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Determining the Empirical Formula of Copper Chloride Purpose of the Experiment To determine the empirical formula of a compound containing only copper.

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Presentation on theme: "Determining the Empirical Formula of Copper Chloride Purpose of the Experiment To determine the empirical formula of a compound containing only copper."— Presentation transcript:

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

2 The mass in grams of 1 mole of a compound is equal to the mass in grams of the components. Molar Mass (Molecular Weight) H 2 OMW = 2(1.008) g AlCl 3 MW = (35.45) g (Note: The Significant Figures.) So the mass in grams of 1 mole of each compound is…

3 The mass in grams of 1 mole of each compound is… Molar Mass (Molecular Weight) H 2 OMW = g AlCl 3 MW= g

4 Percent Composition Defn: The percentages of a compound’s mass that are due to each of the component elements. Mass of C = 2 x g Mass of H = 6 x g Mass of O = 1 x g Mass of 1 mole of C 2 H 5 OH = ?? C 2 H 5 OH

5 Percent Composition Defn: The percentages of a compound’s mass that are due to each of the component elements. Mass of C = g Mass of H = g Mass of O = g Mass of 1 mole of C 2 H 5 OH = = g C 2 H 5 OH Mass percent of C = ??

6 Percent Composition Defn: The percentages of a compound’s mass that are due to each of the component elements. Mass of C = g Mass of H = g Mass of O = g Mass of 1 mole of C 2 H 5 OH = = g C 2 H 5 OH

7 Empirical Formula Represents the simplest whole-number ratio of the various types of atoms in a compound. What is the Empirical Formula for the following compounds? Isoamyl Acetate Scent of Bananas Examples P 4 O 10 CaffeineDrying Agent C 14 H 28 O 4 C 8 H 10 N 4 O 2

8 Empirical Formula Represents the simplest whole-number ratio of the various types of atoms in a compound. What is the Empirical Formula for the following compounds? Examples P 4 O 10 CaffeineDrying Agent C 14 H 28 O 4 C 8 H 10 N 4 O 2 P2O5P2O5 C 7 H 14 O 2 C4H5N2OC4H5N2O Isoamyl Acetate Scent of Bananas

9 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Empirical Formula Example 1 What is the empirical formula? 1 st Step:

10 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. What is the empirical formula? 1 st Step: Because it is a percent we can choose any mass we want and multiply that mass by the percents of the components. What is the easiest mass to choose? Empirical Formula Example 1

11 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. What is the empirical formula? What is the easiest mass to choose? grams 1 st Step: Because it is a percent we can choose any mass we want and multiply that mass by the percents of the components. Empirical Formula Example 1

12 So since we chose g of compound we have 43.64g P & g O. A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Can we compare grams directly? Empirical Formula Example 1 2 nd Step:

13 43.64g P x ??? g O x ??? So since we chose g of compound we have 43.64g P & g O. 2 nd Step: Convert grams to moles. Can we compare grams directly? A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Empirical Formula Example 1

14 43.64g P x (1 mol P / g P) = mol P g O x (1 mol O / g O) = mol O So since we chose g of compound we have 43.64g P & g O. 2 nd Step: Convert grams to moles. Can we compare grams directly? A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Empirical Formula Example 1

15 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. What can we do to compare the moles? So we now have moles of P & moles of O. Empirical Formula Example 1 3 rd Step:

16 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. What can we do to compare the moles? So we now have moles of P & moles of O. Do this by dividing both mole values by the smaller one. O ? and P ?  3 rd Step: Convert the smaller number to one. Empirical Formula Example 1

17 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. What can we do to compare the moles? So we now have moles of P & moles of O. 3 rd Step: Convert the smaller number to one. Do this by dividing both mole values by the smaller one. O and P  Empirical Formula Example 1

18 So by dividing both mole values by the smaller one we deduced: O and P  This yields the formula PO 2.5 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Is this an acceptable empirical formula? Empirical Formula Example 1

19 So by dividing both mole values by the smaller one we deduced: O and P  This yields the formula PO 2.5 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Is this an acceptable empirical formula? Empirical Formula Example 1 No.

