What is the percentage of water in #"MnCO"_3 *8"H"_2"O"#?
1 Answer
Explanation:
You're dealing with a hydrated compound, also known as a hydrate, called manganese(II) carbonate octahydrate,
This hydrate contains manganese(II) carbonate,
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
#M_("M H"_2"O") = "18.015 g mol"^(-1)#
so you know that one mole of water has amass of
#"% 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
Here is a similar lab with analysis conducted using copper (II) sulfate.
Hope this helps!