Plan of action:
This experiment is designed to find out which equation is the correct one. It makes use of the fact that each equation states that copper carbonate produces a different number of moles of gas when heated. A known mass of copper carbonate will be heated, and the volume of gas given off will be measured. From this the number of moles of gas can be calculated, and therefore we can find which equation is correct.
I will need to measure the mass of copper carbonate to use to decompose. I will then measure the amount of gas given off using a gas syringe. The gas syringe I will use can only measure up to 100cm3 so I will need to use a certain amount of copper carbonate that will give slightly less than 100cm3. I will look to collect around 70cm³ of gas as this is far enough from the 100cm³ limit to reduce the risk of the volume being off the scale if too much gas is evolved.
There is also the problem of the gas expanding in the heat, and having a larger volume than the gas syringe. To tackle this problem I will need to let the gas cool before measuring the volume, and may need to cool the gas as it enters the gas syringe.
For my investigation I will use 0.30g of copper carbonate.
For equation 1:
Equation 1: 2CuCO3 (s) Cu2O(s) + 2CO2 + 1/2O2(g)
The ratio of copper carbonate to gases is: 2:21/2
Molar Mass, Mr of copper carbonate = 63.5 + 12 + (16x3) = 123.5
So, in 0.30g of copper carbonate the number of moles will be:
Number of moles = Mass / Mr
n = 0.30/123.5 = 2.429149798x10-3 = 2.43x10-3 mols
So the number of moles of gas will be:
(2.43x10-3) X (21/2 / 2) = 2.43x10-3 X 1.25 = 3.04x10-3 mols
The volume of a gas in cm3 is the number of moles multiplied by 24000 cm3 =
3.04x10-3 X 24000 = 72.96 cm3
For equation 2:
Equation 2: CuCO3 (s) CuO(s) + CO2(g)
The ratio of copper carbonate to gases is: 1:1
Molar Mass, Mr of copper carbonate = 63.5 + 12 + (16x3) = 123.5
So, in 0.30g of copper carbonate the number of moles will be:
Number of moles = Mass / Mr
n = 0.30/123.5 = 2.429149798x10-3 = 2.43x10-3 mols
So the number of moles of gas will be:
(2.43x10-3) X (1) = 2.43x10-3 mols
The volume of a gas in cm3 is the number of moles multiplied by 24000 cm3 =
2.43x10-3 X 24000 = 58.32 cm3
In my investigating I will be using 0.30g of copper carbonate. After decomposition has occurred, if 72.96 cm3 of gas is produced, then equation 1 is correct. If 58.32 cm3 of gas is produced, then equation 2 is correct.
Safety:
Copper carbonate is harmful if swallowed and can irritate the lungs and eyes. Ensure that goggles and a lab coat are worn during the experiment.
Using a Bunsen burner can cause the hazard of burns. When Bunsen is not in use, either keep it of or set onto safety flame.
As we are dealing with some equipment made from glass, this can easily be shattered into pieces if dropped. Broken glass must be dealt with maturely as it can easily cut the skin.
Method:
- Gather all apparatus and chemicals required. Set up both of the clamp with clamp stand.
- Take a Dry Weighing Bottle and place it on the digital weighing scales. Take the weight measurement of this. Retrieve a spatula with the calcium carbonate and add 0.30g of calcium carbonate to the dry weighing bottle.
- Carefully, pour the calcium carbonate into a boiling tube making sure everything has gone in with no spillages and attach a bung with delivery tube.
-
Fix the boiling tube onto a clamp stand and attach the delivery tube to the 100cm3 gas syringe. Place this gas syringe to the other clamp stand. (All attachments made should be airtight)
- Retrieve Bunsen burner and place over heatproof mat and under the boiling tube. Being to heat the calcium carbonate.
- As soon as colour change has turned from green to black (colour of copper oxide), stop heating. Allow all the apparatus to cool, and the gas to contract. When the plunger on the gas syringe has stopped moving, record the volume of gas given off.
References:
-
, this website was used to help me gather information about the hazards of copper carbonate.
[page 3 of this write up]
- ‘AS-Level Chemistry Revision Guide’ by CGP pages 12-13 and 16-17, this helped me with my calculations.
[page 2 & 3 of this write up]