20 What must we do to fix the formula PO 2.5 ? A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers. Empirical Formula Example 1

21 What must we do to fix the formula PO 2.5 ? A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers. 4 th Step: Convert to whole numbers. Do this by multiplying both mole values to get whole numbers. Empirical Formula Example 1

22 What must we do to fix the formula PO 2.5 ? Empirical Formula = P 2 O 5 A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers. 4 th Step: Convert to whole numbers. Do this by multiplying both mole values to get whole numbers. 1 P x 2 = 2 P2.5 O x 2 = 5 O Empirical Formula Example 1

23 Empirical Formula Example 2 Mg (solid – silvery white) + O 2 (gas) → Mg x O y (solid – white) heat (0.353g) limiting reagent (0.585g) resulting mass Atmospheric oxygen is in excess A student burns g of magnesium ribbon in an open crucible. A white powder forms and is found to weigh g. What is the empirical formula? How do we determine the empirical formula?

24 Empirical Formula Example 2 Mg (solid – silvery white) + O 2 (gas) → Mg x O y (solid – white) heat (0.353g) limiting reagent (0.585g) resulting mass Atmospheric oxygen is in excess A student burns g of magnesium ribbon in an open crucible. A white powder forms and is found to weigh g. What is the empirical formula? How do we determine the empirical formula? 1 st Step: Calculate mass percents.

25 Mass of O = total mass of compound – mass of Mg = g Mg = 60.3%andO = 39.7% Empirical Formula Example 2 1 st Step: Calculate mass percents.

26 Formula masses and percent composition of three theoretical compounds of Mg and O Empirical Formula Example 2 2 nd Step: Formula Masses and their Percent Composition. Mg = g/moleO = g/mole Formula of Oxide Mg x O y Formula Weight %Mg%O MgO = MgO (15.999) = Mg 2 O2(24.31) = Mg = 60.3%andO = 39.7%

27 Formula masses and percent composition of three theoretical compounds of Mg and O Empirical Formula Example 2 2 nd Step: Formula Masses and their Percent Composition. Mg = g/moleO = g/mole Formula of Oxide Mg x O y Formula Weight %Mg%O MgO = MgO (15.999) = Mg 2 O2(24.31) = Mg = 60.3%andO = 39.7%

28 Formula masses and percent composition of three theoretical compounds of Mg and O Empirical Formula Example 2 2 nd Step: Formula Masses and their Percent Composition. Mg = g/moleO = g/mole Formula of Oxide Mg x O y Formula Weight %Mg%O MgO = / / MgO (15.999) = / x / Mg 2 O2(24.31) = x 24.31/ / Mg = 60.3%andO = 39.7%

29 Formula masses and percent composition of three theoretical compounds of Mg and O Empirical Formula Example 2 2 nd Step: Formula Masses and their Percent Composition. Mg = g/moleO = g/mole Formula of Oxide Mg x O y Formula Weight %Mg%O MgO = MgO (15.999) = Mg 2 O2(24.31) = Mg = 60.3%andO = 39.7%

30 Formula masses and percent composition of three theoretical compounds of Mg and O Empirical Formula Example 2 2 nd Step: Formula Masses and their Percent Composition. Mg = g/moleO = g/mole Formula of Oxide Mg x O y Formula Weight %Mg%O **MgO = MgO (15.999) = Mg 2 O2(24.31) = Mg = 60.3%andO = 39.7%

31 Today’s Experiment Al(s) + Cu x Cl y (aq)AlCl 3 (aq) + Cu(s) known mass of ~25 ml Limiting reagent (excess) known mass of product (silvery white)(reddish)(blue soln)(gray soln)

32 Ground state electron configuration: [Ar].3d10.4s1 Shell structure: Transition metals may exhibit multiple oxidation states (+1, +2, +3, etc…). These are not easily predicted by position in the periodic table. Transition metal ions in aqueous solutions are frequently brightly colored, this is also due to transitions of electrons in the d orbitals (e.g. Cu ions tend to be blue). Copper is a Transition Metal.

33 Last time we worked with Zinc. Zn and Al are both stronger reducing agents than copper. (Note the redox potentials on next slide.) Because of this either one would work to produce metallic copper from a solution of a copper salt. We chose aluminum to work with because the reaction between Zinc and Copper Chloride is quite exothermic and extreme caution would have had to have been observed to avoid burns.

