A 5.25 gram sample of a cobalt chloride hydrate is heated until dried. The anhydrous sample has a mass of 3.00 gram. What is the percent water by mass of the original hydrate?

1 Answer
Stefan V. · mrpauller.weebly.com
Dec 27, 2015

#42.9%#

Explanation:

So, you're dealing with a #"5.25-g"# sample of a cobalt(II) chloride hydrate, let's say #"CoCl"_2 * color(blue)(n)"H"_2"O"#.

You know that the mass of the anhydrous salt, which is what remains after all the water of crystallization was driven off by heating, is equal to #"3.00 g"#.

Assuming the all the water of crystallization was indeed driven off, you can say that since

#m_"hydrate" = m_"anhydrous salt" + m_"water"#

you will have

#m_"water" = "5.25 g" - "3.00 g" = "2.25 g"#

This means that your initial sample of cobalt(II) chloride hydrate contained #"3.00 g"# of anhydrous cobalt(II) chloride and #"2.25 g"# of water of crystallization.

The percent composition of water in the hydrate will thus be

#(2.25 color(red)(cancel(color(black)("g"))))/(5.25color(red)(cancel(color(black)("g")))) xx 100 = color(green)("42.9% H"_2"O")#

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

Hope this helps!

Related questions
See all questions in Percent Composition
Impact of this question
8439 views around the world
You can reuse this answer
Creative Commons License