A small cube of iron is observed under a microscope. The edge of the cube is $4.00 \times 10^{-6}$ cm long. The atomic mass of iron is 55.9 u, and its density is 7.86 $g/cm^{3}$.
a) Find the mass of the cube.
b) Find the number of iron atoms in the cube.
My work:
$d = \frac{m}{v} \implies m = d \times v$
$m = (6.4 \times 10^{-17} cm^3) \times (7.86 \frac{g}{cm^3})$
$m = 5.0304 \times 10^{-16}$ g
Is this correct? Also how do I find part b?
Your answer to part $(a)$ is correct. For part $(b)$, find the number of moles of iron available with you first:
$$\text{No. of moles}=\dfrac{\text {weight of substance in grams}}{\text{molecular mass of substance in amu}}$$
Hence in the no. of moles($n$) is: $$n=\dfrac{5.0304×10 ^{−16} \text{ g}}{55.9 \text { amu}}$$
$$\implies n\approx 9.0054 \times 10^{-18}$$
$$\implies \text {no. of iron atoms in the cube}=\text {no. of moles}\times \text{Avogadro's number}$$
$$\implies\text {no. of iron atoms in the cube}=n\times 6.023\times 10^{23}$$
$$\implies\text {no. of iron atoms in the cube}\approx 5.424 \times 10^6$$