Question

A chemist is studying small organic compounds for their potential use as an antifreeze. When 0.243 g of a compound is dissolved in 25.0 mL of water, the freezing point of the solution is - 0.201°C.
(b) Analysis shows that the compound is 53.31 mass % C and 11.18 mass % H, the remainder being O. Calculate the empirical and molecular formulas of the compound

Answer 1

Molar mass = 144.2 g/mol.

Calculate Molar mass.

Formula to solve this: ΔT = Kf . m

ΔT → Freezing point of pure solvent - Freezing point of solution

0°C - (-1.94°C) = 1.86 °C/m . m

1.94°C / 1.86 m/°C = m →  1.04 m

1.04 are the moles of solute in 1kg of solvent → molality (mol/kg)

Let's convert the mass of our solvent in kg to determine the moles.

80 g . 1kg / 1000 g = 0.08 kg

Molality . kg of solvent → Solute moles

1.04 mol/kg . 0.080 kg = 0.0832 moles

These are the moles of 12 g of solute. To find the molar mass → g/mol

12 g / 0.0832 mol = 144.2 g/mol

To learn more about Molar mass from the given link:

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