# What is the mass in grams of 2.10 * 10^21 atoms of copper, Cu?

There are ${N}_{A}$ individual copper atoms in a moles of copper metal, and this has a mass of $63.55$ $g$.
So we know that $\text{Avogadro's number} \left({N}_{A}\right)$ of individual copper atoms have a mass of $63.55$ $g$.
Thus $\left(2.10 \times {10}^{21} \text{copper atoms"xx63.55*g*mol^-1)/(6.022140857(74)×10^23 "copper atoms } m o {l}^{-} 1\right)$ $=$ ??g