# What is the number of grams of xenon in 3.958 g of the compound xenon tetrafluoride?

$\text{Moles of } X e {F}_{4}$ $=$ $\frac{3.958 \cdot g}{207.28 \cdot g \cdot m o {l}^{-} 1}$
$\text{Moles of } X e {F}_{4}$ $\cong \frac{1}{50} \cdot m o l$
How many moles of $F$ are present here?