Determining the Correct Equation for the Decomposition of Copper Carbonate
- Pages: 8
- Word count: 1866
- Category: Chemistry
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The task set is to prove or disprove which of the two given equations is correct by thermally decomposing CuCO3 and analysing the two possible oxides formed.
Equation 1: 2CuCO3 ?CuO(s) + 2CO2 (g) + 1/2O2 (g)
Equation 2: CuCO3 ?CuO(s) + CO2 (g)
In order to do this we need to call upon previous knowledge about the mole. The key idea involved in solving this problem is collecting a volume of gas by performing safe and fair test.
My ‘Advanced Chemistry for You’ reinforces my previous knowledge; one mole of gas at room temperature occupies a volume of 24dm3(litres). This is important information as when conducting the experiment in a classroom we have to scale down.
If we tried to prove the equations as they stood above, used one mole of Copper Carbonate and just did the above practical for this experiment, in theory we would collect 24dm of carbon dioxide gas. We would then need to ensure that first and for most we had a big enough vessel to collect the gas in. This won’t work, as too much CO2 is present in both cases!
To begin working out where to begin we need to identify what we are trying to prove and collect information that will be relevant to this.
The first place I look was my textbook – ‘Advanced Chemistry for You’, Laurie Ryan, 2000. It states that
‘At room temperature, 25?C and atmospheric pressure at 1 atmosphere 1 mole of ant gas will occupy a volume of 24dm3.’
This is useful because we have to use this to work out how much Copper carbonate to decompose.
When working this out we also have to use equation found in ‘Foundation Chemistry’:
Moles = mass
You will be shown later how to work this out.
From my research I have also found an equation that when rearrange might be useful in helping us to work out the correct equation. I found this in ‘Advanced Chemistry for You’
It relates to the ideal gas equation, which is:
n=number of moles
We can the go on to replace n with the equation as above:
It then becomes:
PV = mass x RT
We can then rearrange this formula to make the mass the subject:
Mass = R.A.M x PV
This then enables us to work out the mass. If we know that we are using room temperature and pressure, and we know from rearranging:
Moles of a gas = volume of gas
We can get –
Volume of gas = moles x 24dm3
Then we can substitute in:
Mass = R.A.M x P x (moles x 24dm3)
Factors to consider:
* How will you decompose the compound?
I’ve chosen to use the method of heating the solid copper carbonate and collecting the gas formed.
* How are you going to collect the gas?
I have decided to use a gas syringe. I have decided on this method against that of the upturned tube of water in a bath. I have done this because I can reduce the amount of errors that may occur using a gas syringe.
* How are you going to minimise gas losses?
The use of a gas syringe, which is airtight in itself and airtight seals, will also help reduce gas losses to an absolute minimum. Other possible ways of reducing gas losses is to insert the bung in the test tube containing the solid tightly before I begin to heat and decompose the copper carbonate.
* How are you going to make it a fair test?
A) I will make certain that the gas syringe is at 0 before beginning the experiment.
B) I will ensure that all the copper carbonate has been fully decomposed by waiting until the copper carbonate, which is a green colour, has changed fully to a black colour which is the colour of copper oxide. Another method, which I shall also employ, is by looking at the gas syringe. We can tell when the reaction has stopped due to the fact that when the Copper carbonate has finished decomposing gas will stop being produced.
C) I will also wait about a minute before taking a result while the copper oxide cools and all the gas has been released while the copper carbonate decomposes.
D) I will let the gas in the gas syringe cool down to room temperature before I take a reading. This is so it is at room temperature and pressure.
E) I could also repeat the experiments 3-4 times to obtain an accurate average, this would minimise the risk of errors and it would also account for any anomalous results.
1. Heat proof mat
2. Bunsen burner
3. Test tube
5. Glass tubing
6. Gas syringe
7. Accurate weighing scales capable of measuring to 0.01 of a mole.
8. One mole of copper carbonate
9. Safety goggles
10. Laboratory coat
When sorting out the safety for this practical I consulted the ‘School Science Services Hazcards’. These would give me the details about the substance I would be using and what to do in the event of an accident. Basic Copper carbonate is harmful if swallowed. The dust will also irritate the lungs, skin and eyes. This means that when conducting the experiment we will wear goggles and a lab coat. If it gets on our skin we are to wash the area thoroughly.
