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