Step 2. Heat test tube until all of the O2 has evolved Step 3. Record the amount of O2 produced Observations Mass of O2 produced: 3.915576 grams or .1224 moles Mass of KCl produced: 6.083363 grams or .0816 moles Calculations Chemical Reaction 2KClO3-> 2KCl + 3O2 Theoretical Yield (96/245.1)*10 grams= 3.916 grams of O2 Percent Yield 3.915576/3.916= .99989 *100= 99.998 percent Data Actual amount of Oxygen Produced |3.915576 grams | |Theoretical amount of oxygen Produced |3.916 grams of O2 | |The Percent Yield |99.998 percent | | Results The amount of oxygen produced was almost exactly what was expected with a percent yield of 99.998 percent. The Law of conservation of mass was also up held with there being a combined mass of 10 grams from the resulting O2 and KCl. Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium.
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
Acid test= solid + 3M Acetic acid cloudy + 6 ZnO ZnO + Heat yellow Zn Cooled White + ZnCO3 + heat yellow solid. Yellow solid cooled white solid + QUANTITATIVE DATA AND CALUCATIONS Desired Results Calculations Temperature of Water Bath 22.8oC Vapor Pressure (as
StudyBlue Flashcard Printing of Lab Final 2211L UGA http://www.studyblue.com/servlet/printFlashcardDeck?deckId=... In the distillation experiment, the purpose of a fractionating column was to Which liquid would be most easily separated from water by simple distillation? 1. 1-propanol (bp=97 degrees C) 2. 2-propanol (bp=82 degrees C) 3. tetrahydofuran (bp=65 degrees C) 4.
White precipitate shows the presence of chloride (Cl-). Chloride anion equation: HCl(aq) + AgNO3 (aq) → HNO3 (aq) + AgCl(s). The nitrate anion test involves cooling a mixture containing 1 mL of test solution and 3mL 18M H2SO4. 2mL is poured down the inner test tube side and the presence of a brown ring shows nitrate (NO3-) to be present. The carbonate anion test mixes 1 mL of test solution and drops of 6M HCl.
| Observations of Chemicals | Zinc Sulfate | Powder of a white solid | Barium Iodide | Powder of a white solid. | Deionized water | Liquid, transparent. | Trial # | BaI2 | ZnSO4 | Theoretical Yield of ZnI2 | Actual Yield | Percent Yield | 1 | .67g | .45g | .499820g | .52g | 104% | 2 | .67g | .45g | .499820g | .52g | 104% | 3 | .66g | .46g | .493117g | .48g | 97% | Calculations for Cost | Double Replacement | Synthesis | 0.48 grams of Zinc Sulfate - $0.02 | 1.00 gram Granular Zinc - $62.50 | 0.67grams of Barium Iodine Dihydrate - $0.886 | 2.00 gram Iodine - ($74.90 × 2) - $149.80 | 0.52 grams of Zinc Iodide - $0.906 | 1.00 gram zinc - $0.212 | 1000 grams of Zinc Iodide = $1,923.00 | 1000 grams of Zinc Ioidide = $212.30 | Focus Question Should chemists prepare Zinc Iodide, from its Elements or from a Double Replacement Reaction between Barium Iodide and Zinc Sulfate?
Synthesis of Methyl Stearate Post-Lab Submitted by Matthew Sharma TA: Evan Determining the Limiting Reagent Methyl Oleate (MW= 296.49 g/mol) Amount used = 1.141g (1.141g)(1mol / 296.49g) = 0.003848 mol ∴ The limiting reagent for the reaction is methyl oleate because hydrogen is used in excess Theoretical Yield Methyl Stearate (0.003848 mol)(298.504g/mol product) = 1.1486 g methyl stearate Percent Yield Given the 1:1 stoichiometry of the reaction: 100% x (0.2551g product / 1.1486g theoretical yield) = 22.2% Conclusion A 22.2% yield of synthesized methyl stearate was obtained via the catalytic hydrogenation of methyl oleate in the presence of the catalyst 10% palladium on carbon. The product
2) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.43/1.01) x100% = 42.57% That concludes that the percent recovery is 42,57%. 3) Percent recovery for isolation of hydroquinone dimethyl ether % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.16/1.01) x100% = 15.84% That concludes that the percent recovery is 15.84%. Table 2: : Experimental IR peaks compared to literature IR peaks for Benzoic acid Functional groups | Experimental peak (cm-1) | Literature peak (cm-1) | O-H | 3407-2563 | 3400-2564 | C=O | 1689 | 1689 | C-H |
Part B: The graduated pipet’s average density at 22.3 °C was determined to be 0.9785g/mL with a percentage error of 1.89% shows the graduated pipet to be more accurate and precise. Part C: Density of an unknown NaCl solution was measured and a calibration curve used to determine the percentage of NaCl by mass in the solution. y=0.007x + 0.998 which concluded that the concentration of the sodium chloride solution was 3.14%. INTRODUCTION Anything that you can see, touch, taste or smell, occupies space and has mass, it is called matter. Matter can be a gas, a liquid,
Experiment 6: Physical & Chemical Properties Title and Identifiers: Heating of chemical properties 01/23/15 Purpose / Objective: • To investigate the chemical properties of pure chemical substances • To investigate the physical properties of pure chemical substances Procedural Outline: • Matches • Beaker, 100 mL, glass • Burner-fuel • Goggles-Safety • Stirring rod - Glass • Test Tube(5), 13 x 100 mm in Bubble Bag • Test-tube-clamp-holder • Test-tube-cleaning-brush • Well-Plate-24 • Well-Plate-96 • Litmus Paper, Blue • Litmus paper, Red • Copper (II) Carbonate in Vial, 1/2 Full • Copper (II) Nitrate Crystals in Vial • Copper Metal