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

Question

Avery Adams · Accepted Answer

The empirical formula is #"AgNO"_3"#.

Percentages of Ag, N, and O: #"Ag":##63.50%# #"N":##8.25%# #"O":##28.25%# We can assume a 100 g sample because the percentages add up to 100%, 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"# Divide the total number of moles of each element by the minimum number of moles to find the mole ratios. #"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.

Xavier Avery · Answer

undefined In order to get the mole ratio—which is the empirical formula for each element in the compound—you would first need to convert the percentages of each element to moles. Then, you would divide each mole value by the smallest mole value that was obtained.