# How many grams of oxygen are in 6.15times10^23 formula units of (NH_4)_2SO_4?

Aug 1, 2017

$65.4$ $\text{g O}$

#### Explanation:

We're asked to find the mass, in $\text{g}$, of oxygen in $6.15 \times {10}^{23}$ ${\text{formula units (NH"_4")"_2"SO}}_{4}$.

To do this, we recognize that there are $4$ $\text{atoms O}$ per formula unit of ammonium sulfate:

6.15xx10^23cancel("formula units (NH"_4")"_2"SO"_4)((4color(white)(l)"atoms O")/(1cancel("formula unit (NH"_4")"_2"SO"_4)))

$= 2.46 \times {10}^{24}$ $\text{atoms O}$

Now, we use Avogadro's number to find the number of moles of $\text{O}$:

2.46xx10^24cancel("atoms O")((1color(white)(l)"mol O")/(6.022xx10^23cancel("atoms O"))) = ul(4.085color(white)(l)"mol O"

Finally, we use the molar mass of oxygen ($15.999$ $\text{g/mol}$) to convert from moles to grams:

4.085cancel("mol O")((15.999color(white)(l)"g O")/(1cancel("mol O"))) = color(red)(ul(65.4color(white)(l)"g O"