# Question eb51b

Oct 3, 2017

$1.204 \cdot {10}^{23}$

#### Explanation:

$\text{4.032 g}$ divided by the molar mass of $\text{MgO}$, $\text{40.3044 g/mol}$, gives you moles

$\text{4.032 g"/"40.3044 g/mol" = "0.10004 moles}$

Then multiply this by $6.02 \cdot {10}^{23}$ to get the number of formula units, which would be

0.10004 cancel("moles") * (6.02* 10^23"f. units")/(1cancel("mole")) = 0.60224 * 10^(23) $\text{f. units}$.

Since one formula unit of $\text{MgO}$ contains $2$ atoms--one of $\text{Mg}$ and one of $\text{O}$--multiply this by $2$ to get

0.60224 * 10^(23)cancel("f. units") * "2 atoms"/(1cancel("f. unit")) = 1.204 * 10^(23)# $\text{atoms}$