Question

The empirical formula of an imaginary substance called polythrytol has been determined to be C₂H₅O₂. A solution of 2.50 g of polythrytol in 50.0 g water freezes at −0.773°C. What is the molecular formula of polythrytol? The K_f of water is 1.86°C / molal.

Answer 1

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatyle ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

where:

Kf = freezing point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentation.

At the end:

Tf mix = Tf solvent - dTf

now...

-0.733 = -1.86 * (2.5/MW) / (50*10^-3)

MW = 1.86 /0.733 * 2.5 / (50*10^-3)

MW = 126.87 g/mol

MW of empirical = C2H5O2 = 12*2 + 5*1 + 2*16 = 61

ratio --> 126.87/ 61 = 2

then

C2H5O2 x2 = C4H10O4