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?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

Your goal here is to find the 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

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
The mass of a crucible and a hydrated salt was found to be 21.447 g. The mass of the crucible and the anhydrous salt was 20.070 g. The mass of the crucible was 17.985 g. How do you calculate the mass of the hydrate heated?

The question asked you to determine the mass of the hydrate, even though you could calculate the mass of anhydrous salt. And this is simply the intial mass of the...

1 Answers 1 views
A 50.0-kg mixture of salt and water is 15.0% (by mass) salt. What is the mass of salt and water in the mixture?

You know that your mixture is $15.0%$ salt by mass, so make sure that you understand what that means before moving forward. For a component of a mixture, its...

1 Answers 1 views
Lithium and nitrogen react to produce lithium nitride: $6Li(s) + N_2(g) -> 2Li_3N(s)$. How many moles of lithium nitride are produced when 0.710 mol of lithium react in this fashion?

Since the given information is in moles, we need only look at the ratio between the chemicals: Six moles of Li will produce 2 moles of lithium nitride. So, the...

1 Answers 1 views
You weigh a sample of $"CusO"_4 * x"H"_2"O"$ and find that it has a mass of $"6.45 g"$. After you heat the sample, you find that the mass of the anhydrous salt is $"4.82 g"$. What is the value of $x$ ?

The idea here is that the difference between the mass of the hydrate and the mass of the anhydrous salt is equal to the mass of water of hydration present...

1 Answers 1 views
A hydrate containing aluminium sulphate has the formula $"Al"_2 ("SO"_4)_3 * x"H"_2 "O"$ and it contains 11.11% of aluminium by mass. Calculate the value of $x$ in the hydrate formula. ?

Your goal here is to figure out exactly how many moles of water of hydration are present for every mole of hydrate. As you can see, the chemical formula of...

1 Answers 1 views
A copper salt contains $25.5%$ $Cu$, $12.9%$ $S$, $25.6%$ $O$, and $36.0%$ $"water"$...what is the empirical formula of the hydrate?

We ASSUME that there are $100*g$ of copper salt, and work out the molar quantities: $Cu:$ $(25.5*g)/(63.55*g*mol^-1)$ $=$ $0.401*mol$ $Cu$. $S:$ $(12.9*g)/(32.06*g*mol^-1)$ $=$ $0.401*mol$ $S$. $O:$ $(25.6*g)/(15.999*g*mol^-1)$ $=$ $1.60*mol$ $O$...

1 Answers 1 views
An hydrate of $BaCl_2$ contains $0.750*mg$ of the salt, and $0.0965*g$ of water, how many waters in the formula $BaCl_2*n(H_2O)$?

$"Moles of barium chloride"=(0.00075*g)/(208.23*g*mol^-1)=??*mol$ $"Moles of water"=(0.0965*g)/(18.02*g*mol^-1)=7.203xx10^-6*mol$ And you can tell us the molar ratio..........

1 Answers 1 views
If 70.0g nitric acid reacts with 100.0g barium hydroxide what mass of salt may be obtained? If a student collects 138 g of this salt, what is the % yield?

This is an acid-base /titration reaction and the balanced chemical equation is $2HNO_3+Ba(OH)_2->Ba(NO_3)_2+2H_2O$ Since a balanced chemical equation represents ratio in which chemicals react, we now first convert all...

1 Answers 1 views
Inside of a cell suppose there is 2% salt and 98% water. The cell is in a solution containing 1% salt and 99% water. Is the cell hypotonic, hypertonic or isotonic compared to its surroundings?

...I think this is clear. The cell is $"hypertonic"$ with respect to the medium...the prefixes follow the old Greek prefixes... $"hyper "-=" more"$, $"hypo "-=" LESS"$, $"iso "-=" the same...

1 Answers 1 views
A solution is prepared from $25*g$ of salt dissolved in $1*L$ of water. If the salt is an hydrate of sodium thiosulfate, how many water molecules are associated with this salt?

$Na_2S_2O_3*5H_2O$ is the common laboratory reagent, this so-called pentahydrate forms large sugary crystals so it is easy and convenient to handle. The formula mass of this material is $248.18*g*mol^-1$ (i.e....

1 Answers 1 views
X