Start Learning Free

Before You Begin

Question #44670

Difficulty: 8/10 | 🟢 Free

The Question

A gaseous hydrocarbon (A) has an empirical formula of \ce{CH2}. At 300 K300\text{ K} and 100 kPa100\text{ kPa}, the density of the gas is 3.36 kg m−33.36\text{ kg m}^{-3}. Compound (A) exhibits geometrical isomerism and undergoes reductive ozonolysis to yield only one type of organic product (B). What is the molar mass of (A) and the number of α\alpha-hydrogens in (B)? (Given: R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1})

Options

  • A.

    84 g mol−184\text{ g mol}^{-1}, 22

  • B.

    84 g mol−184\text{ g mol}^{-1}, 33

  • C.

    56 g mol−156\text{ g mol}^{-1}, 33

  • D.

    112 g mol−1112\text{ g mol}^{-1}, 22

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Empirical vs Molecular Formula|Density-Volume-Mass Relationship

Why This Question Is Tricky

Common Mistakes

The question tests the dimensional analysis of the ideal gas law combined with organic structural elucidation.

The core trap lies in the unit of density (kg m−3\text{kg m}^{-3}).

When using strictly SI units (PP in Pa, dd in kg m−3\text{kg m}^{-3}, R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}), the molar mass MM is calculated in kg mol−1\text{kg mol}^{-1}.

Students frequently miss this and assume the result is in g mol−1\text{g mol}^{-1}, leading to a massive error, or they incorrectly convert kg m−3\text{kg m}^{-3} to g L−1\text{g L}^{-1} by multiplying by 1000 instead of 1 (since 1 kg m−3=1 g L−11\text{ kg m}^{-3} = 1\text{ g L}^{-1}).

Bypassing this yields M=84 g mol−1M = 84\text{ g mol}^{-1}.

The subsequent chemical trap requires identifying a symmetrical alkene capable of geometrical isomerism (hex-3-ene) and correctly counting the α\alpha-hydrogens in its ozonolysis product (propanal), avoiding confusion with acetaldehyde which would result from a miscalculated mass of 56 g mol−156\text{ g mol}^{-1}.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterSome Basic Concepts of Chemistry - Class 11 Part 1
ConceptEmpirical vs Molecular Formula
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree