1. Review Unit 3 (Chp 1,2,3,4): Stoichiometry & Solutions John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Uncertainty and Error Measuring devices have different uses and different degrees of precision. (uncertainty) % Error = |Accepted – Experimental| x100 Accepted

3. 5.68 1.Nonzero digits are significant. 2.Leading Zeroes are never significant. 3. Trailing Zeroes: SIG, only if a decimal point. 0.0003700400 Significant Figures fewest decimal places fewest significant figures + or – x or ÷ 3.48 + 2.2 = 6.40 x 2.0 = 5.7 12.8 13

4. perchlorateClO 4 – chlorateClO 3 – chloriteClO 2 – hypochloriteClO – CNOF SiPSCl AsSeBr TeI nitrateNO 3 – nitriteNO 2 – InOutIon Name – – “Oxyanion” Names (elbO’s) 4 3 2 1 4 3 ___-ate ___-ite sulfateSO 4 2– sulfiteSO 3 2– phosphatePO 4 3– per-___-ate hypo-___-ite

5. nitrateNO 3 – nitriteNO 2 – sulfateSO 4 2– sulfiteSO 3 2– perchlorateClO 4 – chlorateClO 3 – chloriteClO 2 – hypochloriteClO – Ion Acid add H + Name Acids from these oxyanions: InOutIon NameAcid Name 4– per-___-ate 34 ___-ate 23 ___-ite 1– hypo-___-ite per-___-ic acid ___-ic acid ___-ous acid hypo-___-ous acid Naming Acids

6. Using Moles Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. Mass (grams) Particles (atoms) (molecules) (units) Moles (groups of 6.022x10 23 particles) molar mass # g 1 mol # g Avogadro constant 6.022x10 23 1 mol 6.022x10 23 macro- bridgemicro-

7. 1) Convert grams to moles (if necessary) using the molar mass (from PT) 2) Convert moles (given) to moles (wanted) using the mol ratio (from coefficients) 3) Convert moles to grams using the molar mass (from PT) grams A x 1 mol A. grams A = _ mol B mol A x x grams B 1 mol B 1) molar mass2) mole ratio3) molar mass Stoichiometric Calculations (OR Liters, atoms, …)

8. Percent Composition and Empirical Formulas

9. Percent Composition by Mass % element = (# of atoms)(AW) (FW) x 100 So the percentage of carbon in ethane (C 2 H 6 ) is… %C = (2)(12.01) (30.07) 24.02 30.07 = x 100 = 79.88% C

10. Types of Formulas Empirical formulas: the lowest ratio of atoms of each element in a compound. Molecular formulas: the total number of atoms of each element in a compound. CH 3 C2H6C2H6 C2H4OC2H4O C 6 H 12 O 3 molecular mass = multiple of emp. form. empirical mass

11. Percent to Mass Mass to Mole Divide by Small Times ‘till Whole Steps (rhyme) ÷ moles by smallest to get mole ratio of atoms MM from PT assume 100 g x (if necessary) to get whole numbers of atoms Calculating Empirical Formulas 75 % C75 g C6.2 mol C 25 % H25 g H24.8 mol H 1 C 4 H CH 4 from Mass % Composition

12. A hydrocarbon has is 17.34% H and 82.66% C by mass. Determine its empirical formula. If MM is 58 g mol –1, what is the Molecular Formula? 82.66 g C 17.34 g H 1) Percent to Mass 2) Mass to Mole 82.66 g C x = 6.883 mol C 17.34 g H x = 17.20 mol H 1 mol C 12.01 g C 1 mol H 1.008 g H 6.883 mol 3) Divide by Small 4) Times ’till Whole = 1  1 C = 2.499  2.5 H x 2 = 2 C C2H5C2H5 x 2 = 5 H C 4 H 10 58 29.06 = 2 2 (C 4 H 10 ) = molecular mass empirical mass

13. Hydrocarbons with C and H are analyzed through combustion with O 2 in a chamber.  g C is from the g CO 2 produced  g H is from the g H 2 O produced  g X is found by subtracting (g C + g H) from g sample Combustion Analysis

14. A sample of a chlorohydrocarbon with a mass of 4.599 g, containing C, H and Cl, was combusted in excess oxygen to yield 6.274 g of CO 2 and 3.212 g of H 2 O. Calculate the empirical formula of the compound. If the compound has a MW of 193 g∙mol –1, what is the molecular formula? Combustion Analysis

15. 6.274 g CO 2 x 1 mol CO 2 44.01 g CO 2 1 mol H 2 O 18.02 g H 2 O 1 mol C 1 mol CO 2 x 12.01 g C 1 mol C x 2 mol H 1 mol H 2 O x 1.008 g H 1 mol H x 3.212 g H 2 O x = 1.712 g C = 0.3593 g H 4.599 g sample – (1.712 g C) – (0.3593 g H) = = 2.528 g Cl ? g C ? g H ? g Cl

