posted by Christian .
What percent by mass does the water constitute in Na2SO4*10H2O?
Determine the formula for the hydrate of cobalt (II) acetate from the following data.
Mass of crucible and lid= 28.945g
Mass of crucible, lid, and hydrate= 29.986g
Mass of crucuble, lid, and anhydrous salt= 29.677g
Mass of crucible, lid, and anhydrous salt(2nd)= 29.673g
Surely you know something about how this should b done. Tell us what you don't understand.
I think for the second one I need to to figure out how much the hydrate weighs by itself without the crucible then figure out how much weight the anhydrous salt lost during heating but I am not sure.
For the first question I don't know how to figure it out without knowing the grams so how do I do that?
For the first one, take a mole of the material. Na2SO4*10H2O has a molar mass of ??. And 10 moles H2O has a mass of ??. Then mass 10 moles H2O/mass 1 mole Na2SO4*10H2O and all that times 100 will be the percent water by mass.
For the second one, you are correct.
The mass of water lost = ??
The mass copper acetate converted to moles.
mass of water lost converted to moles.
Then look at the ratio of the moles and determine the number of moles of water per 1 mole copper acetate. Post your work if you get stuck.
you must have a formula to solve any mole because anything in solving a numbers like mole have formula you must search the formula to be easy sor you to finish your work....