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

1 Answer
anor277
Sep 14, 2017

C_2H_5OC2H5O.......Are you sure you quoted the elemental compositon properly?

Explanation:

We assume a 100*g100⋅g mass of compound....

And thus there are (53.31*g)/(12.011*g*mol^-1)=4.44*mol53.31⋅g12.011⋅g⋅mol−1=4.44⋅mol with respect to CC....

And (11.18*g)/(1.00794*g*mol^-1)=11.09*mol11.18⋅g1.00794⋅g⋅mol−1=11.09⋅mol with respect to HH....

And (35.51*g)/(16.00*g*mol^-1)=2.22*mol35.51⋅g16.00⋅g⋅mol−1=2.22⋅mol with respect to OO....

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of C_2H_5OC2H5O.

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_2C4H10O2. The question has not been drawn from actual data, and the values have been interpolated.

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