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.

Molecular formula is: C₁₀H₁₆O

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

the awnser is group 12

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o