# What is the number of moles of beryllium atoms in 36 g of Be?

Mar 14, 2016

Approx. $4$ $\text{moles}$. Thus $4 \times {N}_{A}$ beryllium atoms, where ${N}_{A} = \text{Avogadro's number}$

#### Explanation:

The atomic mass of berylium $=$ $9.012 \cdot g \cdot m o {l}^{-} 1$.

Thus $\text{Avogadro's number}$ of beryllium atoms has a mass of just over $9$ $g$.

Since ${N}_{A} = \text{Avogadro's number}$ $=$ $6.022 \times {10}^{23}$ $m o {l}^{-} 1$, the number of beryllium atoms $=$

(36.0*cancelg)/(9.012*cancelg*cancel(mol^-1) $\times$ $6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1}$ $=$ ??

The answer is a dimensionless number AS REQUIRED.