Decomposition of Copper Carbonate

Calculating the Correct Mass 

The best method would be to take one of the two equations and try to prove or disprove it. Since Equation 2 has only 1 mole of every substance, this would be the easier one to work with in terms of ratio. It can be said that:

"1 mole of CuCO3 will decompose to 1 mole of CuO and 1 mole of CO2"

This can be assumed because the equation is "stoichiometric", meaning that exactly one mole of copper carbonate decomposes to exactly one mole of copper oxide and exactly one mole of carbon dioxide.

So:

CuCO3 (s)           CuO (s) + CO2 (g)

1 mol                   1 mol        1 mol

And because 1 mole of any gas occupies 24dm³ or 24000cm³:

CuCO3 (s)           CuO (s) + CO2 (g)

1 mol                   1 mol      24000cm³

However, 24000cm³ would be far too much gas to be produced in a school laboratory, as that size of apparatus is not available. The largest gas syringe available can hold 100cm³. It would not, however, be sensible to attempt to produce 100cm³ because if Equation 1 turns out to be the correct one, which would mean oxygen is also produced and there would consequently be more gas. Aiming to produce 75cm³ of gas would mean that there is room for error and for the possibility of Equation 1 occurring. It would also be a lot easier to read the measurement.

So, if 1 mole of CuCO3 produces 24000cm³ of CO2, to produce 75cm³ we need:

(1 x 75 moles) / 24000 = 3.125 x 10-3 mol

Mr of CuCO3 = 63.5 + 12 + (16 x 3) = 123.5

Therefore:

Mass = (3.125 x 10-3) x 123.5 = 0.386g (3s.f.)

This will be the mass of copper carbonate that will need to be decomposed.

Now if we are going to decompose 0.386g of copper carbonate, we can then use the information in the second equation to calculate the volume of gas produced in the first.

So:

2CuCO3 (s)                Cu2O (s) + 2CO2 (g) + ½O2 (g)

2 mol                         1 mol           2 mol         ½ mol        

And again:

2CuCO3 (s)                Cu2O (s) + 2CO2 (g) + ½O2 (g)

2 mol                         1 mol         48000cm3    12000cm3

As we have used 0.386g of CuCO3, we have to find its amount of moles.

So:

0.386/123.5 = 0.0031255 mol

If 2CuCO3 is 0.0031255 mol then 2CO2 is 0.0031255 mol as well.

Then to get the volume of gas we multiply 0.0031255 moles with 24000 which we get 75.012 cm3,

If CuCO3 is 0.0031255 mol then ½O2 is (½ x 0.0031255 / 2 = 0.0007814moles.

0.0007814 moles x 24000 = 18.753 cm3 

Therefore, finally we know that:

2CO2 (75.012 cm3) + ½O2 (18.753 cm3) = 93.8cm3 (3s.f.)

This will be the amount of gas produced in equation 1.

Apparatus

Balance scales capable of weighing out 0.01g
Gas syringe (100cm3 max)
Measuring cylinder (250cm3 max)

Test tube
Bung
Glass tubing
Bunsen burner

Heatproof mat
Clamp and clamp stand

0.39g Copper Carbonate

Safety goggles
Lab. coat


Method

1. Place the weighed mass of CuCO3 into the test tube and tightly insert the bung so that no gas can escape. Attach the test tube to the clamp.

2. Set up apparatus as below:

3. Light the Bunsen burner and place it on the heatproof mat, underneath the test tube.

4. Heat until the copper carbonate has entirely decomposed. This will be indicated by the fact that the bubbling will stop, and there will be a colour change from the green of the copper carbonate to the black or red colour of the possible oxides.

5. Take the reading from the gas syringe and note down the volume of gas produced.

6. Repeat these stages a further 3 times, then take an average of the 4 results.

Fair Test

Minimizing gas loss:

Using a gas syringe as opposed to an upturned tube of water in a trough is, in itself, a more careful way of ensuring no gas is lost as the gas syringe has airtight seals.

The bung must also be inserted tightly into the test tube before the decomposition begins.

Keeping measurements accurate:

The gas syringe must be at 0 before the experiment begins.

It must be ensured that the copper carbonate has fully decomposed by the colour change, the lack of bubbling, and the fact the gas syringe has stopped being filled.

The apparatus should be left assembled for a minute before the result is taken so all the gas can be released.

The gas syringe should be left to cool so the gas inside can return to room temperature and pressure.

The experiment should be repeated to obtain an accurate average and highlight anomalous results. If there are a few anomalies, they should be repeated.


Safety

Copper carbonate is harmful. Ingestion may cause nausea and vomiting and inhalation may cause irritation on prolonged contact. Skin contact could cause skin irritation and possible skin discolouration and eye contact may cause mild eye irritation. Care should be taken when dealing with the substance - goggles and a lab. coat must be worn and if contact occurs the area should be rinsed.

Care should also be taken when dealing with the Bunsen burner and the heated test tube.


How to tell which equation is correct 

It was worked out in a previous calculation that 0.39g of CuCO3 should produce 75cm³ of gas, if Equation 2 is correct. If this volume or a little less (to allow for slight gas loss) is obtained, then Equation 2 is correct. If more than this volume has been produced (about 94cm³ from equation 1), it means that another gas, oxygen, has also been produced. This would suggest that the correct equation is Equation 1.

Prediction

I consider Equation 2 to be the most likely. Cu2O is produced by electrolytic or furnace methods whereas CuO is produced by the ignition of suitable salts such as the carbonate, the hydroxide, or the nitrate of copper, or by heating of cuprous oxide. Although the substance was not ignited, copper carbonate was decomposed, which may have a similar effect as ignition would.

References

http://www.britannica.com/eb/article?tocId=81944&query=copper%20oxides&ct=eb

OCR Chemistry 1 – p.20

http://www.iun.edu/~cpanhd/C101webnotes/quantchem/rxnstoich.html

http://www.ctmsupplies.hemscott.net/Copper%20Carbonate.htm