Answer
$Ag_{2}O$
Work Step by Step
Convert the mass of each constituent element into moles by dividing by its molar mass:
$1.651g\ Ag\times\frac{1mol}{107.87g}=0.01530mol\ Ag$
$0.1224g\ O\times\frac{1mol}{16.00g}=0.007630mol\ O$
Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close.
$\frac{0.01530mol\ Ag}{0.007630}=2$
$\frac{0.007630mol\ O}{0.007630}=1$
So the empirical formula for this compound is:
$Ag_{2}O$
