# Question 133bf

Aug 1, 2017

%"H"_2"O" = 36.1%

#### Explanation:

I'll assume the hydrate is copperul(("II") sulfate pentahydrate, as no oxidation number for copper was given (which is indeed a commonly-used hydrate in analyses).

We're asked to find the percentage by mass of $\text{H"_2"O}$ in $\text{CuSO"_4·5"H"_2"O}$.

To do this, what we can do is find the molar mass of $\text{H"_2"O}$ and divide that by the molar mass of the whole hydrate (and then multiply by $100$ to find the percentage):

bb("H"_2"O":

5(overbrace((2)(1.008color(white)(l)"g/mol"))^"hydrogen" + overbrace(15.999color(white)(l)"g/mol")^"oxygen")

= color(red)(ul(90.076color(white)(l)"g/mol")

bb("CuSO"_4·5"H"_2"O":

overbrace(63.546color(white)(l)"g/mol")^"copper" + overbrace(32.066color(white)(l)"g/mol")^"sulfur" + overbrace((4)(15.999color(white)(l)"g/mol"))^"oxygen" + overbrace(color(red)(90.076color(white)(l)"g/mol"))^"water"

= color(green)(ul(249.68color(white)(l)"g/mol"

bb("Mass Percent":

The mass percentage of $\text{H"_2"O}$ in $\text{CuSO"_4·5"H"_2"O}$ is given by the equation

%"H"_2"O" = ("mass H"_2"O")/("mass CuSO"_4·5"H"_2"O") xx 100%#

$= \left(\textcolor{red}{90.076 \textcolor{w h i t e}{l} \text{g/mol"))/(color(green)(249.68color(white)(l)"g/mol")) xx 100% = color(blue)(ulbar(|stackrel(" ")(" "36.1%" }} |\right)$