# How many sodium atoms are there in 6.0 g of Na_3N?

Jun 29, 2016

$\text{Moles of sodium}$ $=$ $3 \times \text{moles of sodium nitride}$

#### Explanation:

$\text{Moles of sodium nitride}$ $=$ $\frac{6.0 \cdot g}{82.99 \cdot g \cdot m o {l}^{-} 1}$

Thus number of sodium atoms (or at least number of sodium ions):

$\frac{6.0 \cdot g}{82.99 \cdot g \cdot m o {l}^{-} 1} \times {N}_{A} \times 3$, where ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

i.e. $3 \times \frac{6.0 \cdot g}{82.99 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$. We get an actual number, without units, as required.