Question #01cd8

1 Answer
Oct 1, 2015

The hydrate contains #color(blue)("10.14 % (m/m) water")#.

Explanation:

The formula of copper(II) sulfate is #"CuSO"_4#.

#1.00 × 10^0 = 1.00 × 1 = 1.00 ≈ 1#

If the formula has 1 water of hydration, the formula is #"CuSO"_4"·H"_2"O"#.

The formula mass of #"H"_2"O"# is

#"2 H" = "2 × 1.008 u" = color(white)(l)"2.016 u"#
#"1 O" = "1× 16.00 u" = "16.00 u"#
#stackrel(————-—————————————)("TOTAL" = color(white)(XXXXll)"18.016 u")#

The formula mass of the compound is

#"1 Cu" = "1 × 63.55 u" = color(white)(Xll)"63.55u"#
#"1 S" = "1 × 32.06 u" = color(white)(XXl)"32.06 u"#
#"4 O" = "4 × 16.00 u" = color(white)(XX)"64.00 u"#
#"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 %#