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Chapter 2 Atoms, Molecules and Ions. History of Chemistry Greeks Alchemy.

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Presentation on theme: "Chapter 2 Atoms, Molecules and Ions. History of Chemistry Greeks Alchemy."— Presentation transcript:

1 Chapter 2 Atoms, Molecules and Ions

2 History of Chemistry Greeks Alchemy

3 Theory vs. Law Theory – human attempts to explain or interpret natural phenomenon Law – summarizes what occurs (observed behavior)

4 Dalton’s Atomic Theory Elements - made up of atoms Same elements, same atoms. Different elements, different atoms. Chemical reactions involve bonding of atoms

5 Periodic Table Rows (Left to Right) - periods Columns (top to bottom) - groups

6 The Atom Made up of: – Protons – (+) charged – Electrons – (-) charged – neutrons

7 Periodic Table Atomic Mass Atomic Number Alkali Metals (Group I) Alkali Earth Metals (Group II) Chalcogens (Group VI) Halogens (Group VII) Noble Gases (Group VIII) Transition Metals

8 Periodic Table Alkaline Metals – Grps. I & II Transition Metals Non-metals Halogens – Group VII Noble Gases –Group VIII - little chemical activity

9 Periodic Table Lanthanides Actinides

10 Periodic Table Atomic Mass - # at bottom how much element weighs Atomic Number - # on top gives # protons = # electrons

11 Periodic Table Atomic Mass – number below the element – not whole numbers because the masses are averages of the masses of the different isotopes of the elements

12 Ions Are charged species Result when elements gain electrons or lose electrons

13 2 Types of Ions Anions – (-) charged Example:F - Cations – (+) charged Example:Na +

14 Highly Important! Gain of electrons makes element (-)=anion Loss of electrons makes element (+)=cation

15 Isotopes Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.

16 Example Isotopes% Abundance 12 C 98.89 % 13 C 1.11 % 14 C 11 C

17 – For example, the mass of C = 12.01 a.m.u is the average of the masses of 12 C, 13 C and 14 C.

18 Determination of Aver. Mass Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

19 Take Note: If there are more than 2 isotopes, then formula has to be re-adjusted

20 Sample Problem 1 Assume that element Uus is synthesized and that it has the following stable isotopes: – 284 Uus (283.4 a.m.u.)34.6 % – 285 Uus (284.7 a.m.u.)21.2 % – 288 Uus (287.8 a.m.u.)44.20 %

21 Solution Ave. Mass of Uus = [ 284 Uus](283.4 a.m.u.)(0.346) [ 285 Uus] +(284.7 a.m.u.)(0.212) [ 288 Uus] +(287.8 a.m.u.)(0.4420) = 97.92 + 60.36 + 127.21 = 285.49 a.m.u (FINAL ANS.)

22 Oxidation Numbers Is the charge of the ions (elements in their ion form) Is a form of electron accounting Compounds have total charge of zero (positive charge equals negative charge)

23 Oxidation States Are the partial charges of the ions. Some ions have more than one oxidation states.

24 Oxidation States - generally depend upon the how the element follows the octet rule Octet Rule – rule allowing elements to follow the noble gas configuration

25 Nomenclature - naming of compounds

26 Periodic Table Rows (Left to Right) - periods Columns (top to bottom) - groups

27 Periodic Table Mendeleev – arranged elements in the (.) table

28 Periodic Table Atomic Mass – number below the element – not whole numbers because the masses are averages of the masses of the different isotopes of the elements

29 – For example, the mass of C = 12.01 a.m.u is the average of the masses of 12 C, 13 C and 14 C.

