Nitration of Naphthalene Wed 2/25/2015 Lab report # 1 Abstract: The purpose of this experiment was to nitrate naphthalene with nitronium ion‚ which is formed at low concentration from a reaction of nitric acid and sulfuric acid. The percent yield from the experiment was 54.4% of the product‚ and the melting point of the possible results were 59 °C for 1-nitronaphthalene‚ and 78°C for 2-nitronaphthalene. Introduction: Polynuclear aromatic hydrocarbons such as naphthalene can be nitrated
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Recrystallization Abstract: Technical grade aniline was reacted with acetic anhydride to give acetanilide a dark brown color‚ due to the presence of impurities. Crystallization of the crude product from water was combined with decolorization with activated carbon‚ furnished pure acetanilide as white flakes‚ melting point 112-115°C‚ yield 70.3%. Experimental: 2 grams of technical grade aniline and 15ml water were placed in a 125ml Erlenmeyer flask. Then‚ 2.5ml of acetic anhydride was
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Gladys Mesta Dr. Mbah December 6‚ 2012 Percent Yield of a Chemical Reaction Introduction Yield is the quantity of product in a chemical reaction‚ the theoretical yield of a reaction can be calculated using mole ratios from the balanced chemical reaction. The actual yield has to be obtained and measured in a laboratory. It may be usual to often find the actual yield to be less than the theoretical yield due to many different factors. This gives rise to the concept of percent yield. Sodium bicarbonate
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4-nitroaniline had a percent recovery of 18.67% and naphthalene had a percent recovery of 80%. Naphthalene had a relatively high percent recovery‚ and the loss of product can be explained by error inherent. The low percent recovery of 4-nitroaniline is mainly due to the fact that anhydrous sodium sulfate was accidently added to the flask. Therefore‚ some product was lost as the aqueous solution with the protonated 4-nitroaniline reacted with the drying agent. However‚ more errors to explain the results
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Although I love science‚ I encountered more problems in this subject than any other. Recently‚ I was assigned a lab. The purpose was to let Copper Sulfate react with Aluminum and obtain Copper. Before the experiment‚ I set up the stoichiometric equation carefully‚ predicting the production of Copper using my assigned mass of Copper Sulfate. Additionally‚ I intentionally made Copper Sulfate an excess in my equation‚ since it would be dissolved in water and I would only have to collect Copper at the
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Purposes of Experiment: To synthesize a transition metal complex‚ potassium tris (oxalate)ferrate(III) trihydrate in a two step process‚ to learn new laboratory techniques such as decantation‚ recrystallization‚ gravity and suction filtration. Also the purpose of the experiment is to determine the actual‚ theoretical‚ and percent yields of product‚ and characterize the final compound by determining the number of waters of hydration by gravimetric analysis List of Observations: After adding
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CHEMICAL REACTIONS OF COPPER AND PERCENT YIELD Objective To gain familiarity with basic laboratory procedures‚ some chemistry of a typical transition element‚ and the concept of percent yield. Apparatus and Chemicals |0.5 g piece of no. 16 or no. 18 copper wire |evaporating dish | |250 mL beaker (2) |weighing paper | |concentrated HNO3 (4 – 6 mL)
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Chapter 3: Stoichiometry 3: Stoichiometry 5: Thermochemistry 8: Covalent Bonding and Molecular Structure 15: Chemical Equilibrium 16: Acids and Bases 3.2 Stoichiometry and Compound Formulas 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis Chapter Summary Chapter Summary Assignment Reference Tools Periodic Table Molarity Calculator Molar Mass Calculator Unit
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Title: Stoichiometry Reaction Objectives: 1. To decompose sodium hydrogen carbonate (sodium bicarbonate) by heating. 2. To accurately measure the degree of completion of the reaction by analysing the solid sodium carbonate product. 3. To calculate amount of product with given amount of reactant. 4. To determine amount of heat release in the reaction. Results: Part 1: Thermal Decomposition of NaHCO3 Materials Mass (g) Clean and dry test tube 15.1632 Clean test tube + NaHCO3 17.1647
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Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards‚ I divided 339.1584 by 84.007 and
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