A chemist using the Dumas method (see previous problem) finds that a pure volati

Question

A chemist using the Dumas method (see previous problem) finds that a pure volatile liquid with a boiling point of 10.8 degree C fills a 250.0-cm^3 bulb at an atmospheric pressure of 0.998 bar. If the weighed condensate has a mass of 635 milligrams, what is the molecular mass of the liquid? Chemical analysis shows that when 1.200 grams of the liquid is combusted, 2.637 grams of CO_2 (g) and 1.439 grams of H_2 O (l) are formed. Determine the molecular formula of the unknown liquid. Assume that the unknown liquid is composed of only the elements carbon, hydrogen, and oxygen.

Explanation / Answer

Duma' method for molar mass determination

moles of liquid = PV/RT

with,

P = 0.998 bar = 0.985 atm

V = 250 cm^3 = 0.250 L

R = gas constant

T = 273 + 10.8 = 283.8 K

we get,

moles of liquid = 0.985 x 0.250/0.08205 x 283.8

                        = 0.0106 mol

molar mass of liquid = 0.635 g/0.0106 mol = 60.05 g/mol

Combustion of liquid,

moles C = 2.637/44 = 0.06 mol

mass of C in sample = 0.06 x 12 = 0.72 g

moles H = 2 x 1.439/18 = 0.16 mol

mass of H in sample = 0.16 x 1 = 0.16 g

mass of O in sample = 1.2 - (0.72 + 0.16) = 0.32 g

moles of O = 0.32/16 = 0.02 mol

Divide all moles with 0.02,

C = 0.06/0.02 = 3

H = 0.16/0.02 = 8

O = 0.02/0.02 = 1

So,

molecular formula of unknown liquid = C3H8O

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