Chem 1212 Jexam HW3 solution 2

Chem 1212 Jexam HW3 solution 2 - x H y N z .) a) Carbon is...

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make, you must divide the actual total mass of compound found by rate of effusion in part c by 65.2kg HW #3, Number 2- Solution A compound contains only C, H, and N. A chemist analyzed it by doing the following experiments: a) Combustion analysis shows that the mass percent of C in the compound is 26.1%. b) The nitrogen content of the compound was analyzed by Dumas method in which the compound was first reacted by passage over hot CuO ( s ) : Compound N 2( g ) + CO 2( g ) + H 2 O ( g ) In a given experiment, a 65.2 mg sample of the compound produced 35.6 mL of dry N 2 at 740.0 torr and 25 °C. c) The effusion rate of the compound as a gas was measured and found to be 24.6 mL/min. The effusion rate of argon gas, under the same conditions, is 26.4 mL/min. What is the molecular formula of the compound? (Enter the formula in the form of C
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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.

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Chem 1212 Jexam HW3 solution 2 - x H y N z .) a) Carbon is...

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