Question

Molar mass of the hydrocarbon (X) which on ozonolysis consumes one mole of O3 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)

Answer 1

Correct Answer: 70

To find the molar mass of hydrocarbon (X) undergoing ozonolysis that yields ethanal and propanone, let's write the balanced reaction equation. Assume the hydrocarbon is C_aH_b. The ozonolysis reaction can be represented as: C_aH_b + O₃ → CH₃CHO + CH₃COCH₃. Given products are ethanal (C₂H₄O) and propanone (C₃H₆O).

The empirical formula of (X) is the sum of the products: CH₃CHO + CH₃COCH₃ = C₅H₁₀O₂. But since one mole of O₃ is used per mole of (X), count C and H only: C₅H₁₀.

Calculate the molar mass: Molar mass of C₅H₁₀ = 5(12 g/mol) + 10(1 g/mol) = 60 g/mol + 10 g/mol = 70 g/mol.

This molar mass falls within the provided range of 70,70 g/mol. Hence, the computed molar mass of hydrocarbon (X) is 70 g/mol.