Question #33b75

1 Answer
Jul 24, 2017

Answer:

We gets a molecular formula......of #C_4H_8O_2#.....

Explanation:

AS for all these sorts of problems, we assume a #100*g# mass of unknown compound, and we access the empirical formula......

#"Moles of C"=(54.5*g)/(12.011*g*mol^-1)=4.54*mol.#

#"Moles of H"=(9.3*g)/(1.00794*g*mol^-1)=9.23*mol.#

#"Moles of O"=(36.2*g)/(15.999*g*mol^-1)=2.26*mol.#

Note that normally #%O# would never be measured OR QUOTED. For a #C_nH_mO# compound, you would assume that the missing percentage i.e. #100%-%C-%H=%O#.

And now we divide thru by the smallest molar quantity, that of oxygen, to get an empirical formula of......

#C:(4.54*mol)/(2.26*mol)=2.00#

#H:(9.23*mol)/(2.26*mol)=4.08#

#O:(2.26*mol)/(2.26*mol)=1.00#

......to get an empirical formula, the simplest whole number ratio defining constituent atoms in a species, of......

#C_2H_4O#.

Now the molecular formula is a whole number mulitple of the empirical formula. And thus........

#"(empirical formula)"xxn="molecular formula"#....but we were quoted a molecular mass, and so.....

#nxx(2xx12.011+4xx1.00794+1xx16.00)*g*mol^-1=88*g*mol^-1.#

Clearly, #n=2#, and so our molecular formula is.....

#2xx{C_2H_4O}=C_4H_8O_2#.....