1. Chapter 3 Molecules and Compounds

2. Molecules and Compounds - Chemical Formulas 1:1 1:2 1:3 2:3 1:4etc CO H 2 O NH 3 Al 2 O 3 CH 4 C + 4 H = CH 4 Molecular model

3. Molecular Models Ethanol

4. Molecular Models

5. Some compounds are IONIC - electrons are TRANSFERRED from a metal to a nonmetal; the compound is held together ELECTROSTATICALLY Coulomb’s Law: force = k (n + )(n - )/d 2

6. Some compounds are COVALENT - electrons are SHARED between two atoms + e-e- + e-e- E. g. carbon dioxide This commonly occurs for two or more NONMETALS

7. The Covalent Bond

8. Ions An atom or group of atoms with a net charge caused by the net loss or gain of electrons CATION = positively charged ion ANION = negatively charged ion

9. Formation of Cations & Anions

10. Predicting Whether an Atom Will Form a Cation or Anion in Order to Make an Ionic Compound Metals LOSE electrons to form CATIONS, nonmetals GAIN electrons to form ANIONS

11. Valence Electrons in Ionic Compounds The A-group (representative) elements follow the OCTET RULE; they obtain an inert gas valence (outer) shell that contains 8 electrons Metals - lose # electrons = group number e.g. Ca  Ca 2+ + 2e - (Ar outer shell) Nonmetals - gain electrons = 8 - group # e.g. N + 3e -  N 3- (Ne outer shell)

12. Valence Electrons for Covalent Compounds Covalent compounds form between two or more nonmetals In this case the nonmetals can either LOSE all of their valence electrons, or - GAIN enough electrons to obtain an OCTET

13. Examples - SO 3 - oxygen (VIA) gains 8-6 = 2 O 2- ion forms - sulfur (VIA) loses all 6 S 6+ ion forms NOTE: There is a rule that states that oxygen is ALWAYS -2. These rules are coming up!

14. Example 3.3 - Predicting Ion Charges When Forming Ionic Compounds Metals lose electrons, Nonmetals gain them Al = group IIIA metal, so LOSES 3 ELECTRONS Al  Al 3+ + 3e - S = group VIA nonmetal, so GAINS 8 - 6 = 2 ELECTRONS S + 2e -  S 2-

15. Polyatomic Ions Contains 2 or more atoms COVALENTLY bonded, and the complete unit contains a net charge, e.g. nitrate, NO 3 -

16. Polyatomic Ion Examples NO 2 -, CO 3 2 -, SO 4 2 -, PO 4 3 - CO 3 2 - ion NO 2 - ion

17. Ionic Compound = Metal + Nonmetal or a Metal + Polyatomic Ion Compound held together electrostatically Very strong forces hold the lattice together, so ionic cmpd’s have very high melting points NaCl crystal lattice m.p. = 800 o C

18. Predicting Formulas of Ionic Compounds Balance positive and negative charges to produce a neutral molecule Ca 2+ + Cl -  Ca 2+ + CO 3 2 -  Ca 2+ + PO 4 3 -  Al 3+ + O 2 - 

19. Oxidation Numbers A number assigned to each element in a compound in order to keep track of the electrons during a reaction Mg 2+ = +2 Cl - = -1 O 2- = -2 N 3- = -3

20. Rules for Assigning Oxidation Numbers (Chap. 5, p. 207) Rules higher up take precedence over lower rules 1The O.N. for an atom in its pure, uncombined state = 0. 2The sum of the O.N.’s for a neutral molecule = 0. For a polyatomic ion, the sum = charge.

21. 3Group IA = +1 Group IIA = +2 4H = +1 UNLESS combined with IA or IIA, then = -1 5Oxygen = -2 6For binary ionic compounds only - Group VA = -3 Group VIA = -2 Group VIIA = -1 Rules cont’d

22. Examples P 4 Al 2 O 3 MnO 4 - NaH Na 2 SO 3 Mg 3 N 2

23. Chemical Nomenclature Examples (More Detail in Lab) Ionic Compounds NaClsodium chloride Al 2 S 3 aluminum sulfide FeSO 4 iron(II) sulfate KClO 3 potassium chlorate Covalent Compounds SO 2 sulfur dioxide P 2 O 5 diphosphorus pentaoxide N 2 Odinitrogen oxide

24. The Mole - The mole is the chemist’s counting unit Avogadro’s Number (N A ) = 6.022 X 10 23 pair = 2Dozen = 12 Gross = 144 Ream = 500

25. By definition, 12 C = 12.000 amu How many particles does it take to have 12.000 grams of 12 C ? N A = 6.022 X 10 23 (as determined by experiment) Where Does Avogadro’s Number Come From?

