1 / 22

Determining the Empirical Formula of Copper Chloride

Determining the Empirical Formula of Copper Chloride. Purpose of the Experiment Determine the empirical formula of a compound containing only copper and chlorine. Molar Mass or Molecular Weight. The molecular weight is the mass, in grams, of one mole of a compound.

kenisha
Download Presentation

Determining the Empirical Formula of Copper Chloride

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Determining the Empirical Formula of Copper Chloride Purpose of the Experiment Determine the empirical formula of a compound containing only copper and chlorine

  2. Molar Mass or Molecular Weight The molecular weight is the mass, in grams, of one mole of a compound. H2O MWt = 18.0148 g/mol AlCl3 MWt = 133.341 g/mol

  3. Percentage Composition The percentage composition is the percent of a compound’s mass that results from each of its constituent elements. For ethanol, C2H5OH, 2  12.011 g/mol C = 24.022 g/mol of C2H5OH 6  1.008 g/mol H = 6.048 g/mol of C2H5OH 1  15.999 g/mol O = 15.999 g/mol of C2H5OH Molecular Weight = the sum = 46.069 g/mol of C2H5OH % C = (24.022 g/mol ÷ 46.069 g/mol) 102 = 52.14 % % H = ( 6.048 g/mol ÷ 46.069 g/mol) 102 = 13.13 % % O = (15.999 g/mol ÷ 46.069 g/mol) 102 = 34.73 % Sum = 100 %

  4. Empirical Formula The empirical formula is the simplest whole- number ratio of the atoms in a compound. Examples: benzene, CH phosphorus pentoxide, P2O5 Molecular Formula The molecular formula is the simplest whole- number ratio of the atoms in a single molecule of a compound. Examples: benzene, C6H6 phosphorus pentoxide, P4O10

  5. Schematic diagram of the combustion device used to analyze substances for carbon and hydrogen CxHy + O2 ----> y/2H2O + xCO2 excess

  6. Determination of the Empirical Formula A white compound has been analyzed and found to contain 43.6 wt. % phosphorous and 56.4 wt. % oxygen. In a separate study it molecular weight has been found to be 284 g/mol. What is the empirical formula of this compound? Assume exactly 100 g of the compound, then one has 43.6 g P ÷ 30.974 g/mol P = 1.408 mol of P 56.4 g O ÷ 15.999 g/mol O = 3.525 mol of O then the molar ratios are 1.408 mol ÷ 1.408 mol = 1.000 mol P 3.525 mol ÷ 1.408 mol = 2.503 mol O and the empirical formula is P2O5.

  7. Determination of the Molecular Formula A white compound has been analyzed and found to contain 43.6 wt. % phosphorous and 56.4 wt. % oxygen. In a separate study it molecular weight has been found to be 284 g/mol. What is the molecular formula of this compound? First determine the empirical weight of P2O5: 2 30.974 g/mol P = 61.948 g/mol P 5 15.999 g/mol O = 79.997 g/mol O The empirical weight of P2O5 = the sum = 141.945 g/mol Molecular weight ÷ Empirical weight = 284 g/mol ÷ 141.945 g/mol = 2.001 and the molecular formula is twice the empirical formula and the molecular formula is P4O10.

  8. Atmospheric oxygen in excess heat Mg(s, silvery-white) + O2(g) MgxOy(s, white) 0.353 g of Mg solid limiting reagent Result: Mass of MgxOy = 0.585 g Determining the Empirical Formula of Magnesium Oxide The goal is to determine x and y in the unknown magnesium oxide, MgxOy, by burning a known mass of Mg in an excess of oxygen.

