Calculation of empirical formula:
`{:("Element","Percentage","At.mass","Relative number of atoms","Simplest ratio"),("Carbon",40.0,12,(40)/(12)=3.33,(3.33)/(3.33)=1),("Hydrogen",6.66,1,(6.66)/(1)=6.66,(6.66)/(3.33)=2),(,53.34,,,),("Oxygen",("by difference"),16,(53.34)/(16)=3.33,(3.33)/(3.33)=1):}`
Empirical formula `= CH_(2)O`
Empirical formula mass `= (12 +2 +16) = 30`
Mol. mass `= 2xx V.D. = 2 xx 30`
`= 60`
`n = ("Mol. mass")/("Emp. mass") = (60)/(30) = 2`
Molecular formula `= 2xx` (Empirical formula)
`= 2xx (CH_(2)O) = C_(2)H_(4)O_(2)`
Alternatively
Number of moles of carbon
`= (% "of carbon")/(100) xx ("Molecular mass")/("Atomic mass of element")`
`= (40)/(100) xx (60)/(12) = 2`
Number of moles of hydrogen
`= (%"of hydrogen")/(100) xx ("Molecular mass")/("Atomic mass of hydrogen")`
`= (6.66)/(100) xx (60)/(1) = 4`
Number of moles of oxygen
`= (%"of oxygen")/(100) = ("Molecular mass")/("Atomic mass of oxygen")`
`= (53.34)/(100) xx(60)/(16) = 2`
Molecular formula `= C_(2)H_(4)O_(2)`
