Question

What is the difference in molar masses of third and fourth homologues of alkane series?

• A. 28 g mol$^{-1}$
• B. 14 g mol$^{-1}$
• C. 15 g mol$^{-1}$
• D. 16 g mol$^{-1}$

Hint:

Successive homologues always differ by 14 mass units (one carbon + two hydrogens).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A homologous series is a series of organic compounds with the same functional group and similar chemical properties in which successive members differ by a fixed unit.
Step 2: Key Formula or Approach:
Successive members in a homologous series differ by a \( -CH_{2} \) (methylene) group.
Difference in molar mass = Atomic mass of C + 2 \( \times \) Atomic mass of H.
Step 3: Detailed Explanation:
For the alkane series (\( C_{n}H_{2n+2} \)):
- 3rd homologue is Propane (\( C_{3}H_{8} \)).
- 4th homologue is Butane (\( C_{4}H_{10} \)).
Difference = \( (C_{4}H_{10}) - (C_{3}H_{8}) = CH_{2} \).
Molar mass of \( CH_{2} = 12 + 2(1) = 14 \text{ g/mol} \).
Step 4: Final Answer:
The difference in molar masses is 14 g mol$^{-1}$.