Free Essay

Stoichiometry

In:

Submitted By bekahmuller
Words 1019
Pages 5
Stoichiometry
Rebekah Muller
Chemistry Lab 101-02
Instructor: Sam Stevenson

Abstract: Stoichiometry is used to figure out how much potassium chlorate in a mixture of potassium chlorate and an inert substance (a substance that does not react chemically). By determining the mass of the product experimentally, the mass percent composition is calculated stoichometrically. After finding the mass percent composition, the final calculation to find is percent error (with the accepted value given). In this experiment, we use a solid that contains potassium chlorate as the reactant and then determine how much potassium chlorate was actually in the solid. To determine the amount of potassium chlorate, we heat up the solid on a Bunsen burner and measure how the solid as changed while the reaction was taking place.
Objective:
There are four objectives to this lab: to calculate the theoretical and percent yields, predict the effect of experimental error on percent yield, calculate the percent composition, and calculate the percent error.
Introduction:
The purpose of this lab is to determine the mass percent composition of potassium chlorate in an unknown substance by using Stoichiometry. Stoichiometry is the calculation of the quantities of chemical elements or compounds involved in chemical reactions. In this lab, the decomposition of the reaction would further the calculations of potassium chlorate in the mixture. Decomposition means the breaking down of compounds or mixtures into the form of elements or smaller substances. The hypothesis trying to be proven is that it is possible is find the mass of a compound can be determined through Stoichiometry and experimentally. Experimentally, the mixture will be heated. If the heating continues for an extended amount of time, the reaction will not be affected because the reaction has already been completed.
Results:
The results data table is connected on another page.
Discussion:
In order to calculate the percent composition, the amount of oxygen after decomposition needed to be measured. Trial one measured .24 grams of oxygen had precipitated. These .24 grams of oxygen were converted to moles of oxygen by the use of molar mass (18 grams per mole). By using the mole to mole ratio for potassium chlorate to oxygen and molar mass of potassium chlorate, the amount of potassium chlorate was calculated to be .613 grams. The percent composition was then found by dividing the found amount of potassium chlorate by the amount of potassium chlorate put into the cylinder before decomposition take place (1.036 for trial one). The quotient of the problem needs to be multiplied 100 percent and the product will be the percent composition of the potassium chlorate for that specific trial. The percent compositions for all three trials were 59.2%, 72.7%, and 70.0% respectively. To find the percent error, the percent composition found for each trial and the given value percent composition was used. The absolute value of the computed value the given, divided by the given. After obtaining this value, the value should be multiplied by 100 percent to get the percent error. Each percent error ended up different for each trial: trial one=15.43%, trial two=3.86% and trial three=.43%. All three trials’ percent error was below 25% meaning that the results from the trials were fairly accurate for each trial. Throughout the experiment there were possible sources of error. Random error is error in measurement that leads to measurable values being inconsistent when repeated measures of a constant attribute or quantity are taken. Systematic error is biases in measurement which lead to the situation where the mean of many separate measurements differs significantly from the actual value of the measured attribute. One source of error would be the drying process. Instead of drying the test tube over the Bunsen burner, drying the test tube with a paper towel could leave behind particles that could potentially change the results of the experiment if the particles catch on fire. Another possible source of error is if the reaction does not completely take place. If the reaction did not completely take place, the amount of oxygen that evolved would decrease because not all of the potassium chlorate reacted during the procedure. These errors could be avoided by taking the time to dry the test tube. Even when the test tube looks dry, take another minute to be cautious of the fact that it may not be completely dry. To avoid the error of not having the reaction fully take place, use pre-caution as before. If the reaction was heated for a longer period of time, there would be certainty that the full reaction had taken place. By using the same amount of the mixture after the oxygen has been combusted then the oxygen amount would stay the same regardless of the amount of time.
Conclusion:
Once completing the experiment, the calculations were carried out. By taking the average of the percent errors from each trial, it can be concluded that with such a low percent error the data is quite accurate.
Experimental:
To begin the experiment, first two test tubes must be thoroughly cleaned with distilled water; and then turn on the Bunsen burner. Use the burner to dry the test tube. Hold the test tube about one inch above the flame. (DO NOT DRY THE TEST TUBE USING A PAPER TOWEL!!!) Once the test tubes dried, the experiment is ready to take place. This experiment required three trials used one gram of the unknown sample for each trial. Determine the weight of the test tube and the amount of the sample before conducting each trial. Put the sample into the test tube and hold the test tube over the Bunsen burner for about five minutes to ensure that the reaction has completely taken place. Then determine the new weight of the test tube with the sample still in the test tube. Collect data in an organized fashion or data table. Conduct three trials to lessen the amount of error. After carrying out three trials, clean up and leave the lab area exactly how it was found.

