# How do you find the number of atoms of oxygen present in 18 g of K_2CO_3?

Sep 19, 2016

$7.83 \times {10}^{22}$ atoms

#### Explanation:

We have: $18$ $g$ of ${K}_{2} C {O}_{3}$

In order to determine the number of atoms of oxygen, we first need to determine the number of moles of ${K}_{2} C {O}_{3}$:

$\implies n = \frac{m}{M}$

$\implies n \left({K}_{2} C {O}_{3}\right) = \left(\frac{m \left({K}_{2} C {O}_{3}\right)}{M \left({K}_{2} C {O}_{3}\right)}\right)$ $g$ $m o {l}^{- 1}$

$\implies n \left({K}_{2} C {O}_{3}\right) = \left(\frac{18}{\left(2 \times 39.10\right) + \left(12.01\right) + \left(3 \times 16.00\right)}\right)$ $g$ $m o {l}^{- 1}$

$\implies n \left({K}_{2} C {O}_{3}\right) = \left(\frac{18}{78.2 + 12.01 + 48.00}\right)$ $g$ $m o {l}^{- 1}$

$\implies n \left({K}_{2} C {O}_{3}\right) = \left(\frac{18}{138.21}\right)$ $g$ $m o {l}^{- 1}$

$\implies n \left({K}_{2} C {O}_{3}\right) = 0.130$ $g$ $m o {l}^{- 1}$

Now, let's determine the number of atoms of $O$:

$\implies n = \frac{N}{L}$

$\implies N = n L$

$\implies N \left(O\right) = \left(n \left({K}_{2} C {O}_{3}\right) \times 6.022 \times {10}^{23}\right)$ atoms

$\implies N \left(O\right) = \left(0.130 \times 6.022 \times {10}^{23}\right)$ atoms

$\implies N \left(O\right) = 7.83 \times {10}^{22}$ atoms