What is the molar quantity of oxygen given a 3.42*mol quantity of Co(ClO_4)_2*6H_2O?

Feb 16, 2017

You don't need to calculate formula mass.

Explanation:

Clearly, in one formula unit of $C o {\left(C l {O}_{4}\right)}_{2}$, there are $8$ oxygen atoms. Agreed? In one formula unit of $C o {\left(C l {O}_{4}\right)}_{2} \cdot 6 {H}_{2} O$, there are $14$ oxygen atoms. Equivalently there are 14 oxygen atoms per formula unit of the solvate.

And in a $1 \cdot m o l$ quantity of $C o {\left(C l {O}_{4}\right)}_{2} \cdot 6 {H}_{2} O$ there are thus clearly $14 \times 1 \cdot m o l$ oxygen atoms. I use the mole as a number, just as I would use a $\text{dozen}$ or a $\text{score}$ or a $\text{100}$, and of course the $\text{mole}$ IS a number, i.e. ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus number of oxygen atoms in a $\text{mole}$ of $C o {\left(C l {O}_{4}\right)}_{2} \cdot 6 {H}_{2} O$,

$= 14 \cdot m o l \times {N}_{A} = 14 \cdot \cancel{m o l} \times 6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1}$

$\cong 9 \times {10}^{24} \text{ oxygen atoms, a number as required. }$

And in $3.42 \cdot m o l$, there are $\cong 3.1 \times {10}^{25} \text{ oxygen atoms.}$