*no references were used in the writing of this paper*

Similar Documents

Premium Essay

Stoichiometry

...Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction Objectives: • Observe the reaction between solutions of sodium carbonate and calcium chloride. • Determine which of the reactants is the limiting reactant and which is the excess reactant. • Determine the theoretical mass of precipitate that should form. • Compare the actual mass with the theoretical mass of precipitate and calculate the percent yield. Materials: Balance 0.70 M sodium carbonate solution, Na2CO3(aq) Graduated cylinder 0.50 M calcium chloride solution, CaCl2(aq) Beaker (250 mL) Wash Bottle (distilled H2O) Filter paper Funnel Iron ring Ring stand Procedure: Part I: The Precipitation Reaction (Day 1) 1. Obtain two clean, dry 25 mL graduated cylinders and one 250 mL beaker. 2. In one of the graduated cylinders, measure 25 mL of the Na2CO3 solution. In the other graduated cylinder, measure 25 mL of the CaCl2 solution. Record these volumes in your data table. 3. Pour the contents of both graduated cylinders into the 250 mL beaker and observe the results. Record these qualitative observations in your observations table. Allow the contents of the beaker to sit undisturbed for approximately 5 minutes to see what happens to the suspended solid particles. Meanwhile, proceed to step 4. 4. Obtain a piece of filter paper and put your initials and your partner’s initials on it using a pencil. Measure and record the mass of the filter paper, then use...

Words: 821 - Pages: 4

Free Essay

Lab 4: Stoichiometry

...Braidon Berry Lab 4: Stoichiometry of a Precipitation Reaction Background: Stoichiometry is a challenging yet essential part of learning chemistry. By utilitizing the resources of this course, as well as the lab work. Learning this difficult section of the course can be done with hard work and sharp attention. Purpose: The purpose of this lab was to learn how to connect calculations with real life examples. Doing this lab while doing the calculations for the molarity and molar masses, made it more challenging, however it did help in seating the knowledge and connecting it to the course. Procedure: To start, the dry Calcium Chloride Dihydrate, was weighed out and mixed with a small amount of distilled water. After that step the molarity of the CaCl2 was determined and logged in the data sheet. Similar steps were followed in terms of measuring dry material, weighing it, and then mixing it with distilled water. This will be necessary in order to observe the reaction and calculate more. Then the water was now filtered in the paper filter, this filter would later be dried in the sunlight and weighed to measure the mass of the product. After the filtering was complete, the percent yield could be calculated, as well as the actual mass of the precipitate. Data: |Initial: CaCl2*2H20 (g) |1.0 g | |Initial: CaCl2*2H20 (moles) ...

