And a #"chlorite"# is an ionic chloride oxide with a #+III# oxidation state with respect to chlorine, so we have the salt formed from #Mg^(2+)# and #ClO_2^(-)#.
We can write a Lewis description of chlorite as #""^(-)O-Cl=O#; around each atom from the LEFT, there are 3, 2, and 2 lone pairs of electrons, hence the charge distribution of #-1#, #0#, and #0#.
And so we got #Mg(ClO_2)_2#, and ONE mole of this stuff contains 1 mole each of magnesium, 2 moles of chlorine, and FOUR moles of oxygen....
We will be requiring #8.47xx10^23*"oxygen atoms"#, and so we take the quotient....
#(8.47xx10^23*"oxygen atoms")/(4xx6.022xx10^23*"oxygen atoms"*mol^-1)=3.52xx10^-4*mol#, and this is in respect to moles of #"magnesium chlorite..."#
This is not easy, (at least I don't think it is) so please review my reasoning, and calculation. Good luck.
Let's check the calculation....
WE have a mass of #3.52xx10^-4*molxx159.2*g*mol^-1=0.0560*g# with respect to the salt.....
....and with respect to oxygen there is #3.52xx10^-4*molxx4*O*"atoms"xx16*g*mol^-1=0.023*g#
....and with respect to magnesium metal there is #3.52xx10^-4*molxx1*Mg*"atoms"xx24.31*g*mol^-1=8.56xx10^-3*g#.
....and with respect to chlorine there is #3.52xx10^-4*molxx2*Cl*"atoms"xx35.45*g*mol^-1=2.50xx10^-2*g#.
We add the masses up to get....#(0.023+8.56xx10^-3+2.50xx10^-2)*g=0.0560*g#
#=(0.0565*g)/(159.2*g*mol^-1)=3.55xx10^-4*mol#...
