Anhydrous lithium perchlorate (4.78 g) was dissolved in water and re-crystallized. Care was taken to isolate all the lithium perchlorate as its hydrate. The mass of the hydrate salt obtained was 7.21 g. What hydrate is it?

1 Answer
Oct 29, 2015

Answer:

Lithium perchlorate trihydrate

Explanation:

Your goal here is to find the mole ratio that exists between the anhydrous salt lithium perchlorate, #"LiClO"_4#, and the water of crystalization, #"H"_2"O"#.

Now, you know that the anhydrous salt is rehydrated to form the hydrate. The difference bwteen the mass of the hydrate and the mass of the anhydrous salt will be the water of crystaliztion.

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

#m_"water" = m_"hydrate" - m_"anhydrous"#

In your case, you will have

#m_"water" = "7.21 g" - "4.78 g" = "2.43"#

So, you know that your hydrate sample contains

  • #"4.78 g " -># anhydrous lithium perchlorate
  • #"2.43 g " -># water of crystalization

To determine how many moles of each you get in this sample, use their respective molar masses

#4.78color(red)(cancel(color(black)("g"))) * "1 mole LiClO"_4/(106.39color(red)(cancel(color(black)("g")))) ="0.04493 moles LiClO"_4#

and

#2.43color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) ="0.1349 moles H"_2"O"#

To get their mole ratio, divide both values by the smallest one

#"For LiClO"_4: " "(0.04493color(red)(cancel(color(black)("moles"))))/(0.04492color(red)(cancel(color(black)("moles")))) = 1#

#"For H"_2"O: " (0.1349color(red)(cancel(color(black)("moles"))))/(0.04492color(red)(cancel(color(black)("moles")))) = 3.0024 ~~ 3#

This means that the hydrate will be

#"LIClO"_4 * 3"H"_2"O" -># lithium perchlorate trihydrate