Question

The osmotic pressure of a solution containing 1.44 g of an unknown compound dissolved in 175.0 mL of solution at 25 ∘C is 1.93 atm . The combustion of 32.88 g of the unknown compound produced 55.60 gCO2 and 22.76 gH2O . What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)?

Answer 1

we know that

osmotic pressure = C x R x T

so

1.93 = C x 0.0821 x 298

C = 0.07888

now

concentration = moles x 1000 / volume (ml)

so

0.07888 = moles x 1000 / 175

moles = 0.0138

now

moles = mass / molar mass

so

0.0138 = 1.44 / molar mass

molar mass = 104.3

so

molar mass of the compound is 104.3

now consider the combustion reaction

we know that

moles = mass / molar mass

so

moles of compound taken = 32.88 / 104.3 = 0.315

moles of C02 obtained = 55.6 / 44 = 1.2636

moles of H20 obtained = 22.76 / 28 = 1.26444

now

let the compound be CaHbOc

the combustion reaction will be

CaHbOc + ( a+ 0.5b-0.5c) 02 ---> a C02 + b H20

now

moles of C02 obtained = a x moles of CaHbOc taken

so

1.2636 = a x 0.315

a = 4

now

moles of H20 obtained = b x moles of CaHb0C taken

so

1.2644 = b x 0.315

b = 4

now

the compound becomes C4H4Oc

so

molar mass of compound = ( 4 x 12 ) + ( 4 x 1) + ( c x 16)

so

52 + 16c = 104

c = 3.25

c = 3

so

the compound is C4H403