# How many oxygen atoms are present in a 120*g mass of "nitric acid"?

Feb 24, 2017

Approx. $3.5 \times {10}^{24}$ $\text{oxygen atoms......}$

#### Explanation:

$\text{Moles of nitric acid} \equiv \frac{120 \cdot g}{63.01 \cdot g \cdot m o {l}^{-} 1} = 1.90 \cdot m o l$.

And thus, given a formula of $H N {O}_{3}$, there are $3 \times 1.90 \cdot m o l$ $\text{oxygen atoms}$ in the given mass of nitric acid.

And since $1 \cdot m o l$ specifies a quantity of $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, there are:

$3 \times 1.90 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $\text{oxygen atoms}$

=?? $\text{oxygen atoms............}$

Feb 24, 2017

Depends on the concentration.....

#### Explanation:

If you are thinking of concentrated nitric acid, then it is typically 68% $H N {O}_{3}$, 32% water. Fuming nitric acid will have $H N {O}_{3}$ content of 86%, and only 14% water.

Oxygen atoms are present in both components, so you'd need to work out the number of moles of both.......

If you take standard conc. nitric acid, then 120 g contains 81.6 g of $H N {O}_{3}$ and 38.4 g of water. This gives you (81.6 / 63.01) = 1.295 moles of $H N {O}_{3}$ and (38.4 / 18.02) = 2.131 moles of water.

Each mole of $H N {O}_{3}$ contains 6.02 x ${10}^{23}$ molecules each containing an oxygen atom. The same applies to water.

So you have (6.02 x ${10}^{23}$ x 1.295) + (6.02 x ${10}^{23}$ x 2.131) = 2.062 x ${10}^{24}$ oxygen atoms in total.