Hemoglobin contains $$0.34\%$$ of iron by mass. The number of Fe atoms in $$3.3 \text{ g}$$ of hemoglobin is (Given: Atomic mass of Fe is $$56u$$, $$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$$)

Since hemoglobin contains $$0.34\%$$ iron by mass and the sample has a mass of $$3.3 \text{ g}$$, with the atomic mass of Fe being $$56 \text{ u}$$ and $$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$$, we first calculate the mass of iron present.

Substituting the given percentage into the formula $$m_{Fe} = \frac{0.34}{100} \times 3.3$$ gives $$m_{Fe} = 0.01122 \text{ g}$$.

Next, we determine the number of moles of iron by using $$n_{Fe} = \frac{m_{Fe}}{M_{Fe}} = \frac{0.01122}{56}$$, which yields $$n_{Fe} = 2.004 \times 10^{-4} \text{ mol}$$.

Finally, multiplying the moles of iron by Avogadro’s number via $$N = n_{Fe} \times N_A = 2.004 \times 10^{-4} \times 6.022 \times 10^{23}$$ results in $$N = 1.21 \times 10^{20}$$ atoms.

Therefore, the number of Fe atoms in 3.3 g of hemoglobin is $$1.21 \times 10^{20}$$, which corresponds to Option C.