# Calculate the percentage of water of hydration in barium chloride crystals (the atomic weight of Ba is about 137.4 g/mol)?

Jul 15, 2018

14.75%

#### Explanation:

You'll need to start off by knowing the formula of the barium chloride crystals, which is barium chloride dihydrate, $B a C {l}_{2} \cdot 2 {H}_{2} O$. From the formula, you'll notice that there are 2 moles of ${H}_{2} O$ for every mole of the hydrate, $B a C {l}_{2} \cdot 2 {H}_{2} O$.

When the 1 mole of hydrate crystals are heated, you will be left with 1 mole of the anhydrous crystal ($B a C {l}_{2}$) and 2 moles of ${H}_{2} O$, like this:

$B a C {l}_{2} \cdot 2 {H}_{2} O \to B a C {l}_{2} + 2 {H}_{2} O$

To calculate the percentage of water of hydration, we'll need to find the molar mass of water (${H}_{2} O$) and hydrate crystal ($B a C {l}_{2} \cdot 2 {H}_{2} O$).

$\text{molar mass " H_2O = 2(1.01) + 16.00 = 18.02 " g/mol}$

$\text{molar mass " BaCl_2 *2H_2O = 137.4 + 2(35.45) + 2[2(1.01) + 16.00] = 244.34 " g/mol}$

We can then proceed to calculating the percentage of water of hydration in barium chloride crystals:
"Percentage of water of hydration" =(2 xx 18.02 ) / (244.34)*100%=14.75%