26. Significance of the Mole Mass in amu’s Mass in grams/mole N A of carbon atoms weighs N A of iron atoms weighs

27. Molar Mass - the mass in grams of one mole of any element Molar mass of sodium (Na) = mass of 1 mol of Na atoms = 22.99 g/mol = mass of 6.022 X 10 23 Na atoms Molar mass of lead (Pb) = mass of 1 mol of Pb atoms = 207.2 g/mol = mass of 6.022 X 10 23 Pb atoms

28. Mass Moles Conversion Moles to Mass moles grams = grams 1 mole Molar mass Mass to Moles grams 1 mole = moles grams 1 / Molar mass

29. Example 3.6 - Mass to Moles How many moles are represented by 125 g of silicon, an element used in semiconductors?

30. Example 3.7 - Moles to Mass What mass, in grams, is equivalent to 2.50 mol of lead (Pb) ?

31. Mole Calculation Using Density The graduated cylinder in the photograph contains 25.0 cm 3 of Hg. If the density of Hg = 13.534 g/cm 3 at 25 o C, how many moles of Hg are in the cylinder? How many atoms of Hg are there?

32. Molar Mass of a Compound Sum up the molar masses of each atom in the compound HC 2 H 3 O 2

33. Example 3.9 - Molar Mass & Moles You have 16.5 g of the common compound oxalic acid, H 2 C 2 O 4. Calculate - 1. The number of moles 2. The number of molecules 3. The number of C atoms 4. The mass of one molecule

34. Other Fun Stuff 1 molecule contains - 2 carbon atoms 1 oxygen atom 6 hydrogen atoms 1 mole contains - 2 moles of carbon atoms 1 mole of oxygen atoms 6 moles hydrogen atoms 46.07 g contains - 2(12.01) = 24.02 g of carbon 1(16.00) = 16.00 g of oxygen 6(1.008) = 6.05 g of hydrogen C 2 H 5 OH MW = 46.07

35. Conversion factors for C 2 H 5 OH - 2(12.01) g C/ 46.07 g C 2 H 5 OH OR 24.02 g C/ 46.07 g C 2 H 5 OH 6 moles H/ mole C 2 H 5 OH 1 mole oxygen/ 2 moles C

36. More Problems - How many grams of Na are there in 200. g of Na 2 CO 3 ? How many moles of oxygen are there in 25.0 mol of SO 2 ?

37. More Problems - How many aluminum atoms are there in 150. g of Al 2 O 3 ? How many oxygen atoms are there in 500. mL of a 30.0 % solution of H 2 SO 4 with a density of 1.250 g/cm 3 ? (MW = 98.1)

38. Percent Composition from a Known Formula NH 3 MW = 17.03 g/mol % N = % H =

39. Empirical & Molecular Formulas Empirical = simplest ratio of atoms in the molecules Molecular = actual ratio

40. Calculating Empirical Formulas Formulas of unknown compounds are determined from the percent composition of each element by mass. Assume 100 g and divide by atomic weight Divide by fewest number of moles

41. Calculating Molecular Formulas The molecular weight must be known. It is obtained from a separate experiment Benzene empirical formula = CH formula weight = 12.01 + 1.008 = 13.018 If the MW = 78.11, then what is the molecular formula?

42. Example 3.10 Eugenol is the active component of oil of cloves. It has a MW of 164.2 g/mol and is 73.14 %C and 7.37 %H; the remainder is oxygen. What are the empirical and molecular formulas?

43. Another example - Vanillin is a common flavoring agent. It has a molar mass of 152 g/mol and is 63.15 %C and 5.30 %H; the rest is oxygen. What are the empirical and molecular formulas?