Question

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.

Answer 1

There are `9xx0.203*molxx6.022xx10^23*mol^-1` oxygen atoms in a `47.6*g` mass of `Al_2(CO_3)_3`.

Explanation:

In the one mole of `Al_2(CO_3)_3` there are `N_A` formula units, i.e. `6.022xx10^23` formula units.

`"Moles of aluminum carbonate"` `=` `"Mass"/"Molar mass"`

`=` `(47.6*g)/(233.99*g*mol^-1)` `=` `0.203*mol`

But in each mole of the carbonate there are clearly `9` `mol` `O` atoms.

And thus there are `9xx0.203*mol` `O` atoms in the given quantity.