Question

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?

Answer 1

`"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`