# The formula for the illegal drug cocaine is C_17H_21NO_4, which is 303.39 g/mol. What is the percentage of carbon in the compound?

$\text{%Carbon}$ $=$ $\text{Mass of carbon"/"Mass of molecule}$ xx100%
$=$ $\frac{17 \times 12.011 \cdot g \cdot m o {l}^{-} 1}{303.39 \cdot g \cdot m o {l}^{-} 1}$ xx100% $=$ ??%