Question

Complete combustion of 3.500 g of a compound of carbon, hydrogen, and oxygen yielded 4.441 g CO₂ and 1.212 g H₂O. When 15.10 g of the compound was dissolved in 316 g of water, the freezing point of the solution was found to be -0.855 °C. For water, K_fp = 1.86 °C/m. What is the molecular formula of the compound?

Answer 1

step 1: find molar mass of compound

we have below equation to be used:
delta Tf = Kf*mb
0.855 = 1.86 *mb
mb= 0.4597 molal


mass of solvent = 316 g
= 0.316 kg     [using conversion 1 Kg = 1000 g]

we have below equation to be used:
number of mol,
n = Molality * mass of solvent in Kg
= (0.4597 mol/Kg)*(0.316 Kg)
= 0.1453 mol

mass of solute = 15.10 g


we have below equation to be used:
number of mol = mass / molar mass
0.1453 mol = (15.1 g)/molar mass
molar mass = 104 g/mol

step 2: find empirical formula and then molecular formula

Number of moles of CO2 = mass of CO2 / molar mass CO2
= 4.441/44
= 0.1009

Number of moles of H2O = mass of H2O / molar mass H2O
= 1.212/18
= 0.0673

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.1009

Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*0.0673 = 0.1347

Molar mass of O = 16 g/mol


mass O = total mass - mass of C and H
= 3.5 - 0.1009*12 - 0.1347*1
= 2.1542

number of mol of O = mass of O / molar mass of O
= 2.1542/16.0
= 0.1346
Divide by smallest :
C: 0.1009/0.1009 = 1
H: 0.1347/0.1009 = 4/3
O: 0.1346/0.1009 = 4/3

multiply by 3 to get simplest whole number ratio:
C : 1*3 = 3
H : 4/3*3 = 4
O : 4/3*3 = 4

So empirical formula is:C3H4O4

Molar mass of C3H4O4 = 3*MM(C) + 4*MM(H) + 4*MM(O)
= 3*12.01 + 4*1.008 + 4*16.0
= 104.062 g/mol

Now we have:
Molar mass = 104.0 g/mol
Empirical formula mass = 104.062 g/mol
Multiplying factor = molar mass / empirical formula mass
= 104.0/104.062
= 1

So molecular formula is:C₃H₄O₄
Answer: molecular formula is:C₃H₄O₄

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know