A research chemist isolates a new compound with an empirical formula C₃H₃O₂. Dissolving 2.51 g of the compound in 100. g of water produces a solution with a freezing point of -0.325^$\circ$C. What is the molecular formula of the compound (For water, K_f = 1.86^$\circ$C/m.) A) C₃H₃O₂ B) C₆H₆O₄ C) C₉H₉O₆ D) C₃H₁₂O₄ E) None of the above

Question

Quizplus · Accepted Answer

1) Calculate the change in freezing point:
ΔTf = Kf * m
m = molality = moles of solute / mass of solvent (in kg)
mass of solvent = 100 g / 1000 = 0.1 kg
molality = (2.51 g / molar mass) / 0.1 kg
molar mass of empirical formula = (1*12.01) + (18*1.01) + 13 + (12*32.07) + 14 + (19*79.9) + 16 = 2165.26 g/mol
molality = (2.51 g / 2165.26 g/mol) / 0.1 kg = 0.001157 mol/kg
ΔTf = (1.86°C/m) * (0.001157 mol/kg) = 0.00215202°C
2) Calculate the van't Hoff factor, i:
i = (ΔTf_calculated / ΔTf_expected)
ΔTf_expected = Kf * m * i
ΔTf_expected = (1.86°C/m) * (0.001157 mol/kg) * i = 0.00215202°C
i = ΔTf_calculated / ΔTf_expected = -0.325°C / 0.00215202°C = -150.72
The van't Hoff factor should be a whole number, indicating the number of particles formed by the solute in solution. A negative value indicates an error in the calculation or that the solute is not a strong electrolyte. 
3) Since all of the subscripts in the empirical formula are whole numbers and there are no fractions, we can assume that the compound does not dissociate into ions in solution, and that i = 1.
4) Calculate the molecular formula:
The empirical formula has a molar mass of 2165.26 g/mol.
The molecular formula must have a molar mass that is a whole number multiple of the empirical formula molar mass. 
Possible molecular formulas are CS16U16B4S4U4B0H4S16U16B0O4S48U48B0, CS8U8B2S2U2B0H2S8U8B0O2S24U24B0, CS4U4B1S1U1B0HS4U4B0OS4S12U12B0, CS2U2B0.5S0.5U0.5B0H0.5S2U2B0O0.5S6U6B0, and CS1U1B0.25S0.25U1B0H0.25S1U1B0O0.25S3U3B0.
The molecular mass of the first formula is 8661.04 g/mol, which is too large.
The molecular mass of the second formula is 4330.52 g/mol, which is also too large.
The molecular mass of the third formula is 2165.26 g/mol, which is the correct molar mass.
Therefore, the molecular formula of the compound is CS1U1B13S1U1B0HS1U1B13S1U1B0OS1U1B14S1U1B0.

A Research Chemist Isolates a New Compound with an Empirical ∘\circ∘

A Research Chemist Isolates a New Compound with an Empirical $\circ$