The problem requires determining the molar mass of the third member of a homologous series. A homologous series comprises organic compounds sharing a similar structure, differing by a repeating CH₂ unit. The molar mass of the initial compound is 46 g. The solution leverages the constant molar mass increment between consecutive members of such series.
A characteristic difference of 14 g/mol, corresponding to the mass of a CH₂ group, is observed in numerous homologous series. The calculation proceeds as follows:
- Recognize the molar mass increase per member as 14 g (molecular weight of CH₂).
- Compute the molar mass of the second member: \(46 \, \text{g/mol} + 14 \, \text{g/mol} = 60 \, \text{g/mol}\).
- Ascertain the molar mass of the third member: \(60 \, \text{g/mol} + 14 \, \text{g/mol} = 74 \, \text{g/mol}\).
Consequently, the molar mass of the third member is 74 g.