# What is the mass of 4.50 * 10^22 atoms of gold, Au?

May 4, 2016

$\left(4.50 \times {10}^{22} \text{ gold atoms")/(6.022xx10^23" gold atoms per mole}\right) \times 196.97 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$
One mole of gold has a mass of $196.97 \cdot g \cdot m o {l}^{-} 1$. By definition there are $\text{Avogadro's number}$ of gold atoms that constitute this mass, i.e. ${N}_{A} = 6.022 \times {10}^{23}$.
$\left(4.50 \times {10}^{22} \text{ gold atoms")/(6.022xx10^23" gold atoms per mole}\right) \times 196.97 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$