A solution contains 11.10 g of unknown compound (non-electrolyte) dissolved in 5

Question

A solution contains 11.10 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.49 C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O.

What is the molecular formula of the compound?

A solution contains 11.10 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.49 C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O.

What is the molecular formula of the compound?

Explanation / Answer

I am using freezing point data to calculate the molar mass

delta t = m x Kc x 1 (assume a molecular compound with 1 particle in solution).

m = 3.49 / 1.86 = 1.87 moles of solute / kg of water

we also have 11.10 g of the compound in 50 g of water or 11.10 x 1000 / 50 g / kg of water.

====> 222 g/kg of water.

so in 1 liter we have 1.87 moles = 222 g

mass of 1 mole = 222 / 1.87 = 118.7g / mol

in 100 g of compound we have 60.98 / 12 moles of C = 5.1 moles

moles of H = 11.94 / 1 = 11.94 mols of H

mass of Oxygen in 100g = 100-(60.98+11.94)

                                    = 27 g

                                    = 27/16 = 1.70

molar ratio of C:H:O = 5.1: 11.9:1.7

molecular formula would be C3H7O (mass= 59g)

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