The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of "Ag"Ag to "O"O.
"Mass of Ag = 7.96 g"Mass of Ag = 7.96 g
"Mass of silver oxide = mass of Ag + mass of O"Mass of silver oxide = mass of Ag + mass of O
"8.55 g = 7.96 g + mass of O"8.55 g = 7.96 g + mass of O
"Mass of O = (8.55 – 7.96) g = 0.59 g"Mass of O = (8.55 – 7.96) g = 0.59 g
"Moles of Ag" = 7.96 color(red)(cancel(color(black)("g Ag"))) × "1 mol Ag"/(107.9color(red)(cancel(color(black)( "g Ag")))) = "0.073 77 mol Ag"Moles of Ag=7.96g Ag×1 mol Ag107.9g Ag=0.073 77 mol Ag
"Moles of O "= 0.59 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.0369 mol O"Moles of O =0.59g O×1 mol O16.00g O=0.0369 mol O
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
"Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mll)"Integers"ElementAgMass/gXMolesXllRatiomllIntegers
stackrel(—————————————————-———)(color(white)(m)"Ag" color(white)(XXXm)7.96 color(white)(Xm)0.073 77
color(white)(Xll)2.00color(white)(mmm)2)
color(white)(ml)"O" color(white)(XXXXl)0.59 color(white)(mm)"0.0369 color(white)(Xml)1 color(white)(mmmml)1
There are 2 mol of "Ag" for 1 mol of "O".
The empirical formula of silver oxide is "Ag"_2"O".
Here is a video that illustrates how to determine an empirical formula.
