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# Thermal decomposition of copper (II) carbonate watch

1. You are given basic copper (II) carbonates of the form Cu(OH)2.xCuCO3. The value of x can be determined by gravimetric analysis, i.e. by measuring mass changes.

Thermal decomposition equations:

CuCO3 --> CuO + CO2
Cu(OH)2 --> CuO + H2O

Question: Explain how you would process your data to determine the value of x. NB this is a planning exercise; no data are actually obtained or given.

Mark scheme:
Work out Mr for CuCO3 =123.5, Cu(OH)2=97.5 and CuO = 79.5
Relative mass loss for CuCO3 = 35.6%; that for Cu(OH)2 = 18.5%.
(% mass loss - % mass loss for Cu(OH)2) : (% mass loss - % mass loss for CuCO3) gives ratio CuCO3:Cu(OH)2
Divide by value of Cu(OH)2 to obtain x.

I don't understand the bold bit. If both copper compounds decompose together, how can one work out the composition?

Ref: 9791_y08_sp1.pdf
Cambridge Pre-U Specimen for use 2008 onwards
2. (Original post by bbrain)
You are given basic copper (II) carbonates of the form Cu(OH)2.xCuCO3. The value of x can be determined by gravimetric analysis, i.e. by measuring mass changes.

Thermal decomposition equations:

CuCO3 --> CuO + CO2
Cu(OH)2 --> CuO + H2O

Question: Explain how you would process your data to determine the value of x. NB this is a planning exercise; no data are actually obtained or given.

Mark scheme:
Work out Mr for CuCO3 =123.5, Cu(OH)2=97.5 and CuO = 79.5
Relative mass loss for CuCO3 = 35.6%; that for Cu(OH)2 = 18.5%.
(% mass loss - % mass loss for Cu(OH)2) : (% mass loss - % mass loss for CuCO3) gives ratio CuCO3:Cu(OH)2
Divide by value of Cu(OH)2 to obtain x.

I don't understand the bold bit. If both copper compounds decompose together, how can one work out the composition?

Ref: 9791_y08_sp1.pdf
Cambridge Pre-U Specimen for use 2008 onwards
You can solve this algebraically.

For 100g of mixture
If you let mass of carbonate in mixture = x, moles of carbonate = x/Mr = x/123.5
Mass of hydroxide = (100 - x), moles of hydroxide = (100 - x)/97.5

after decomposition moles of CuO formed = mol carbonate + mol hydroxide

mol CuO = x/123.5 + (100 - x)/97.5

mass CuO = [x/123.5 + (100 - x)/97.5]*79.5

From this you can work out x ...

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