Words: 357 - Pages: 2

Premium Essay

Stoiochiometry of a Precipitation of Reaction

...Lab: STOIOCHIOMETRY OF A PRECIPITATION OF REACTION By: Ruth Pierre 6/1/15 PURPOSE Stoichiometry is a section of chemistry that involves using relationships between reactants and products in a chemical reaction to determine desired quantitative data. Stoichiometry means the measure of elements. In order to use stoichiometry it is important to understand the relationships that exist between products and reactants and why they exist, which require understanding how to balanced reactions. The purpose of this experiment is to predict the amount of product produced in a precipitation reaction using stoichiometry. This experiment will also to determine the actual yield versus the theoretical yield in order to calculate the percent yield. MATERIALS * Distilled water * Paper towels * Small paper cup * Coffee cup or mug * Beaker 100 ml glass * Funnel * Cylinder 25 ml * Goggles for safety * Scale Digital-500g * Weighing boat plastic * CaCl2·2 H2O-Calcium Chloride, Dihydrate - 2.5 g * Filter Paper 12.5-cm, Na2CO3 -Sodium Carbonate - 2 g PROCUDURE 1. Put on your goggles. 2. Weigh out 1.0 g of CaCl2·2H2O and put it into the 100-mL beaker. Add 25 mL of distilled water and stir to form the calcium chloride solution. Use only distilled water since tap water may have impurities that interfere with the experiment. 3. Use stoichiometry to determine how much Na2CO3 you will need for a full reaction. 4. Weigh the calculated...

Words: 1172 - Pages: 5

Free Essay

Tutorial

...UNIVERSITI TUNKU ABDUL RAHMAN Centre Course Year/ Trimester Lecturer : : : : Centre for Foundation Studies Foundation In Science Year 1 Trimester 1 Unit Code Unit Title Session : : : FHSC1114 Physical Chemistry 2014/05 Ms Azlina Banu, Mr Tam Yun Hong, Ms Chong Pui Kuan, Ms Amelia Chiang Kar Mun, Ms Farhanah, Ms Wong Jing Tyng, Ms Jamie Anne, Ms Lau Mei Chien, Mr Ng Sweet Kin, Ms Phang Ying Ning, Ms Precilla, Ms Rachel Tham, Ms Rajalakshmi, Mr Tan Jun Bin, Ms Tan Lee Siew Tutorial 3 : Chapter 3 Stoichiometry and Solution Concentration 1. Balance the following equations: (a) Al(s) + NH4ClO4(s) → Al2O3(s) + AlCl3(s) + NO(g) + H2O(g) (b) GaBr3(aq) + Na2SO3(aq) → Ga2(SO3)3(aq) + NaBr(aq) 2. Ethanol, C2H5OH, is a liquid with a density of 0.789 g ml-1 at 25 °C. Calculate the molarity of ethanol solution made by dissolving 20.00 mL of ethanol at 25 °C in enough water to make 250.0 ml of solution. [Ans: 1.37 mol L-1] 3. Copper sulfate is widely used as a dietary supplement for animal feed. A lab technician prepares a “stock” solution of CuSO4 by dissolving 79.80 g of CuSO4 in enough water to make 500.0 mL of solution. (a) Determine the molarity of the CuSO4 “stock” solution prepared by the technician. [Ans: 1 mol L-1] (b) Calculate the volume of CuSO4 “stock” solution that should be diluted to give 2.5 L of [Ans: 0.25 L] 0.1 M CuSO4. 4. Aluminum is a limiting reactant in the reaction with sulfur gas to form aluminum sulfide. Initially, 1.18 mol of aluminum and 2.25 mol of...

Words: 447 - Pages: 2

Free Essay

Lab Chapter 3

...CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS This chapter reviews the mole concept, balancing chemical equations, and stoichiometry. The topics covered in this chapter are: • Atomic mass and average atomic mass • A vogadro’ s number, mole, and molar mass • Percent composition calculations • Empirical and molecular formula determinations • Chemical equations, amount of reactant and product calculations • Limiting reagents and reaction yield calculations Take Note: It is absolutely essential that you master the mole concept to do well on the quantitative aspects of AP Chemistry!! When solving quantitative problems on the Free Response section of the AP exam, supporting work must be shown to receive credit. Using dimensional analysis is a very powerful technique in solving problems. Be sure to report your answer to the correct number of significant figures (see Chapter 1 in this review book). Atomic mass and average atomic mass Atomic mass is the mass of an atom in atomic mass units (amu). One amu is defined as 1/12 of one C-12 atom. The C-12 isotope has a mass of exactly 12.000 amu. The C-12 isotope provides the relative scale for the masses of the other elements. Average atomic mass is the value reported on the periodic table, which takes into account the various isotopes of an element and their respective frequencies. To calculate the average atomic mass of an element, add up the different masses of the isotopes (using amu) multiplied by each isotope’s abundance...

