A volatilization gravimetry experiment was conducted on an unknown, but nominall

Question

A volatilization gravimetry experiment was conducted on an unknown, but nominally organic liquid (presumed to only contain carbon, hydrogen, and oxygen). The liquid was massed and then combusted in an excess of oxygen. The exhaust gas was passed through a trap containing drierite and then through a trap containing ascarite. The starting and ending masses of the drierite and ascarite were used to obtain the data below.

Mass Unknown Mass CO2 Mass H2O
12.4370 17.0822 13.9849
12.5590 17.2497 14.1221
12.2230 16.7882 13.7443

A.) Calculate the mass percent of carbon, hydrogen, and oxygen in the unknown liquid.

B.) Calculate the empirical formula of the unknown compound.

C.) (extra credit for the organically inclined) What is the unknown liquid (most
probably)? Why does this seem to be the only reasonable choice?

Explanation / Answer

A)

CxHyOz + (x+y/2-z/2)O2 --> xCO2 + y/2H2O

Mass Unknown Mass CO2 Mass H2O

12.4370 17.0822 13.9849

Moles of CO2 =17.0822/44=0.388

Moles of H2O =13.9849/18=0.777

Moles of C atom =0.388

Moles of H atom = 2*0.777=1.554

Total mass=17.0822+ 13.9849=31.0671

Mass % of C = 0.388*12/31.0671*100=15

Mass % of H = 1.554/31.0671*100=5

Mass % of O = 100-(15+5)=80

B)

Empirical formula

Moles of O atom = 0.8*31.0671/16= 1.554 moles

Moles of C atom =0.388

Moles of H atom = 2*0.777=1.554

Least mole = 0.388

Mole ratio

C =1

H 1.554/.388=4

O 1.554/.388= 4

Empirical formula CH4O4

C)

Orthocarbonic acid C(OH)4

(CH4O4)n only possible is for n=1

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