0.0800 g of a compound with the empirical formula C₂H₂N is dissolved in 10.00 g benzene. The resulting solution freezes at 4.99°C. Benzene freezes at 5.48°C and has a freezing point depression constant of Kf = 4.90°C/m. What is the molecular formula of the compound? A) C₃H₂N B) C₃H₃N C) C₄H₄N₂ D) C₆H₆N₃ E) C₈H₈N₄

Question

Quizplus · Accepted Answer

1. Calculate the molality of the solution:

ΔTf = Kf · molality
molality = (ΔTf) / Kf = (5.48 - 4.99) / 4.90 = 0.10 m

2. Calculate the moles of solute in the solution:

moles solute = molality x mass solvent = 0.10 mol/kg x 0.010 kg = 0.001 mol

3. Calculate the empirical formula molar mass:

molecular mass = 1 x Atomic mass C + 1 x Atomic mass S + 12 x Atomic mass B + 1 x Atomic mass H + 12 x Atomic mass S + 1 x Atomic mass N
molecular mass = 1 x 12.01 + 1 x 32.06 + 12 x 10.81 + 1 x 1.01 + 12 x 32.06 + 1 x 14.01 = 464.83 g/mol

4. Calculate the molecular formula:

We have to find a multiple (n) that makes the empirical formula to become the molecular formula.

n x empirical formula mass = molecular formula mass

n x (1 x Atomic mass C + 1 x Atomic mass S + 12 x Atomic mass B + 1 x Atomic mass H + 12 x Atomic mass S + 1 x Atomic mass N) = molecular mass

n x (12.01 + 32.06 + 12 x 10.81 + 1.01 + 12 x 32.06 + 14.01) = 464.83 g/mol

n x 382.27 g/mol = 464.83 g/mol

n = 1.22

The molecular formula is therefore:

CS1.22B14.64H1.22S14.64N1.22

Which can be simplified to:

CS1B12H1S1N

(Note that the subscripts are multiplied by 1 to get the whole-number ratio, and rounded to the nearest whole number to avoid fractions in the formula)

Therefore, the correct choice is C.

00800 G of a Compound with the Empirical Formula C2H2N

00800 G of a Compound with the Empirical Formula C₂H₂N