Words: 5108 - Pages: 21

Free Essay

Science

...UNIVERSITI TUNKU ABDUL RAHMAN Centre Course Year/ Trimester Lecturer : : : Centre for Foundation Studies Foundation In Science Year 1 Trimester 1 Unit Code Unit Title Session : : : FHSC1114 Physical Chemistry 2015/05 : Ms. Amelia Chiang, Ms. Azlina Banu, Ms. Farhanah, Ms.Gurpreet, Ms. Jamie, Ms. Lau Mei Chien, Ms. Lily Lee, Ms. Nabilah, Mr. Ng Sweet Kin, Ms. Phang Ying Ning, Ms. Precilla, Ms. Rachel Tham, Ms. Rajalakshmi, Mr. Sivabalan, Ms. Tan Lee Siew Tutorial 3: Chapter 3 Stoichiometry and Solution Concentration 1. Balance the following equations: (a) (b) 2. V2O5(s) + CaS(s)  CaO(s) + V2S5(s) GaBr3(aq) + Na2SO3(aq)  Ga2(SO3)3(aq) + NaBr(aq) 316.0 g of aluminum sulfide, Al2S3 reacts with 493.0 g of water, H2O. Given the unbalanced equation as below: Al2S3(s) + H2O(l) → Al(OH)3(s) + H2S(g) (a) Find the excess reactant. (Ans: H2O) (b) Find the mole of the excess reactant that remains after the reaction. (Ans: 14.742 mole) [Sep 2014] 3. Consider the reaction below: 2Al(s) + 3I2(s)  2AlI3(s) (a) Determine the limiting reagent and the theoretical yield of the product if 1.20 moles of aluminium and 2.40 moles of iodine are used. (Ans: 489.218 g) (b) Calculate the percentage yield of the product if 450 g of AlI3 is obtained. (Ans: 91.98%) 4. A salt solution is produced when 2.9 g of sodium chloride, NaCl dissolved in 200 ml of water. Calculate the molality (m) of the NaCl solution...

Words: 365 - Pages: 2

Free Essay

Yolo

...enjoy with our cookies. | 5 | Percentage Composition | Read pgs 258-261Chewing Gum LabDry Lab pg 260Pg 260 # 1-10 (even)Pg. 262 # 1-6 | 6 | Empirical Formula | Nut boltide labRead pgs 268-270Pg 270 # 7, 9, 12Read pgs 271-276Do # 31-40 (even) pg 275 | 7 | Combustion Analysis type problem and Hydrate Lab Prep | Pg 277 # 16, 17, 18 (Pre-lab) + data table for 3 hydrates | 8 | Mole Airlines flight 1023 | Work in groups of 4 and hand in at the end of the period with full calculations and ID chart filled outPg 225 # 41-47 (odd)Pg 278 # 51-60 (even)Pg 279 # 1-16 (all) | 9 | Problem Set on Empirical Formula and % Composition | Work in pairs and set up a solutions gallery with chart paper solutions in review for their quizQUIZ | 10 | Stoichiometry | Bicycle Activity – IntroRead pgs 296-299Pg 298 # 1-10 (odd)Pg 299 # 1-6 (even)Pg 300 # 11-20 (odd) | 11 | Chalk Lab | Activtiy 7.1 pg 301Mole MethodRead pgs 301-304 Pg 304 #21-30 (even)Pg 305 # 1, 4, 6, 7,10, 11 | 12 | Limiting Reagent Problems | Socratic – full step by step with a problemRead pgs 306-308Pg 309Do # # 31-39 (even) (practice)Do # 8, 10,...

