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Unformatted text preview: x H y N z .) a) Carbon is 26.1% b) n= PV/(RT) n= (.974atm)(.0356L)/(.0821x 298.15K) = .00142 mols N 2 Look for grams of 1 mol of Nitrogen: .00142 mols N 2 x (2 mol N/ 1 mol N 2 ) x (14.01g/ 1 mol N) = .0398g N Find the percent by mass of Nitrogen: (46.0g/mol) = 705.5 times .0652g So… .0398g N x 705.5 times = 28.07g N <- total mass in compound N%= total mass of N/ total mass of compound = 28.07g/46.0g = 61.0% of N H%= 100%- (61.0%+ 26.1%) = 12.9% c) Rate of Effusion= 24.6 ml/min = ( . / ) 39 95 m1 26.4 ml/min M 1 = 46.0g/mol (this is the total molar mass of compound) CONCLUSIVELY: We now know 26.1% of C -> 12.01g x (1 mol C/ 12.01 g C) = ~ 1 12.9% of H -> 5.934g x (1 mol H/ 1.008 g H) = ~ 6 61.0% of N -> 28.07g x (1 mol N/ 14.01g N) = ~ 2 Soooo…. The compound is C 1 H 6 N 2...
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This note was uploaded on 03/17/2009 for the course CHEM 1212 taught by Professor Suggs during the Spring '08 term at University of Georgia Athens.
- Spring '08