11 Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 3A – STOICHIOMETRY: CHEMICAL CALCULATIONS Empir. & molec. formulas and related stoichiometry Benzene and acetylene are different substances with the same empirical formulas. Which of the following is true? A. Their molar masses will have a simple integer ratio. B. Their molecular formulas are the same. C. Their structural formulas are similar. D. They have similar chemical properties. E. They have similar physical properties. Benzene is C6H6; its empirical formula is C1H1or CH. Acetylene is C2H2; its empirical formula is C1H1or CH. A. True. The molar mass of C6H6is 78; the molar mass of C2H2is 26. Their molar masses have a simple integer ratio of 3:1. Even if the student didn’t know the molecular formulas of benzene and acetylene the student would still get “A” as the answer. Let’s say that the student thought that benzene was C6H6, but that acetylene was C4H4. The molar masses would be 78 and 52, respectively, which is 3:2, a simple integer ratio. B. False. Their molecular formulas are different, C6H6and C2H2. C. H H | | C ═C H ―C C ―H C ―C | | H H H ─C ≡C ─H False. Their structural formulas are different. D. False. The molecules are completely different and therefore have different reactions with varied reagents, e.g., bases. E. False. The molecules are completely different and therefore have different physical properties, e.g., boiling point and melting point. A
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22 Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 5 - GASES Reaction stoichiometry 5.00 grams of alum, KAl(SO4)2.xH2O, is heated in an oven, and the water vapor driven off is collected and found to exert a pressure of 1.33 atm in a 3.00 Liter container at 111 oC. Calculate the value of “x” in the hydrate. A. 2 B. 4 C. 5 D. 7 E. 12 KAl(SO4)2.xH2O →KAl(SO4)2+ xH2O 5.00 g 1.33 atm 3.00L 111 + 273K Plan: V,P,TH2O →molH2O →gH2O →g KAl(SO4)2 →molKAl(SO4)2→mol ratio PV = nRT n = PV/RT n = (1.33 atm x 3.00 L)/(0.08205Latmdeg-1mol-1x (273 + 111)) = 0.1266 mol H2O 0.1266 mol H2O x 18.02gH2O/mol = 2.28g H2O 5.00g alum – 2.28g H2O = 2.72g KAl(SO4)2 2.72gKAl(SO4)2x (1/(39.098+26.982+(96.02 x 2))) = 0.01054 mol KAl(SO4)2X = mol H2O/molKAl(SO4)2 = 0.1266/0.01054 = 12.01E