How do you convert #5.28xx10^23# formula units of iron (III) carbonate to grams?

Check what's given first
#Fe_2(CO_3)_3#
Number of molecules
Alright so firstly, we need to calculate the number of moles in #5.28 xx 10^23# molecules of Iron (III) Carbonate

To calculate the number of moles we will use the formula
Number of particles = number of moles x Avogadro's Contant
Thus, since we need the number of moles, rearrange this equation to-
Number of moles = number of particles / Avogadro's constant
Thus we have
#5.28 xx 10^23//6.02 xx 10^23#
#0.87#
Thus we have 0.87 moles of Iron(III) Carbonate

Now that we have the number of moles, we can find the mass in grams using the equation (Moles = Mass / Molar Mass)
So since we dont have the molar mass of Iron(III) Carbonate, lets find it out
Firstly, the equation is #Fe_2(CO_3)_3#
The molar mass of Fe - #2(55.845)#
The molar mass of C - #3(12.01)#
The molar mass of O- #9(16)#
Add em up all together and you have #291.72g# of #Fe_2(CO_3)_3#
Thus, rearranging the equation of Moles = Mass/ Molar Mass to find Mass, we have
Mass = Moles x Molar Mass
Mass =#0.87 xx 291.72#
Mass = #253.8#

So #5.28 xx 10^23# molecules of Iron(III) Carbonate is #253.8g#

Hope this helps