This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
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
 Suggs
 Chemistry

Click to edit the document details