Answer
$C_{18}H_{24}O_{2}$
Work Step by Step
Assume 100g of the compound. Convert the mass of each constituent element into moles by dividing by its molar mass:
$79.37g\ C\times\frac{1mol}{12.01g}=6.609mol\ C$
$8.88g\ H\times\frac{1mol}{1.01g}=8.79mol\ H$
$11.75g\ O\times\frac{1mol}{16.00g}=0.7344mol\ O$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close.
$\frac{6.609mol\ C}{0.7344}=9$
$\frac{8.79mol\ H}{0.7344}=12$
$\frac{0.7344mol\ O}{0.7344}=1$
So the empirical formula for this compound is:
$C_{9}H_{12}O$
Divide the molar mass of the compound by the molar mass of the empirical formula and then multiply the subscripts of the empirical formula by this number.
$272.37\frac{g}{mol}\div136.12\frac{g}{mol}= 2$
So the molecular formula is $C_{18}H_{24}O_{2}$.
