# A "140.5-g" sample of "NiSO"_4 * x "H"_2"O" was heated until no further decrease in mass was observed. The mass of the anhydrous (dry) salt is "77.5 g". Determine the value of x in the formula?

Feb 26, 2018

#### Explanation:

Feb 26, 2018

$x = 7$....

#### Explanation:

$N i S {O}_{4} \cdot n \left({H}_{2} O\right) \left(s\right) + \Delta \rightarrow N i S {O}_{4} \left(s\right) + n {H}_{2} O \left(g\right) \uparrow$

We heat the nickel hydrate such that we are left with nickel sulfate, and the mass loss represents the moles of water...

$\text{Moles of nickel sulfate (anhydrous)} = \frac{77.5 \cdot g}{154.75 \cdot g \cdot m o {l}^{-} 1} = 0.501 \cdot m o l$...

But the mass LOST represented the water present....

$\text{Moles of water lost} = \frac{140.5 \cdot g - 77.5 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 3.50 \cdot m o l$...and so we divide thru by the LOWEST molar quantity to get...

${\left(N i S {O}_{4}\right)}_{\frac{0.501 \cdot m o l}{0.501 \cdot m o l}} \cdot {\left(O {H}_{2}\right)}_{\frac{3.50 \cdot m o l}{0.501 \cdot m o l}} \equiv \left(N i S {O}_{4}\right) \cdot {\left(O {H}_{2}\right)}_{7}$..

And so we got...$N i S {O}_{4} \cdot {\left(O {H}_{2}\right)}_{7}$...a salt of nickel sulfate with SEVEN waters of crystallization....