# What is the mass of 6xx10^22 "zinc atoms"?

Jun 8, 2017

Well, we use the molar quantity, ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, and gets approx. a mass of $6.5 \cdot g$.
We know that ${N}_{A}$ $\text{zinc atoms}$, i.e. $6.022 \times {10}^{23}$ individual zinc atoms, have a mass of $65.39 \cdot g$ precisely. How do we know this? Well, because we has access to a Periodic Table, and this EXPLICITLY tells us the mass of a molar quantity.
And so we simply have to take the equivalent mass of zinc metal, i.e. (6xx10^22*"zinc atoms")/(6.022xx10^23*"zinc atoms"*mol^-1)xx65.38*g*mol^-1=??g.