Question

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 by adding the missing subscripts.

Answer 1

P = 0.328atm
T= 27.0 oC
= (27.0+273) K
= 300 K


use:
P = C*R*T
0.328 = C*0.0821*300.0
C =0.0133 M

volume , V = 75.0 mL
= 7.5*10^-2 L


use:
number of mol,
n = Molarity * Volume
= 0.0133*0.075
= 9.988*10^-4 mol

mass(solute)= 152 mg
= 0.152 g


use:
number of mol = mass / molar mass
9.988*10^-4 mol = (0.152 g)/molar mass
molar mass = 152 g/mol

we have mass of each elements as:
C: 78.9 g
H: 10.59 g
O: 10.51 g


Divide by molar mass to get number of moles of each:
C: 78.9/12.01 = 6.5695
H: 10.59/1.008 = 10.506
O: 10.51/16.0 = 0.6569


Divide by smallest to get simplest whole number ratio:
C: 6.5695/0.6569 = 10
H: 10.506/0.6569 = 16
O: 0.6569/0.6569 = 1


So empirical formula is:C10H16O

Molar mass of C10H16O,
MM = 10*MM(C) + 16*MM(H) + 1*MM(O)
= 10*12.01 + 16*1.008 + 1*16.0
= 152.228 g/mol

Now we have:
Molar mass = 152.0 g/mol
Empirical formula mass = 152.228 g/mol
Multiplying factor = molar mass / empirical formula mass
= 152.0/152.228
= 1

So molecular formula is:C₁₀H₁₆O