# A 0.49*g mass of water is LOST from a mass of 3.75*g Fe(NO_3)_3*2H_2O upon prolonged heating. What is the percentage composition of water with respect to the starting salt?

Sep 16, 2017

Approx. $3 \cdot g$..........

#### Explanation:

You gots $F e {\left(N {O}_{3}\right)}_{3} \cdot 2 {H}_{2} O$, which has a formula mass of $277.88 \cdot g \cdot m o {l}^{-} 1$.

And so you got a molar quantity of $\frac{3.75 \cdot g}{277.88 \cdot g \cdot m o {l}^{-} 1} = 0.0135 \cdot m o l$......

ANd now you subject this to drying to remove the water.....

$F e {\left(N {O}_{3}\right)}_{3} \cdot 2 {H}_{2} O + \Delta \rightarrow F e {\left(N {O}_{3}\right)}_{3} \left(s\right) + 2 {H}_{2} O \left(g\right) \uparrow$

And now you still have a $0.0135$ molar quantity WITH RESPECT to $F e {\left(N {O}_{3}\right)}_{3}$, i.e. $0.0135 \cdot m o l \times 241.86 \cdot g \cdot m o {l}^{-} 1 = 3.26 \cdot g$...

And so in fact you have lost a mass of $0.49 \cdot g$ with respect to water....which is necessarily (why?) TWICE the molar quantity of starting hydrate salt.....

Finally we address the percentage of water by mass in the $\text{iron(II) hydrate}$, and this is simpy "mass of water"/"mass of hydrate salt"xx100%...

=(0.49*g)/(3.75*g)xx100%=13.1%. Agreed?