A compounds analyzes to be 61.7% Cl, 26.72% P, and 12.11% N. 1.2952 g of the com

Question

A compounds analyzes to be 61.7% Cl, 26.72% P, and 12.11% N. 1.2952 g of the compound was dissolved in 15.00 ml of benzene (density of benzene=.879 g/ml). Producing a solution freezing at 4.03 degress celcius. pure benzene has a freezing point of 5.48 degrees celcius with a Kf of 5.12Co/molal. What is the empirical and molecular formula for this compound. 

Explanation / Answer

61.7 g Cl *(1 mol /35.5g ) = 1.73802817 mol Cl 26.72 g P *(1 mol / 31 g) = 0.862 mol P 12.11 g N *( 1mol / 14 g) = 0.865 mol N Divide by 0.865 => Cl2P1N1 (empiricalformula) T =Kf*m           m= moles compound/kg solvent = (mass compound / MW) / kg solvent MW = Kf (mass)/[T * kg solvent] MW = (5.12 C *kg/mol)*(1.2952 g) / [[5.48-4.03]C * 15 mL *[0.879e-3kg/mL]] MW = 347 g/mol empirial weight = 2*35.5 + 30 + 14 = 115 molecular weight = 347/115 = 3.006 Thus Cl6P3N3 (molecular formula)

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