Question

A solution contains 10.35 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.26 ∘C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O.

What is the molecular formula of the compound?

Answer 1

Freezing point depression DTf = 3.26 deg C

Molaity = DTf/Kf = 3.26/1.86 = 1.75 m


Now the mass of water = volume x density = 50.0 x 1.00 = 50.0 g

= 0.05 kg


Moles of compound = molality x mass of water in kg

= 1.75 x 0.05 = 0.0875 mol


Molar mass = mass/moles of compound

= 10.35/0.0875 = 118.28 g/mol


Since moles = mass/molar mass

Moles of C : H : O = 60.98/12.01 : 11.94/1.008 : (100 - 60.98 - 11.944)/16.00

= 5.07 :11.84 : 1.69

= 5 : 12 : 2


The empirical formula is C2H6O


Let the molecular formula be C2aH3aOa

Molar mass = 12.01 x 2a + 1.008 x 3a + 16.00 x a = 118.28

43.044a = 118.28 => a = 2.74 let it be 3


The molecular formula is C6H18O3