Determine Number of Glycine Units in Decapeptide Hydrolysis

To find the number of glycine units present in the decapeptide, we analyze the hydrolysis reaction of the decapeptide.

A decapeptide consists of 10 amino acid residues linked together by 9 peptide bonds. The complete hydrolysis of 1 mole of a decapeptide requires 9 moles of water ($\text{H}_2\text{O}$) to break all peptide bonds and produce individual amino acids:

$\text{Decapeptide} + 9\text{H}_2\text{O} \longrightarrow \text{Hydrolysed Products (Amino acids)}$

Given:

• Molecular weight of the decapeptide = $796\text{ g mol}^{-1}$
• Molar mass of water ($\text{H}_2\text{O}$) = $18\text{ g mol}^{-1}$
• Molecular weight of glycine = $75\text{ g mol}^{-1}$

The total molar mass of the hydrolysed products obtained from 1 mole of the decapeptide is:

$\text{Total weight of products} = M_{\text{decapeptide}} + 9 \times M_{\text{H}_2\text{O}}$
$\text{Total weight of products} = 796 + 9 \times 18 = 796 + 162 = 958\text{ g mol}^{-1}$

It is given that glycine contributes $47.0\%$ to the total weight of the hydrolysed products. Thus, the total mass of glycine produced per mole of decapeptide is:

$\text{Mass of glycine} = \frac{47.0}{100} \times 958 = 450.26\text{ g mol}^{-1}$

Let $n$ be the number of glycine units present in the decapeptide. The mass of glycine is also given by:

$\text{Mass of glycine} = n \times M_{\text{glycine}} = n \times 75\text{ g mol}^{-1}$

Equating the two values of the mass of glycine:

$75 \times n = 450.26$
$n = \frac{450.26}{75} \approx 6.003$

Since $n$ must be an integer, the number of glycine units in the decapeptide is $6$.