# How do you convert 27.6 g Ar to atoms?

Jun 6, 2016

Well, I know there are $\text{Avogadro's number}$ of $\text{Ar}$ atoms in a mass of $39.95 \cdot g$.

#### Explanation:

Thus $\text{Argon atoms}$ $=$ $\frac{27.6 \cdot g}{39.95 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \text{ argon atoms } m o {l}^{-} 1$ $=$ ??" argon atoms"

Jun 6, 2016

$4.15 x {10}^{23}$ atoms of Argon

#### Explanation:

To change 27.6 g of Argon to atoms we begin by converting the grams of Argon to moles using a conversion factor of 1 mole equal to the mass of Argon from the periodic table.

$27.6 \cancel{g} \left(\frac{1 m o l}{39.95 \cancel{g}}\right) = 0.69 m o l$

Now convert moles to atoms using Avogadro's number of $6.02 x {10}^{23}$

$0.69 \cancel{m o l} \left(\frac{6.02 x {10}^{23} a t m s}{1 \cancel{m o l}}\right) =$

$4.15 x {10}^{23}$ atoms of Argon