Question

Question #133bf

Answer 1

`%"H"_2"O" = 36.1%`

Explanation:

I'll assume the hydrate is copper`ul(("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 `"H"_2"O"` in `"CuSO"_4·5"H"_2"O"`.

To do this, what we can do is find the molar mass of `"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 `"H"_2"O"` in `"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%`

`= (color(red)(90.076color(white)(l)"g/mol"))/(color(green)(249.68color(white)(l)"g/mol")) xx 100% = color(blue)(ulbar(|stackrel(" ")(" "36.1%" ")|)`