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

Step-by-step solution

The general formula for alkanes is C_nH_ 2n+2 . The third homologue corresponds to n = 3 , yielding propane C₃H₈ . The molar mass is calculated as (3 12) + (8 1) = 44 g/mol. For the fourth homologue at n = 4 , the formula is butane C₄H₁₀ , with molar mass (4 12) + (10 1) = 58 g/mol. The difference is 58 - 44 = 14 g/mol. This matches the mass of a CH₂ unit ( 12 + 2 = 14 g/mol), which is the constant increment between successive homologues in any homologous series. Final answer: 14 g/mol