Answer to Question #157321 in Physical Chemistry for Joshua

Question

unknown compound consists of 33.81% C, 1.42% H, 45.05%O, and 19.72% N  by mass. A 1.505 g sample of the compound when dissolved in 50.00 ml benzene  (d=0.879g/ml), lowers the freezing point of benzene to 4.70o C. What is the  molecular formula of the compound?

Answer 1

Solution.

$\Delta T = K \times m$

$m = \frac{m(com.)}{M(com). \times m(benzene)}$

$m(benzene) = 0.879 \times 50.00 = 43.95 \ g = 0.044 \ kg$

$m = -\frac{4.70-5.53}{K} = 0.1596 \ \frac{mol}{kg}$

$M(com.) = \frac{m(com.)}{m \times m(benzene)}$

M(com.) = 214.56 g/mol

n(C) = 6.045 mol

n(H) = 3.047 mol

n(O) = 6.041 mol

n(N) = 3.022 mol

$n(C):n(H):n(O):n(N) = 6:3:6:3$

C6H3N3O6

Answer:

C6H3N3O6