What is the mass in grams of 3.7 x 1019 particles of iron(III) carbonate?

Question

This is really confusing to me.

Answer 1

Find molar mass of iron(III) carbonate; Fe₂(CO₃)₃

molar mass = 291.7 g/mol

Find the number of moles of Fe₂(CO₃)₃ in 3.7x10¹⁹ particles

3.7x10¹⁹ particles x 1 mol / 6.02x10²³ particles = 6.15x10^-5 moles

Convert moles to gram using the molar mass

6.15x10^-5 moles x 291.7 g/mol = 0.01793 g = 0.018 grams (2 sig. figs.)