Step 1:
use:
Δ Tf = Kf*mb
26.4 = 40.0 *mb
mb= 0.66 molal
m(solvent)= 17.25 g
= 1.725*10^-2 kg
use:
number of mol,
n = Molality * mass of solvent in Kg
= (0.66 mol/Kg)*(1.725*10^-2 Kg)
= 1.138*10^-2 mol
mass(solute)= 3.236 g
use:
number of mol = mass / molar mass
1.138*10^-2 mol = (3.236 g)/molar mass
molar mass = 284.2 g/mol
Step 2:
let in compound number of moles of C, H and O be x, y and z
respectively
Number of moles of CO2 = mass of CO2 / molar mass CO2
= 0.01812/44
= 4.118*10^-4
Number of moles of H2O = mass of H2O / molar mass H2O
= 0.00198/18
= 1.1*10^-4
Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 4.118*10^-4
so, x = 4.118*10^-4
Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*1.1*10^-4 = 2.2*10^-4
Molar mass of O = 16 g/mol
mass O = total mass - mass of C and H
= 0.0078 - 4.118*10^-4*12 - 2.2*10^-4*1
= 2.638*10^-3
number of mol of O = mass of O / molar mass of O
= 2.638*10^-3/16.0
= 1.649*10^-4
so, z = 1.649*10^-4
Divide by smallest:
C: 4.118*10^-4/1.649*10^-4 = 5/2
H: 2.2*10^-4/1.649*10^-4 = 4/3
O: 1.649*10^-4/1.649*10^-4 = 1
Multiply by 6 to get simplest whole number ratio:
C: 5/2*6 = 15
H: 4/3*6 = 8
O: 1*6 = 6
So empirical formula is:C15H8O6
Molar mass of C15H8O6,
MM = 15*MM(C) + 8*MM(H) + 6*MM(O)
= 15*12.01 + 8*1.008 + 6*16.0
= 284.214 g/mol
Now we have:
Molar mass = 284.2 g/mol
Empirical formula mass = 284.214 g/mol
Multiplying factor = molar mass / empirical formula mass
= 284.2/284.214
= 1
So molecular formula is:C15H8O6
Answer: C15H8O6
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