[bbco88]

When I searched I found a problem exactly like mine (must be tricky!) but I was still confused so here is my similar problem....

A mixture of CaCO3 and CaO weighing 0.692 g was heated to produce CO2. After heating, the remaining solid weighed 0.488 g. Assuming that all the CaCO3 broke down to CaO and CO2, calculate the mass percent of CaCO3 in the original mixture.

This is what I have done... is this correct so far...?
CaCO3--> CaO + CO2
so first I found the mass of CO2
.692 g / 100.09 g CaCO3= .0069 mol CaCO3 same as .0069 mol CO2, then multiply it by molar mass of CO2 44.01g to get .303669 g CO2
Then find mass of CaO
.488g/ 100.09 g CaCo3= .0049 mol CaCO3 same as .0049 mol CaO, then multiply it by molar mass of CaO 56.08 to get .2748 g CaO
Then add both together to get .578461 divided by the mass of CaCO3 100.09
and I got .00578 I don't think Im doing this right at all... can someone please work me through this I would really appreciate that!

[Borek]

No idea what you are doing 

Sample got lighter - why? There is a difference in masss - why? Where is the missing mass?

[bbco88]

Yeah I know I have no idea what I am doing... can walk me through the steps?

[macman104]

When I searched I found a problem exactly like mine (must be tricky!) but I was still confused so here is my similar problem....

A mixture of CaCO3 and CaO weighing 0.692 g was heated to produce CO2. After heating, the remaining solid weighed 0.488 g. Assuming that all the CaCO3 broke down to CaO and CO2, calculate the mass percent of CaCO3 in the original mixture.

Ah, yes.  Students always come to the resource center at our school with this problem.  And the first time a student came to me for help, I was similarly stumped.  Only upon working backwards and thinking about it did I get it.  You'll probably go "ah ha" once it's solved.

This is what I have done... is this correct so far...?
CaCO3--> CaO + CO2
so first I found the mass of CO2
.692 g / 100.09 g CaCO3= .0069 mol CaCO3

.692 g is not the weight of CaCO3, it is CaCO3 & CaO.  The starting material is a mix of CaCO3 and CaO.  When you heat it, the CaCO3 breaks down into CaO and CO2. 

The key to the problem is realizing the all of the moles of carbon released as CO2 (which is the difference in weight between the starting material and the product) came ONLY from the CaCO3.  So the moles of carbon in CO2 is going to equal your moles of CaCO3.  Once you have your moles of CaCO3 you can convert that to grams and get the mass percent of the original mixture.

Once you get that initial jump of the carbon relationship it becomes a fairly trivial problem.

[Borek]

The key to the problem is realizing the all of the moles of carbon released as CO2 (which is the difference in weight between the starting material and the product) came ONLY from the CaCO3.

The key to the problem is to answer questions I have asked earlier

[macman104]

The key to the problem is to answer questions I have asked earlier

Apologies, to be honest, I didn't actually pick up that those were leading questions, I thought for whatever reason you were asking for clarification.  Sorry.

[bbco88]

It's okay thanks for both of your help, but Im still a little confused so I understand that you have to find the moles of CO2 which equals moles of CaCO3 then convert that into g, but I have to weights in the equation so Im not sure how to use them in this question....

[macman104]

It's okay thanks for both of your help, but Im still a little confused so I understand that you have to find the moles of CO2 which equals moles of CaCO3 then convert that into g, but I have to weights in the equation so Im not sure how to use them in this question....

You'll have to be a little more clear, I'm not sure what you mean.  Maybe, can you write out the calculations you have so far with the new info, and we can go from there.

[AWK]

set two equation:
mass(CaCO3)+ mass(CaO)_in_sample =0.692
moles(CO2)+moles(CaO)_in_sample =final_moles(CaO)

since moles(CO2) = moles(CaCO3)

[Borek]

It's okay thanks for both of your help, but Im still a little confused so I understand that you have to find the moles of CO2 which equals moles of CaCO3 then convert that into g, but I have to weights in the equation so Im not sure how to use them in this question....

Fact that moles of CO₂ equals moles of CaCO₃ is direct application of balanced reaction equation for this problem, if that's what you are asking about.

Note: while AWK is right about setting two equations system, there is no need for that - you can easily calculate how much CaCO₃ there was initially, then CaO is just the rest of the sample.

[bbco88]

Okay so I started by writing the balanced equation again...
CaCO3 + CaO --> 2CaO + CO2 then I found how many moles of CO2 would come from .656 g.
.692g/ 44.01 g CO2 = .015724 moles of CO2 which would also be moles of CaCO3
so now I have .015724 moles of CaCO3 which I convert to grams by multiplying it by the mass of CaCO3 100.09 to get 1.5738 g of CaCO3, is this the orginal mass percent... am I finished.... I dont feel this is right, how come I didn't even use the second weight of the solid sample?

[Borek]

You are doing exactly the same mistake as before. Please reread macman104 posts and try to answer questions I have asked. Don't bother with the reaction equation for now, just concentrate on these questions:

Sample got lighter - why? There is a difference in masss - why? Where is the missing mass?

[bbco88]

Yeah still don't understand, I have been working on this problem for DAYS, soooo I GIVE UP!!!! Chemistry is ruining my life!

[DrCMS]

Yeah still don't understand, I have been working on this problem for DAYS, soooo I GIVE UP!!!! Chemistry is ruining my life!

Stop being such a loser.  You've been given lots of good advice and help you just need to look at it again and read the question to get the answer.

A mixture of CaCO₃ and CaO is heated and gives off CO₂ gas. 
The CaCO₃ in the mixture breaks down to CaO and CO₂ gas.
The CaO part of the mixture is unchanged.

You're given the weights before and after heating.  The rest is simple maths.