Calculate Hydrated Calcium Formula

[Winzer]

Homework Statement

I have .021 moles of H2o
and .0137 CaSO4 annhydrous.

What is the formula for hydrated calcium?

Homework Equations

CaSo4*H20-->CaSo4 +H2o

The Attempt at a Solution

we divide .0137 by.021 and get .65.

We then multipy that by three and get:

2CaSo4*1H20-->2CaSo4 +1H2o
Right?

 

[Winzer]

oops that should be
2CaSo4*3H20-->2CaSo4 +3H2o

 

[Winzer]

Here is another Calculation I have to solve.

What mass due to waters of crystallization is present in a 50g sample of:
CuSo4*5H20?

1 mole of CuSo4=159.61g
5 moles H20= 90.08g

Do I just take the ratio of h20 to cuso4 an multply by 50g to obtain mass of h20?

 

[lalbatros]

To my best knowledge, there are 2 types of hydrates:

gypsum or dihydrate: CaSO4.2H2O
hemihydrate: CaSO4.(0.5)H2O

But the question is really not clear.
Can you make it clearer.
What do you mean by this:

"I have .021 moles of H2o and .0137 CaSO4 annhydrous."

What does "I have" mean ?

What is the context of this question?
Is it a textbook exercice, then reproduce it completely.
Is it a experimental question, then explain more about you procedure.

 

[chemisttree]

If the question is related to an analysis of an unknown hydrate and the values you gave are for the constituent components, the solution is fairly easy.

Find the number of moles of each constituent and place these numbers as subscripts for each component. Divide both by the smallest number (of the subscript). This should yield the empirical formula. In the case where fractional numbers result, multiply both subscripts by an integer to remove the fractional value.

eg.


n moles of X and m moles of Y.

first attempt $$(X)_n(Y)_m$$
Using some numbers, 0.03 for n and 0.045 for m this becomes:

$$(X)_0._0_3(Y)_0._0_4_5$$

Dividing both subscripts by the smallest number (0.03):

$$(X)_1(Y)_1._5$$

multiply both subscripts by '2' to yield:

$$X_2Y_3$$ (the correct empirical formula)

 

[Kushal]

for the second problem... you have to think in terms of moles...
1 mol of CuSO4.5H2O contains 1 mol CuSO4 and 5 Mol H2O
so 50 g of CuSO4.5H2O, which is (50/250) mol = 0.2 mol will contain 0.2 mol CuSO4 and (5*0.2)mol H2O

you nneed the mass of H2O present in the 50g...
this corresponds to (5*0.2*mass of 1 mol H2O) = 18 g

it's all a matter of proportion...

 

[chemisttree]

for the second problem... you have to think in terms of moles...
1 mol of CuSO4.5H2O contains 1 mol CuSO4 and 5 Mol H2O
so 50 g of CuSO4.5H2O, which is (50/250) mol = 0.2 mol will contain 0.2 mol CuSO4 and (5*0.2)mol H2O

you nneed the mass of H2O present in the 50g...
this corresponds to (5*0.2*mass of 1 mol H2O) = 18 g

it's all a matter of proportion...

AARRRGGHHHH!

 

[Kushal]

hey chemisttree... is there a problem in my way of proceeding with this calculation? then why you said... AARRRGGHHHH!

 

[steven10137]

lol Kushal, I think he was just sarcastiaclly annoyed at the fact that what you wrote is exactly what he said lol ...

 

[Kushal]

aaawww... sry then... i may b stupid sometimes...lool