Question

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C6H6 (d = 0.879 g/mL), lowers the freezing point from 5.53 to 1.37 °C. What is the molecular formula of this compound?

Answer 1

i dont know what im doing wrong! i found the empirical formula to be C3H2NO2 with M = 84.054 g/mol. And then i calculated the molarmass like this: molality = (mass/Molarmass) /(Volume (L) * density (Kg/L)) and got M = 180.62 g/mol .... it does not match .. 180.62/84.054 = 2.149 when it is should be 2

Answer 2

C6H4N2O4