# How many atoms of Chlorine are present in 0.072g of FeCl3?

Sep 3, 2017

Approx. $8 \times {10}^{20}$ individual chlorine atoms...........

#### Explanation:

We (i) take the ration of the mass of the sample to the formula mass of $F e C {l}_{3}$ to get the number of moles of $F e C {l}_{3}$.

$\frac{0.072 \cdot g}{\left(55.85 + 3 \times 35.46\right) \cdot g \cdot m o {l}^{-} 1} = 4.44 \times {10}^{-} 4 \cdot m o l$.

And thus we have a molar quantity with respect to $F e C {l}_{3}$...the molar quantity with respect to $C l$ is CLEARLY three times this figure, i.e. $4.44 \times {10}^{-} 4 \cdot m o l \times 3 = 1.33 \times {10}^{-} 3 \cdot m o l$.....

But (ii) a mole specifies $\text{Avogadro's number}$ of individual items of stuff, i.e. ${N}_{A} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And so ..........

"number of chlorine atoms"=6.022xx10^23*mol^-1xx1.33xx10^-3*mol=??