Chem 111 1st Edition Exam # 1 Study GuideSee info sheet we are given on moodleOWL Chapters 1-4.3 (not 4.4 or 4.5)Multiple ChoicePractice eExam gives you an extra 5% on exam!Tables in OWL book to memorize:2.3.2, 2.3.3, 2.3.5, 2.3.6, 2.4.2, 2.4.1, 4.2.1Use scientific notation and SI unitsScientific Notation is X.YZ x 10^WSignificant Figure RulesAddition or Subtraction: Least number of decimal places controlsEx. 1.3468+1.2=2.5 (2 sig figs)Multiplication and Division: least number of decimal controlsEx. 1.3468x1.21=1.63 (3 sig figs)Inside an AtomProton-positive chargeElectron-negative chargeNeutron-no chargeWriting it OutNumber of protons determine what element it isZ= protons=atomic numberThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Mass Number= A= Z+(number of neutrons)❑ZACIsotopes-differ number of neutrons (weight) Ionic Bonds-electron is transferredCovalent Bond-electron is sharedFormulasEmpricial: Smallest whole number ratio of atomsMolecular: Shows actual number of each atom in a moleculeStructural: 2D drawingCondensed Structural Formula: Shows common functional groups and indicates structureFinding % composition from a formula1. Find the molar mass using the periodic table2. Calculate percentages of each element3. Example: Calculate MW of C2H6O(2x12.011u)(6x1.0079u)+(1x15.9999u)46.068 u%C=(2x12.011u/46.068)x100%=52.145%%H=(6x1.0079u/46.068)x100%=13.127% %O=(1x15.999u/46.068)x100%=34.729%Avogadro’s Number=6.0221415E23A mole of anything is 6.0221415E23 of anything!Molar Mass=mass of 1 mole of “the formula”Calculate empirical formula given percent composition1. Assume weight of 100 g2. Convert grams to moles3. Find the smallest ratios-divide by the smallest numberGiven: N=30.45%O=69.55%N=60.45gO=69.55g30.45gN x (1mol/14.007g)=2.17 mol69.55gO x (1 mol/15.999g)=4.35 molN2.17O4.35N: 2.17/2.17=1O: 4.35/2.17=2Anwer : N O2A balanced reaction-Every atom on the left side (reactant) MUST end up on the right side (product)Calculating weight of product 1. Write balanced equation2. Convert units of weight to moles3. Determine # of moles in product produced4. Convert back to original unitsC4H10+O2−→ C O2+H2OC4H10+6.5 O2−→ 4 C O2+5 H2O8lbs x (453.9 g/lb) x (mol/58.123g)=62.47 moles butaneC4H10+6.5 O2−−→ 4 C O2+5 H2O62.47 moles C4H10x(44.01 gmol)=249.9 moles C O262.47 moles C4H10x(44.01 gmol)x(lb453.9 g)=24.23 lb of C O2Determine the limiting reagent1. Write a balanced equation2. Convert to moles3. Calculate amount of product with limiting reagentTypes of ReactionsCombination ReactionsA+BABDecomposition ReactionsABA+BSingle Replacement ReactionsAB+C AC+BDouble Replacement ReactionsAB+CDAD+CBPercent Yield1. Write balanced equation2. Convert to moles3. Calculate theoretical 100% yield4. Compare actual to theoretical yieldElectrolyteAny substance that increases the number of ions in water when the substance is dissolved in waterCan be strong or weakStrong ElectrolyteWhen dissolved in water is or becomes almost entirely ions in solutionWeak ElectrolyteWhen dissolved in water results in relatively few ions All weak acids and weak bases are weak electrolytesAcidIncreases H^+ when dissolved in waterStrong AcidStrong electrolytesBaseIncreases OH^- when dissolved in waterTake protons from acidsStrong BaseStrong