But we must interrogate the #"empirical formula"#, and then use the quoted molecular mass to calculate the #"molecular formula..."#
And thus for the given mass of compound we gots....
#(0.590*g)/(1.00794*g*mol^-1)=0.585*mol# with respect to hydrogen....
and #(9.40*g)/(15.00*g*mol^-1)=0.585*mol# with respect to oxygen....
And thus we get the #"empirical formula,"# the simplest whole number ratio representing constituent atoms in a species of ...
#H_((0.585*mol)/(0.585*mol))O_((0.585*mol)/(0.585*mol))=HO#...
But it is a fact that the #"molecular formula"# is a simple whole number multiple of the #"empirical formula."#.
And thus in terms of the given #"molecular mass...."#
#34.00*g*mol^-1=nxx{1.00794+15.999}*g*mol^-1#...where of course #1.00794*g*mol^-1# and #15.999*g*mol^-1# are the respective ATOMIC masses of hydrogen and oxygen....
And so #n=(34.00*g*mol^-1)/({1.00794+15.999}*g*mol^-1)#
Clealry, #n=2#, and the #"molecular formula"-=H_2O_2#, #"hydrogen peroxide"# as we anticipated.....