Words: 448 - Pages: 2

Premium Essay

Study Guide

...CHEMISTRY 1031 (CHILDS) – STUDY GUIDE FOR EXAM II Tuesday October 14 BE 160 Exam II will cover the material I lectured on from chapters 4 and 5. You are also expected to retain material from chapters 1, 2, and 3. Everything covered in lecture is fair game for the exam. You will be provided with a periodic table, solubility rules, conversion factors and constants (similar to those at the back of the chapter.) The exam will contain a mixture of multiple choice questions similar to those assigned as homework and at the end of each chapter in the textbook. (The Practice Examples are within the chapter, the problems appear at the end of the chapter.) Students must have their Temple photo ID card during lecture exams. TU-ID Numbers must be correctly and completely filled in on exams to ensure your score is properly recorded. Students are not allowed to have cell phones and other PDA devices on their persons while taking an exam. Cell phones/PDAs are to be placed in bags/backpacks which will be kept at the back of the lecture hall. Students may leave cell phones/PDAs on the front bench of the lecture hall. Students who are caught using cell phones/PDAs during an exam will be ejected from the exam and will receive a score of zero. This violation will be reported to the Dean and the Vice-Provost who will then take the appropriate disciplinary action. Only pens/pencils and scientific calculators (non-programmable and non-graphing) are allowed to be with students during...

Words: 1433 - Pages: 6

Premium Essay

Evaluation of L – Proline as a Catalyst for an Asymmetric Aldol Reaction

...Evaluation of L – Proline as a Catalyst for an Asymmetric Aldol Reaction Abstract This reaction is divided into two parts. In the first part acetone, L – proline and 4 – nitrobenzaldehyde are reacted to give (R)-4-hydroxy-4-(4-nitrophenyl)butan-2-one as the major product along with (S)-4-hydroxy-4-(4-nitrophenyl)butan-2-one. The identity of the product is confirmed by IR spectra of the product which gives peaks at 1073.94 cm-1, 1330 cm-1, 1515.05, 1600.13 cm-1, 1708.25 cm-1, 2930.82 cm-1and broad peak at 3418.10 cm-1. The mass of the product is 0.013 grams which gives a percentage yield of 29.81%. The melting point of the product is not taken due to minimal product. In the second part of the reaction excess reagents are used and the synthesized product is in very small quantities. The product synthesized is diastereomers of 1-(4-nitrophenyl)-3-oxobutyl 3,3,3-trifluoro-2-methoxy-2-phenylpropanoate. The identity of this compound is confirmed by the following peaks seen in 1H NMR: 8.19 and 7.62 ppm, 5.47 ppm, 2.90, 2.13 ppm, 3.30 ppm, 7.36 – 7.38 ppm. The melting point, IR spectrum, theoretical yield or percentage yield is not found since all of it used in 1H NMR analysis. The product 1H NMR shows a mix of both the diastereomers, but it is difficult to ascertain which one is in excess. Introduction The aldol reaction that was experienced in this lab is the nucleophillic addition of an enolate to a carbonyl group to form a B-hydroxycarbonyl. This reaction is a very powerful...

Words: 2273 - Pages: 10

Free Essay

Acetaminophen Lab Report

...Randall Elrod 12/6/2013 Acetaminophen Title Acetaminophen Purpose To prepare acetaminophen. Theory Acetaminophen is a synthetic, nonopiate, centrally acting analgesic derived from p-aminophenol.1 The full chemical name is N-acetyl-p-aminophenol.1 The pKa of acetaminophen is 9.51 at 25°C.1 Acetaminophen is used to relieve mild to moderate pain from headaches, muscle aches, menstrual periods, colds and sore throats, toothaches, backaches, and reactions to vaccinations (shots), and to reduce fever.2 Acetaminophen may also be used to relieve the pain of osteoarthritis (arthritis caused by the breakdown of the lining of the joints).2 Acetaminophen is in a class of medications called analgesics (pain relievers) and antipyretics (fever reducers).2 It works by changing the way the body senses pain and by cooling the body.2 Acetaminophen is one of the most commonly used medicines in the United States.3 When used according to the label directions, it has a well-established record of safety and efficacy.3 Although acetaminophen overdose is very rare in the context of its broad usage, overdose can be toxic and lead to acute liver failure.3 Preparation of acetaminophen involves treating an amine with an acid anhydride to form an amide.4 In this experiment, p-aminophenol (the amine) is treated with acetic anhydride to form acetaminophen (p-acetamidophenol), the amide. The chemical reaction for acetaminophen is as follows: 4 Procedure Procedure is as taken from Pavia,...

