Which equation is correct - Equation 1: 2CuCo3(s) ® Cu2O(s) + 2Co2 (g) + 1/2O2(g)Equation 2:CuCo3(s) ® CuO(s) + Co2(g)

Equation 1: 2CuCo3(s) → Cu2O(s) + 2Co2 (g) + 1/2O2(g)

Equation 2:CuCo3(s)   → CuO(s) + Co2(g)

Variables
There are several variables, which could affect the outcome of this investigation. In order to make it a fair test we will attempt to keep the mass of copper carbonate constant (0.4g), we call this independent variable.

Amount of Metal Carbonate:
The more metal carbonate we use the more products we will get.

Our dependent variable is the amount of products we will get from the reaction.

 Prediction

From my lecture and the preliminary experiment I have done, I can say that equation two is correct. From my preliminary experiment, when I heated the copper carbonate, it changed from green to black, I got black powder and I know that CuO is black in colour. There was a gas given off when I heated it, and I tested what gas is given off by mixing it with calcium hydroxide. From this I observe the calcium hydroxide turned into milky that is lime water.

Equipments:

• Weight bottle
• Balance
• 2 Stand
• Spatula
• Gas syringe
• Boiling tube
• 2 Clamp
• Bunsen burner
• Delivery tube
• CoCo3  (0.40g)
• Tong
• Goggle
• Lab coat

Diagram

 

Method

1. Put the weighing bottle in the balance and tear it and take 0.4g of CuCo3 in the weighing bottle.
2. Transfer 0.4g copper carbonate in the boiling tube.
3. Heat it gently using Bunsen burner.
4. Observe the colour change. I carry out the experiment after I saw the copper carbonate change from green to black.
5.  I set up the clamp stands, gas syringe and Bunsen burner as shown in the diagram.
6. Weigh 0.4g of copper carbonate and transfer it into the boiling tube.
7. Fix the boiling tube and the gas syringe in to two different stands shown in the figure above.
8. Close the boiling tube with bung and connect the boiling tube and syringe by delivery tube.
9.  Turn on the gas and light the Bunsen burner and heat the copper carbonate in the boiling tube. Keeps burning till all the metal turns black and the volume of carbon dioxide in the syringe stops moving.
10. Record the final volume of carbon dioxide.
11. After the boiling tube cold down remove the bang from the boiling tube and measure the mass of the black product we get after burning copper carbonate.
12. Repeat the procedure to get a better result.

Preliminary test result

 

• Mass of copper carbonate=0.4g
• Mass of theoretical copper oxide=0.3g
• Initial volume of gas in the gas syringe = 0.01dm3
• Final volume of gas in the gas syringe = 0.062dm3
• volume gas produced during the reaction = 0.061dm3

 

The equation of the reaction will be:

CuCo3  → Cuo + Co2

Let’s say that

Number of mole of CuCo3 = n1

   “                 “          Cuo = n2

   “                           “                Co2  = n3

mass of copper oxide                       =  m2

mass of copper carbonate               =  m1

volume of Co2                 = V3

volume of gas at room temperature = 24dm3

From stochiometry we know that n1=n2=n3

Calculation

Actual mole of copper carbonate

n1 = m

      M

n1 = 0.4g

       124

n1= 0.0032mol =

 
n1=n2=n3

expected volume carbon dioxide

 

n3 = V3

        Vo

V3 = n3 x Vo

V3 = 0.0032 x 24

V3 = 0.077dm3

Theoretical volume of carbon dioxide = 0.061dm3

Actual mass of CuO

 

m1  =     m2

M1         M2

m2= m1 x M2

           M1

m2 = 80 x 0.4

             124

m2= 0.258g

Theoretical mass of copper oxide = 0.3g

conclusion

Error

1. mass of CuO

  error %  = theoretical value - actual value x 100%

theoretical value

  error  = 0.3g  - 0.258g x 100%

0.258g

  error  = 16.3%

2. error in CO2

error %  = theoretical value - actual value x 100%

theoretical value

error  =  0.077dm 3 - 0.061 dm3 x 100%

0.061dm3

error  = 26.2 %

• from my experiment I was expecting to get 0.258g of copper oxide but I get 0.3g. this shows I get  0.042g more carbon dioxide than I it should be, this might be due to:

• weighing error.
• transfer error, when I transfer the copper carbonate into the boiling tube.
• Measurement error.
• Errors of transferring the black powder from the boiling tube.
• Calculation error.

• I was expecting to get 0.077dm3 of gas from the experiment but I get 0.061dm3 . in this case I get less volume of gas than I should be getting. This error might be due to one or two of the following reasons:

• I took the final volume before the copper carbonate burns completely.
•  Problems of fixing the bung into the boiling tube.

• From the above preliminary experimental  result I got, I can say that equation 1 is correct. To confirm that the mass of copper carbonate I used in the beginning of the experiment has to be equal to the total mass of copper oxide and the volume of carbon dioxide.

CuCO3(s) → CuO(s) + CO2(ga)

n3 =   n1

Vo = m1 = m2 

V3    M1     M2

• m3 = n3 x M3

m3 = 0.0032 x 44

      m3 = 0.1408g

• m1 = m2 + m3

m1 = 0.258g + 0.1408g

m1 = 0.3988g

bibliography:

Chemistry 1 Endorsed by OCR

The rocks and minerals of the world by Charles A. Sorrel and George F. Sandstorm