Given :
A compound has a molar mass of 129 g/mol .
Empirical formula of compound is C₂H₅N .
To Find :
The molecular formula of the compound.
Solution :
Empirical mass of compound :
![M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + ( 1 \times 14 )\\\\M_e = 43\ gram/mol](https://tex.z-dn.net/?f=M_e%20%3D%20%28%202%20%5Ctimes%2012%20%29%20%2B%20%28%205%20%5Ctimes%201%20%29%20%2B%20%28%20%201%20%20%5Ctimes%2014%20%29%5C%5C%5C%5CM_e%20%3D%2043%5C%20gram%2Fmol)
Now, n-factor is :
![n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3](https://tex.z-dn.net/?f=n%20%3D%20%5Cdfrac%7BM%7D%7BM_e%7D%5C%5C%5C%5Cn%20%3D%20%5Cdfrac%7B129%7D%7B43%7D%5C%5C%5C%5Cn%20%3D%203)
Multiplying each atom in the formula by 3 , we get :
Molecular Formula, C₆H₁₅N₃
Not 100% sure but I believe it is C.
The answer is a conjugate acid.