Answer
$BeSO_{4}$
Work Step by Step
Convert the mass of each constituent element into moles by dividing by its molar mass:
$2.128g\ Be\times\frac{1mol}{9.0122g}=0.2361mol\ Be$
$7.557g\ S\times\frac{1mol}{32.06g}=0.2357mol\ S$
$15.107\ O\times\frac{1mol}{16.00g}=0.944mol\ O$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close. Then multiply by whatever factor makes them all a whole number or very close to a whole number.
$\frac{0.2361mol\ Be}{0.2357}=1$
$\frac{0.2357mol\ S}{0.2357}=1$
$\frac{0.944mol\ O}{0.2357}=4$
So the empirical formula for this compound is:
$BeSO_{4}$
