1. Focusing on Molecules and Ions Today

2. Molecules, Formulas, Structures Molecular StructureMolecular FormulaEmpirical Formula H - HH2H2 H O = S = OSO 2 C3H8C3H8 C3H8C3H8 C2H6C2H6 CH 3 N2H4N2H4 NH 2

3. Chemical Bonds Molecular Compounds = Covalent Compounds –Formed by sharing electrons –Between nonmetals –Shared electrons are shown by shared lines Ionic Compounds –Formed by ions held together by electrostatic charges

4. Ions When atoms lose or gain electrons, they become ions. –Cations - positive ions are formed by elements on the left side of the periodic chart. Mg Mg +2 + 2e - –Anions - negative ions are formed by elements on the right side of the periodic chart. F + e - F -

5. Cations vs Anions How do I know if an element will become a cation or an anion? –Metals generally lose electrons during reactions (rxns) –Nonmetals generally gain electrons during rxns –So… metals tend to be cations and nonmetals tend to be anions

6. Trends in Ions +1 +2 +3 -4-3-2

7. Monatomic Ions Single atoms that have gained or lost electron –Metals Groups 1A-3A form positive charges according to the group number Transition metals (Group B) –Form cations –Not as easy to predict as group A metals –Often times they have multiple charges –Nonmetals Often form negative charges according to the group number (group A) with the exception of group 8A… noble gases –Hydrogen –can gain or lose electrons

8. Polyatomic Ions Made up of two or more atoms that have a collective charge (either positive or negative) Common Polyatomic Ions (pg 74) IonsNames NH 4 + Ammonium NO 3 - Nitrate CO 3 -2 Carbonate OH - Hydroxide SO 4 -2 Sulfate PO 4 -3 Phosphate C2H3O2-C2H3O2- Acetate CN - Cyanide

9. Ionic Bonds Ionic compounds are generally formed between a metal and a nonmetal

10. Ionic Bonds - Writing Formulas When ions combine to form compounds, the result must be neutral in charge. You can determine the formula of a compound in 2 ways: 1.Use math to make sure the total charge of all ions add up to zero. 2.Use the “criss-cross method” oThe charge on the cation becomes the subscript on the anion. The charge on the anion becomes the subscript on the cation. If these subscripts are not in the lowest whole-number ratio, divide them by the greatest common factor.

11. Writing Formulas

12. Predict the formula of the compound: Magnesium + Chlorine Mg + Cl Mg +2 + Cl - MgCl 2 Rubidium + Oxygen Rb + O Rb + + O -2 Rb 2 O

13. Writing Formulas Calcium + Carbonate –Ca + CO 3 CaCO 3 Ammonium + Phosphate –NH 4 + PO 4 (NH 4 ) 3 PO 4

14. Naming Ionic Compounds 1.Name the cation a.Monatomic cation - name the metal then add “ion” i.Ca +2 = Calcium Ion b.If the cation has more than one charge possibility (think transition metal), indicate the charge using roman numerals in parentheses immediately following the ion name i.Fe +2 = Iron(II)

15. Naming Ionic Compounds 2.Name the anion a.Monatomic - Use the suffix “-ide” i.Cl - = Chloride b.Polyatomic - memorize the common one (we’ll go through the naming procedure in a bit) 3.Name the cation then the anion

16. Naming Ionic Compounds (Practice) NiS Li 3 N Al(OH) 3 PbCl 2 Nickel Sulfide Lithium Nitride Aluminum Hydroxide Lead (II) Chloride

17. Properties of Ionic Compounds Held together by electrostatic forces Force of attraction between ions is given by Coulomb’s Law Insert Naming Here… Insert Naming of Covalent as well

18. MOLE A mole is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope 6.0221415 x 10 23 things Avagadro’s number –He had the basic idea… but never determined the number… “false advertising”!

19. MOLE Is the cornerstone of quantitave chemistry

20. MOLE Here’s a fun little fact for you… AMU’s… where exactly did they get that number? –1 amu = 1.66054 x 10 -24 g –1g = 6.02214 x 10 23 amu

21. MOLE Practice How many moles of glass beads are present in 425.6g of beads? (100 beads = 8.24g) 425.6g x (100beads/8.24g) = 5165.0 beads 5156.0 beads x (1mol/6.02214x10 23 beads) = 8.58 x 10-21 mole of beads

22. Molar Mass (M) The mass in grams of one mole of an element. Unit = g/mol An element’s molar mass is the quantity in grams equal to it’s atomic weight

23. Molar Mass to Moles Consider a 62.02 gram sample of K 2 SO 4. How many moles of K 2 SO 4 are present? –62.02g x (1mol/174.3g) = 0.3558 mol K 2 SO 4 How many molecules of K 2 SO 4 are present? –0.3558 mol x (6.022x10 23 molecules/ 1 mol) = 2.143 X 10 23 molecules

24. Describing Compound Formulas Any sample of a pure substance always consists of the same elements combined in the proportion by mass –We know this… law of definite proportion (law of constant composition)

25. Describing Compound Formulas Can express molecular composition at least 3 ways: 1.Molecular Formula 2.in terms of the mass of each element per mole of compound 3.Mass of each element in the compound relative to the total mass of the compound –mass percent

26. Percent Composition What is the mass percent of each element in NH 3 ? 1 mole NH 3 = 17.031 g %N = 14.007g N / 17.031g NH 3 x 100% = 82.244% %H = 3.0237g H / 17.031g NH 3 x 100% = 17.755%

27. Empirical and Molecular Formulas from Percent Composition Consider a substance with the following composition: 59.99 % C, 8.05 % H, and 31.96 % O Can you find the empirical formula? YES!!!!

28. Empirical and Molecular Formulas from Percent Composition Three Steps: 1.Convert the percent to mass If I assume a 100g sample, then the percent will equal the mass 2.Convert mass of each element to moles 3.Find the mole ratio of elements

29. Empirical and Molecular Formulas from Percent Composition 1.Convert the percent to mass 59.99% C = 59.99g C 8.05% H = 8.05g H 31.96% O = 31.96g O

30. Empirical and Molecular Formulas from Percent Composition 2.Convert the mass to moles 59.99g C x ( 1mol / 12.011g ) = 5 mol C 8.05g H x ( 1mol / 1.008g ) = 7.99 mol H 31.96g O x ( 1mol / 15.999g ) = 1.998 mol O

31. Empirical and Molecular Formulas from Percent Composition 3.Convert the mass to moles C / O = ( 5 / 1.998 ) = 2.5 H / O = ( 7.99 / 1.998 ) = 4 O / O = ( 1.998 / 1.998 ) = 1

32. Empirical and Molecular Formulas from Percent Composition Empirical Formula: C 2.5 H 4 O 1 What’s the problem? Subscripts need to be whole numbers! C5H8O2C5H8O2

33. Empirical and Molecular Formulas from Percent Composition What is the molecular formula? –Need to know the molecular weight! –The molecular weight = 200 g/mol Find the empirical mass C 5 H 8 O 2 = 100.1g/mol Divide the molecular weight by the empirical mass : 200/100.1 = 2 Molecular formula = C 10 H 16 O 4