Words: 713 - Pages: 3

Free Essay

Bromination Report

...Radical Bromination of Name Gabriel Greene a Hydrocarbon Date 6/26/2012 |Equation [structural and balanced] | | [pic] + 2Br2 → [pic]+2 HBr | Stoichiometric Data |Compounds |Mole Wt. |Mass |Moles | | | |or volume | | |Reactant(s)# | |(measured) |(actual) | |Bibenzyl |182.3g/mol |.267g |.00146 mol | |10% bromine-dichloromethane |159.8g/mol |2.00 ml |.00390 mol | |Product | |(theoretical) |(theoretical) | |1,2-dibromo-1,2-diphenylethane |340.1g/mol |.496g |.00146 mol | #Only those part of the stoichiometic equation. |Product Data ...

Words: 349 - Pages: 2

Premium Essay

Chemistry

...NAME: AKINWANDE OLUFISAYO PARNTNER: COURTNEY DATE: JULY 8TH 2014. ALDOL CONDENSATION TO MAKE VANILIDENEACCETONE. DATA: REACTION: TABLE 1: DATA FROM EXPERIMENT | Vanillin | Acetone | Vanillindeneacetone | Chemical formula | C8H803 | C3H60 | C11H1203 | Molecular weight | 152.25 g/mol | 58.08 g/mol | 192.21g/mol | Density (g/cm3) | 1.06 | 0.788 | | Amount used | | | | Total moles | | | | Theoretical yield | | | | Percentage yield | | | | TABLE 2: DATA FROM EXPERIMENT: REAGENTS | Sodium Hydroxide | Hydrochloric Acid | Chemical formula | NaOH | HCl | Molecular weight | 39.997 g/mol | 36.46 | Denisty | 2.13 | 1.18 | Amount used | 3mL | 7mL | Total moles | 0.i598 | 0.227 | TABLE 3 : IR SPECTRUM Peaks (cm-1) | Functional group | 3699.8 | OH group | 3267.6 | Alkene | 1634.79 | Aromatic | CALCULATIONS: Total moles of Vanillin = 0.5g/152.15 = 0.0033 moles of vanillin. Total moles of acetone = (4mL * 0.788)/ 158.08 = 0.0199 moles of Acetone. Theoretical yields: Vanillin = 0.5g * 1mole * 1mole * 192.21 = 0.632g 152.15 1mole 1mole Acetone = 4ml * 0.788 * 1mole * 1mole * 192.21 = 3.832g 1mole 158.08 1mole 1mole Therefore, vanillin is the limiting...

Words: 487 - Pages: 2

Premium Essay

Chem Test2

...Question 1 Which of the following is NOT a strong electrolyte? | | LiOH | | | CaCl2 | | | MgCO3 | | | NaC2H3O2 | | | Li2SO4 | 1 points   Question 2 Which of the following compounds is soluble in water? | | CaS | | | MgCO3 | | | PbCl2 | | | BaSO4 | | | None of these compounds is soluble in water. | 1 points   Question 3 Which of the following compounds is insoluble in water? | | Hg2I2 | | | MgSO4 | | | (NH4)2CO3 | | | BaS | | | All of these compounds are soluble in water. | 1 points   Question 4 Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of Na2CO3 and HCl are mixed. | | 2 H+(aq) + CO32-(aq) H2CO3(s) | | | 2 Na+(aq) + CO32-(aq) + 2 H+(aq) + 2 Cl-(aq) H2CO3(s) + 2 NaCl(aq) | | | 2 H+(aq) + CO32-(aq) H2O(l) + CO2(g) | | | 2 Na+(aq) + CO32-(aq) + 2 H+(aq) + 2 Cl-(aq) H2CO3(s) + 2 Na+(aq) + 2 Cl-(aq) | | | No reaction occurs. | 1 points   Question 5 Write balanced complete ionic and net ionic equations for the following reactions: AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq) | | Ag+(aq) + NO3-(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)Ag+(aq) + Cl-(aq) --> AgCl(s) | | | Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) --> AgCl(s) + Na+(aq) + NO3-(aq)Ag+(aq) + Cl-(aq) + NO3-(aq) + Na+(aq)--> AgCl(s) | | | Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) --> AgCl(s) + Na+(aq) + NO3-(aq)Ag+(aq) + Cl-(aq) --> AgCl(s) | | |...

