As with all these types of problems it is useful to assume #100*g# of compound, and address the empirical formula on this basis.
And so #"moles of potassium"=(26.56*g)/(39.10*g*mol^-1)=0.679*mol#
And #"moles of chromium"=(35.43*g)/(52.00*g*mol^-1)=0.681*mol#
And #"moles of oxygen"=(38.01*g)/(16.00*g*mol^-1)=2.38*mol#.
And now we divide thru by the LOWEST molar quantity to get an empirical formula of....
#K_((0.679*mol)/(0.679*mol))Cr_((0.681*mol)/(0.679*mol))O_((2.38*mol)/(0.679*mol))-=KCrO_(3.50)#...but because we desire WHOLE numbers we double this trial formula to get #K_2Cr_2O_7#, i.e. #"potassium dichromate"#, a deep orange salt that is a common oxidant that is reduced to #Cr^(3+)#....
Now that when we send something for analysis, typically we get #%C#, #%H#, #%N#...an analyst will rarely give you #%O# because it is a bit difficult to measure. The difference between the quoted percentages, and 100% is often assumed to be #%O#, as indeed it is assumed to be here.
