What is the percent yield of oxygen in a sample of 115g of oxygen produced by heating 400g of potassium chlorate

The %  yield   of oxygen in  a sample  of 115g of oxygen  produced by  heating  400g  of  potassium  chlorate  is  73.37%

         calculation

%   yield   = actual  yield/ theoretical  yield  x100

the   actual  yield =  115 g

find  the  theoretical  yield  by
  
  first write the equation    for decomposition  of  KClO3
that  is  2KClO3  =  2KCl  +3 O2

 then find the moles  of  KClO3 = mass/molar  mass

=400g/122.5 g/mol = 3.265  moles
by use of mole  ratio  between  KClO3 to O2   which is   2:3 the moles of  O2=  3.265  x3/2 =4.898  moles

the  theoretical mass  of O2  =moles   of O2  x molar mass   oF  O2

=4.898   x 32 g/mol = 156.74   grams
%    yield  is therefore  =115/ 156.74 x100 = 73.37%
  

They are held together by chemical bonds. i am pretty sure.

What do you mean by that? like the contents? the thc level?