Lecture 5 Professor Hicks General Chemistry II (CHE132) Percent Composition (aka percent by mass) % by mass component 1 = mass component 1 mass sample 100% sample component 1 100 g sample component 1 component 2 component 3 etc. conservation of mass requires all % mass add up to 100% % by mass is an intensive property Percent Composition Prince Rupert s Metal (brass) Calculate the percentage of copper in Prince Rupert s Metal, a form of brass. A sample of it contains 15 g of copper and 5 grams of zinc. % by mass component 1 = percentage copper = mass component 1 mass sample mass copper mass brass 100% 100% percentage copper = 15 grams 20 grams 100% copper zinc = 75% copper 5 grams 15 grams 1
percent composition from formula mass 1 formula unit of CaCl 2 40.078 amu Ca 2+ Cl - Cl - 35.45 amu 35.45 amu Cl - Ca 2+ formula mass = 110.98 amu % by mass Cl = 2 x 35.45 amu x 100% 110.98 amu = 63.88 % Cl mass 2 Cl - Cl - Cl - x 100% Ca mass CaCl 2+ 2 formula unit Cl - Cl - 110.98 amu sample CaCl 2 2 x 35.45 amu Cl - Cl - Ca 2+ 40.078 amu Cl - 2
Molarity (M) unit of concentration molarity = moles liter conversion factor from volume to moles moles liter x liters = moles molarity x volume = # moles in that volume molarity multiply by volume liters M x V = # moles mass compound multiply molar mass divide molar mass moles of formula units, molecules, atoms, or ions divide Avogadro's number multiply Avogadro's number number formula units, molecules, atoms, or ions Example: What is the molarity of glucose (C 6 H 12 O 6 ) in a solution made by dissolving 100.0 grams of glucose in enough water to make 250.0 ml of solution 1) molarity has units of liters convert ml liters 250.0 ml x 10-3 liter 1.0 ml = 0.2500 lit 2) Convert mass C 6 H 12 O 6 to moles 100.0 grams x 1 mole 180.16 grams = 0.5551 moles Molarity = # moles #liters = 0.5551 moles 0.2500 lit = 2.220 M glucose 3
Example: How many moles of NaCl are in 50.5 ml of a 0.250 M NaCl solution? 1) molarity has units of liters convert ml liters 50.5 ml x 10-3 liter 1.0 ml = 0.0505 lit 0.250 M NaClx 0.0505 liters = 0.0126 moles NaCl or if you write out the base units of molarity 0.250 moles NaCl liter solution x 0.0505 liters solution = 0.0126 moles NaCl 4
dilution formula 1) add 10 ml HCl stock V conc = 0.010 lit M conc x V conc = # moles HCl 12 M x 0.010 lit added = 0.12 moles HCl added M conc V conc = M dil V dil used to prepare a dilute solution from a concentrated stock solution # moles HCl = # moles HCl taken from stock after adding water M dil = M conc V conc 12 x 0.010 = M = 0.24 M adding water did not change # moles V 0.50 dil HCl but it did change molarity of HCl 2) add water to 500 ml mark V dil = 0.50 lit water concentrated stock solution M conc = 12.0 M HCl 500 ml flask M dil = 0.24 M 5
Acids H + + anion H+ - anion Acids are molecular compound because they can dissolve without dissociating into ions Ionic compounds must separate into ions to dissolve Weak acids have a small percentage of molecules separated into H + and an anion, the rest stay together as one particle HF ~ 95% H + and F - 5% Strong acids separate 100% into H+ and anion in water HCl ~ 0 % H + and Cl - ~ 100% Hydrogen acts as a non-metal Strong acids HCl HNO 3 H 2 SO 4 H + Cl - H + NO - 3 H + SO 2-4 hydrochloric acid nitric acid sulfuric acid Weak acids HC 2 H 3 O 2 HF H + C 2 H 3 O - 2 H + F - acetic acid hydrofluoric acid 6
molecular compounds ionic compounds all other molecular compounds HF ~ 95% (molecules) Acids are molecular compounds dissolve dissolve acids HCl H + + Cl - ~ 100% dissolve H + + F - ~5% separated ions (separated ions) Dissolved molecules cations (+ ions) anions (- ions) Strong Bases - ionic compounds with hydroxide ion (OH-) any hydroxide cation ion + cation OH - Examples: NaOH, LiOH, KOH, Ca(OH) 2, Ba(OH) 2, Sr(OH) 2, Acid-Base Neutralization Reactions acid H+ - anion metathesis reactions ions change partners + cation OH - strong base acid + base H 2 O (l) + salt HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) 7
Weak Bases react with water to produce OH - Base + H 2 O Base-H + + OH - NH 3 (aq) + H 2 O (l) NH 4+ (aq) + OH - (aq) React with acids neutralizing H + NH 3 (aq) + H + (aq) NH 4+ (aq) many compounds that contain nitrogen are weak bases like CH 3 NH 3 Equivalents are a quantity that is sort of like Moles mass sample / molar mass = moles in sample mass sample / equivalent mass = equivalents in sample molar mass equivalent mass = # of H + or OH Example Molar mass for HCl = equivalent mass HCl = 36.45 grams per mole or equivalent Equivalents are sort of like Moles Another example molar mass for H 2 SO 4 = 98 grams per mol equivalent mass H 2 SO 4 = 98/2 grams per eq = 49 grams per eq Yet Another example molar mass for Ca(OH) 2 = 74 grams per mol equivalent mass H 2 SO 4 = 74/2 grams per eq = 37 grams per eq 8
Normality is sort of like Molarity Normality = # equivalents per liter of solution Example: Calculate the normality of an H 2 SO 4 solution containing 25 grams of H 2 SO 4 in 0.150 L equivalent mass H 2 SO 4 = 49 grams per eq equivalents of H 2 SO 4 = 25/49 eq = 0.51 eq Normality = equivalents / liters =.51/ 0.150 = 3.4 N In a titration Normality and Equivalents are VERY useful at the Equivalence Point exact amounts of reactants have been combined so that neither or both are the limiting reactant # of equivalents of each reactant is equal Note: moles of each reactant may not be equal if balancing numbers are not all 1 N 1 V 1 = N 2 V 2 equivalents reactant 1 = equivalents reactant 2 9
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