An unknown compound (152 mg) was dissolved in water to make 75.0 mL of solution. The solution did not conduct electricity and had an osmotic pressure of 0.328 atm at 27°C. Elemental analysis revealed the substance to be 78.90% C, 10.59% H, and 10.51% O. Determine the molecular formula of this compound.

The molecular formula of the compound is C10H16O.

Explanation:

Based on the given information, the mass of an unknown compound is 152 mg or 0.152 g, which was dissolved to produce 75 ml of solution.

The osmotic pressure of the solution is 0.328 atm at 27 degrees C or 300 K.

The formula of osmotic pressure is,

P = CRT

Now putting the values in the formula we get,

0.328 = C*0.0821*300

C = 0.013317

C = (mass/molecular mass) * (1000/volume of solution in ml)

0.013317 = (0.152/Molecular mass) * (1000/75)

Molecular mass = 152.186

C mass = 152.186 * 0.789 = 120.07 grams

C mole = 120.07/12 = 10

H mass = 152.186 * 0.1059 = 16.11 grams

H mole = 16.11/1 = 16

O mass = 152.186 - 120.07 - 16.116 = 16 grams

O mole = 16/16 = 1

Thus, the molecular formula of the compound will be,

C10H16O

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