

49. The computer attached to the detector gives a readout like this:

50. The locations tell the masses – one group of Cl had a mass of 35 amu’s and the other had a mass of 37 amu’s.