# Chemistry

posted by James

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% Carbon and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20 celsius is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g of water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a non electrolyte.

1. DrBob222

C = 31.57%
H = 5.30%
Therefore, O = 100%-31.57-5.30 = about 63%.

Take a 100g sample which will give you
31.57g C atoms.
5.30g H atoms.
63g O atoms.

Convert to mols and I'll estimate.
31.57/12 = about 2.63 mols C
5.30/1 = 5.30 mols H
63/16 = about 3.94 mols O

Now find the ratio to each other with the smallest being 1.0. The easy way to do that is to divide the smallest number by itself and follow suit with the other numbers.
2.63/2.63 = 1.0 mols C
5.30/2.63 = 2.01 mols H
3.94/2.63 = 1.49 mols O

Normally one would round these off to the nearest whole number but the O is too much to round. We can see that is about 1.5 so the ratio is 2C, 4H, 3O and the empirical formula is C2H4O3 (I just multilied the 1, 2, and 1.5 by 2 to get whole numbers). That makes the mass of the empirical formula about 76.

For the molar mass we use the freezing point data.
delta T = Kf*m
5.20 = 1.86*m
Solve for m = molality

Then m = mols/kg solvent. You know m and kg solvent (0.025g), solve for mols.

Then mols = grams/molar mass. You know mols and grams, solve for molar mass
I approx about 150 for molar mass.

empirical formula mass x ? = 150
76 x ? = 150 so ? (a whole number) = 150/76 = about 1.97 which rounds to 2.0 as a whole number so the molecular formula is
(C2H4O3)2 or C4H8O6 with a molar mass of 152. The 150 determined by the freezing point method is only CLOSE to the right answer and it allows you to find that whole number of 2 which then gives you the correct molecular formula and correct molar mass.

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