Question

A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16

Hint:

Use simultaneous equations with given molar masses and percentage composition to find numbers of monosaccharide units in polysaccharides.

Answer 1

Correct Answer: 2

Step 1: Let the number of ribose units be \( x \), 2-deoxyribose units be \( y \), and glucose units be \( z \). Total units: \[ x + y + z = 8 \]
Step 2: Total molar mass: \[ 150x + 134y + 180z = 1024 \]
Step 3: Given 2-deoxyribose is 58.26% by weight of total monosaccharides: \[ \frac{134y}{150x + 134y + 180z} = 0.5826 \] Multiply both sides by denominator: \[ 134y = 0.5826 (150x + 134y + 180z) \] \[ 134y = 87.39x + 78.1y + 104.87z \] Rearranged: \[ 134y - 78.1y = 87.39x + 104.87z \] \[ 55.9 y = 87.39 x + 104.87 z \] Since \( z = 8 - x - y \), substitute: \[ 55.9 y = 87.39 x + 104.87 (8 - x - y) \] \[ 55.9 y = 87.39 x + 839 - 104.87 x - 104.87 y \] \[ 55.9 y + 104.87 y = 87.39 x - 104.87 x + 839 \] \[ 160.77 y = -17.48 x + 839 \] Rewrite: \[ 160.77 y + 17.48 x = 839 \]
Step 4: Try integer values of \( y \) and solve for \( x \): For \( y = 5 \), \[ 160.77 \times 5 + 17.48 x = 839 \] \[ 803.85 + 17.48 x = 839 \implies 17.48 x = 35.15 \implies x = 2.01 \approx 2 \] Therefore, \[ x = 2, \quad y = 5, \quad z = 8 - 2 - 5 = 1 \] Step 5: Number of ribose units \( x = 2 \).