16. 0.07131 mol 1.712 g C x 1 mol C 12.01 g C 1 mol H 1.008 g H 0.1425 mol C = 0.3564 mol H = 0.3593 g H x = = 2 C 5 H 2.528 g Cl x 1 mol Cl 35.45 g Cl 0.07131 mol Cl = =1 Cl C 2 H 5 Cl If the compound has a MW of 193 g∙mol –1, what is the molecular formula? MW EW 193 64.51 = 3 C 6 H 15 Cl 3

17. Before 2 H 2 + O 2 2 H 2 O Initial: ? mol ? mol ? mol Change: End: 10 7 0 0 mol 2 mol 10 mol –10 –5+10 H2H2 O2O2 O 2 is in smallest amount, but… H 2 is in smallest “stoichiometric” amount Does limiting mean smallest amount of reactant?No! After Limiting Reactant

18. Solid aluminum metal is reacted with aqueous copper(II) chloride in solution. If 0.030 g Al are reacted with 0.0060 mol CuCI 2, which is the limiting reactant? How much product can be produced? Limiting Reactant Al (s) + CuCl 2 (aq) Cu (s) + AlCl 3 (aq) 2332

19. 2 Al (s) + 3 CuCl 2 (aq) 3 Cu (s) + 2 AlCl 3 (aq) Limiting Reactant 0.030 g Al 0.0060 mol CuCl 2 1 mol Al 26.98 g Al 3 mol Cu 3 mol CuCl 2 x x x 3 mol Cu 2 mol Al = 0.0017 mol Cu = 0.0060 mol Cu Al is limiting

20. Theoretical Yield theoretical yield: the maximum mass of product that can be formed –calculated by stoichiometry (using LR only) This is different from the actual yield, the amount one actually produces and measures (or experimental) 0.030 g Al 1 mol Al 26.98 g Al x x 3 mol Cu 2 mol Al = 0.0017 mol Cu

21. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make %Yield = x 100 Actual Theoretical (calculate using the LR only) % Error = |Accepted – Experimental| x100 Accepted NOT % Error:

22. 101.96 g Al 2 O 3 1 mol Al 2 O 3 4 Al + 3 O 2 2 Al 2 O 3 Percent Yield 23.4 g Al 1mol Al 26.98 g Al x 2 mol Al 2 O 3 4 mol Al x = 44.2 g Al 2 O 3 x %Yield = x 100 = 39.3 g 44.2 g 88.9 % When 23.4 g of Al are allowed to burn in excess oxygen, 39.3 g of aluminum oxide are formed. What is the percentage yield? WS MC #6-11 FR #2

23. Strong Acids: Only 6 strong acids: Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Perchloric (HClO 4 ) proton (H + ) donors HI + H 2 O  H 3 O + + I –

24. Strong Bases: The strong bases are soluble hydroxides (OH – ) of… Group 1 (Li,Na,K) CBS (Ca, Ba, Sr) Mg(OH) 2 not as soluble proton (H + ) acceptors OH – + H 3 O +  H 2 O + H 2 O ase

25. Salts: Ionic Solids: (metal-nonmetal) dissociate (dissolve) by separation into ions Electrolytes: ions in solution that conduct electricity

26. Electrolytes: Strong, Weak, or Non? Compound Ionic STRONG Molecular Acid (H____) STRONG (6) WEAK (& NH 3 ) Not Acid NON nonmetals (Covalent)metal-nonmetal C 11 H 22 O 11 C 2 H 5 OH H 2 O CH 3 COOH HNO 2 HF KBr CaI 2 FeCl 3 NaOH Ca(OH) 2 (strong bases) HCl, HBr, HI HNO 3 H 2 SO 4 HClO 4 (ions conduct electricity)

27. ALWAYS Soluble Ions: Na +, K +, etc. group I (alkali metals) NH 4 + ammonium NO 3 – nitrate HCO 3 – bicarbonate C 2 H 3 O 2 – (CH 3 COO – )acetate (ethanoate) ClO 3 –, ClO 4 – chlorate, perchlorate Solubility Rules Common Precipitates form with:examples Ag +, Pb 2+, Hg 2+ (AP/H)AgCl, PbI 2 OH – (hydroxide)Cu(OH) 2 CO 3 2 – (carbonate)CaCO 3 ******

28. Preparing a Solution 1-mass solute 2-add solvent, swirl to dissolve 3-add solvent to mark

29. Dilution M 1 V 1 = M 2 V 2 1-calc M 1 V 1 =M 2 V 2 2-pipet V 1 from concentrated 3-fill to mark w DI

30. Dilution M 1 V 1 = M 2 V 2 What volume of a stock solution of concentration 8.0 M CuSO 4 is needed to prepare 2.0 L of 3.0 M CuSO 4 ?

31. g A L of A g B mol BL of B g A 1 mol A mol A 1 L g B 1 mol B mol B 1 L molar mass A molar mass B molarity A (M) molarity B (M) mol-to-mol ratio mol A Rxn: A (aq) + 2 B (aq)  C (aq) + 2 D (aq) Solution Stoichiometry

32. Transition Metal Ion Colors Color from e – movement in unfilled d orbitals. Filled or Empty d sublevel has no e – movement so ion is colorless. WS MC #1-5, 12-15 FR #3