An organic sample has composition by mass, #C:53.31%;H:11.18%;#, and the balance was oxygen. What is its empirical formula?

1 Answer
Sep 14, 2017

Answer:

#C_2H_5O#.......Are you sure you quoted the elemental compositon properly?

Explanation:

We assume a #100*g# mass of compound....

And thus there are #(53.31*g)/(12.011*g*mol^-1)=4.44*mol# with respect to #C#....

And #(11.18*g)/(1.00794*g*mol^-1)=11.09*mol# with respect to #H#....

And #(35.51*g)/(16.00*g*mol^-1)=2.22*mol# with respect to #O#....

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of #C_2H_5O#.

This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is #C_4H_10O_2#. The question has not been drawn from actual data, and the values have been interpolated.