Potassium Iodide Lab

Lindsay Zackeroff Mr. Mardulier/4 Chemistry Honors 2/27/08 I. Title: Finding the Formula for Lead Nitrate II. Purpose: This experiment was to test the different concentration levels of specified alkali metals to determine the greatest mass of lead nitrate. III. Background Information: Potassium Iodide Potassium Iodide is a crystallized, white salt but known to turn a bright yellow when exposed to prolonged moisture such as mixing with water. It is a simple iodine salt. In its natural state it is mostly colorless and odorless. If tasted, it would be like saline and extremely bitter and is has a relatively low level of hazard. Its main use is in photography but also used in table salt to "iodize" food and can be used in expectorants for lung congestion. It can also be used to protect the thyroid from radioactive iodine Potassium iodide IUPAC name Potassium iodide Other names Kalium iodide, knollide, potide Identifiers CAS number [7681-11-0] RTECS number TT2975000 Properties Molecular formula KI Molar mass 66.00 g/mol Lead Nitrate Lead Nitrate is a hazardous colorless crystal or white powder. It has a long history of uses. Until 1974, when the dangers of lead were realized, it was in a variety of products. Lead(II) nitrate IUPAC name Lead(II) nitrate Other names Lead nitrate Plumbous nitrate Lead dinitrate Plumb dulcis Identifiers CAS

  • Word count: 1403
  • Level: International Baccalaureate
  • Subject: Chemistry
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Aspirin Lab Report

Ximena Delgado Ms. Petty Chemistry HL-Pd. 5 8/04/11 Aspirin Lab Report Introduction Aspirin is an acid. In order to make an aspirin there is a mix of salicylic acid and acetic anhydride with H2SO4.The synthesis of aspirin is an esterification, in other words it's when an acid and an alcohol mix together in order to form a product which in this case it's an aspirin. However esterification is a reversible reaction. Different strengths of aspirin are based on the amount of active ingredients that they contain. Titration is a way to determine how much acid is in a solution by adding just enough base of a known concentration to neutralize the acid. In neutralization, the number of moles of acid, H+, are combined with an equal number of moles of base, OH-. The aspirin will be titrated against a standard solution of base, 0.100 M NaOH. Base will be dispensed from a biuret into a beaker containing the dissolved (in ethanol) acid and phenolphthalein indicator, which will show a faint pink color in basic solutions. This range of colors determines the color change interval and it expresses the pH range. Phenolphthalein indicator shows acid form when the color is clear and a base form when the color is pink. The approximate range of change in color is when the pH is 8 to 9.8. Purpose The purpose of this lab is to experimentally determine the molar mass of pure aspirin(acetyl

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  • Level: International Baccalaureate
  • Subject: Chemistry
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Preparation of Aspirin Lab

The Preparation of Aspirin Lab Calculating the Percent Yield of Aspirin Mass of salicylic acid +/-0.0001 g Volume of Acetic Anhydride +/- 0.05 mL Mass of Filter Paper +/- 0.0001 g Mass of Filter Paper & Aspirin +/- 0.0001 g Mass of aspirin +/- 0.0002 g Relative Molecular Mass of Salicylic Acid +/- 0.0000 g/mol Relative Molecular Mass of Aspirin +/- 0.0000 g/mol Theoretical Yield of Aspirin +/- 0.000001 g Percent Yield of Aspirin +/- 0.0002 % .0167 2.08 0.6499 .2305 0.5806 38.1216 80.1582 .326127 43.7816 Observations: After sulfuric acid was added, the solution fumed and heat up. After all of the reactants had reacted, a colorless solution was formed. After the solution was allowed to cool down and condense, the solid product that was filtered was white and in powder form. The aqueous solution left over was colorless. Sample Calculations: Relative Molecular Mass of Salicylic Acid: Mass of carbon: 12.011 g/mol Mass of oxygen: 15.999 g/mol Mass of hydrogen: 1.0079 g/mol Mass of Salicylic acid: 7 (12.011) + 3 (15.999) + 6 (1.0079) = 138.1216 +/- 0.0000 g/mol Relative Molecular Mass of Aspirin: Mass of carbon: 12.011 g/mol Mass of oxygen: 15.999 g/mol Mass of hydrogen: 1.0079 g/mol Mass of Aspirin: 9 (12.011) + 4 (15.999) + 8 (1.0079) = 180.1582 +/- 0.0000 g/mol Theoretical Yield of Aspirin: Mass of Salicylic Acid: 1.0167 +/- 0.0001 g Relative

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  • Level: International Baccalaureate
  • Subject: Chemistry
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Crystallization - Rock candy is collection of sugar crystal.

