Question #133bf

1 Answer
Aug 1, 2017

Answer:

#%"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%" ")|)#