Peptide Bond Formation

From the given reaction sequence, the two acyclic monomers obtained are
X : NH2‒CH2‒COOH  (glycine, an α-amino acid)
Y : NH2‒(CH2)5‒COOH  (6-aminocaproic acid, another α-amino acid)

On heating, these amino-acids undergo condensation (peptide) polymerisation with loss of water to give the biodegradable, linear (acyclic) copolymer nylon-2-nylon-6, Z.

Step-1 : Molar masses of the monomers
Using the given atomic masses (H = 1, C = 12, N = 14, O = 16):
Glycine, NH2CH2COOH → C2H5NO2
$$M_{\text{gly}} = 2(12) + 5(1) + 14 + 2(16) = 24 + 5 + 14 + 32 = 75\ \text{g mol}^{-1}$$
6-Aminocaproic acid, NH2(CH2)5COOH → C6H13NO2
$$M_{\text{cap}} = 6(12) + 13(1) + 14 + 2(16) = 72 + 13 + 14 + 32 = 131\ \text{g mol}^{-1}$$

Step-2 : Stoichiometry of the polymerisation
500 mol of X react with 500 mol of Y, so the copolymer chain contains
500 glycine residues and 500 caproic-acid residues → total 1000 residues.

Step-3 : Mass of water eliminated
For an ordinary linear polyamide, the number of peptide (-CO-NH-) links formed is one less than the number of residues, i.e.
$$\text{No.\ of links} = 1000 - 1 = 999$$
Each link eliminates one molecule of water (18 g mol-1).
$$m_{\text{water}} = 999 \times 18 = 17\,982\ \text{g}$$

Step-4 : Mass of the single mole of polymer Z
Total initial mass of the two monomers
$$m_{\text{initial}} = 500(75) + 500(131) = 500 \times 206 = 103\,000\ \text{g}$$
After loss of water,
$$m_{\text{polymer}} = m_{\text{initial}} - m_{\text{water}}$$ $$m_{\text{polymer}} = 103\,000 - 17\,982 = 85\,018\ \text{g}$$

Because the entire 500 mol of each monomer gives exactly 1 mol of copolymer Z, the molar mass of Z is 85 018 g mol-1.

Hence, the amount of Z formed is 85018.00 g.