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

Feb 23, 2016

Check what's given first
$F {e}_{2} {\left(C {O}_{3}\right)}_{3}$
Number of molecules
Alright so firstly, we need to calculate the number of moles in $5.28 \times {10}^{23}$ 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
$5.28 \times {10}^{23} / 6.02 \times {10}^{23}$
$0.87$
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 $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$
The molar mass of Fe - $2 \left(55.845\right)$
The molar mass of C - $3 \left(12.01\right)$
The molar mass of O- $9 \left(16\right)$
Add em up all together and you have $291.72 g$ of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$
Thus, rearranging the equation of Moles = Mass/ Molar Mass to find Mass, we have
Mass = Moles x Molar Mass
Mass =$0.87 \times 291.72$
Mass = $253.8$

So $5.28 \times {10}^{23}$ molecules of Iron(III) Carbonate is $253.8 g$

Hope this helps