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Chapter 3: Lecture Notes

by: Caroline Hess

Chapter 3: Lecture Notes 110.0

Marketplace > University of Pittsburgh > Chemistry > 110.0 > Chapter 3 Lecture Notes
Caroline Hess
GPA 4.0

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These are the Chapter 3 notes for Dr. Vine's CHEM 0110 course. The notes follow her lecture outline and her PowerPoints.
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This 8 page Class Notes was uploaded by Caroline Hess on Wednesday October 14, 2015. The Class Notes belongs to 110.0 at University of Pittsburgh taught by Paul,Stanley in Fall 2015. Since its upload, it has received 21 views. For similar materials see GENERAL CHEMISTRY 1 in Chemistry at University of Pittsburgh.


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Date Created: 10/14/15
Chapter 3 Chemical Formulas and Equations 31 Molecular and Formal Weight 0 Molecular Weight The sum of all the atomic weights of all the atoms in a molecule of the substance Example C6H12O6 Glucose Carbon 12011 6 72066 amu Hydrogen 10079 12 Total Oxygen 16000 6 18016 amu Formula Weight The sum of all atomic weights of all the atoms in a formula unit of the compound Example MgSO4 7 H20 Epsom salt Magnesium 243050 1 243050 Sul de 32066 1 3 Total Oxygen 16000 11 gt 1 2464765 Hydrogen 10079 14 14112 32 The Mole Concept Mole The quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon12 12C 12000 amu 12C 1 mole 12 g of atoms of 12C 1200 g 1 12C atom weights 1992648 103923 12 g 1992648 10 23 26022 10 23 Avogrado s Number 6022103923 Molar Mass The mass of one mole of the substance 0 For all substances the molar mass in grams per mole is numerically equal to the formula weight in amu Example How much does one molecule of glucose weigh C6H12O6 1 l C H 0 1m0le0fglucose18016 g X moe 6 12 6 2991023gm0lecule quot1016 6022 10 23 molecules Example Grams to atoms How many atoms of Carbon are in 1803 grams of glucose 23 1803gx1m 0162001m0lXM6022X 1021x 18016 gmol 1 mole 1 molecule M236 X1022Cat0ms Example Molecules to grams If we have 4221022 molecules of H20 in a sample of Epsom salt MgSO47H20 how many grams of MgSO47H20 do we have 1m0leH20 422 x 1022 H 20 molecules x 23 6022 x 10 700 X 10 2 moles of H20 lmolMg 504 700 X 10 2moles of H2 0 X 001 mole of Mg 5047 H20 7molH20 001 mol Mg 5047H20 X 24164818 247gMg 5047 H20 m0 33 Mass Percentage from the Formula Percentage Composition 0 Percentage composition The mass percentage of each element in the compound 0 Mass percentage A as the parts of A per hundred parts of the total by mass 0 mass of AEthe whole X 100 Mass A Of A massof total Example Which compound contains the highest of Nitrogen a NO3 2259 b N02 3044 d NH3 8225 35 Determining Formulas Empirical Formula The formula of a substance written with the smallest integer subscripts 0 Example C1H1 Molecular Formula A multiple n of its empirical formula o N Molecular weight Empirical Formula 0 Example C6H6 78gm0l6 0 Example 13gm01 Example A certain organic molecule contains Carbon Hydrogen and Oxygen It contains 400 carbon and 673 hydrogen Determine the empirical formula 10040 60 673 5327 of Oxygen Assume you have 100 g 40 gof C x 112730111 2333 molC 333 mol 1 lfH 121 moo 7 H 666 lH333 l 6 3g0f X 10079g m0 m0 rato lmolofO 5327 0 333 10333 l 30f X 10079g m0 m 34 Elemental Analysis Percentage of C H and O 0 Unknown compound CxHyOz Known simple mass cxHyoz 02 n H20 co2 All carbon ends up in C02 0 All hydrogen ends up as H20 Combustion reaction Example A 08223 9 sample of C H and O was combusted in excess 02 to produce 2445 g of C02 and 06003 g of H20 were obtained What s the empirical formula 1 mol lmol C l 1211 7 44018xlmolCO2 005556m0 Cx 0 g 066 3g 244 gofCOzx 1m0lH20 ZmolH X 1802g 1m0l H20 06003 g H2 0 x 006663 mol H x 10073 2 006715 g lmol 8223 06673 006715 008882 2 005551 1 g X 16000 g quot 0 00550 1 0 006663 12 0005551 0005551 C10H120 36 Molar Interpretation of a Chemical Equation 2H202D2H20 o 2 molecules of H2 1 molecule of 02 l 2 molecules Of H20 0 2 moles of H2 1 moles of 02 l 2 moles of H20 0 2202 g of H2 320 g of 02 2 1802 g of H20 Stoichiometry The calculation of the quantities of reactants and products involved in a chemical reaction 0 Compare moletomole ratio 37 Amounts of Substances in Chemical Reaction 1 Write equation 2 Balance an equation 3 Change to moles 4 Compare molemoe ratio stoichiometry 5 Moles to grams Example How much 02 in grams is used in the combustion of 1 tank 8 gallons of gas octane C8H18 4 C8H18 I 25 02 9 I 16 C02 9 18 H20 9 8 gallons Liters l Grams 8 gallons x 3789 L x 1000 mL 2 30280mL 1 gal L 30280 mL x 221289 g mL 1 mol 21289 g X 11423g 1864 moles 25 mol 02 1864 mol X 2330 mole 02 mol C8H18 320 3 m0 2330 mol x 74560 3 02 38 Limiting Reactant Theoretical and Percentage Yields Writing equation Balance equation Change to moles ICE table Determine limiting reactant Moles to grams Yields Theoretical and Percentages NewewNH Example Bromine and sodium chloride solution can be prepared from a reaction between chlorine and sodium bromide solution How many grams of bromine are formed if 250 g of chlorine is reacted with 500 g of sodium bromide 2 NaBr aq Clz g 2 NaCl aq Brz l 0486 NaBr 0352 Oz 0 NaCl 0 Brz 0486 0486 10486 0243 0 0109 0486 0243 lmol 250 g Clzx 70906g 0352 mol Cl2 500 g NaBr x 1327180918 20486 mol NaBr C2 as limiting 2m01NaBr Can t be limiting 0352m01XW 0706moleNaBr because it gives NaBr as limiting 0486NaBrx 117101012 202431710102 Is limiting because 0 2molNaBr matter IS left in end Theoretical yield The expected amount of product 0 Actual yield The measured amount of product 0 Percent yield Theoretical Yield 1 Actual Yield X 00


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