Words: 1680 - Pages: 7

Premium Essay

Methyl Ester Lab Report

...oil mole ratio (1:6 to 1:24). The synthesis of methyl esters were achieved by an autoclave nitrogen pressurized reactor equipped with a magnetic stirrer and a thermocouple.42 A maximum yield (97%) of methyl ester was obtained at a temperature of 180 °C, 1:15 molar ratio of methanol and oil for a reaction period of 2 hours. The 3 wt% of catalyst amount and 350-360 min-1 agitation speed were fixed respectively to obtain good yields. 2.13. Esterification of free fatty acids in crude palm oil The production of biodiesel from the crude palm oil by the conventional method, using the homogeneous base catalyst contains large amounts of free fatty acid and it also results in the soap formation and thereby reducing the yield of biodiesel. To overcome these problems the free fatty acids need to be esterified to their esters by using an acid catalyst prior to alkaline catalysed transesterification. The sulfated metal oxide was found to be the best catalyst for this purpose due to its high acidic strength. The sulfated tin oxide solid acid catalysts were doped with aluminium and used in the transesterification reaction of free fatty acids present in the crude palm oil.43 A number of aluminium doped sulfated tin oxide catalyts were synthesised by using different aluminium presursors/starting materials. From the results, it is observed that the different aluminium precursors gave different acitivities of catalysts. The esterification reaction of free fatty acids with methanol using these catalysts...

Words: 1187 - Pages: 5

Premium Essay

Hypochlorite Oxidation of Methyl Ketones by the Haloform Reaction: Benzoic Acid

...Experiment 34A: Hypochlorite Oxidation of Methyl Ketones by the Haloform Reaction: Benzoic Acid Objective: To practice the Haloform reaction as synthetic rout to form benzoic acid To understand the mechanis that provide chorine from NaOH To obtain the melting point and FTIR spectrum of benzoic acid Experimental Procedure: The procedure was followed as described in Mayo, pages 405 -406, with the modifications given on Blackboard. Additional modifications to the procedure included the following: * The filtered product was left on the Hirsch funnel for 1 5 min * The amount of Na2SO3 was increased from 15 to35 mg * The amount of Acetophenone was increased from 15 to35 mg * The amount of Na2SO3 was increased from 90 μL, from 60 μL. * The amount of Sodium hypoclorite was increased to 3.2 mg Reaction scheme Data and Results: Physical properties Compound | Formula | MW | Amount | mmol | Mp 0C | Bp 0C | d | nD | Acetophenone | C₆H₅COCH₃ | 120.16 | 90 μL. | 77 | 20.7 | 2.2.6 | 1.03 | 1.5372 | Aquous NaOCl | NaOCl | 74.442 | 3.2mL | | 18 | 101 | 1.11 | | Sodium sulfite | Na2SO3 | 120.6 | 35mg | | | 33.4 | | | Benzoic acid | C6H5COOH | 122.12 | 94mg | 77 | 121.0 | 249 | | | Chloroform | CHCl3 | 119.38 | 92mg | | | | | | Water | H2O | 18.015 | 1.5mL | | 32 | 100 | 1.00 | | Observations: * Acetophenone and NaOCl were clear liquids * Sodium Sulfite was a whote powder * A yellow greasy liquid was found at the...

Words: 893 - Pages: 4