CuCO3 Lab Answer Key

CuCO3 LAB ANSWER KEY Purpose: to determine the percentage of CO2 (g) in a given sample of CuCO3. Procedure: 1) An empty test tube was weighed and its mass recorded. 2) A scoop full of CuCO3 was placed in the test tube. 3) The test tube was weighed again. 4) The test tube was heated with a Bunsen Burner. Observations were recorded before, during, and after heating. 5) SAFETY: When heating test tubes over an open flame, always hold the test tube with tongs, at a 45 degree angle, and point mouth of test tube AWAY from others!! 6) The test tube was weighed again once the reaction was complete and the tube was cooled down to room temperature. Observations: BEFORE HEATING Full thorough observations of the green powder before it was heated DURING HEATING Full thorough Descriptions of how the powder was changing and moving as well as any vapour that was leaving the test tube DATA: Balance #: __________________ AFTER HEATING Full thorough of What the powder looked like after the Bunsen Burner was turned off before heating: at end of heating: Mass of empty tube = 19.87 g Mass of tube + CuCO3 = Mass of tube + contents 24.16 g = 22.32 g Mass of CuCO3 = Mass of contents = Mass of tube + CuCO3 Mass of tube + contents – Mass of empty tube – Mass of empty tube = 24.16 – 19.87 = 4.29 g = 22.32 – 19.87 g = 2.45 g CALCULATIONS: We want to determine the % of CO2 in a CuCO3 molecule. % of CO2 = mass of Carbon dioxide mass of Copper carbonate X 100 Mass of Carbon Dioxide = mass of tube and green contents before heating – mass of tube and black contents after heating Mass of Carbon Dioxide = 24.16 – 22.32 = Note that this also equals: 1.84 g 4.29 g – 2.45 g= 1.84 g So now, % of CO2 = mass of CO2 that escaped X 100 original mass of CuCO3 sample = 1.84 g X 100 4.29 g = 42.89% Therefore 42.89% of our copper carbonate (CuCO3) sample was carbon dioxide. (CO2) Questions: 1) Was this a chemical or physical change? This was a CHEMICAL CHANGE. 2) What is your evidence or proof of your answer to question #1? (In other words, explain, based on what you saw in this experiment). 1) We saw a substance leaving the test tube while it was being heated on the Bunsen Burner. You could visibly see the vapour coming out of the tube. 2) The original tube contained a green, fine powdery sample of CuCO3. This remaining substance is called copper oxide and is a black, coarse powder. The black powder never turns back into the green powder, even after the test tube cools down. Therefore it is not reversible and that means it was a chemical change. 3) Also, Mrs. Toombs gave us a hint when she told us the black powder had a new chemical name. 3) How does your explanation (in #2) correlate with the definition of chemical or physical change? This molecule called copper carbonate (CuCO3) is made up of I atom of copper, I atom of carbon, and 3 atoms of oxygen. The Copper (Cu) has a “value” or “charge” of +2 The Carbonate ion (which is actually in a grouping of CO3) has a charge of -2 This is why molecule forms together to make CuCO3 When you remove CO2 from a CuCO3 molecule, you are left with CuO. That means a new molecule is formed (called copper oxide, CuO), and one molecule called carbon dioxide, CO2, has been permanently removed) 4) Why was it necessary to heat the tube? In order to remove CO2 from a CuCO3 molecule, you have to increase the kinetic energy of the CuCO3 molecules so that they are moving fast enough to start to shake apart their bonds. This allows a chunk of the molecule, in the form of CO2 to leave the sample. This CO2 molecule is a gas and is moving fast enough to float right out of the tube. The remaining atoms are the CuO that stay in the form of a course, heavy black powder that stay at the bottom of the test tube. Please note that we did NOT burn the CuCO3. We heated it until it split into two new molecules.