34 * * * The oxidation of aluminum is more likely than the oxidation of zinc. These potentials indicate the relative thermodynamic tendency for the indicated half-reaction to occur. Al Al e – E = volts Zn Zn e – E = volts Cu Cu e – E = volts

35 2Al(s) + 6HCl(aq) --> 2AlCl 3 (aq) + 3H 2 (g) Cu(s) + n HCl(aq) -x-> No Reaction Removal of Excess Reducing Agent Once you have ensured that the aluminum is in excess (some aluminum will be floating on the surface), then it is time to proceed with the next step: removal of the aluminum. This is accomplished by adding HCl in excess. The HCl reacts with the aluminum and not the copper.

36 Reagents in Lab Cu x Cl y solution in 4L spigot jugs - Take ~25 ml for each run (using 100 ml Grad Cylinder) Checkout 3 pc Al foil – Add small pieces. Do not fold. Add in excess. If there is any remaining unused Al, give to TA. 1 - pair of Beaker Tongs Return to stockroom when you are done. Record data: ( g Cu x Cl y /ml, d=1.074 g/ml) 10% HCl in 1L wash bottles - Take 5-10 ml for each run (using 10 ml Grad Cylinder) (N.B. solid NaHCO 3 is to be used for acid spills.) Record on page 87.

37 Flow Chart for Procedure Casserole 1 – Weigh to the nearest 0.001g. Record mass [3]. 100 ml Graduated Cylinder - Acquire ~25 ml Cu x Cl y solution. Casserole 1 with ~25 ml Cu x Cl y soln – Weigh to the nearest 0.001g. Record mass [4]. Pour Cu x Cl y soln into casserole Casserole 1 with ~25 ml Cu x Cl y soln Obtain Al foil. Tear Al foil into ~1 cm 2 pieces. Add Al foil to Cu x Cl y soln. Submerge foil with glass stirring rod. Do NOT stir. When solution turns green – start adding ~ 1 / 4 cm 2 pieces. When solution is grey stop. Are there small pieces of Al floating on the surface? Yes.No. Add more Al.Decant AlCl 3 soln into waste beaker.

38 Casserole 1 with solid Cu. Beaker 1 - Have TA verify that all Cu has been removed from solution. Decant AlCl 3 soln into waste beaker. Rinse twice with distilled water. Decant rinsings into waste beaker. Obtain ~10 ml of HCl. Pour HCl over Cu in casserole. Stir gently. Are there small pieces of Al visible on the surface? Are there bubbles still forming on the surface? Yes. Add more Al. No. Pour in proper waste container. Add more HCl. Yes. No. Yes. Casserole 1 with damp solid Cu. Beaker 1 with waste AlCl 3 solution. Pour in proper waste container. Flow Chart for Procedure

39 Casserole 1 with damp solid Cu. Place on hotplate (~100 o C) for 5 min.* Take off hotplate. Cool to room temp. Weigh to the nearest 0.001g. Record mass [8] under 1 st heating. *Repeat until 2 successive readings are within 0.010g. (Note: Decrease heating time each successive heating and break apart solid each time with a microspatula to increase surface area.) Is the mass within 0.010g of the previous mass? No.Yes. Stop. Pour solid Cu into “used solids” container. Continue with next heating. Flow Chart for Procedure

40 Highlights of Procedure Notes  Record all weights to 0.001g.  Weigh ~25 ml of Cu x Cl y solution, use exact density to calculate exact volume, then calculate the mass of Cu x Cl y.  Do not use the metal microspatulas when stirring solution.  Add Al foil until blue color is gone, allow excess foil to dissolve also.  Allow container to cool before weighing.  Speed up cooling by placing under hoods.  Break apart solid Cu between weighing with microspatula.

41 Hazards  10% HCl - strong acid, corrosive  Cu x Cl y solution - heavy metal, irritant  Hot surfaces - hotplates, glassware (Be Careful: Hot glassware looks just like cold glassware.) Waste  Liquid Waste: Al +3 / HCl  Used Solids: Cu

42 Homework Assignment – Due Feb Complete: Oxidation / Reduction (Packet pages 75-78) For February Read: Ternary Mixture (Text pp ) Turn-In: Empirical Formula (Text pp ) along with a Calculations Page* (explained on p.87) & Oxidation / Reduction (Packet pp 75-78). *Note that on the bottom of page 87 it states that the calculations which need to be shown on a separate sheet of paper are indicated by asterisks (*).


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