Chemistry - 64 - molecular formula?

Anthraquinone contains only carbon, hydrogen, and oxygen and has an empirical formula of C7H4O. The freezing point of camphor is lowered by 22.3C when 1.32g anthraquinone is dissolved in 11.4g camphor. Determine the molecular formula of anthraquinone.

Answer:
C14H8O2

The freezing point of pure camphor is 179.8 ° C and Kf = 39.7 ° C/mol.

The change in freezing point = von't Hoff factor * Kf * activity.

The change in freezing point = 22.3 ° C.
The von't Hoff factor = 1 (substance is covalently bonded)
So solve for the activity (22.3/39.7 = 0.562)

Use the fact that the MM of camphor is 152.23 g/mol and you used 11.4 grams to get 0.07489 moles of camphor (grams/MM). 1.32 grams of anthraquinone gives a molality which points to a Molecular Mass of just over 200 g/mol. The molecular formula is therefore twice the empirical formula.
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