Determination of Ribose Units in Octasaccharide Hydrolysis

To find the number of ribose units present in one molecule of the octasaccharide, we perform the following steps:

1. Mass Balance of the Hydrolysis Reaction: A linear octasaccharide consists of $8$ monosaccharide units joined by $7$ glycosidic linkages. Complete hydrolysis of $1\text{ mol}$ of the linear octasaccharide requires $7\text{ mol}$ of $\text{H}_2\text{O}$.

   The molar mass of $\text{H}_2\text{O}$ is: $M_{\text{H}_2\text{O}} = 2 \times 1 + 16 = 18\text{ g mol}^{-1}$

   Therefore, the total mass of the monosaccharides produced from $1\text{ mol}$ of the octasaccharide is: $\text{Total mass of products} = M_{\text{octasaccharide}} + 7 \times M_{\text{H}_2\text{O}}$ $\text{Total mass of products} = 1024\text{ g} + 7 \times 18\text{ g} = 1024\text{ g} + 126\text{ g} = 1150\text{ g}$

2. Determination of 2-Deoxyribose Units ($y$): Let $x$, $y$, and $z$ be the number of units of ribose, 2-deoxyribose, and glucose present in one molecule of octasaccharide, respectively.

   Since the molecule is an octasaccharide: $x + y + z = 8$

   The problem states that the amount of 2-deoxyribose formed is $58.26\%\text{ (w/w)}$ of the total hydrolyzed products. $\text{Mass of 2-deoxyribose} = y \times 134\text{ g}$ $\frac{y \times 134}{1150} \times 100 = 58.26$ $134y = \frac{58.26 \times 1150}{100} = 669.99 \approx 670$ $y = \frac{670}{134} = 5$

3. Determination of Ribose Units ($x$): Substituting $y = 5$ into the total monosaccharide count: $x + z = 8 - 5 = 3 \implies z = 3 - x$

   The sum of the molar masses of all individual monosaccharides in $1\text{ mol}$ of the product is: $x M_{\text{ribose}} + y M_{\text{deoxyribose}} + z M_{\text{glucose}} = 1150$ $150x + 134(5) + 180z = 1150$ $150x + 670 + 180z = 1150$ $150x + 180z = 480$

   Dividing the equation by $30$: $5x + 6z = 16$

   Substituting $z = 3 - x$: $5x + 6(3 - x) = 16$ $5x + 18 - 6x = 16$ $-x = -2 \implies x = 2$

Thus, the number of ribose units present in one molecule of octasaccharide is 2.