Empirical formula = C2H5NO
Molecular mass = 2 x (vapour density)
= 2 x 29.5
= 59.0
Empirical formula mass of C4H5NO = 24 + 5 + 14 + 16
= 59
n = (Molecular mass) / (Empirical formula mass)
= 59/59 = 1
Empirical formula mass = 59
\(\therefore\) Molecular formula = (Empirical formula)n
= (C2H5NO)1
Molecular formula = C2H5NO