Question

ANALO387: Percent Water in a Hydrate

Answer 1

Answer:

$\large \boxed{37.77\,\%}}$

Explanation:

(1) Mass of hydrate

$\begin{array}{rcr}\text{Mass of crucible + hydrate} & = & \text{21.447 g}\\-\text{Mass of crucible} & = & \text{17.985 g}\\\text{Mass of hydrate} & = & \textbf{3.462 g}\\\end{array}$

(2) Mass of water

$\begin{array}{rcr}\text{Mass of crucible + hydrate} & = & \text{21.447 g}\\-\text{(Mass of crucible + anhydrous salt)} & = & \text{20.070 g}\\\text{Mass of water} & = & \textbf{1.377 g}\\\end{array}$

(3) Percent of water

$\begin{array}{rcl}\text{Percent of water}&=&\dfrac{\text{Mass of water}}{\text{Mass of hydrate}} \times 100 \%\\\\& = &\dfrac{\text{1.377 g}}{\text{3.462 g}} \times 100 \% \\\\& = &\mathbf{37.77 \%}\\\end{array}\\\text{The mass percent of water is \large \boxed{\mathbf{37.77 \%}} }$