• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

determine the correct equation

Extracts from this document...


Aim- To determines the correct equation of the thermal decomposition of copper carbonate. The two possible equations are shown below: Equation 1: 2CuCO� (s) --> Cu�O (s) + 2CO� (g) + 1/2O� (g) Equation 2: CuCO� (s) --> CuO (s) + CO� (g) Apparatus * Heat mat * Clamp stand * 100ml gas syringe * Conical flask * Bunsen burner * Tripod * Tube/hose with bung attachment * Top-pan balance * Gauze Chemicals Powdered copper carbonate Calculations I expected the volume of gas would be 80cm�. This is because the bigger the volume of gas, the smaller the percentage error for the gas syringe as it is 100ml which has small graduations and big scale. V/24000 = n 80/24000=3.33 x10-� m/Mr = n Mr x n =m Mr=molar mass n=mole m= mass R.M.M of copper carbonate = 63.5 + 12 + (16 x 3) = 123.5 grams 123.5 x 3.33 x10-� =0.4g So in this experiment, I decide to use 0.4g of copper carbonate. ...read more.


Using 0.1 grams of copper carbonate: Moles of copper carbonate used = mass / relative molecular mass Mass = 0.4 grams R.M.M of copper carbonate = 63.5 + 12 + (16 x 3) = 123.5 grams Moles = 0.4 / 123.5 = 3.24 x 10-3 If we now compare the moles of copper carbonate to the moles of gas (carbon dioxide) given off with find that the molar ratio of the equation is 1:1. As the equation is a 1:1 equation then the moles of gas are the same as the moles of carbon carbonate. I shall now again use the equation above to calculate the gas produced to actual room temperature and pressure. n=v/24000 n= mole v=volume of gas Moles = 3.24 x 10-3moles 3.24 x 10-3=v/24000 Then rearrange the formula to n x 24000=v 3.24 x 10-3x 24000=77.76 cm� My two expected volumes of gas are 77.76 cm3 and 97.2 cm3, These are big enough for me to be able to use the more accurate, syringe, method of collecting gas, this should make my experiment more accurate. ...read more.


7. Wait 20 minutes until the gas has returned to room temperature as when it is hot it will expand and take up more volume. So to acquire a correct volume of gas we must wait for it too return to room temperature, approx 20oC, so that the volume is correct for a temperature of 20oC. 8. After 20 minutes record volume of gas. Safety Remember as we are using a Bunsen burner to wear goggle and a safety apron at all time and leave Bunsen burner on a safety flame when not in use Result Compare the recorded gas produced to out theoretical values. Expected gas produced for equation 1= 97.2cm� If the gas given off was between 95.00 - 100.00cm� then equation 1 was the correct equation for the decomposition of copper carbonate. Expected gas produced for equation 2 = 77.76 cm� If the gas given off was between 75.00 - 80.00 cm� then equation 2 was the correct equation for the decomposing of copper carbonate. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level War Poetry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level War Poetry essays

  1. Which Equation is Correct?

    This number, Avogadro's number, is 6.022x1023. The mass in grams of one mole of a compound is equal to the molecular weight of the compound in atomic mass units. One mole of a compound contains 6.022x1023 molecules of the compound.

  2. Which Equation is Correct?

    * This suitable mass should not produce more than 100cm³ volume of gas. * This can then be decomposed by the heat from the Bunsen burner. * We will continuously heat the copper carbonate until no more gas is produced.

  1. Free essay

    Which equation is correct?

    Therefore this shows that if I have 0.15 grams of copper carbonate, it will release and produce 30cm� of gas if equation 2 is correct. Equation 1: 2 moles of CuCO3 =2 and a half moles of gas This is telling me that 2 moles of copper carbonate will produce 2 and a half moles of gas.

  2. I need to produce a marketing strategy for a new or existing product. I ...

    My questionnaire results show that 70% of the people surveyed were British Gas customers. Initially they thought that the main age group that uses their services was 21-30. The results of the survey show that people in the age groups 31-40 and 41-50 are the higher users of the services.

  1. Determination of the Value of the Gas Constant and the Molar Volume of Oxygen ...

    = 0.108 ? 0.004g --> ? 100/27 % or ? 3.70 % M(O2) = 32.00 (assume effectively no error) VF(O2) = 8.9 x 10-5 m3 ? 5 x 10-7 m3 --> ? 50/89 % or ? 0.562 % Temp. = 294.5 K ?

  2. Determine which of two equations below is the correct equation of the thermal decomposition ...

    Use test-tube holders to hold boiling tube with copper carbonate above a Bunsen burner on full power 5. When copper carbonate has gone completely black (completely decomposed) and bubbling has stopped 6. Wait for the gas syringe to cool down before measuring and noting down the volume of gas obtained.

  1. The Thermal Decompasition Of Copper Carbonate

    Repeat these stages a further 2 times, then take an average of the 3 results. In this experiment it is imperative that the test is kept fair, this can be done in many ways. Firstly by making sure that the weighing of the CuCO3 is correct, This is important as

  2. Determining the Correct Equation for the Decomposition of Copper Carbonate.

    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: PV=nRT P=pressure V=volume n=number of moles R=gas constant T=temperature

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work