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!

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