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Colligative Properties

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An unknown organic compound is a nonelectrolyte known to be composed of C, N, O, and H. Combustion analysis showed that it contained 67.3% C, 4.62% N, and 6.93% H. A solution made by dissolving 1.20 g of the unknown compound in 25.0 g of benzene had a freezing point of 4.7 degrees Celsius. Determine:
a) the empirical formula
b) the molar mass
c) the molecular formula of the unknown compound

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% combination by weight Atomic weights: No of g-atoms in 100 gms =
C= 67.30% 12.011 5.6032 =67.3/12.011
N= 4.62% 14.007 0.3298 =4.62/14.007
O= 21.15% 15.999 1.3220 =21.15/15.999
H= 6.93% 1.008 6.8750 =6.93/1.008

Therefore C,N,O, H are in the ratio 5.6032:0.3298:1.3220:6.8750

If we divide by the smallest number ie 0.3298 we get
C= 16.99
N= 1
O= 4.01
H= 20.85

Therefore C N O H are in ...

Solution Summary

The solution calculates the empirical formula, the molar mass and the molecular formula of the unknown compound

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1. Anthraquinone contains only carbon, hydrogen and oxygen and has an empirical formula of C7H4O. The freezing point of camphor is lowered by 22.3oC when 1.32 g of antraquinone is dissolved in 11.4 g of camphor. Determine the molecular formula of anthraquinone. ( Kf for camphor is 40.0).

2. A solution is prepared by mixing 25 ml of C5H12 ( density = 0.63 g/ml) and 45 ml of C6H14 ( density = 0.66 g/ml). Assume that the volumes add on mixing, calculate for C5H12
a) mass %
b) mole fraction
c) molality
d) molaroty.

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