# How many N_2O_4 molecules are contained in 76.3 gN_2O_4?

We work out (i) the number of moles of ${N}_{2} {O}_{4}$, and (ii) convert this to molecules.
$\text{Moles of dinitrogen tetroxide}$ $=$ $\frac{76.3 \cdot g}{92.011 \cdot g \cdot m o {l}^{-} 1}$. This gives an answer in $\text{moles}$. Now it is a fact that in $1$ $\text{mole}$ of stuff there are $6.022 \times {10}^{23}$ individual items of that stuff:
$\text{Number of "N_2O_4" molecules}$ $=$ $6.022 \times {10}^{23} m o {l}^{-} 1 \times \frac{76.3 \cdot g}{92.011 \cdot g \cdot m o {l}^{-} 1}$ $=$ ??" molecules."