# Question #fa394

Feb 28, 2017

$\text{In one mole of ferric oxide?}$

#### Explanation:

A $\text{mole}$ specifies a quantity of $6.022 \times {10}^{23}$ individual items of stuff. We use it in the same way as we would use a $\text{dozen}$, or a $\text{bakers' dozen}$, or a $\text{gross}$. It is simply another (admittedly large) collective number.

We have $1 \cdot m o l$, $6.022 \times {10}^{23}$ formula units of $F {e}_{2} {O}_{3}$, i.e. $2 \times {N}_{A}$ $\text{iron atoms}$, and $3 \times {N}_{A}$ $\text{oxygen atoms}$, where ${N}_{A} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$. And thus, there are $2 \times 6.022 \times {10}^{23}$ iron atoms in such a molar quantity.

Now I know ${N}_{A}$ iron atoms have a mass of $55.8 \cdot g$, and ${N}_{A}$ oxygen atoms have a mass of $15.999 \cdot g$. How do I know this?

And thus $1 \cdot m o l$ of $\text{ferric oxide}$ has a mass of $1 \cdot m o l \times \left(2 \times 55.8 \cdot g \cdot m o {l}^{-} 1 + 3 \times 16.00 \cdot g \cdot m o {l}^{-} 1\right) = 159.60 \cdot g$.

Do you agree?