Question

How many formula units make up 13.0 g of magnesium chloride (MgCl2)?

Answer 1

$N_{f.u.}=\frac{m}{M} \cdot N_{A}$

$N_{f.u.}$ - number of formula units

$m$ - mass

$M$ - molar mass

$N_{A} \approx 6.02 \cdot 10^{23} \; mol^{-1}$ - Avogadro's number

$M(MgCl_{2})=1 \cdot 24.305+2 \cdot 35.453=95.211 \; g/mol$

$N_{f.u.}=\frac{13}{95.211} \cdot 6.02 \cdot 10^{23} \approx 8.22 \cdot 10^{22}$