Answer
$C_{4}H_{5}N_{2}O$
Work Step by Step
Assume 100g of the compound. Convert the mass of each constituent element into moles by dividing by its molar mass:
$49.48g\ C\times\frac{1mol}{12.01g}=4.120mol\ C$
$5.19g\ H\times\frac{1mol}{1.01g}=5.13mol\ H$
$28.85g\ N\times\frac{1mol}{14.01g}=2.059mol\ N$
$16.48g\ O\times\frac{1mol}{16.00g}=1.030mol\ O$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close.
$\frac{4.120mol\ C}{1.030}=4$
$\frac{5.13mol\ H}{1.030}=5$
$\frac{2.059mol\ N}{1.030}=2$
$\frac{1.030mol\ O}{1.030}=1$
So the empirical formula for this compound is:
$C_{4}H_{5}N_{2}O$
