CHEM 1441 1st EditionExam # 2 Study Guide Lectures: 8 - 15Lecture 8 (February 11) Empirical Formula Problem Example:Menthol (MW[molecular weight]= 156.3), the strong smelling substance in many cough drops, is a compound of carbon, hydrogen and oxygen. When 0.1595 g of Methanol was burned in a combustion apparatus, 0.449 g of CO2 and 0.184 g of H2O formed. What is menthol’s molecular formula? First write the formula and given measurements: CxHyOz + O2  CO2 + H2O 0.1595 g 0.449 g 0.184 gFind the mols of every element:C from CO2: 0. 44 9 g CO2 x 1 mol CO2 x 1 mol C = 0.01020 mol C 44 g CO2 1 mol CO2H from H2O: 0. 18 4 g H2O x 1 mol H2O x 2 mol H = 0.02044 mol H 18 g H2O 1 mol H2OWhat about mols of Oxygen? We know the total mass of menthol is 0.1595 g. So if we subtract the masses of Carbon and Hydrogen from total menthol mass, we will get the mass of Oxygen: CxHyOz = mass C + mass H + mass OFirst we find mass of C and H by using molar mass:Mol stuff  mass stuff0.01020 mol C x 12.011 C = 0.1224 g C 1 mol C0.02044 mol H x 1.008 H = 0.02061 g H 1 mol H0.1595 = 0.1224 + 0.02061 + mass OMass O = 0.01649 gNow convert to mols:0.01649 g O x 1 mol O = 0.00131 mol O 15.999 g OCalculate empirical formula:C: 0.01020 mol C C: 9.98H: 0.02044 mol H divide all of them by the smallest H: 19.82O: 0.001031 mol O O: 1On Exam!Make first guess: C10H20O1Check the molecular weight (same thing as molar mass) 120+20+16 = 156This matches up with MW we were given in the problem. The empirical formula is also the molecular formula for menthol.Chapter 4: Major Classes of Chemical Reactions Stoichiometry Map for aqueous reactions and solidsAqueous (MA, VL) (MB, VL) Mol A Mol BSolids mass A × ÷ × ÷ mass B Stuff A Stuff B M stands for Molarity which is a concentration term Molarity= mols of solute Remember this equation: M = ɳ Volume (in Liters) of solution VLMolarity is always paired with a volume. It is either given or derived; Molarity can’t be given by itself. Example 1:Calculate the molarity of a solution made by adding 141 g C2H3O2H (MW 60.1 g/mol) to enough water to make a 750 mL solution.what we are given what we want141 g C2H3O2H find Molarity of C2H3O2H750 ml solution Convert mL to LVL= 0.750 L Calculate MMass C2H3O2H  mol C2H3O2H  MC2H3OOH,VL141 g C2H3O2H x 1 mol C2H3O2H = 2.35 mol C2H3O2H; M = 2.35 mol C2H3O2H = 3.13 M 60.1 g C2H3O2H 0.750 Liters Example 2:A chemist wants to prepare 250 mL of 2.00 M NaOH (MW 40.0 g/mol). How many grams of NaOH should be used?what we are given what we wantNaOH M = 2.00 M find mass of NaOH250 mL NaOH by Avogadro’s numberX molar mass÷ molar massConvert ml to LVL=0.250 LCalculate mass in grams(M,V)  mol NaOH  mass NaOH2.00 x 0.250 = 0.50 mol NaOH x (use molar mass or the MW) 40.0 g NaOH = 20.0 g 1 mol NaOH Concentration by Dilution the keyword here is dilution! For dilutions, use M1V1 =M 2V2  Example 1:Commercial Ammonia is 15.0 M NH3. How can a solution of 500 mL of 6.0 M aqueous ammonia be prepared? stock diluteM1V1 = M2V2(15 M)(V1) = (6.0 M)(500 mL)V1=200 mL or 0.2 L of stock NH3 is needed Example 2:Prepare 3.0 M H2SO4 at 500 mL, given that the concentrated bottle is 18.0 M. How much stock solution is needed for the dilution? Notice the keyword dilution! That means you can use MV=MVstock diluteMV = MV(18 M)( V ) = (3 M)(500 mL)V=83.3 mL or 0.0833 L of solutionLecture 9 (February 13)  Concept Check Aqueous Reactions How many L of a 0.1 M HCl are required to completely react with 13.0 g of MgCO3 in the following reaction?MgCO3 (s) + 2 HCl (aq)  MgCl2 (aq) + H2Cl (l) + CO2 (g)given want13.0 g MgCO3 volume of HCl 0.1 M HClMass MgCO3  mol MgCO3  mol HCl13.0 g MgCO3 x 1 mol MgCO3 x 2 mol HCl = 0.3083 mol HCl NaOHɳM V 84.32 g MgCO3 1 mol MgCO3 Use Molarity equation to find volume VL=ɳ/M 0.3083 mol HCl = 3.08 0.1 M Electrolytes Electrolytes are ionic compounds that are made up of a metal and nonmetal or a polyatomic cation and polyatomic anion. Ionic compounds are salts. When salts are put into water 2 processes occur: 1. The salt dissolves 2. Salts dissociate into their ions  For example: if NaCl (s) is put into water, it will dissolve and dissociate into Na+ (aq) and Cl- (aq) ions Some molecular compunds also dissociate, for example acids:HCl (aq)  H+ (aq) + Cl- (aq) The arrows used to write equations have different meanings One way arrow means the reaction has gone to completion; it is a complete dissociation. Compounds that undergo complete dissociations are strong electrolytes. Two arrows that show that the reaction is only partially complete; it is an incomplete reaction. Compounds that undergo incomplete reactions are weak electrolytes. This is not actually a reaction arrow. This is a resonance arrow. How to tell if something if strong or weak to write Net Ionic Reactions: MEMORIZE SOLUBILITY RULES Soluble