Potassium chlorate, #"KClO"_3"#, is an ionic compound. The particles of an ionic compound are called formula units. There are #6.022xx10^23# formula units in one mole of an ionic substance.
Start with a balanced equation:
#"2KClO"_3##rarr##"2KCl + 3O"_2#
Determine the molar mass, #M#, of #"O"_2"# .
#M"O"_2:##(2xx"15.999 g/mol O"_2)="31.998 g/mol O"_2"#
Determine mol #"O"_2"#
Multiply given mass of #"O"_2"# by its molar mass so that mol is in the numerator and mass is in the denominator, #"mol"/"grams"#.
#8.14color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="0.25439 mol O"_2"#
Determine mol #"KClO"_3"#
Multiply mol #"O"_2# by the mol ratio from the balanced equation,
#"2 mol KClO"_4/("3 mol O"_2")# with #"O"_2"# in the denominator so that it cancels.
#0.25439color(red)cancel(color(black)("mol O"_2))xx(2"mol KClO"_3)/(3color(red)cancel(color(black)("mol O"_2)))="0.16959 mol KClO"_3"#
Determine the number of formula units of #"KClO"_3"#.
Multiply mol #"KClO"_3"# by #6.022xx10^23# formula units/mol.
#0.16959color(red)cancel(color(black)("mol KClO"_3))xx(6.022xx10^23color(white)(.)"formula units KClO"_3)/(1color(red)cancel(color(black)("mol KClO"_3")))=1.02xx10^23##"formula units KClO"_3"# (rounded to three significant figures)
Note: I kept some extra digits during the process to reduce rounding errors, but the final answer was rounded to three significant figures due to #"8.14 g O"_2"#.
