Cu2O is a red crystalline material, which is produce by the electrolytic or furnace method. CuO is a black powder prepared by the ignition of suitable salts such as carbonate (1)
The mole is the mass of substance that has the same number of particles as atoms in exactly 12g of carbon-12. One mole of any substance contains Avogadro’s number 6.02×10²³ molˉ1 (2).Avogadro discovered that at room temperature (25°C) and pressure (1 atm), all gases occupy the same volume. 1 mole of any gas will occupy a volume of 24dm³.
CALCULATING THE NEEDED MASS
The best method is 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 easiest to work out in terms of ratio. This assumption is made because the equation is “stoichiometric” (3) meaning 1 mole of CuCOз decomposes to exactly 1 mole of CuO and 1 mole of carbon dioxide. However, 24000cm³ would be too much gas to produce in a school lab, as that size of apparatus is not available. Gas syringe will be used but has a maximum of 100cm³, so it would not be sensible to produce 100cm³ of gas because if equation 1 turns out to be correct (which would mean oxygen will also be produced), and there would consequently be more gas. Aiming to produce 75cm³ of gas would leave room for error and the possibility of equation 1 occurring.
Equation 2: CuCOз ➔ CuO + CO2
If 1 mole of CuCOз produces 24000cm³ of CO2, to produce 75cm³ (for equation 2) we need:
Moles in gas = volume / 24000
= 75 / 24000
Moles in gas = 0.003125 moles
Ratio = CO2 : CuCOз
1 : 1
0.003125 : 0.003125
Moles of CuCOз = 0.003125 moles
R.A.M of CuCOз = 63.5 + 12 + (16 × 3)
Mass of CuCOз = 0.003125 × 123.5
Mass of CuCOз needed to be decomposed = 0.39g (2 s.f)
Equation 1: 2CuCOз ➔ Cu2O + 2CO2 + ½O2
For equation 1, volume of gas that will be produced using 0.39g of CuCOз will be:
Moles of 2CuCOз = mass / R.A.M
= 0.39 / 123.5
Moles of 2CuCOз = 0.003157894 moles
Ratio = 2CuCOз : 2CO2
= 2 : 2
= 0.00315784 : 0.00315784
Moles of 2CO2 = 0.00315784 moles
Volume of 2CO2 = moles × 24000
= 0.00315784 × 24000
Volume of 2CO2 = 76cm³
0.39g of CuCOз
Heat proof mat
Scales (capable of weighing out 0.01g)
Collect and set up apparatus as shown in the drawing
Ensure the gas syringe is at 0 before the experiment begins.
Place the weighed mass of CuCOз into the conical flask and insert the bung tightly so that no gas escapes.
Light the Bunsen burner and place it underneath the test tube (as shown in the drawing).
Heat the conical flask until the CuCOз has fully decomposed. (This will be indicated by the fact that the bubbling will stop, and there will be a significant colour change of the CuCOз form bluish green to black or red).
Continue to heat even after the bubbling has stopped for about 1 minute. This will ensure that the reaction has completely ended
Leave the apparatus unattended to for a few seconds to ensure all gas has been released.
Then record the reading from the gas syringe which shows the volume of gas produced.
Repeat theses stages thrice, to get a more accurate result. Then take the average for the results.
WHICH EQUATION IS CORRECT?
In previous calculation, 0.39g of CuCO³ should produce 75cm³ of gas, if equation 2 is correct. If this volume of gas or a slightly less volume (to allow for slight gas loss) is obtained during the experiment, then equation 2 is correct. But if more than 75cm³ of gas is produced, then it means that another gas has been produced which will be oxygen. This would then mean that equation 1 is correct.
Also, after the decomposition has fully occurred, you can try to guess which equation is happening through looking at the colour of the substance after decomposition. If it’s a red compound, it may be equation 1 occurring due to the Cu2O which is a red crystalline. Whereas if it’s a black compound then it may be equation 2 occurring as CuO is a black powder.
CuCOз is harmful; ingestion may cause nausea or vomiting, and inhalation may cause irritation (4).
Care must be taken when dealing with Bunsen burner to avoid been burnt.
(2) OCR Chemistry 1 – p20