# How many molecules, and how many ATOMS, are present in a 1*g mass of nitrogen gas?

Jul 23, 2017

Of dinitrogen, there are $\frac{1}{28} \times {N}_{A}$ molecules..., where ${N}_{A} = \text{Avogadro's Number}$, i.e. ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

#### Explanation:

And the number of nitrogen atoms is $\frac{1}{14} \cdot {N}_{A}$.

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

Why use such an absurdly large number? Well, if we got ${N}_{A}$ $\text{hydrogen atoms}$ we have a mass of $1 \cdot g$ of hydrogen precisely. We got $1 \cdot g$ of dinitrogen.......

And thus we got..........

(1*g)/(28.0*g*mol^-1)xxN_A=??"nitrogen molecules".....

(1*g)/(14.0*g*mol^-1)xxN_A=??"nitrogen atoms"