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

Let the octasaccharide contain $$r$$ ribose units, $$d$$ 2-deoxyribose units and $$g$$ glucose units.

Because it is an octasaccharide, the total number of monosaccharide residues is
$$r + d + g = 8 \qquad -(1)$$

During hydrolysis all the glycosidic bonds break. In a linear chain of 8 units there are $$8-1 = 7$$ bonds, each bond having been formed by the loss of one water molecule (molar mass $$18\text{ g mol}^{-1}$$).
Hence,

total mass of isolated monosaccharides, $$M = (\text{molar mass of octasaccharide}) + 7 \times 18$$
$$M = 1024 + 126 = 1150\text{ g mol}^{-1} \qquad -(2)$$

The masses contributed by the individual monosaccharides are

$$M = 150r + 134d + 180g \qquad -(3)$$

The question states that 2-deoxyribose accounts for 58.26 % of the total mass of the hydrolysis products, i.e.

$$\frac{134d}{M} = 0.5826 \qquad -(4)$$

Substituting $$M = 1150$$ from (2) into (4):
$$134d = 0.5826 \times 1150 = 669.99 \approx 670$$
$$d = \frac{670}{134} = 5 \qquad -(5)$$

Insert $$d = 5$$ into (1):
$$r + g = 8 - 5 = 3 \qquad -(6)$$

Insert $$d = 5$$ into (3) and use $$M = 1150$$:
$$150r + 134(5) + 180g = 1150$$
$$150r + 670 + 180g = 1150$$
$$150r + 180g = 480$$
Divide by 30:
$$5r + 6g = 16 \qquad -(7)$$

Solve simultaneous equations (6) and (7).
From (6): $$r = 3 - g$$
Substitute in (7):
$$5(3 - g) + 6g = 16$$
$$15 - 5g + 6g = 16$$
$$15 + g = 16$$
$$g = 1$$
Therefore $$r = 3 - 1 = 2$$.

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