# What is the mass of 6.02 * 10^23 atoms of calcium?

In $1$ $m o l$ of $C a$ $\text{atoms}$, there are ${N}_{A}$ individual calcium atoms.
How does this help you? Well, it happens that ${N}_{A}$, $\text{Avogadro's Number}$ $\cong$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
And it is stated on your trusty Periodic Table that the molar mass of $C a$ is $40.08 \cdot g \cdot m o {l}^{-} 1$. So, if we specify a molar quantity, we specify a precise number of atoms and molecules that can be approximated by mass.