Number of o atoms in 7.3 x10^-3 g of caso4 x 2h2o

The number of O atoms in 7.3 x10 ^-3 g of CaSO4 x 2H2O  are = 1.535 x10^20  atoms
calculation

step 1: calculate the moles of CaSO4 . 2H2O

moles = mass/molar mass molar mass is of CaSO4.H2O=Mw of Ca + S +O +H2O= 40+32 +( 4x16)  +2( (1x2)+16)=  172 g/mol
moles is therefore= 7.3 x10^-3 g/172 g/mol=   4.244  x10^-5 moles find the moles of Osince there are 6 atoms of O the moles of O=   6 x ( 4.244 x10^-5)= 0.000255 moles
by use  of Avogadro's law constant find the number of O atomthat is  1 mole     =       6.02 x10 ^23 atoms

                      0.000255 moles=? atoms

by cross multiplication  = 0.000255 moles  x(6.02 x10^23 atoms)/ 1 mole=  1.535  x10^20 atoms

Inot sure what it is i sorry : ((