Question

In a compound `C, H, N` atoms are present in `9:1:3.5` by weight. Molecular weight of compound is `108`. Its molecular formula is:
A. `C_(2)H_(6)N_(2)`
B. `C_(3)H_(4)N`
C. `C_(9)H_(12)N_`
D. `C_(6)H_(8)N_(2)`

Answer 1

Correct Answer - D
Ratio of masses `m_(C ) : m_(H) : m_(N) = 9:1:3.5`
Ratio of moles
`= n_(C ):n_(H):n_(N) = (9)/(12):(1)/(1):(3.5)/(14)`
`= (0.75)/(0.25):(1)/(0.25):(0.25)/(0.25)`
`= 3:4:1`
`:.` Empirical formula of `o .c. = C_(3)H_(4)N`
Empirical formula mass `= (3 xx 12 + 4 xx 1 + 1 xx 14) u`
`= 54 u`
`n = ("Molecular mass")/("Empirical formula mass") = (108 u)/(54 u) = 2`
Molecular formula `= n xx` Empirical formula
`= 2 xx C_(3)H_(4)N`
`= C_(6)H_(8)N_(2)`
Shortcut method:
Find the molecular masses using the given formulas. Only `C_(6)H_(8)N_(2) (6 xx 12 + 8 xx 1 + 2 xx 14)` gives `108 u`.