Interpret the ozonolysis information

Ozonolysis of an alkene breaks one $\mathrm{C=C}$ bond using one mole of $\mathrm{O_3}$ per mole of alkene.

Since $1$ mole of $\mathrm{O_3}$ is consumed per mole of hydrocarbon $X$, the compound $X$ contains one double bond.

Identify the alkene from the products

The ozonolysis products are:

• one mole of ethanal: $\mathrm{CH_3CHO}$
• one mole of propanone: $\mathrm{CH_3COCH_3}$

In ozonolysis, each carbon of the double bond becomes a carbonyl carbon.

• To get ethanal, one alkene carbon must have had substituents: $\mathrm{CH_3}$ and $\mathrm{H}$
• To get propanone, the other alkene carbon must have had substituents: $\mathrm{CH_3}$ and $\mathrm{CH_3}$

Therefore, the alkene is: $\mathrm{CH_3-CH=C(CH_3)_2}$

Write molecular formula

Count atoms in $\mathrm{CH_3-CH=C(CH_3)_2}$:

• Carbons: $5$
• Hydrogens: $10$

So the molecular formula is: $\mathrm{C_5H_{10}}$

Calculate molar mass

$M = 5(12) + 10(1) = 60 + 10 = 70\ \mathrm{g\ mol^{-1}}$

Compare with stored answer

Derived answer = $70$

Stored correct answer = $70$

Hence, they agree.