A linear octasaccharide (molar mass = 1024 g mol^-1) on complete hydrolysis produces three monosacch

Q: Why do we add the mass of 7 water molecules?

Because hydrolysis involves the addition of water to break bonds. An octasaccharide has 8 units connected by 7 bonds, so 7 water molecules are incorporated into the final products.

Q: How did we get 1150 g/mol?

It's the mass of the octasaccharide (1024) plus the mass of 7 water molecules ($7 \times 18 = 126$). $1024 + 126 = 1150$.

Q: Why is the number of 2-deoxyribose units exactly 5?

When you calculate $58.26\%$ of 1150, you get $669.99$. Dividing this by the molar mass of 2-deoxyribose (134) gives $4.999$, which is practically exactly 5. In chemistry problems, these numbers are designed to be integers.

Q: Where did the equation $x + y = 3$ come from?

An octasaccharide has 8 units in total. We found that 5 of them are 2-deoxyribose. That leaves $8 - 5 = 3$ units, which must be a combination of ribose ($x$) and glucose ($y$).

Q: Could there be 0 units of glucose?

If $y=0$, then $x=3$. The mass would be $3 \times 150 = 450$, but we need the remaining mass to be 480. So $y$ cannot be 0. The math perfectly dictates $x=2$ and $y=1$.

Question

A linear octasaccharide (molar mass = $1024 	ext{ g mol}^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is $58.26\% 	ext{ (w/w)}$ 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 $	ext{g mol}^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180;
Atomic mass (in amu): H = 1, O = 16

Sigma Percentile · Accepted Answer

The problem of hydrolyzing an octasaccharide is a beautiful intersection of chemical stoichiometry and algebraic puzzle-solving. It might look intimidating at first glance, with its percentages and multiple unknown sugars, but once we break it down using the fundamental law of conservation of mass, it unravels elegantly.

### The Anatomy of an Octasaccharide

Imagine an octasaccharide as a train with 8 cars. These cars are the monosaccharide units (ribose, 2-deoxyribose, and glucose). To connect 8 cars, you need exactly 7 couplings. In chemistry, these couplings are glycosidic bonds.

When we perform complete hydrolysis, we are essentially breaking every single one of these 7 couplings. Breaking a glycosidic bond requires the addition of one water molecule ($	ext{H}_2	ext{O}$). Therefore, to completely dismantle our 8-car train into individual cars, we must consume exactly 7 water molecules.

### The Law of Conservation of Mass

This is where the magic happens. The law of conservation of mass dictates that the total mass of our final products must equal the mass of everything we started with.

We started with one mole of the octasaccharide, which weighs $1024 	ext{ g}$. But we also added 7 moles of water! Since each water molecule has a molar mass of $18 	ext{ g/mol}$, the total mass of water added is $7 	imes 18 = 126 	ext{ g}$.

Adding these together gives us the total mass of our 8 individual monosaccharide units:
$$ M_{	ext{total}} = 1024 + 126 = 1150 	ext{ g/mol} $$

**This step is the most common trap.** Many students forget to add the mass of the water and try to calculate the percentages based on the $1024 	ext{ g}$ alone. Always remember: hydrolysis adds mass!

### Decoding the Percentages

The problem generously tells us that $58.26\%$ of this total product mass is composed of 2-deoxyribose. Let's translate this percentage into a concrete mass:
$$ M_{	ext{deoxy}} = \frac{58.26}{100} 	imes 1150 \approx 670 	ext{ g} $$

Now, we know that a single unit of 2-deoxyribose has a molar mass of $134 	ext{ g/mol}$. To find out how many units of 2-deoxyribose we have, we simply divide the total mass of 2-deoxyribose by the mass of one unit:
$$ n_{	ext{deoxy}} = \frac{670}{134} = 5 $$

Perfect! Out of our 8 total units, exactly 5 are 2-deoxyribose.

### The Final Algebraic Puzzle

We have 8 units in total, and we've identified 5 of them. This leaves us with $8 - 5 = 3$ units. These remaining 3 units must be a combination of ribose and glucose.

Let's define two variables:
- Let $x$ be the number of ribose units.
- Let $y$ be the number of glucose units.

Our first equation is simply the count of the remaining units:
$$ x + y = 3 $$

Now, let's look at the remaining mass. The total mass was $1150 	ext{ g}$, and the 5 units of 2-deoxyribose accounted for $670 	ext{ g}$. The mass left for ribose and glucose is:
$$ 1150 - 670 = 480 	ext{ g} $$

Since each ribose unit weighs $150 	ext{ g}$ and each glucose unit weighs $180 	ext{ g}$, we can write our second equation based on mass:
$$ 150x + 180y = 480 $$

We now have a neat system of linear equations. Let's simplify the mass equation by dividing everything by 30:
$$ 5x + 6y = 16 $$

From our first equation, we know that $y = 3 - x$. Substituting this into our simplified mass equation gives:
$$ 5x + 6(3 - x) = 16 $$
$$ 5x + 18 - 6x = 16 $$
$$ -x = -2 \implies x = 2 $$

And there we have it! The variable $x$ represents the number of ribose units, which perfectly evaluates to **2**. This means our original octasaccharide was composed of 5 units of 2-deoxyribose, 2 units of ribose, and 1 unit of glucose.

Enter Numerical Value:

Visualized Solution

$Reaction Setup$

$Total Mass Calculation$

$Mass of 2-Deoxyribose$

$Units of 2-Deoxyribose$

$Unit Balance Equation$

$Mass Balance Equation$

$Solving for x$

$Final Result$

$Conclusion$

The Sigma Insight: Biomolecules

The Anatomy of an Octasaccharide

The Law of Conservation of Mass

Decoding the Percentages

The Final Algebraic Puzzle

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