Title: Crystallization Name: Anas Mansour Group members: Shaun, Raina, Michelle. Date: 10/5/2008 Introduction: Have you ever eaten a rock candy? Rock candy is collection of sugar crystal. It looks and tastes good and dissolves fairly well when placed in a cup of tea or coffee. Firstly I will be describing the shape of crystals. To understand the solid shape we need to have some insight into the structure of simple crystals and the forces that hold them together, Single-crystal X-ray diffraction is one of the ways of determining the structure of the crystal. A crystal acts as a three dimensional diffraction grating to a beam of X-ray. To have a closed packed structure in three dimensions we must add a second layer of atoms. The spheres of the second layer sit in half of the hollows of the first layer. If we write this sequence we would build up the layers ABABABA..... Etc. This is known as a hexagonal close-packing (hcp). As shown is the diagram below. Side view of two anion layers, Mg3 (OH) 6, structure. However there are still possibilities to add a third layer which would go directly over the first layer. The third layer which we will label C will have the sequence ABCABCABC.... etc. This is known as a cubic close-packing (ccp). Close-packing represents the most efficient use of space when packing identical spheres, it is said that the packing efficiency will

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  • Level: International Baccalaureate
  • Subject: Chemistry
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Melting and Freezing point of naphthalene

Melting and Freezing point of Naphthalene Research Question . What is the specific temperature for the melting point and freezing point of naphthalene? Background During the lab experiment, we are going to place the Naphthalene on the gauze mat which will be kept above the Bunsen burner using a retort stand. A matter can be classified as solid, liquid, or gas and they are known as the three states of matter. We know that particles during the melting process move faster than particles that are in the solid state due to their high kinetic energy. Particles also move slower as the matter changes back into solid due to solid's low amount of kinetic energy. Kinetic energy is the energy that makes the particles move. When the matter is in its solid state, the particles are stuck together in such a way that they do not move; however, when matter is in its liquid state the particles slide over each other. For gas, the particles move freely in the container it's in. Solid is a definite shape while liquid and gas are indefinite. Solids stay the same shape while liquids and gasses take the shape of the container it's in. Hypothesis The melting point of Naphthalene is 80°and the freezing point of Naphthalene is 60°. Materials and Equipment * A 500ml beaker * Tap water * Ring Clamp * Wire Gauze * Bunsen Burner * Naphthalene * Test tube holder * Stopwatch *

  • Word count: 1074
  • Level: International Baccalaureate
  • Subject: Chemistry
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im To determine the relative molecular mass of chloroacetic acid ClCH2COOH

Aim To determine the relative molecular mass of chloroacetic acid (ClCH2COOH). Data Collection Table (1): masses of items used in the experiment. Item Mass (g) (±0.05g) Beaker 45.80 Beaker + Chloroacetic acid 47.80 Table (2): volume of solutions used in the experiment. Solution Volume (cm3) Sodium hydroxide (0.1M) 20.8 ± 0.1 Chloroacetic acid (total) 00.00 ± 0.20 Chloroacetic acid (reacted) 0.00 ± 0.06 * Two drops of phenolphthalein indicator were added to the chloroacetic acid solution to enable us to determine when the solution is neutralized. Phenolphthalein is pink in basic and neutral solutions and colorless in acidic solutions. * When 20.8 cm3 ± 0.1cm3 of sodium hydroxide were titrated into the flask containing 10.00cm3 ± 0.06 cm3 chloracetic acid and two drops of phenolphthalein indicator, the color of the solution changed from colorless to pink, which indicates that the solution has been neutralized. Data Processing and Presentation . Number of moles of sodium hydroxide reacted Conversion factor cm3 = 10-3 dm3 20.8 cm3 = 20.8 * 10-3 dm3 = 2.08 * 10-2 dm3 Number of moles (n) = concentration (c) * volume (v) = (2.08 * 10-2) (0.10) = 2.1 * 10-3 moles of sodium hydroxide reacted 2. Number of moles of chloroacetic acid reacted Chloroacetic acid (ClCH2COOH) reacts with sodium hydroxide (NaOH) to produce sodium chloroacetate

