CHEM 1441 1st Edition Exam 2 Study Guide Lectures 8 15 Lecture 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 On Exam 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 g Find the mols of every element C from CO2 0 449 g CO2 x 1 mol CO2 x 1 mol C 0 01020 mol C 44 g CO2 1 mol CO2 H from H2O 0 184 g H2O x 1 mol H2O x 2 mol H 0 02044 mol H 18 g H2O 1 mol H2O What 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 O First we find mass of C and H by using molar mass Mol stuff mass stuff 0 01020 mol C x 12 011 C 0 1224 g C 1 mol C 0 02044 mol H x 1 008 H 0 02061 g H 1 mol H 0 1595 0 1224 0 02061 mass O Mass O 0 01649 g Now convert to mols 0 01649 g O x 1 mol O 0 00131 mol O 15 999 g O Calculate empirical formula C 0 01020 mol C H 0 02044 mol H divide all of them by the smallest O 0 001031 mol O C 9 98 H 19 82 O 1 Make first guess C10H20O1 Check the molecular weight same thing as molar mass 120 20 16 156 This 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 solids Aqueous MA VL MB VL Mol A Mol B X molar mass Solids mass A molar mass mass B by Avogadro s number Stuff A Stuff B M stands for Molarity which is a concentration term Molarity mols of solute Remember this equation M 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 141 g C2H3O2H 750 ml solution what we want find Molarity of C2H3O2H Convert mL to L VL 0 750 L Calculate M Mass C2H3O2H mol C2H3O2H MC2H3OOH VL 141 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 NaOH M 2 00 M 250 mL NaOH what we want find mass of NaOH Convert ml to L VL 0 250 L Calculate mass in grams M V mol NaOH mass NaOH 2 00 x 0 250 0 50 mol NaOH x use molar mass or the MW 40 0 g NaOH 20 0 g NaOH 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 dilute M 1 V1 M 2 V2 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 MV stock dilute MV MV 18 M V 3 M 500 mL V 83 3 mL or 0 0833 L of solution Lecture 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 MgCO 3 in the following reaction MgCO3 s 2 HCl aq MgCl2 aq H2Cl l CO2 g given 13 0 g MgCO3 0 1 M HCl want volume of HCl Mass MgCO3 mol MgCO3 mol HCl 13 0 g MgCO3 x 1 mol MgCO3 x 2 mol HCl 0 3083 mol HCl 84 32 g MgCO3 1 mol MgCO3 Use Molarity equation to find volume VL M 0 3083 mol HCl 3 08 L 0 1 M M V 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 Compounds strong Exceptions weak Group 1 cations Li Na K none All NH 4 NO3 C2H3O2 ClO4 none Halides Cl Br I Ag Pb Ca Hg Flourides F Pb Group 2 Sulfates SO4 Ca Sr Ba Ag Pb Insoluble Compounds strong Exceptions weak Metal hydroxides OH Group 1 Ca Sr Ba NH 4 Carbonates and Phosphates CO3 PO4 Group 1 NH4 Sulfides S Group 1 Group 2 Or look at this chart http swhaftrnotes blogspot com 2012 02 ap chemistrysolubility rules html Practice Example 1 Write out net inonic reation of the following AgNO3 aq NaCl aq First we have to figure out if these compounds are ionic Yes they are Then find out which species are weak or strong using solubility rules Weak species will stick together Strong species will break into ions If there is at least one strong species then the reaction will go to completion Use the one way arrow Since both AgNO3 and NaCl are strong they will dissociate AgNO3 aq NaCl aq Ag aq NO3 Na Cl For writing products on the other side pair the opposite positive and negative species with the other Ag aq NO3 Na Cl AgCl s NaNO3 …