A quantity of 7.480 g of an organic compound is dissolved in water to make 300.0 mL of solution. The solution has an osmotic pressure of 1.43 atm at 27°C. The analysis of this compound shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent O, and 16.3 percent N. Calculate the molecular formula of the compound

C5H6N2O3

Explanation:

First the empirical formulas

C= 41.8÷12= 2.48

H= 4.7÷1= 4.7

O= 37.3÷ 16= 2.33

N= 16.3÷14 = 1.16

Divide by the smallest

C= 3.48/1.16=3

H= 4.7/1.16=4.1

O= 2.33/1.16=2

N= 1.16/1.16=1

Therefore empirical formula = C3H4NO2

To calculate molecular formula for osmotic pressure,

π= cRT

Or

π=cgRT/M where cg is in g/liter & T is temperature in Kelvin. Thus

π= (7.480*0.0821*300)/ M

M= 184.23/1.43

M= 128.83

To find molecular Formula

Molecular Mass= (empirical mass)n

128.83= (C3H4NO2)n

128.83= 86n

n= 1.5

Therefore the molecular formula

(C3H4NO2)1.5

= C4.5H6N1.5O3

Approximately

C5H6N2O3