First let us find the molar mass of the compound using freezing point depression

Question

First let us find the molar mass of the compound using freezing point depression calculation

t = i Kf m

Molecular weight of benzene = 50 x 0.879 = 43.95 gm

(5.53-1.37) °C = (1) (5.12 °C kg mol-1) (x / 0.04395 kg)

X = 0.0357 Moles

Molar mass of the compound = 6g / 0.0357 mol = 168.02 gm/mol

Number of carbons in the molecule = 42.9 x 168.02 / 100 x 12 = 6 carbons

Number of Hydrogens in the molecule = 2.4 x 168.02 / 100 x 1.008 = 4 Hydrogens

Number of nitrogens in the molecule = 16.7 x 168.02 / 100 x 14 = 2 Nitrogens

Number of Oxygen in the molecule = 38.1 x 168.02 / 100 x 16 = 4 Oxygen

Hence the molecular formula of the compound = C6H4N2O4  

Explanation / Answer

please explain understandable

previous | 21 of 21 I returm to assignment Exercise 14.73 se Home bus gnments Part A A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% o, by mass. Addition of 6.00 g of this What is the molecular formula of this compound? compound to 50.0 mL benzene, CeHe Express your answer as a chemical formula. Enter the elements in the order: C, H, N, O (d 0.879 g/mL), lowers the freezing point from 5.53 to 1.37 °C, rse Tools xt dy Area My Answers Give Up er Settings Provide Feedback Continue nment 18.12 F10 F12 96 5 6

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