Question

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

Answer 1

Answer:

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