# Question #8e917

Jan 4, 2018

$2 \cdot {10}^{23}$

#### Explanation:

The trick here is to realize that the molar mass of oxygen gas tells you the mass of $6.022 \cdot {10}^{23}$ molecules of oxygen gas.

Since oxygen gas has a molar mass of about ${\text{32.0 g mol}}^{- 1}$, you can say that $6.022 \cdot {10}^{23}$ molecules of oxygen gas will have a mass of $\text{32.0 g}$.

This is the case because $1$ mole of oxygen gas must contain a number of molecules of oxygen gas equal to Avogadro's constant.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 mole O"_2 = 6.022 * 10^(23)color(white)(.)"molecules O}}_{2}}}}$

So if $1$ mole of oxygen gas has a mass of $\text{32.0 g}$, you can say that $6.022 \cdot {10}^{23}$ molecules of oxygen gas will have mass of $\text{32.0 g}$.

This means that your sample contains

$8 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * (6.022 * 10^(23)color(white)(.)"molecules O"_2)/(32.0color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(2 * 10^(23)color(white)(.)"molecules O}}_{2}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of oxygen gas.