Determining the correct equation for the composition of Copper Carbonate

6) Spatular

7) Pastette

Saftey equipment

> Goggles

> Lab coat.

* remember to tie back any long hair or loose clothing.

Method

) Put on your lab coat and collect the apparatus that you require then rinse out the glassware with the de-ionnised water.

2) Set up the apparatus as in the diagra below....

3) Put on your goggles

4) Place the conical flask on to the digital scales and press the tear button (T) to total the scales back to 0. Then add 1g of CuCO3 using the spatular. Replace the bung in the conical flask.

5) Place the conical flask on to gauze on the tripod.

6) Check that the water level within the measuring cylinder is at 250cm3 (remember to measure to the bottom of the miniscus), if it is not then take the pastette and bubble some air up the cylinder to displace the water to 250Cm3.

7) Now make sure that the delivery tube is directley underneath the measuring cylinder. If it is proceed on to point 7 if not rectify the situation before moving on.

8) Turn on and light the bunsen burner. Turn the air hole at the bottom of the bunsen to make a blue flame, but not roaring.

9) Record the amount of gas given off by the amount of water that had been displaced.

0) Now repeat the whole experiment again, but replace the CuCO3 with 2CuCO3 and record yet again the amount of water displaced/gas given off.

Calculations

To work out which equation is correct ideas about the mole must be applied, as so....

CuCO3 --> CuCO + CO2

mass volume

moles moles

In step 1 the equation n = m

M

must be applied......

g

23.5gmol-1 = 8.097 x 10-3 moles

In step 2 you must times by the ratio of each part in this case it is 1:1 so the moles stay the same 8.097x10-3

In step 3 the number of moles must be times by 24dm3 to achieve the volume...

8.097x10-3 x 24dm3 = 0.19433dm3

x 1000 = 194.33cm3

94cm3 is how much water should be displaced from the measuring cylinder if 1g of CuCO3 is used.

So if you used 1g of 2CuCO3 you carry out the calculations in the same way...

. 1g

247gmol-1 = 4.048x10-3moles

2. 4.048x10-3mols x 1.25

1 = 5.06x10-3moles

3. 5.06x10-3mols x 48dm3 = 0.243dm3

4. 0.243dm3 x 1000 = 243cm3

This is because...

CuCO3 --> CuCO + CO2

mol --> 1mol

mol --> 24dm3

2mol --> 48dm3

OR

2CuCO3 --> Cu2O + 2CO2 + 1/2 O2

2 mol --> 2 1/2 mol

2mol --> 60dm3

So after carrying out the experiment you will be able to look at the amount of gas given off and decide from that "which equation is correct."

If the water displaced is around 243cm3 it is going to be the 2nd equation that is correct

2CuCO3 --> Cu2O + 2CO2 + 1/2 O2

If the water displaced is around 194.33cm3 it is the 1st equation that is correct

CuCO --> CuO + CO2

* note some of the volumes cannot be read as accurately, as given in these calculations, off of the measuring cylinder. The nearest number should be taken.

Bibliography

> Internet

> Student intranet

> Chemistry 1 - Cambridge

> Personal class notes.

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Sam Granshaw Chemistry