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Place your recovered materials in the appropriate labeled collection 3. Cleaning

ID: 505413 • Letter: P

Question

Place your recovered materials in the appropriate labeled collection 3. Cleaning Up containers as directed by your laboratory instructor. Clean your glassware with soap or detergent. CAUTION wash your hands thoroughly with soap or detergent before leaving the on MICRO SCALE ALDO L COND ENSATI ON Equipment 2 beakers, 100-mL melting point capillary tubes 2 conical vials, 5-mL 10-100 HL micropipet 10-mL Erlenmeyer flask. 3 Pasteur pipets, with latex bulb 25-mL filter flask, with tubing 2 pipets, 1.0-mL filter paper product vial glass stirring rod red litmus paper Hirsch funnel, with adapter support stand hot plate 2 test tubes, 13 x 100-mm labels thermometer, -10 to 260 C magnetic stir bar utility clamp magnetic stirrer 2 watch glasses marking pen "or 10mL reaction vial Reagents and Properties quantity (g/mol) mp (C) by d (g/ml.) 30 mg 58.08 56 0.79 benzaldehyde 105 mg 106.12 178 1.044 dibenzalacetone" ethanol 2.8 mL sodium hydroxide 1.0 mL 40.00 Preview Prepare an ice-water bath Place ethanol and NaOH solution into a beaker Cool the ethanol-NaOH solution to 20 C Prepare a mixture of benzaldehyde and acetone Add the benzaldehyde acetone mixture to the ethanol-NaOH solution Stir the mixture for 30 min Cool the mixture in an ice-water bath Rinse the crystals with water

Explanation / Answer

Ans. Balanced reaction: 1 Acetone + 2 Benzaldehyde ---------> 1 Dibenzalacetone

Stoichiometry: 1 mol acetone reacts with 2 mol benzaldehyde to form 1 mol dibenzalacetone.

            Theoretical molar ratio of reactant = Acetone : benzaldehyde :: 1: 2

Given:

            I. Mass of acetone = 30.0 mg = 0.030 g                           ; [1 g= 1000 mg]

                        Moles of acetone = Mass/ molar mass

                                                = 0.030 g/ (58.08 g/ mol) = 5.165 x 10-4 mol

            II. Mass of benzaldehyde = 105 mg = 0.105 g

                        Moles of benzaldehyde = 0.105 g / (106.121 g/ mol) = 9.894 x 10-4 mol

Experimental molar ratio of reactant =

Acetone : benzaldehyde :: 5.165 x 10-4 : 9.894 x 10-4

                                                = 5.165 : 9.984

Divide the ratio by 5.165 so that moles of acetone become 1. Note that multiplying or dividing a ratio with any factor does not affect the ration.

            So, Acetone : benzaldehyde :: (5.165 : 9.984) / 5.165 = 1: 1.91

So, experimental molar ration = Acetone : benzaldehyde :: 1: 1.191

Now, compare the theoretical and experimental molar ratio of reactions. The experimental moles of benzaldehyde is less than theoretical value while keeping moles of acetone constant.

That is, experimental moles of benzaldehyde < theoretical moles of benzaldehyde    

Thus, benzaldehyde is the limiting reagent.

#2. According to the stoichiometry of balanced reaction, 2 mol benzaldehyde (limiting reagent) forms 1 mol Dibenzalacetone.

So,

Theoretical moles of Dibenzalacetone produced = (1/2) x moles of benzaldehyde

Or, Theoretical moles of Dibenzalacetone produced = (1/2) x 9.894 x 10-4 mol

                                                = 4.947 x 10-4 mol

Theoretical mass of Dibenzalacetone produced = Moles x Molar mass

                                                = 4.947 x 10-4 mol x (224 g/ mol)

                                                = 1108.128 x 10-4 g

                                                = 0.1108128 g

                                                = 110.81 mg

Therefore, theoretical yield of the experiment = 110.81 mg