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Topic: Loss on ignition  (Read 5765 times)

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Offline verygreen

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Loss on ignition
« on: January 19, 2014, 11:09:37 AM »

I don't even know where to start, any help or direction would be greatly appreciated.

 A student combusted a sample at 1050 C for several hours and obtained the loss on ignition value of 15 wt%.   

Assume that the sample did not contain any hydrous minerals and the loss is due to release of CO2.  The petrographic observation indicates similar amounts of calcite, CaCO3, and dolomite, (CaMg)(CO3)2.  Calculate the contents of calcite and dolomite in the sample and give the answers in weight percent.

Offline sjb

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Re: Loss on ignition
« Reply #1 on: January 19, 2014, 12:43:51 PM »

I don't even know where to start, any help or direction would be greatly appreciated.

 A student combusted a sample at 1050 C for several hours and obtained the loss on ignition value of 15 wt%.   

Assume that the sample did not contain any hydrous minerals and the loss is due to release of CO2.  The petrographic observation indicates similar amounts of calcite, CaCO3, and dolomite, (CaMg)(CO3)2.  Calculate the contents of calcite and dolomite in the sample and give the answers in weight percent.

So, for instance, if you started with 100g of sample, and x g was calcite, how much was dolomite? What did the mass go down to?

Offline verygreen

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Re: Loss on ignition
« Reply #2 on: January 19, 2014, 12:56:38 PM »

  Ok, so starting with 100g sample, it would go down 15wt% so it would go down 15g to 85g.

I am just not sure how to tie in the CO2 with dolomite and calcite, the 15g is loss of CO2 from both, I don't know how to differentiate how much was from each respective mineral...

Offline sjb

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Re: Loss on ignition
« Reply #3 on: January 19, 2014, 01:00:44 PM »

  Ok, so starting with 100g sample, it would go down 15wt% so it would go down 15g to 85g.

I am just not sure how to tie in the CO2 with dolomite and calcite, the 15g is loss of CO2 from both, I don't know how to differentiate how much was from each respective mineral...

That's a good start. So, if you had x g of calcite and y g of dolomite initially, what does x + y equal?

Can you write separate balanced equations for the loss of carbon dioxide from both?

Offline verygreen

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Re: Loss on ignition
« Reply #4 on: January 19, 2014, 02:17:24 PM »
 Thanks for your assistance!
 
 So

   X=calcite    y= dolomite

calcite decomposition

   CaCO3-> heat -> CaO + CO2

Dolomite decomposition

   (CaMg)(Co3)2-> CaO + MgO+ 2CO2


so the CO2 from both equal 15g.
so twice as many mols of CO2 is coming from dolomite, so can I just calculate the # of mols of 15g of CO2 and determine how much CO2 came from calcite and how much from dolomite? How could i determine x and y then?

Offline verygreen

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Re: Loss on ignition
« Reply #5 on: January 19, 2014, 02:25:00 PM »
or wait

 10%/10g would be from the dolomite and 5%/5g would be from the CO# of calcite..

Offline sjb

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Re: Loss on ignition
« Reply #6 on: January 19, 2014, 05:08:51 PM »
or wait

 10%/10g would be from the dolomite and 5%/5g would be from the CO# of calcite..

No, this would only be the case if there were equimolar quantities of the two minerals. All you know is that a certain amount of CO2 was given off. 2 moles per mole of dolomite, and 1 mole per mole of calcite.

How many moles are there in x moles of calcite? And in your dolomite?

Offline verygreen

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Re: Loss on ignition
« Reply #7 on: January 19, 2014, 05:30:08 PM »

I am not sure if I understand correctly but

  Is it x moles of CO2 in x moles of calcite and 2y moles of CO2 in Y moles of dolomite?

Offline verygreen

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Re: Loss on ignition
« Reply #8 on: January 19, 2014, 05:32:45 PM »


aka

    1 mole of CO2 in Calcite and 2 moles of CO2 in dolomite?

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