Answer
$C_{5}H_{7}N$
Work Step by Step
Assume 100g of the compound. Convert the mass of each constituent element into moles by dividing by its molar mass:
$74.03g\ C\times\frac{1mol}{12.01g}=6.164mol\ C$
$8.70g\ H\times\frac{1mol}{1.01g}=8.61mol\ H$
$17.27g\ N\times\frac{1mol}{14.01g}=1.233mol\ N$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close.
$\frac{6.164mol\ C}{1.233}=5$
$\frac{8.61mol\ H}{1.233}=7$
$\frac{1.233mol\ N}{1.233}=1$
So the empirical formula for this compound is:
$C_{5}H_{7}N$
