Final answer:
A 3.07 g sample of Fe contains approximately 3.31 x 10^22 atoms of iron (Fe), calculated using the molar mass of Fe and Avogadro's number.
Step-by-step explanation:
To find out how many atoms of Fe are contained in a 3.07 g sample of Fe, we use the molar mass of Fe and Avogadro's number. The molar mass of Fe, or iron, is approximately 55.85 g/mol. To convert the mass of Fe to moles, we divide the mass of the Fe sample by its molar mass:
3.07 g Fe * (1 mol Fe / 55.85 g Fe) = 0.055 mol Fe
Next, we use Avogadro's number, which is 6.022 x 10^23 atoms/mol, to find the number of atoms:
0.055 mol Fe * (6.022 x 10^23 atoms/mol) = approximately 3.31 x 10^22 atoms of Fe
Therefore, a 3.07 g sample of Fe contains approximately 3.31 x 10^22 atoms of Fe.