# 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 24, 2017

$1.10 \cdot {10}^{24}$

#### Explanation:

The first thing that you need to do here is to convert the mass of aluminium carbonate to moles by using the compound's molar mass.

In this case, you know that aluminium carbonate has a molar mass of ${\text{233.99 g mol}}^{- 1}$, which means that $1$ mole of aluminium carbonate has a mass of $\text{233.99 g}$.

You can thus say that your sample will contain

47.6 color(red)(cancel(color(black)("g"))) * ("1 mole Al"_2("CO"_3)_3)/(233.99color(red)(cancel(color(black)("g")))) = "0.2034 moles Al"_2("CO"_3)_3

Now, the chemical formula of aluminium carbonate tells you that every mole of this salt contains

• two moles of aluminium, $2 \times \text{Al}$
• three moles of carbon, $3 \times \text{C}$
• nine moles of oxygen, $3 \times 3 \times \text{O}$

This means that your sample will contain

0.2034 color(red)(cancel(color(black)("moles Al"_2("CO"_3)_3))) * "9 moles O"/(1color(red)(cancel(color(black)("mole Al"_2("CO"_3)_3)))) = "1.831 moles O"

Finally, to convert this to atoms of oxygen, use Avogadro's constant, which tells you that in order to have $1$ mole of atoms of oxygen, you need to have $6.022 \cdot {10}^{23}$ atoms of oxygen.

$1.831 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"))) * (6.022 * 10^(23)color(white)(.)"atoms O")/(1color(red)(cancel(color(black)("mole O")))) = color(darkgreen)(ul(color(black)(1.10 * 10^(24)color(white)(.)"atoms O}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of aluminium carbonate.