Finding CaCl2 percentage in a 4.22g CaCl2 and NaCl mixture

[dracolnyte]

Homework Statement

a 4.22g mixture of CaCl₂ and NaCl was dissolved, and then treated to precipitate all of the Ca as CaCO₃, which was then heated to drive off CO₂, leaving a sample of pure CaO which was found to weigh 0.969g. What percentage of the original mixture was CaCl[/SUB]2[/SUB]?

I think I know how to solve this question if I just knew the balanced chemical equation for this.

The Attempt at a Solution

If i knew the balanced chemical equation, i can just calculate the moles for CaO, then do ratios to find the moles of CaCl2, convert it back to grams and divide over 4.22g to find percentage.

 

[Bystander]

You have/are given two equations in two unknowns: the solution for the second equation is trivial, 0.969/MW(CaO). The 4.22 g mixture is xMW(NaCl) + (1-x)MW(CaCl₂).

 

[epenguin]

Homework Statement

a 4.22g mixture of CaCl₂ and NaCl was dissolved, and then treated to precipitate all of the Ca as CaCO₃, which was then heated to drive off CO₂, leaving a sample of pure CaO which was found to weigh 0.969g. What percentage of the original mixture was CaCl[/SUB]2[/SUB]?

I think I know how to solve this question if I just knew the balanced chemical equation for this.

The Attempt at a Solution

If i knew the balanced chemical equation, i can just calculate the moles for CaO, then do ratios to find the moles of CaCl2, convert it back to grams and divide over 4.22g to find percentage.

You don't need any 'balanced equation' whatever that would be - you have got the formulae that give you the composition of the substances stated right there in the problem. Using atomic masses of Ca and O you can find the number of moles of CaO which then is the same as the number of moles of Ca which is in turn the same as the number of moles of CaCl₂.
You don't even need to calculate the number of moles really - you can just calculate the grams of Ca in Ca0 by atomic or molecular mass. Then that's the number of grams of Ca in the CaCl₂ it came from vand from that, again by proportions, you calculate what was the number of grams of CaCl₂.

This sort of chemistry is basically arithmetic of proportions.

It seems to throw some students. They have forgotten the arithmetic of proportions.

May be another reason is the formulae look like algebra. If it's any help, try drawing these molecules with the appropriate number of blobs, Little white blob for a Ca atom, a black one for a Cl atom, and a red one for an O atom, and write the atomic mass just next to each blob; you might latch on to the idea more easily.

 

[epenguin]

Note to self: remember to look at the dates of original postings - there has been more than time to get a complete chemical education since this was posted.

 

[Bystander]

Note to self: remember to look at the dates of original postings

Some people collect stamps --- on the other hand, I browse the "practice forum," so when you see me/my response(s), check the original date.