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Unformatted text preview: CHE 178 Problem Set #2 Solutions 1 1. Example on “reverse engineering” Number average molecular weight: ∑ = i i n M w M 1 (Eq. 1.6, pg. 13 Y&L) Weight average molecular weight: ( ) ∑ = i i n M w M (Eq. 1.7, pg. 13 Y&L) Company A ω 1 = 0.5 M 1 = 20,000 g/mol ω 2 = 0.5 M 2 = 100,000 g/mol g/mol 3 . 333 , 33 000 , 100 5 . 000 , 20 5 . 1 = ⇒ + = n n M M ( ) ( ) g/mol 000 , 60 100,000 0.5 000 , 20 5 . = ⇒ + = w w M M Company A ω 1 = x M 1 = 20,000 g/mol ω 2 = x M 2 = 100,000 g/mol ω 3 = 12x M 3 = 1,000,000 g/mol 1 58 000 , 000 , 1 000 , 000 , 1 2x 1 10x 50x 1 000 , 000 , 1 2 1 000 , 100 000 , 20 1 + = ⇒ + + = + + = x M x x x M n n ( ) ( ) ( ) x M x x x M w w 1,880,000 1,000,000 1,000,000 2 1 100,000 000 , 20 = ⇒ + + = (a) 5% increase in M n : 1.05 x 33,333 = 35,000 g/mol x M w 1,880,000 1,000,000 = 4754 . 1 58 000 , 000 , 1 000 , 35 = ⇒ + = x x CHE 178 Problem Set #2 Solutions 2 ( ) 0493 . 4754 . 2 1 3 = = ω (b) 5% increase in M w 4984 . 1,880,000 1,000,000 0000 , 63 = ⇒ = x x ( ) 0032 . 4984 . 2 1 3 = = ω (c) I would make the M w measurements since it is more sensitive to the component of interest. M w allows for the determination of a much smaller weight fraction of the 1,000,000 g/mol polymer (0.0032 << 0.0493). If the amount were less than ~5 wt%, it would not be detected by the M n measurement. CHE 178 Problem Set #2 Solutions 3 2a. Problem 2.2 from Y&L Monomer: HO(CH 2 ) 14 COOH M n = 24,000 g/mol n HORCOOH HO RCOO H where R = (CH 2 ) 14 Molecular weight of the monomer unit: (CH 2 ) 14COO = 240 g/mol => M o 100 1 1 100 240 000 , 24 = = = = = p x M M x n w n n 99 . = p CHE 178 Problem Set #2 Solutions 4 2b. Problem 2.3 Y&L Hexamethylene diamine (AA): H 2 N(CH 2 ) 6 NH 2 M AA = 116 g/mol AA mol 0795 ....
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This note was uploaded on 12/09/2011 for the course CHEM ENG 178 taught by Professor Segalman during the Fall '08 term at Berkeley.
 Fall '08
 SEGALMAN

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