Combustion analysis of a 12.01 g sample of tartaric acts - which contains only carbon, hydrogen, and oxygen - produced 14.08 g CO2 and 4.32 g H2O. What is the empirical formula for tartaric acid?

1 Answer
Jun 9, 2018

Answer:

#"Empirical formula of tartaric acid"-=C_2H_3O_3#

Explanation:

We address the molar quantities of carbon, and hydrogen .... convert these into masses, and then address the empirical formula.

#"Moles of carbon dioxide"=(14.01*g)/(44.01*g*mol^-1)=0.3183*mol#

And thus mass of carbon...#-=0.3183*molxx12.011*g*mol^-1=3.824*g#

#"Moles of water"=(4.32*g)/(18.01*g*mol^-1)=0.2399*mol#

And thus mass of hydrogen...#-=2xx0.2399*molxx1.00794*g*mol^-1=0.4835*g#

The balance of the mass....#12.01*g-0.4835*g-3.824*g=7.702*g#...represents the mass of oxygen....a molar quantity of #(7.702*g)/(16.00*g*mol^-1)=0.4814*mol#

And so we divide thru by the LEAST molar quantity, that of OXYGEN, to get a trial empirical formula of...

#C_((0.3183*mol)/(0.2399*mol))H_((0.4814*mol)/(0.2399*mol))O_((0.4814*mol)/(0.2399*mol))# #-=# #C_(1.33)H_(2.00)O_(2)#..

But by definition, the #"empirical formula"# #"IS THE SIMPLEST WHOLE NUMBER ratio..."# and so we multiply the calculated empirical formula by a factor of THREE...

#3xx{C_(1.33)H_(2.00)O_(2)}-=C_4H_6O_6-=C_2H_3O_3#