Question

How many atoms are in 56.7 grams Mg3P2?

Answer 1

$N=\frac{m}{M} \cdot N_A$

where $N$ - number of atoms; $m$ - mass, g; $M$ - molar mass, g/mol; $N_A$ - Avogadro's number, $N_A \approx 6.022 \cdot 10^{23} \; mol^{-1}$.

$M(Mg_3P_2)=3 \cdot 24.305+2 \cdot 30.974=134.863 \; g/mol$

$N=\frac{56.7}{134.863} \cdot 6.022 \cdot 10^{23} \approx 2.53 \cdot 10^{23}$