As with all these problems, we assume a #100*g# mass, and interrogate the molar quantities in order to assess the empirical formula:
#"Moles of carbon"=(74.97*g)/(12.011*g*mol^-1)=6.24*mol#
#"Moles of hydrogen"=(7.4*g)/(1.00794*g*mol^-1)=7.34*mol#
#"Moles of oxygen"=(17.6*g)/(16.00*g*mol^-1)=1.10*mol#
Note that normally, you would NOT be given #%O#...oxygen is a difficult element to analyze... We divide thru each molar quantity by the LOWEST molar quantity...i.e. divide thru by moles of oxygen to get a trial empirical formula of...
#C_((6.24*mol)/(1.10*mol))H_((7.34*mol)/(1.10*mol))O_((1.10*mol)/(1.10*mol))-=C_(5.67)H_(6.67)O#...but we want integral numbers... and so we multiply by THREE...
#3xxC_(5.67)H_(6.67)O-=C_17H_20O_3#
Now the #"molecular formula"# is a #"WHOLE NUMBER MULTIPLE"# of the #"empirical formula"#...and clearly the multiple is one...i.e. the molecular mass of #C_17H_20O_3-=272.2*g*mol^-1#
