A mixture which contains 0.550g of camphor and 0.045g of an organic solute freezes at 157^(@)C . The solute contains 93.46% of C and 6.54% of H by weight . What is the molecular formula of the compound ? (Freezing point of camphor =178.4^(@)C and K_(f)=37.70)

A mixture which contains 0.550g of camphor and 0.045g of an organic so

1.	A mixture which contains 0.550g of camphor and 0.045g of an organic solute freezes at 157^(@)C . The solute contains 93.46% of C and 6.54% of H by weight . What is the molecular formula of the compound ? (Freezing point of camphor =178.4^(@)C and K_(f)=37.70)
Answer» Solution :Molality `=(DeltaT_(f))/(K_(f))=(178.4-157)/(37.70)=(21.4)/(37.70)` If the molecular wt. of SOLUTE is M then molality `=("moles of solute")/("wt.of solvent in grams")xx1000` `=(0.045//M)/(0.550)xx1000=(4500)/(55M)` Thus, `(4500)/(55M)=(21.4)/(37.70)`, `M=144.14` Now, from the given weight `%` of C and H , we get ` {:("moles of " C=(93.46)/(12)=7.79),("moles of "H=(6.54)/(1)=6.54):}}`(supposing 100g of the solute) `:.` molar ratio of C and `H=7.79 : 6.54`, i.e. `1.2 : 1`, i.e. `12 : 10` `:.` EMPIRICAL FORMULA is `C_(12)H_(10)` Empirical-formula weight `=12xx12+1xx10=154` The empirical-formula weight is slightly greater than the molecular weight which MIGHT be due to some molecular change of the solute in the solvent. However, the molecular formula will be `C_(12)H_(10)`.