# A 1.59-g sample of a metal chloride, MCl_2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.60 g. How do you calculate the molar mass of M?

Dec 12, 2016

$\text{Molar mass of metal chloride}$ $=$ $126.2 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

We need a stoichiometric equation:

$M C {l}_{2} \left(a q\right) + 2 A g N {O}_{3} \left(a q\right) \rightarrow M {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 A g C l \left(s\right) \downarrow$

$\text{Moles of silver chloride:}$
$=$ $\frac{3.60 \cdot g}{143.32 \cdot g \cdot m o {l}^{-} 1} = 2.51 \times {10}^{-} 2 \cdot m o l .$

Given the stoichiometry of the precipitation reaction, there were $\frac{2.51 \times {10}^{-} 2 \cdot m o l}{2}$ $=$ $1.26 \times {10}^{-} 2 \cdot m o l$ $M C {l}_{2}$.

So we have a molar quantity, and a given mass, and the $\text{molecular mass}$ is the $\text{quotient}$, $\text{mass"/"molar quantity}$

$= \frac{1.59 \cdot g}{1.26 \times {10}^{-} 2 \cdot m o l}$ $=$ $126.2 \cdot g \cdot m o {l}^{-} 1$.