# How many formula units in 5.88*mol of CaO_2?

Jul 20, 2017

Well how many eggs in 5 dozen eggs.........? And you mean on $C a O$ not $C a {O}_{2}$.

#### Explanation:

The given question is precisely the same as that I asked in the opening, except that here I deal with MOLES not DOZENS.....

$1 \cdot m o l$ of stuff unequivocally specifies $6.022 \times {10}^{23}$ individual items of that stuff, where $6.022 \times {10}^{23}$ is Old Avogadro's number, symbolized by ${N}_{A}$.

Now in each formula unit of $C a O$, there are TWO atoms.......

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

$\text{number of atoms"-=2*"atoms} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 5.88 \cdot m o l \cong 30 \cdot {N}_{A}$

What is the mass of this quantity of $C a O$?

Jul 20, 2017

A $5.88$ mole sample of $\text{CaO}$ contains $3.54 \times {10}^{24}$ formula units of $\text{CaO}$.
There is no compound with the formula $\text{Ca"_2"O}$. The formula for calcium oxide is $\text{CaO}$. One mole of anything is $6.022 \times {10}^{23}$ of anything, including formula units. In order to determine the number of formula units in a $5.88$ mole sample of $\text{CaO}$, multiply the given moles by $6.022 \times {10}^{23}$ formula units/mol.
5.88color(red)cancel(color(black)("mol CaO"))xx(6.022xx10^23 "formula units CaO")/(1color(red)cancel(color(black)("mol CaO")))=3.54xx10^24" formula units of CaO"#