# What is the number of atoms in 23.227 g of Cr?

$6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times \frac{23.227 \cdot g}{51.996 \cdot g \cdot m o {l}^{-} 1}$ $=$ ??" atoms"
It is a fact that $\text{Avogadro's number}$, $6.022 \times {10}^{23}$ $\text{chromium atoms}$ have a mass of $51.996 \cdot g$. From where did I get this mass?
So all we have to do is divide the given mass by the molar mass, and mulitply this number by $\text{Avogadro's number}$, $6.022 \times {10}^{23}$ to give the number of chromium atoms.
We get approx. $3 \times {10}^{23} \text{ chromium atoms.}$