# How many ATOMS in a 0.400*mol quantity of N_2O_5?

Mar 27, 2017

Do you mean $\text{how many molecules}$, or $\text{how many atoms?}$

#### Explanation:

Clearly, there are ${N}_{A} \times 0.400$ ${N}_{2} {O}_{5}$ molecules in such a molar quantity. Here we use ${N}_{A}$, $\text{Avogadro's number}$, just as we would use any other collective number, like a $\text{dozen}$, or a $\text{score}$, or a $\text{gross}$.

And so there are $0.400 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, approx. $2.40 \times {10}^{23}$ individual ${N}_{2} {O}_{5}$ molecules. What is the mass of this molar quantity?

And as regard to $\text{atoms}$ there are $0.400 \cdot m o l \times 7 \cdot \text{atoms} \cdot m o {l}^{-} 1$ $=$ $2.8 \times {N}_{A}$ atoms.

The Lewis structure of ${N}_{2} {O}_{5}$ is non-trivial, and charge separation with quaternized nitrogen occurs to give: ""^(-)O(O=)N^(+)-O-^(+)N(=O)O^-. What is the overall charge on this molecular representation?