Question

freezing point is found to be -64.221°C for solution consisting of 10.5 g of an unknown organic compound in 342 g of chloroform. If the freezing point of pure chloroform is -63.5°C empirical formula of the unknown compound is found to be C5H802 by combustion analysis, what is the molecular formula of the unknown?

analysis, what is 11. 86. The freezing point is found to be -64.221°C f 64 221°C for a solution nic compound and sure chloroformis inknown compound of 10.50 g of an unknown organic compo 242 g or chloroform. If the freezing point of pure chla -63.5°C and the empirical formula of the unknown is found to be C3H2O2 by combustion analysis, what is molecular formula of the unknown?

Answer 1

of unknown Ans : Let us take the molecular formula organic compound is (Cg Hs O2) . Given that, the empirical formula of compound is cstte og the unknown Molecular weight of the compound (C6H8O2) = {(1245) + (exe) + (1642) germa? = 100ng met Now, 342 g of chloroform contain 10.5 g of the Unknown compound Cu CHCl3 contain 1005 g loong most of of the the unkne unknown (imprux CHCl contain 10.5 x 1000 ob mol of unknown 100nx 342 compound 0.31 mole of the compound = So molah concentration of the solution = 031 0 Now we know, A1 = k*m (0) For choroform, f = 4.68 °C kg mot alma 0.31 mal kigt 444 = (-63.5 °C ) - (-69221c) = 0.721 °C

Now from (1) no. equation we get- 0.721% = 4.68 °C kg molt x 0.31 mol kg n = 0012 o So the compound molecular formula of is (Cs HQ Oz) or the unknown Go H1604