Chapter 15, Problem 15.116P

(a)

Interpretation Introduction

Interpretation:

The hydrolysis of a 100.00-g sample of peptide forms the given individual amounts of amino acids (molar mass in g/mol are given in parentheses).

  $\begin{array}{l}\text{3}\text{.00 g of glycine (75}\text{.07) 0}\text{.90 g of alanine (89}\text{.10)}\\ \\ \text{3}\text{.70 g of valine (117}\text{.15) 6}\text{.90 g of proline (115}\text{.13)}\\ \\ \text{7}\text{.30 g of serine (105}\text{.10) 86}\text{.00 g of arginine (174}\text{.21)}\end{array}$

Why the total mass of amino acids exceeds the mass of peptide has to be given.

Explanation of Solution

The peptides are formed by the amino acid units.  Amino acids combines together to form the amide bonds of peptides.  The amide bond is formed by the loss of water molecule from amine group of one amino acid and carboxylic acid group of another amino acid.

As the loss of water molecule occurs in the formation of peptide, the mass of peptide is less than that of total mass of amino acids.

(b)

Interpretation Introduction

Interpretation:

The hydrolysis of a 100.00-g sample of peptide forms the given individual amounts of amino acids (molar mass in g/mol are given in parentheses).

  $\begin{array}{l}\text{3}\text{.00 g of glycine (75}\text{.07) 0}\text{.90 g of alanine (89}\text{.10)}\\ \\ \text{3}\text{.70 g of valine (117}\text{.15) 6}\text{.90 g of proline (115}\text{.13)}\\ \\ \text{7}\text{.30 g of serine (105}\text{.10) 86}\text{.00 g of arginine (174}\text{.21)}\end{array}$

The relative numbers of amino acids in the peptide has to be given.

Concept Introduction:

The relative numbers can be calculated by converting the mass in to moles and by dividing the moles with smallest number.

Explanation of Solution

The given amounts of amino acids are

  $\begin{array}{l}\text{3}\text{.00 g of glycine (75}\text{.07) 0}\text{.90 g of alanine (89}\text{.10)}\\ \\ \text{3}\text{.70 g of valine (117}\text{.15) 6}\text{.90 g of proline (115}\text{.13)}\\ \\ \text{7}\text{.30 g of serine (105}\text{.10) 86}\text{.00 g of arginine (174}\text{.21)}\end{array}$

The mass of amino acids is converted to moles by using the equation

  $\text{moles = }\frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

Glycine:

  $\begin{array}{l}\text{moles = }\frac{\text{3}\text{.00 g}}{\text{75}\text{.07 g/mol}}\\ \\ \text{ = 0}\text{.0399627 mol }\end{array}$

Alanine:

  $\begin{array}{l}\text{moles = }\frac{\text{0}\text{.90 g}}{\text{89}\text{.10 g/mol}}\\ \\ \text{ = 0}\text{.010101010 mol }\end{array}$

Valine:

  $\begin{array}{l}\text{moles = }\frac{\text{3}\text{.70 g}}{\text{117}\text{.15 g/mol}}\\ \\ \text{ = 0}\text{.0315834 mol }\end{array}$

Proline:

  $\begin{array}{l}\text{moles = }\frac{\text{6}\text{.90 g}}{\text{115}\text{.13 g/mol}}\\ \\ \text{ = 0}\text{.0599323 mol }\end{array}$

Serine:

  $\begin{array}{l}\text{moles = }\frac{\text{7}\text{.30 g}}{\text{105}\text{.10 g/mol}}\\ \\ \text{ = 0}\text{.0694577 mol }\end{array}$

Arginine:

  $\begin{array}{l}\text{moles = }\frac{\text{86}\text{.00 g}}{\text{174}\text{.21 g/mol}}\\ \\ \text{ = 0}\text{.493657 mol }\end{array}$

The moles obtained can be divided by the smallest number to get the relative number.

The smallest number is $\text{0}\text{.010101010 mol of alanine}$.

  $\begin{array}{l}\text{Glycine : }\frac{\text{0}\text{.0399627 mol}}{\text{0}\text{.010101010 mol}}\text{ = 4}\\ \\ \text{Alanine : }\frac{\text{0}\text{.010101010 mol}}{\text{0}\text{.010101010mol}}\text{ =1}\\ \\ \text{Valine : }\frac{\text{0}\text{.0315834 mol}}{\text{0}\text{.010101010 mol}}\text{ =3}\\ \\ \text{Proline : }\frac{\text{0}\text{.0599323 mol}}{\text{0}\text{.010101010 mol}}\text{ =6}\\ \\ \text{Serine : }\frac{\text{0}\text{.0694577 mol}}{\text{0}\text{.010101010 mol}}\text{ =7}\\ \\ \text{Arginine : }\frac{\text{0}\text{.493657 mol}}{\text{0}\text{.010101010 mol}}\text{ =49}\end{array}$

The relative numbers of amino acids are calculated.

(c)

Interpretation Introduction

Interpretation:

The hydrolysis of a 100.00-g sample of peptide forms the given individual amounts of amino acids (molar mass in g/mol are given in parentheses).

  $\begin{array}{l}\text{3}\text{.00 g of glycine (75}\text{.07) 0}\text{.90 g of alanine (89}\text{.10)}\\ \\ \text{3}\text{.70 g of valine (117}\text{.15) 6}\text{.90 g of proline (115}\text{.13)}\\ \\ \text{7}\text{.30 g of serine (105}\text{.10) 86}\text{.00 g of arginine (174}\text{.21)}\end{array}$

The minimum molar mass of the peptide has to be given.

Concept Introduction:

The minimum molar mass of peptide can be calculated by adding the products of moles of each amino acid with their molar mass to get the amount of amino acids and the water molecules mass is eliminated from the mass of amino acids to get the molar mass of peptide.

Explanation of Solution

The minimum molar mass of the peptide is calculated as

  $\begin{array}{l}\text{mass of 70 moles of amino acids = }\left(\text{4×75}\text{.07 g/mol}\right)\text{+}\left(\text{1×89}\text{.09 g/mol}\right)\text{+}\left(\text{3×117}\text{.15 g/mol}\right)\\ \text{ +}\left(\text{6×115}\text{.13g/mol}\right)\text{+}\left(\text{7×105}\text{.09 g/mol}\right)\text{+}\left(\text{49×174}\text{.20 g/mol}\right)\\ \\ \text{ = 10,703}\text{.59 g}\end{array}$

$\text{70}$ Moles of amino acids are linked by $\text{69}$ peptide bonds.  For the formation of one peptide bond, one water molecule is removed.  Thus, for $\text{69}$ peptide bonds $\text{69}$ water molecules are removed.

  $\begin{array}{l}{\text{mass of H}}_{\text{2}}\text{O eliminated = }\left({\text{69 mol H}}_{\text{2}}\text{O}\right)\left(\frac{\text{18}{\text{.02 g H}}_{\text{2}}\text{O}}{{\text{1 mol H}}_{\text{2}}\text{O}}\right)\\ \\ \text{ = 1243}\text{.38 g}\end{array}$

The minimum molar mass of peptide bond is calculated by subtracting the mass of water molecules from the mass of amino acids.

  $\begin{array}{l}\text{minimum mass of peptide = 10,703}\text{.59g - 1243}\text{.38g}\\ \\ \text{ = 9,460 g/mol }\end{array}$

The minimum mass of peptide is $\text{9,460 g/mol}$.