Microanalysis of an organic compound containing carbon, hydrogen, and oxygen, gives #62.6%*C#, and #4.21%*H#. What is the empirical formula of the compound?

1 Answer
Sep 7, 2017

Answer:

#C_5H_4O_2# is the #"empirical formula....."#

Explanation:

We assume a #100*g# mass of compound, and thus....

#"Moles of carbon"=(62.6*g)/(12.011*g*mol^-1)=5.21*mol*C#

#"Moles of hydrogen"=(4.21*g)/(1.00794*g*mol^-1)=4.16*mol*H#

#"Moles of oxygen"=(33.3*g)/(15.999*g*mol^-1)=2.08*mol*H#

In each case, we have divided the mass of element by the ATOMIC mass of each element to give an elemental ratio.... How did I know that the compound was #33.3%# by mass with respect to oxygen?

And now we divide thru by the LOWEST molar quantity, that of oxygen to get a trial empirical formula of #C_(2.50)H_2O#. But by definition, the empirical formula is the lowest WHOLE number ratio defining constituent atoms in a species...so we gots #C_5H_4O_2#.

We cannot access the molecular formula inasmuch as we need a measurement of molecular mass. See here.

#"Molecular formula"="empirical formula"xxn#....