# How many oxygen atoms in a mass of 132*g with respect to carbon dioxide?

Apr 17, 2016

There are $3$ $m o l$ of $C {O}_{2}$ in such a mass. Thus there are $6 \times {N}_{A}$ individual oxygen atoms.

#### Explanation:

What is ${N}_{A}$? It is simply $\text{Avogadro's number}$, $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

If I have $6.022 \times {10}^{23}$ individual items of stuff, I have a mole of that stuff. $1$ $m o l$ of $C {O}_{2}$ is equivalent to a mass $44.0 \cdot g$; $12.01$ $g$ of $C$, and $32.0$ $g$ of $O$.

Apr 17, 2016

$36.132 \times {10}^{23}$ oxygen atoms

#### Explanation:

If we work out the relative formula mass of $C {O}_{2}$

Relative atomic mass, ${A}_{r}$ of $C = 12$
Relative atomic mass, ${A}_{r}$ of $O = 16$, so....

${M}_{r}$ of $C {O}_{2} = 44$

The number of moles of a substance is given by :

$\frac{\text{mass in grams}}{{M}_{r}}$

So in this case:

$\frac{132}{44} = 3 \cdot \text{moles}$

One mole contains $6.022 \times {10}^{23}$ molecules (Avogadro's number).

So 3 moles contains $18.066 \times {10}^{23}$ molecules.

Each molecule contains 2 oxygen atoms, so there will be $36.132 \times {10}^{23}$oxygen atoms