Question

Name: iron(III) carbonate (ferric carbonate).

Calculation:
Molar mass Fe2(CO3)3 ≈ 2(55.85) + 3[12.01 + 3(16.00)] = 291.7 g·mol−1.
Mass Fe per mole = 2(55.85) = 111.7 g.
Mass percent Fe = 111.7 / 291.7 = 0.3829 = 38.29%.
Fe recovered from 115.0 g = 0.3829 × 115.0 ≈ 44.0 g Fe.

Answer 1

Your calculation is correct. Using more precise atomic masses:

- M(Fe2(CO3)3) = 2(55.845) + 3[12.011 + 3(15.999)] = 291.714 g·mol−1
- Mass Fe per mole = 2(55.845) = 111.690 g
- Mass % Fe = 111.690 / 291.714 = 0.38278 = 38.278%

Fe recovered from 115.0 g = 0.38278 × 115.0 = 43.988 g ≈ 44.0 g Fe (rounded).