What is the percentage of water in #"MnCO"_3 *8"H"_2"O"#?

1 Answer

#"55.63% H"_2"O"#


You're dealing with a hydrated compound, also known as a hydrate, called manganese(II) carbonate octahydrate, #"MnCO"_3 * 8"H"_2"O"#.

This hydrate contains manganese(II) carbonate, #"MnCO"_3#, as the anhydrous salt. Each formula unit of the hydrate also contains #8# molecules of water of crystallization, hence the prefix octa added to the word hydrate in the name of the compound.

Now, the idea here is that you can look at the chemical formula of the hydrate and say two things

  • one formula unit of the hydrate contains one formula unit of manganese(II) carbonate
  • one formula unit of the hydrate contains eight molecules of water

This implies that one mole of manganese(II) carbonate octahydrate will contain

  • one mole of anhydrous salt
  • eight moles of water

Look up the molar mass of manganese(II) carbonate

#M_("M MnCO"_3) = "114.947 g mol"^(-1)#

This tells you that one mole of anhydrous manganese(II) carbonate has a mass of #"114.947 g"#. Water has a molar mass of

#M_("M H"_2"O") = "18.015 g mol"^(-1)#

so you know that one mole of water has amass of #"18.015 g"#. You can thus say that the percent composition of water in the hydrate is

#"% H"_2"O" = (8 xx 18.015 color(red)(cancel(color(black)("g"))))/((114.947 + 8 xx 18.015)color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(55.63%)color(white)(a/a)|)))#

This tells you that for every #"100 g"# of manganese(II) carbonate octahydrate, you get #"55.63 g"# of water.

Here is a similar lab with analysis conducted using copper (II) sulfate.

Hope this helps!