30 Determination of Aver. Mass Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

31 Take Note: If there are more than 2 isotopes, then formula has to be re-adjusted

32 Sample Problem 1 Assume that element Uus is synthesized and that it has the following stable isotopes: – 284 Uus (283.4 a.m.u.)34.6 % – 285 Uus (284.7 a.m.u.)21.2 % – 288 Uus (287.8 a.m.u.)44.20 %

33 Solution Ave. Mass of Uus = [ 284 Uus](283.4 a.m.u.)(0.346) [ 285 Uus] +(284.7 a.m.u.)(0.212) [ 288 Uus] +(287.8 a.m.u.)(0.4420) = 97.92 + 60.36 + 127.21 = 285.49 a.m.u (FINAL ANS.)

34 Periodic Table Atomic Mass – number below the element – not whole numbers because the masses are averages of the masses of the different isotopes of the elements

35 STOICHIOMETRY – For example, the mass of C = 12.01 a.m.u is the average of the masses of 12 C, 13 C and 14 C.

36 Formula Weight & Molecular Weight The FORMULA WEIGHT of a compound equals the SUM of the atomic masses of the atoms in a formula. If the formula is a molecular formula, the formula weight is also called the MOLECULAR WEIGHT.

37 The MOLE Amount of substance that contains an Avogadro’s number (6.02 x 10 23 )of formula units.

38 The MOLE The mass of 1 mole of atoms, molecules or ions = the formula weight of that element or compound in grams. Ex. Mass of 1 mole of water is 18 grams so molar mass of water is 18 grams/mole.

39 Formula for Mole Mole = mass of element formula weight of element

40 Sample Mole Calculations 1 mole of C = 12.011 grams » 12.011 gm/mol 0.5 mole of C = 6.055 grams » 12.011 gm/mol

41 Avogadro’s Number Way of counting atoms Avogadro’s number = 6.02 x 10 23

42 Point to Remember One mole of anything is 6.02 x 10 23 units of that substance.

43 1 gram = 6.02 x 10 23 a.m.u

44 Avogadro’s Number and the Mole If one mole of anything is 6.02 x 10 23 units of that substance, then: 1 mole of oranges = 6.02 x 10 23 oranges

45 And…….. 1 mole of C has the same number of atoms as one mole of any element

46 Also….. 1 mole of sand = 6.02 x 10 23 particles

47 An Even Better Analogy….. 1 dozen = 12 entities a dozen apples has the same number of entities as a dozen oranges

48 Formula Weight & Molecular Weight The FORMULA WEIGHT of a compound equals the SUM of the atomic masses of the atoms in a formula. If the formula is a molecular formula, the formula weight is also called the MOLECULAR WEIGHT.

49 Summary Avogadro’s Number gives the number of particles or atoms in a given number of moles 1 mole of anything = 6.02 x 10 23 atoms or particles

50 Sample Problem 2 Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.

51 Solution PART I: Formula for Mole: – Mole = mass of element atomic mass of element

52 Solution (cont.) Part II:To determine # of atoms # atoms = moles x Avogadro’s number

53 Problem # 2 A diamond contains 5.0 x 10 21 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?

54 Molar Mass Often referred to as molecular mass – Unit = gm/mole Definition: – mass in grams of 1 mole of the compound

55 Example Problem Determine the Molar Mass of C 6 H 12 O 6

56 Solution Mass of 6 mole C = 6 x 12.01 = 72.06 g Mass of 12 mole H = 12 x 1.008 = 12.096 g Mass of 6 mole O = 6 x 16 = 96 g Mass of 1 mole C 6 H 12 O 6 = 180.156 g

57 Problem #3 What is the molar mass of (NH 4 ) 3 (PO 4 )?

58 Molar Mass Often referred to as molecular mass – Unit = gm/mole Definition: – mass in grams of 1 mole of the compound

59 Sample Problem Given 75.99 grams of (NH 4 ) 3 (PO 4 ), determine the ff: 1. Molar mass of the compound 2. # of moles of the compound 3. # of molecules of the compound 4. # of moles of N 5. # of moles of H 6. # of moles of O 7. # of atoms of N 8. # of atoms of H 9. # of atoms of O

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