Answer
$C_{6}H_{12}O_{6}$
Work Step by Step
Assume 100g of the compound. Convert the mass of each constituent element into moles by dividing by its molar mass:
$40.00g\ C\times\frac{1mol}{12.01g}=3.331mol\ C$
$6.72g\ H\times\frac{1mol}{1.01g}=6.65mol\ H$
$53.28g\ O\times\frac{1mol}{16.00g}=3.330mol\ O$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close.
$\frac{3.331mol\ C}{3.330}=1$
$\frac{6.65mol\ H}{3.330}=2$
$\frac{3.330mol\ O}{3.330}=1$
So the empirical formula for this compound is:
$CH_{2}O$
Divide the molar mass of the compound by the molar mass of the empirical formula then multiply the subscripts of the empirical formula by this number.
$180.16\frac{g}{mol}\div30.03\frac{g}{mol}=6$
So the molecular formula is $C_{6}H_{12}O_{6}$.
