# Question 01cd8

Oct 1, 2015

The hydrate contains $\textcolor{b l u e}{\text{10.14 % (m/m) water}}$.

#### Explanation:

The formula of copper(II) sulfate is ${\text{CuSO}}_{4}$.

1.00 × 10^0 = 1.00 × 1 = 1.00 ≈ 1

If the formula has 1 water of hydration, the formula is $\text{CuSO"_4"·H"_2"O}$.

The formula mass of $\text{H"_2"O}$ is

$\text{2 H" = "2 × 1.008 u" = color(white)(l)"2.016 u}$
$\text{1 O" = "1× 16.00 u" = "16.00 u}$
stackrel(————-—————————————)("TOTAL" = color(white)(XXXXll)"18.016 u")

The formula mass of the compound is

$\text{1 Cu" = "1 × 63.55 u" = color(white)(Xll)"63.55u}$
$\text{1 S" = "1 × 32.06 u" = color(white)(XXl)"32.06 u}$
$\text{4 O" = "4 × 16.00 u" = color(white)(XX)"64.00 u}$
$\text{1 H"_2"O" = "1 × 18.016 u" = "18.016 u}$
stackrel(—————————————————————)("TOTAL" =color(white)(XXXXXXl)"177.626 u")

This tells you that there are 18.016 u of water in 177.626 u of the hydrate.

"% of water" = "mass of water"/"total mass" × 100 % = (18.016 color(red)(cancel(color(black)("u"))))/(177.626 color(red)(cancel(color(black)("u")))) × 100 % = 10.14 %#