JEE MainChemistry

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Molar Mass Determination via Reductive Ozonolysis Products

Q: Can we directly add the atoms of the products and subtract two oxygens?

Yes, during reductive ozonolysis, each C=O oxygen comes from O_3. Thus, combining all atoms from ethanal (C_2H_4O) and propanone (C_3H_6O) and removing the two oxygen atoms gives C_5H_10.

Molar mass of the hydrocarbon (X) which on ozonolysis consumes one mole of O_3 per mole of (X) and gives one mole each of ethanal and propanone is g mol^(-1) (Molar mass of C: 12 g mol^(-1), H: 1 g mol^(-1))

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Correct answer

The hydrocarbon (X) is 2-methylbut-2-ene with the molecular formula $C_{5} H_{10}$ , giving a molar mass of $70 {g mol}^{- 1}$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

1 mole of hydrocarbon (X) consumes 1 mole of $O_{3}$ to yield 1 mole of ethanal ( $C H_{3} C H O$ ) and 1 mole of propanone ( $C H_{3} C O C H_{3}$ ). Atomic masses: $C = 12 g/mol$ , $H = 1 g/mol$ .

02Find

Molar mass of hydrocarbon (X) in $g \cdot m o l^{- 1}$ .

03Visualise

Ozonolysis cleaves a $C = C$ double bond and replaces it with two $C = O$ bonds. Reversing the ozonolysis: remove the two carbonyl oxygens and join the carbonyl carbons with a double bond: $C H_{3} - C H = C (C H_{3})_{2}$ (2-methylbut-2-ene).

04Strategise

Determine the molecular formula of (X). Ethanal ( $C_{2} H_{4} O$ ) + Propanone ( $C_{3} H_{6} O$ ) gives a combined carbon count of $2 + 3 = 5$ and hydrogen count of $4 + 6 = 10$ . Hydrocarbon (X) is $C_{5} H_{10}$ . Then compute its molar mass: $M = 5 \times 12 + 10 \times 1$ .

05Execute

Molar mass of $C_{5} H_{10} = 5 \times 12 + 10 \times 1 = 60 + 10 = 70 g/mol$ .

✓Verify

$C_{5} H_{10}$ has 1 degree of unsaturation (matching consumption of 1 mole $O_{3}$ ), and gives 2-methylbut-2-ene which produces $C H_{3} C H O$ and $C H_{3} C O C H_{3}$ . Molar mass is exactly 70.

Hints that build this answer step by step

What is the structure of hydrocarbon (X) if its reductive ozonolysis produces one mole of ethanal ( ${C H}_{3} C H O$ ) and one mole of propanone ( ${C H}_{3} {C O C H}_{3}$ )?
${C H}_{3} C H = C ({C H}_{3})_{2}$ (2-methylbut-2-ene)
What is the molar mass of hydrocarbon (X) with the molecular formula $C_{5} H_{10}$ ?
$70 {g mol}^{- 1}$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Can we directly add the atoms of the products and subtract two oxygens?

Yes, during reductive ozonolysis, each $C = O$ oxygen comes from $O_{3}$ . Thus, combining all atoms from ethanal ( $C_{2} H_{4} O$ ) and propanone ( $C_{3} H_{6} O$ ) and removing the two oxygen atoms gives $C_{5} H_{10}$ .

Answer

Accepted answer: 70 g mol^{-1}

The hydrocarbon (X) is 2-methylbut-2-ene with the molecular formula $C_{5} H_{10}$ , giving a molar mass of $70 {g mol}^{- 1}$ .

Why each option works or fails

Step-by-step solution

Given: 1 mole of hydrocarbon (X) consumes 1 mole of $O_{3}$ to yield 1 mole of ethanal ( $C H_{3} C H O$ ) and 1 mole of propanone ( $C H_{3} C O C H_{3}$ ). Atomic masses: $C = 12 g/mol$ , $H = 1 g/mol$ .
Find: Molar mass of hydrocarbon (X) in $g \cdot m o l^{- 1}$ .
Visualise: Ozonolysis cleaves a $C = C$ double bond and replaces it with two $C = O$ bonds. Reversing the ozonolysis: remove the two carbonyl oxygens and join the carbonyl carbons with a double bond: $C H_{3} - C H = C (C H_{3})_{2}$ (2-methylbut-2-ene).
Strategise: Determine the molecular formula of (X). Ethanal ( $C_{2} H_{4} O$ ) + Propanone ( $C_{3} H_{6} O$ ) gives a combined carbon count of $2 + 3 = 5$ and hydrogen count of $4 + 6 = 10$ . Hydrocarbon (X) is $C_{5} H_{10}$ . Then compute its molar mass: $M = 5 \times 12 + 10 \times 1$ .
Execute: Molar mass of $C_{5} H_{10} = 5 \times 12 + 10 \times 1 = 60 + 10 = 70 g/mol$ .
Verify: $C_{5} H_{10}$ has 1 degree of unsaturation (matching consumption of 1 mole $O_{3}$ ), and gives 2-methylbut-2-ene which produces $C H_{3} C H O$ and $C H_{3} C O C H_{3}$ . Molar mass is exactly 70.

Molar Mass Determination via Reductive Ozonolysis Products

What feels right?

The hydrocarbon (X) is 2-methylbut-2-ene with the molecular formula $C_{5} H_{10}$ , giving a molar mass of $70 {g mol}^{- 1}$ .

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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