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

Molecular formula is: C₁₀H₁₆O

Step-by-step explanation:

152 mg = 0.152 g

Let's apply the formula for osmotic pressure

π = M . R . T . i

where M is molarity ( mol / L)

R, the Universal Gases constant

T, aboslute T°

i, Van't Hoff factor. As the solution does not conduct electricity, i = 1

0.328 atm = M . 0.082 L.atm/mol .K . 300K

(0.328 atm / 0.082 L.atm/mol .K . 300K) = n

0.0133 M = Mol/L

If I have 0.0133 moles in 1000 mL

in 75mL, I will have 0.0133 mol/L . 0.075L = 9.97x10⁻⁴ moles

Molar mass of compound = 0.152 g / 9.97x10⁻⁴ moles = 152.38 g/m

Let's apply the percents:

100 g of compound have 78.9 g C ___ 10.59g H ____ 10.51 O

152.38 g of compound have:

(152.38 g . 78.9g)/100 g = 120 g C→ 10 moles

(152.38 g . 10.59g)/100 g = 16 g H → 16 moles

(152.38 g . 10.51g)/100 g = 16 g O → 1 mol

Molecular formula is: C₁₀H₁₆O