Chapter 23, Problem 65A

Interpretation Introduction

Interpretation:

The molecular weight of stachyose should be calculated.

Concept introduction:

A biological molecule which consists of carbon, hydrogen and oxygen is known as saccharide or carbohydrate. The general formula of saccharide is

Saccharides (sugar) are classified as: monosaccharides, disaccharides and polysaccharides.

Monosaccharides are defined as a simpler carbohydrate which contains one sugar molecule and can’t hydrolyze into smaller carbohydrate.

Disaccharides are defined as a saccharide which contains two sugar molecules. It is formed when two monosaccharides are linked by glycosidic bonds. It is a double ring structure.

Tetrasaccharides are defined as a saccharide which contains four sugar molecules. It is formed when four monosaccharides are linked by glycosidic bonds.

Polysaccharides are defined as a saccharide which contains many units of sugar molecules that is more than two monosaccharides are linked by glycosidic bonds.

Explanation of Solution

It is given that the Stachyose is a tetrasaccahride which contains two D-galactose units, D-glucose unit and one D-fructose unit.

Molecular weight of each sugar molecule = 180 g/mole

Since, it is given that four sugar molecules are present.

Thus,

Total molecular weight =

=

There are four sugar units implies three linkages are present, thus, there water molecules are released.

Now,

Molar mass of water = 18 g/mole

Total molecular weight =

=

Therefore,

Molecular mass of Stachyose =

=