Using end group analysis to find molecular weight, 6.50 moles of ethylenediamine

Question

Using end group analysis to find molecular weight, 6.50 moles of ethylenediamine (H2N(CH2)2NH2) were reacted with 6.50 moles of adipic acid (HOOC(CH2)4COOH). After the reaction, a titration was performed using NaOH to determine the number of acid groups remaining free after the polymerization. If 73.0 mL of 3.0 M NaOH were required to neutralize the polymer, what is the number average molecular weight of the polymer?

30 Using end group ana lysis to find molecular weight, 650 moles of ethylenediamine adipic acid (H00cICHUdco0H. After the IHNUCHal:NHI were reacted with 650 moles titration was parformed using Na0H to determine the number of acid groups remaining free after the If 30mL of 30 M NaOH were required to neutralize the polymer, what isthenumber average moleoular weight of the polymer?

Explanation / Answer

One mole of base neutralizes one mole of acid

The moles of acid left = Molarity of naOH X volume of NaOH used

The moles of acid left = 3 X 0.073 = 0.219 moles

Number of moles of COOH present = 2 X 6.5 = 13 moles

number of moles of COOH reacted = 13-0.219 = 12.781 moles

The fraction reacted = p = number of COOH reacted / Number of COOH initially present = 12.781 / 13 = 0.983

Degree of polymerisaton = 1 / (1-p) = 1/(1-0.983) = 58.8

So here 58.8/ 2 is adipic acnd 58.8/ 2 is diamine = 29.4

The repeat unit is : C8N2O2H14 with molecular weight = 170

So number average molecular weight = 29.4 X 170 = 4998 g / gmol

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