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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

##### Solution Summary

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

##### Solution Preview

% 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
100.00%

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 ...

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