Question

The empirical formula of an organic compound is $CH_{2}$. The molar mass of the compound is $56g~mol^{-1}$. The organic compound is ________.

• A. n-Butane
• B. Propene
• C. Propane
• D. 2-Methylpropane
• E. Cyclobutane

Hint:

Molecular Formula = $n \times$ (Empirical Formula).

Answer 1

Answer: (E)

Step 1: Understanding the Concept
The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule. The molecular formula is always an integer multiple of the empirical formula. We can find this integer by comparing the molar mass to the empirical formula mass.
Step 2: Key Formula or Approach
1. Calculate the mass corresponding to the empirical formula (empirical formula mass). 2. Find the integer multiplier, \(n\), where \(n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}\). 3. Determine the molecular formula: Molecular Formula = (Empirical Formula)\(_n\). 4. Identify the compound from the options that matches the molecular formula.
Step 3: Detailed Explanation
1. Calculate the empirical formula mass.
The empirical formula is \(CH_2\). Using atomic masses C=12 and H=1: Empirical Formula Mass = \(1 \times 12 + 2 \times 1 = 14 \text{ g/mol}\).
2. Find the integer multiplier (\(n\)).
The given molar mass of the compound is 56 g/mol. \[ n = \frac{56 \text{ g/mol}}{14 \text{ g/mol}} = 4 \] 3. Determine the molecular formula.
The molecular formula is the empirical formula multiplied by \(n\). Molecular Formula = \((CH_2)_4 = C_4H_8\).
4. Identify the compound from the options.
We need to find the compound with the formula \(C_4H_8\). - (A) n-Butane: An alkane (\(C_nH_{2n+2}\)), formula is \(C_4H_{10}\). - (B) Propene: An alkene (\(C_nH_{2n}\)) with 3 carbons, formula is \(C_3H_6\). - (C) Propane: An alkane (\(C_nH_{2n+2}\)) with 3 carbons, formula is \(C_3H_8\). - (D) 2-Methylpropane: An isomer of butane (alkane), formula is \(C_4H_{10}\). - (E) Cyclobutane: A cycloalkane (\(C_nH_{2n}\)) with 4 carbons, formula is \(C_4H_8\). This matches. (Note: Butene also has the formula \(C_4H_8\), but is not an option).
Step 4: Final Answer
The organic compound is Cyclobutane.