# How many atoms of oxygen are contained in 47.6 g of Al_2(CO_3)_3? The molar mass of Al_2(CO_3)_3 is 233.99 g/mol.

Sep 20, 2016

There are $9 \times 0.203 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ oxygen atoms in a $47.6 \cdot g$ mass of $A {l}_{2} {\left(C {O}_{3}\right)}_{3}$.

#### Explanation:

In the one mole of $A {l}_{2} {\left(C {O}_{3}\right)}_{3}$ there are ${N}_{A}$ formula units, i.e. $6.022 \times {10}^{23}$ formula units.

$\text{Moles of aluminum carbonate}$ $=$ $\text{Mass"/"Molar mass}$

$=$ $\frac{47.6 \cdot g}{233.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.203 \cdot m o l$

But in each mole of the carbonate there are clearly $9$ $m o l$ $O$ atoms.

And thus there are $9 \times 0.203 \cdot m o l$ $O$ atoms in the given quantity.