Somewhat difficult chemistry problem, one that I do not know how to solve. Please show work so I may learn.?

Question

A compund is 42.4% C, 2.4% H, 16.6% N, and 37.8% O by mass. The addition of 6.45 g of this compound to 50.0 mL benzene, C(6)H(6) [Density=0.879 g/cm cubed], lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound? Show work, and provide a balanced equation.

Answer 1

42.4 / 12.011 = 3.53 ( C )

2.4 / 1.0079 = 2.38 ( H)

16.6 / 14.0067 = 1.18 (N)

37.8 / 15.9994 = 2.36 ( O )

We divide each number for the smallest and we get

C3 H2 N O2 (mass = 84 )

From the density of benzene we get 50.00 mL >>

50.00 mL x 0.879 g/mL = 43.95 g = 0.04395 Kg

delta T = 5.53 - 1.37 = 4.16 °C

kf for benzene = 5.12

delta T = m kf

4.16 / 5.12 = 0.8125 = m

m = moles /Kg benzene

0.8125 = moles / 0.04395

moles = 0.0357

molecular weight = 6.45 g / 0.0357 = 180. g / mol

180 / 84 = 2

now we multiply each number of the formula by 2 and we have

C6 H4 N2 O4

Answer 2

You need to give us the freezing point constant for benzene. Since that is in degrees per mole/liter or mole/kg of benzene, assuming that the compound is NOT ionic, you can determine the mol wt of the compound. To do this, you restate the problem as adding 129 g of the compound to 1 Liter of benzene (if the constant is in mol/kg, then you play with the density).
The 4.16 deg C depression of freeze pt/molal freeze point tells us the number of moles. That times 129 gives us the mol wt of the compound. Multiply that by each component atoms percent, and round off the numbers to the nearest integers will give you the empirical formula. There is no equation involved.