How would you determine the empirical formula of a compound found to contain 63.50% silver, 8.25% nitrogen, and the remainder oxygen.?

Percentages of Ag, N, and O:
#"Ag":##63.50%#
#"N":##8.25%#
#"O":##28.25%#

Since the percentages add up to 100%, we can assume a 100 g sample, and we can rewrite the percentages as mass in grams.

#"Ag":##"63.50 g"#
#"N":##"8.25 g"#
#"O":##"28.25 g"#

We need to determine the number of moles of each element using each element's molar mass, which is the #color(red)"atomic weight (relative atomic mass) on the periodic table in grams/mole (g/mol)"#. The molar mass is the mass of one mole of the element.

#63.50cancel"g Ag"xx(color(red)(1"mol Ag"))/(color(red)(107.8682cancel"g Ag"))="0.5887 mol Ag"#

#8.25cancel("g N")xx(color(red)(1"mol N"))/color(red)(14.007cancel"g N")="0.589 mol N"#

#28.25cancel"g O"xx(color(red)(1"mol O"))/(color(red)(15.999cancel"g O"))="1.766 mol O"#

Determine the mole ratios by dividing the number of moles of each element by the least number of moles.

#"Ag":##(0.5887"mol Ag")/(0.589"mol")=0.999~~1 "mol Ag"#

#"N":##(0.589"mol N")/(0.589"mol")="1.00 mol N"#

#"O":##(1.766"mol O")/(0.589"mol")="3.00 mol O"#

The empirical formula is #"AgNO"_3"#. This compound is silver nitrate.