1. + Synthesis/Combination Decomposition Single-Replacement  Activity Series Double-Replacement  Solubility Rules Combustion O2O2 ++ hydrocarbon CO 2 H2OH2OO2O2

2. Metals Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron Nickel Tin Lead Hydrogen* Copper Mercury Silver Platinum Gold Halogens Fluorine Chlorine Bromine Iodine Decreasing Activity

4. 3.28 a) What is the mass of 2.50 x 10 -2 mol of MgCl 2 ? FW(MgCl 2 ) = Mg + 2 Cl = 24.305 g/mol + 2(35.453 g/mol) FW(MgCl 2 ) = 95.211 g/mol Mass = (2.50 x 10 -2 mol)(95.211 g/mol) = 2.38 g b) How many moles of NH 4 Cl are there in 75.6 g of this substance? FW(NH 4 Cl) = 53.491 g/mol Moles = mass/FW = 76.5 g/(53.491 g/mol) = 1.43 mol FW = Formula Weightgfm = gram formula massMW = Molar Weight These are all equivalent terms for the molar mass of a substance:

5. c) How many molecules are there in 0.0772 mol HCHO 2 ? #m’cules = (6.022 x 10 23 )(0.0772 mol) = 4.65 x 10 22 m’cules d) How many nitrate ions, NO 3 , are there in 4.88 x 10  3 mol Al(NO 3 ) 3 ? #f.u. = (6.022 x 10 23 )(4.88 x 10  3 mol) = 2.939 x 10 21 f.u. #ions = (2.939 x 10 21 f.u.)(3 NO 3  /f.u.) = 8.82 x 10 21 ions Remember that ionic compounds don’t exist as discrete molecules, but are instead an extended lattice of ions. Hence, they are described in terms of the number of formula units (f.u.’s) that are present. In one f.u. of aluminum nitrate, there are three nitrate ions.

6. Perchlorate Perbromate Periodate ClO 4  BrO 4  IO 4  NitrateNO 3  Chlorate Bromate Iodate ClO 3  BrO 3  IO 3  Sulfate Carbonate SO 4 2  CO 3 2  NitriteNO 2  Chlorite Bromite Iodite ClO 2  BrO 2  IO 2  SulfiteSO 3 2  Hypochlorite Hypobromite Hypoiodite ClO  BrO  IO  Phosphate Arsenate PO 4 3- AsO 4 3- Polyatomic ions that you need to know are: The ‘ides’ Peroxide O 2 2  Hydroxide OH  Cyanide CN  Acetate C 2 H 3 O 2  OR CH 3 COO  The halogenates Permanganate MnO 4  Dichromate Cr 2 O 7 2  Chromate CrO 4 2  Form colored solutions The only cation Ammonium NH 4 +

7. BrO 4 , +7 SO 4 2 , +6 BrO 3 , +5 BrO 2 , +3 BrO , +1 SO 3 2 , +4 S 2 O 3 2 , +6 ClO 4 , +7 ClO 3 , +5 ClO 2 , +3 ClO , +1 IO 4 , +7 IO 3 , +5 IO 2 , +3 IO , +1 NO 3 , +5 NO 2 , +3 PO 4 3 , +5 PO 3 3 , +3 NH 4 +, -3 CrO 4 2 , +6 Cr 2 O 7 2 , +6 CO 3 2 , +4 C2H3O2C2H3O2

8. Combustion Analysis 3.48 (a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of CO 2 and 2.58 mg of H 2 O. What is the empirical formula of the compound (element order is C H O)? C x H y O z + O 2 CO 2 + H 2 O 2.78 mg6.32 mg2.58 mg Step 1: Determine the number of moles of C and H atoms FW(CO 2 ) = 44.010 mg/mmolFW(H 2 O) = 18.015 mg/mmol mmoles C = (6.32 mg CO 2 )/(44.010 mg/mmol) = 0.1436 mmol C mmoles H 2 O = (2.58 mg H 2 O)/(18.015 mg/mmol) = 0.1432 mmol mmoles H = (0.1432 mmoles H 2 O)(2 H atoms/H 2 O) = 0.2864 mmol H

9. Step 2: Convert the mmoles of C and H to mg of the two elements so can determine how much oxygen was in the compound. mg C = (0.1436 mmol)(12.011 mg/mmol) = 1.7248 mg C mg H = (0.2864 mmol)(1.00794 mg/mmol) = 0.2886 mg H mg O = 2.78 mg – (1.7248 mg + 0.2886 mg) = 0.766 mg O mmol O = (0.766 mg)/(15.9994 mg/mmol) = 0.0479 mmol Step 3: Divide through by the smallest molar amount to determine the empirical formula. C: 0.1436 mmol/0.0479 mmol = 3 H: 0.2864 mmol/0.0479 mmol = 6 O: 0.0479 mmol/0.0479 mmol = 1 The empirical formula is C 3 H 6 O

10. Stoichiometry 3.56 The fermentation of glucose (C 6 H 12 O 6 ) produces ethyl alcohol (C 2 H 5 OH) and CO 2. (a) How many moles of CO 2 are produced when 0.400 mol of C 6 H 12 O 6 reacts in this fashion? C 6 H 12 O 6 (aq)  2 C 2 H 5 OH (aq) + 2 CO 2 (g) 0.400 mol #mol x2x 0.800 mol (b) How many grams of C 6 H 12 O 6 are needed to form 7.50 g of C 2 H 5 OH? mol C 2 H 5 OH = (7.50 g)/(46.069 g/mol) = 0.1628 mol FW(C 2 H 5 OH) = 46.069 g/molFW(C 6 H 12 O 6 ) = 180.158 g/mol 0.1628 mol0.08140 mol Mass glucose = (0.08140 mol)(180.158 g/mol) = 14.7 g

11. LR: 3.70 Aluminum hydroxide reacts with sulfuric acid as shown below. (b) How many moles of Al 2 (SO 4 ) 3 form when 0.450 mol Al(OH) 3 and 0.550 mol H 2 SO 4 are allowed to react ? Limiting Reactants/Theoretical Yield 2 Al(OH) 3 (s) + 3 H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 6 H 2 O (l) 0.450 mol 0.550 mol 2x3xx 0.225 mol 0.183 mol The H 2 SO 4 is the limiting reactant because it limits how much aluminum sulfate will be formed. The theoretical (calculated) yield for this reaction is 0.183 mol of aluminum sulfate and 1.10 mol of water. What is the percent yield for this reaction if only 0.750 mol of water are collected? %yield = (0.750 mol)/(1.10 mol)*100 = 68.2% 6x