When dealing with the busen burner we must also take care not to burn ourselves, by either touching the burner while hot or touching and of the equipment it has heated.
Calculating the Correct Mass:
I have decided that it will be easier to prove or disprove Equation 2. this is because it has the easiest reaction ratio. We can say that:
‘1 mole of CuCO3 will decompose to give you 1 mole of CuO and 1 mole of CO2’
We know this because it is a stoichyometric equation.
CuCO3 ?CuO(s) + CO2 (g)
1 ? 1 + 1
And we also know that
‘1mole of any gas will occupy 24dm3 or 24000cm3’
There fore if…
1 mole is 24000cm3 then 1 is 1cm3
No. Of Moles = mass
* Mass = moles x R.A.M
M = 1 x CuCO3
(16 x 3) = 48
(1 X 12) = 12
(1 X 64) = 64
Mass = 1 x 124
? 124g of CuCO3 will produce 24dm3
Which then means if we scale down….
Mass = 1 x 124
? If it is a 1:1 ratio…..
124g : 24000cm3
? .00517g : 1 cm3
So to get the ideal measurement of gas, we have to look back at the available size of gas syringe. In the lab we have access to the biggest we have is able to hold 100cm3. so if aim to get our gas to fill three quarters it will be an amount of gas that we shall be able to read easily.
So if we…
.00517g : 1cm3
x 70 : x 70
then we shall have…..
.36g : 70cm3
1. Collect and set up equipment
2. Weigh copper carbonate to 0.01 mole. The need to measure 0.01 of a mole instead of 1 mole is because; one mole of copper carbonate will produce 24 litres of carbon dioxide. In order to do the experiment in school where the gas syringes we have only hold they hold 100ml , the amount of copper carbonate decomposed needs to be reduced so that the amount of carbon dioxide produced is reduced.
I used ‘Foundation Chemistry’ to find the following equations;
Moles = mass/R.A.M
Moles of gas = volume of gas (cm) / 24,000
Using these equations I calculated the following;
Cu = 64 C = 12 O = 16
R.M.M of copper carbonate = 64 + 12 + (3 x 16)
(CuCO3) = 124g
Mass = moles x R.A.M
0.01 x 124g
1.24g/124g x 24l = 0.24
As a result in theory I will only collect 0.24 litres of gas in the syringe at room temperature.
3. Place the weighed quantity of copper carbonate in a test tube and insert a bung tightly to ensure that it is air tight and gas loss is kept to an absolute minimum. This also guarantees that the results obtained are accurate.
4. Light Bunsen burner and place under tube.
5. Wait until the copper carbonate has completely decomposed. This will be indicated by a colour change, and the fact that the copper carbonate would have stopped bubbling.
Copper carbonate is a green solid. Copper oxide is a jet black solid.
6. When you are sure the copper carbonate has completely decomposed, measure and note down the volume of gas obtained.
7. Repeat stages 1 – 6 three to four times, then take the average. This reduces the risk of error in your results. Also ensure that all variables capable of influencing the results are kept constant. For example, the same amount of copper carbonate is used in each experiment and using the same apparatus again helps reduce the risk of error.
How to tell which equation is correct?
If for example we completely decomposed one mole of copper carbonate, which is not possible in the school laboratories, but in theory, we would produce 24 litres of carbon dioxide. If any more gas was obtained, this would indicate that another gas was being produced, as is the case in option 1. If a little less or exactly 24 litres of gas was collected this would indicate that equation 2 is correct. Collecting less than 24 litres would just suggest there was an error in the method or a gas leakage
I conclude that the most probable equation that is correct is equation 2. Previous knowledge may also help to support this prediction as when anything combusts oxygen gas is never given off as a product because the oxygen is needed for the reaction.
Advanced Chemistry for You – RYAN Laurie, 2000: Page 117
Foundation Chemistry – Cambridge, 1999: Page 22-3
The Complete A-Z Chemistry Handbook – HUNT Andrew, 2000: Page 241-3
A Level Chemistry 4th Edition – RAMSDEN E.N, 2000: Page 151-6