  9. Determining the Empirical Formula of Magnesium Oxide Note: mass of O = total mass – mass Mg = 0.585 g – 0.353 g = 0.232 g then % Mg = (mass Mg ÷ total mass MgxOy) 102 = (0.353 g ÷ 0.585 g) 102 = 60.3 % % O = (mass O ÷ total mass MgxOy) 102 = (0.232 g ÷ 0.585 g) 102 = 39.7 % Mg = 60.3 % and O = 39.7 %

  10. Mg = 60.3% and O = 39.7% Stoichiometry and the percentage composition of three possible magnesium oxides Possible Oxides MgxOy Mg O 60.3 % 39.7 % MgO 43.2 % 56.8 % MgO2 Mg2O 75.2 % 24.8 % and the unknown oxide is MgO

  11. Today’sExperiment Original experiment Zn(s, silvery white) + CuxCly(aq, blue) ZnCl2(aq) + Cu(s, reddish) Modified experiment Al(s, silvery white) + CuxCly(aq, blue) AlCl3(aq) + Cu(s, reddish) ~0.3 g ea. strip (excess) 25 mL Limiting reagent known mass known mass

  12. From: http://wine1.sb.fsu.edu/chm1045/tables/period/PT_large.jpg

  13. Copper is a transition metal with partially filled d orbitals. A transition metal may exhibit multiple oxidation states, such as +1, +2, or +3, oxidation states which are not easily predicted by its position in the periodic table. Because of the partially filled d orbitals, a transition metal ion in aqueous solution frequently is brightly colored, e.g., copper ions are blue in aqueous solution. Zinc and aluminum are both stronger reducing agents than copper, see their redox potentials, so either will yield metallic copper from a solution of a copper salt.

  14. These potentials indicate the relative thermodynamic tendency for the indicated half-reaction to occur.

  15. Other Reactions in the Procedure: Removal of Excess Reducing Agent Zn(s) + 2 HCl(aq) ----> ZnCl2(aq) + H2(g) 2 Al(s) + 6 HCl(aq) ----> 2 AlCl3(aq) + 3 H2(g) Cu(s) + n HCl(aq) --x--> no reaction Note: Cu is below H (0.00 V ref) in table, so It will not react with acids H+(aq) to form H2(g) Metals above H in the table will react with acids.

  16. Checkout 2 piece Al foil, ~0.3 g ea 1 pr Beaker Tongs Reagents in Lab CuClx solution in 4 L spigot jugs, use ~25 ml for each analysis Record data: 0.08067 g CuClx/ml, d = 1.074 g/ml 10 % HCl in 1 L wash bottles, use ~5 ml Use solid NaHCO3 on acid spills

  17. Add Al foil 25 mL copper chloride, weigh and use exact density to get mass of CuClx Stir (takes about 5 min) Add 5-10 drops of 10 % HCl and stir (HCl will dissolve excess Al) Decant the supernatant liquid Cu Flow Chart for Procedure waste

  18. Flow Chart for Procedure Cu Wash with distilled water to remove aluminum chloride Transfer Cu residue to a pre-heated and pre-weighed casserole waste CAUTION: Do not overheat to avoid oxidation heat Determine the mass of Cu waste

  19. Procedure Notes • Record all weights to 0.001g • Weigh 25 ml of CuClx solution, use exact density to calculate exact volume, then calculate the mass of CuClx • Do not use metal forceps or spatulas • Add Al foil until blue color is gone, allow excess foil to dissolve • Allow container to cool before weighing • Speed up cooling by placing flask in front of hood sash raised 4-6” • The second beaker does not have to be 150 mL • A casserole will also work as an evaporating dish

  20. Hazards • 10 % HCl is a corrosive strong acid • CuClx solution-heavy metal, irritant • Hot surfaces - hotplates, glassware Waste • Liquid waste: Al+3 solution and HCl • Cu solids

  21. Summary of Data & Calculations Collected data Results (calculations) Mass percent of Cu Mass of Cl Mass percent of Cl Empirical formula Mass of CuClx Mass of Cu

  22. Next Week’s Experiment: Antacid Analysis (packet) Additional background reading for Antacid Analysis/Titrations: Atkins, “Chemical Principles”, 3rd ed, pp. F67-F72, F85-F88, 415-425

More Related