Combustion of 27.44 g of a compound containing only carbon, hydrogen, and oxygen produces 53.60 gCO2 and 27.43 gH2O. What is the empirical formula of the compound?

I know that the answer will be some variation of C_a H_b O_c.
I have determined the mols of O (0.7894), H (0.1808), and C (1.218) in the products through calculation but I do not know what to do with them now.

1 Answer
May 14, 2018

Well, let us see...I make in #C_2H_5O#...

Explanation:

We got #27.44*g# of an organic compound...#C_nH_mO_l#...

Combustion yields #53.60*g# #CO_2#...a molar quantity of ...

#(53.60*g)/(44.01*g*mol^-1)=1.218*mol#...and so there were #1.218*molxx12.011*g*mol^-1=14.63*g# WITH RESPECT TO CARBON in the initial sample....

#(27.43*g)/(18.01*g*mol^-1)=1.523*mol#...and so there were #3.046*mol# with respect to hydrogen, a mass of #3.07*g# WITH RESPECT TO HYDROGEN in the initial sample....why did I double the molar quantity of water?

What mass is unaccounted? That of oxygen in the initial sample...there are few ways to analyze for oxygen content... On the other hand, carbon, as carbon dioxide, and hydrogen, as WATER, may be routinely and accurately analyzed on a gas chromatograph; and these are the unspoken premises of the question.

#"Mass of oxygen"=(27.44-14.63-3.07)*g=9.740*g#...a molar quantity of #0.6087*mol#...

And so we get a trial empirical formula of #C_(1.218)H_(3.046)O_(0.6087)#. CLEARLY we gots to normalize this...and we divide thru by the LEAST molar quantity, that of oxygen to give the #"empirical formula"# of ...

#C_((1.218)/(0.6087))H_((3.046)/(0.6087))O_((0.6087)/(0.6087))# #-=# #C_2H_5O#

This is probably a first year problem....but all I have done is to assume COMPLETE combustion of a hydrocarbon...i.e.

#C_nH_mO_y + (n+m/2-y/2)O_2 rarr nCO_2(g)+m/2H_2O#

The rest is arithmetic....which is quite laborious...and I think you get this....anyway if you are unclear as to the steps ask for qualification and someone will help.