# Question a348e

Nov 2, 2017

$9.022 \cdot {10}^{23}$ atoms

#### Explanation:

As you can infer from the formula, one $N {a}_{2} C {O}_{3}$ molecule contains 2 atoms of sodium, 1 atom of carbon, and 3 molecules of oxygen. You can use this as a conversion factor, therefore, to infer that for every mole of $N {a}_{2} C {O}_{3}$ you have, you will have 3 moles of oxygen.

However, you don't have $1$ mol of $N {a}_{2} C {O}_{3}$, you have $0.5$. However, this is easily remedied using a simple conversion factor:

0.5 cancel(" mol " Na_2CO_3) * ("3 mol oxygen")/("1" cancel("mol" Na_2CO_3)) = 1.5 "mol oxygen"#

However, we're not done yet! We have 1.5 moles of oxygen, not atoms. However, we do know that $\text{1 mole} = 6.022 \cdot {10}^{23}$ atoms. Hence, we can use that as another conversion factor:

$1.5 \cancel{\text{ mol oxygen") * (6.022* 10^23 " atoms of oxygen")/cancel(1 " mol oxygen}}$

This leaves $9.022 \cdot {10}^{23}$ atoms of oxygen.

If you need more help on converting between moles and atoms and mass, here's some useful resources:

Handy Graphic:

Really useful video:

Hope that helped :)