[Johan]

I'm having some issues with this chapter.. first i'll show you guys the whole problem.

A sample containing a mixture of CaCl₂ and MgCl₂ weighs a total of 2.000grams. The mixture was added to aqueous H₂S0₄ and 0.736 grams of CaSO₄ precipitated out.

(a) How many moles and how many grams of CaCl₂ were in the original 2.000g sample?
(b) How many moles and how many grams of MgCl₂ were in the original 2.000g sample?
(c) What is the percentage by mass of CaCl₂ in the original 2.000g sample?

hint: always being with a balanced equation.

Alright, now my actual problem lies within setting up the balanced equation.. i think after i have that foothold i can do the rest, but i'm not even sure on that as of now. I have done all the homework problems in the book and none of them have been set up like this at all, so i'm really confused. Sad

This is what i'm thinking *un-balanced*

CaCl₂+MgCl₂+H₂S0₄  --------> CaS0₄+ MgS0₄+ HCl

The reason why i'm confused.. i'm assuming that the acid will react with the magnesium as well.. which would leave H and Cl ions floating around.. i'm not positive but since H was acting as the cation and Cl as the anion i'm assuming they would attract each other by default and form HCl.

Obviously the reaction equation is important to figure out the mole ratios for the rest of the problem. T.T

So.. if anyone could help me out i'd be so great full.

[Yggdrasil]

Your reaction equation looks good to me.  It may also be useful to think of two separate reactions happening:

CaCl₂ + H₂SO₄ --> CaSO₄ + 2HCl
MgCl₂ + H₂SO₄ --> MgSO₄ + 2HCl

Also while acid will react with magnesium, acid won't react with magnesium ions.  There is an important difference between Mg and Mg²⁺.

[Johan]

After some time spent scratching my head, i've realized that the net ionic equation is:

Ca²⁺(aq)  +  SO₄^2-(aq) -->  CaSO₄(s)

assuming all of the Ca precipitates out with the SO₄.. i came up with a 29.4% ratio of Ca:Formula Weight. So 29.4% of .736g CaSO₄ = .217g per 2.000g of sample.

That now leaves me with trying to find out how many grams of Cl and Mg were in the original 2.000g sample.. and now i'm completely lost and even wondering if the above works out at all.

[Yggdrasil]

How many moles of calcium is that?  How would you relate that to the amount of CaCl₂?

[Johan]

.217g Ca X 1mol Ca / 40.1g Ca = .005mol Ca

then.. i see there being 1 Ca: 2 Cl

so.. .005 x 2 = .01 mole Cl x 35.5g/1 mol Cl = .355g Cl

.355g Cl + .217g Ca = .572g CaCl₂

2.000 - .572g = 1.478g MgCl₂

If the above is correct, now i must figure out the moles of Mg and Cl... from the compound.. which is something else i've been struggling with the last two weeks.
Going from grams of a compound all the way to grams of each individual element. My teacher gave us a "strategy" chart, but its more confusing than helpful to me. And the book doesn't really get into multiple stepped stoichiometry problems.

thanks a ton for the help already given , i really appreciate it.

[Yggdrasil]

I'll show you an example calculation that may help you see the process in action:

You have 1.478g of MgCl₂ and you want to find the number of moles of Mg and Cl.  First figure out the molecular weight of MgCl₂

MW = 24.3g/mol + 2(35.5g/mol) = 95.3g/mol

Using the MW of MgCl₂, we can figure out how many moles of magnesium chloride are in 1.478g of MgCl₂:

1.478g / (95.3g/mol) = 0.0155mol

From the formula, MgCl₂, you know that one mole of MgCl₂ contains 1 mole of magnesium and two moles of chlorine:

moles of Mg = 0.0155mol * 1 = 0.0155 mol
moles of Cl = 0.0155mol * 2 = 0.0310 mol

[Johan]

Hmm.. well

1.478g MgCl₂ X 1 mol MgCl₂/95.3g/mol= 0.0155mol MgCl₂

.572g CaCl₂ X 1 mol CaCl₂/111.1g = .00515mol CaCl₂

Now.. lol.. if thats right.. one of the handout periodic tables gives us the AMU's to 1 decimal place.. so because the Molar Mass has 3SigFigs the answer should have 3 as well being the 1,5,5.. am i correct?

I know the rules for the most part, but the book is also very unclear on calculations when they result with zero's in the front. Now i know zero's in the front are not significant, but when reporting your answer i get a little confused.

So looking back up to the main question.. i think i just realized they only want Moles/Grams of CaCl₂ and Moles/Grams of MgCl₂. For the percentage by mass.. im fairly sure i just take Mass of CaCl₂ in the 2.000g sample / Mass of 2.000g total sample size = 28.6%

So if everything looks alright to anyone reading it, i think i finally understand for the most part what is going on. And can finally get to my other 3 problems with more confidence.
Once again thanks a ton!