1. Find molecular formula of hydrocarbon $X$

Given molar mass of $X = 80$. Since $X$ is a hydrocarbon, let its formula be $\mathrm{C_nH_m}$.

For an acyclic alkene/alkadiene, a very likely formula near molar mass 80 is:

$12n + m = 80$

We will use the hydrogenation data to determine degree of unsaturation.

2. Use hydrogenation data

Mass of $X$ taken $= 10.0\,\text{mg} = 0.0100\,\text{g}$

Moles of $X$:

$n_X = \frac{0.0100}{80} = 1.25 \times 10^{-4}\,\text{mol}$

Hydrogen consumed $= 8.40\,\text{mL}$ at STP

At STP,

$1\,\text{mol gas} = 22400\,\text{mL}$

So moles of $H_2$ consumed:

$n_{H_2} = \frac{8.40}{22400} = 3.75 \times 10^{-4}\,\text{mol}$

Thus,

$\frac{n_{H_2}}{n_X} = \frac{3.75\times 10^{-4}}{1.25\times 10^{-4}} = 3$

So 1 mole of $X$ consumes 3 moles of $H_2$, meaning $X$ has 3 pi bonds.

Since $X$ is a hydrocarbon and ozonolysis is mentioned, these are most reasonably 3 double bonds.

For an acyclic hydrocarbon with 3 double bonds:

$\mathrm{C_nH_{2n+2-2\times 3}} = \mathrm{C_nH_{2n-4}}$

Now use molar mass:

$12n + (2n-4) = 80$ $14n - 4 = 80$ $14n = 84$ $n=6$

Hence,

$X = \mathrm{C_6H_8}$

So $X$ is an acyclic triene.

3. Use ozonolysis clue

Ozonolysis gives only formaldehyde and a dialdehyde.

• Formation of formaldehyde ($\mathrm{HCHO}$) indicates a terminal alkene unit of type $\mathrm{CH_2=}$.
• Since only formaldehyde and a dialdehyde are formed, the structure must cleave into exactly these kinds of fragments.

A 6-carbon triene fitting this is:

$\mathrm{CH_2=CH-CH=CH-CH=CH_2}$

This is 1,3,5-hexatriene.

Check ozonolysis:

• Cleavage at the first terminal double bond gives one $\mathrm{HCHO}$ from the left end.
• Cleavage at the last terminal double bond gives one $\mathrm{HCHO}$ from the right end.
• The internal carbon framework breaks into dialdehyde fragments.

Let us count fragments by cutting all three double bonds.

Write carbon sequence:

$\mathrm{C_1=C_2-C_3=C_4-C_5=C_6}$

Ozonolysis cleaves at:

• $\mathrm{C_1=C_2}$
• $\mathrm{C_3=C_4}$
• $\mathrm{C_5=C_6}$

This produces segments:

• $\mathrm{C_1}$ as $\mathrm{HCHO}$
• $\mathrm{C_2-C_3}$ as ethanedial $\mathrm{OHC-CHO}$
• $\mathrm{C_4-C_5}$ as ethanedial $\mathrm{OHC-CHO}$
• $\mathrm{C_6}$ as $\mathrm{HCHO}$

So the products are:

$\mathrm{HCHO + OHC-CHO + OHC-CHO + HCHO}$

Thus total number of molecules/fragments produced is:

$4$

4. Final answer

The total number of fragments formed on ozonolysis is

$\boxed{4}$

5. Comparison with stored answer

Stored correct answer = $4$

My derived answer = $4$

They agree.