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  • Level: International Baccalaureate
  • Subject: Chemistry
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The Molar Volume of a Gas

LAB REPORT - THE MOLAR VOLUME OF A GAS Aim: To try the Avagadro's hypothesis of molar volume of a gas under STP. Introduction: Avagadro stated that the volume of one mole of any gas is 22.4 dm3 under STP. In the experiment, we measured the molar volume of H2. The formula for it is: Mg(s) + 2HCl (l) --> MgCl2 (aq) + H2 (g) A certain mass of magnesium ribbon is used to react with HCl and the volume of the H2 gas produced is measured. Then the conditions of the lab is measured and we calculated the value at standart temperature and pressure. Raw Data: Lenght of the whole Magnesium Ribbon: 64 cm. Mass of the whole Magnesum Ribbon: 0.81 g. Temperature of the lab: 220 C (295 K) Vapor pressure at 220 C : 2.6447 Trial 1: Lenght of Magnesium ribbon used: 2.2 cm Mass of Magnesium Ribbon used: 0.02784375 g. Volume of H2(after the reaction): 29.5 cm3 Volume of MgCl2(aq): 20.5 cm3 Volume of H2(in the big beaker, 1 atm) : 29.1 cm3 Volume of MgCl2(aq)(in the big beaker): 20.9 cm3 Trial 2: Lenght of Magnesium ribbon used: 2.2 cm Mass of Magnesium Ribbon used: 0.02784375 g. Volume of H2(after the reaction): 29.0 cm3 Volume of MgCl2(aq): 21.0cm3 Volume of H2(in the big beaker, 1 atm) : 29.4 cm3 Volume of MgCl2(aq)(in the big beaker): 20.6 cm3 Trial 3: Lenght of Magnesium ribbon used: 3.6 cm Mass of Magnesium Ribbon used: 0.0455625 g. Volume of H2(after the

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  • Level: International Baccalaureate
  • Subject: Chemistry
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DETERMINATION OF WATER OF CRYSTALLIZATION

DETERMINATION OF WATER OF CRYSTALLIZATION Student: Vu Cuong Nguyen School: Auckland International College Due date: Wednesday 15th August 2007 I. Aim: To determine the formula of the hydrate of copper (II) sulfate. II. Hypothesis: Hydrates are ionic compounds that have a definite amount of water as part of their structure. The water is chemically combined with the salt in a definite ratio. Ratios vary in different hydrates but are specific for any given hydrate. In this experiment, the hydrate of copper sulfate has the formula CuSO4.xH20. When it is heated, the water is released as vapor and we can describe the reaction as below: Hydrate Anhydrous salt + Salt CuSO4.xH20 CuSO4 + xH20 The amount of water in the hydrate is determined by taking the difference between the mass of the hydrate before and after heated. Then, the percentage composition of water can be calculated basing on its mass. According to usual theory, hydrated copper sulfate has the following formula CuSO4.5H2O. Therefore, the result collected from the experiment should be x = 5. III. Equipment/Chemical: . Equipment: * 1 x Evaporating dish * 1 x Pipe-clay triangle * Some crucible tongs * 1 x Pressed fiber pad * 1 x Digital balance (uncertainty: ±0.001g) * 1 x Spatula * 1 x Glass stirring rod * 1 x Bunsen burner * 1 x Tripod stand 2. Chemical: Hydrated

  • Word count: 703
  • Level: International Baccalaureate
  • Subject: Chemistry
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rate of evaporation

RESEARCH QUESTION How does different type of liquids affect the rate of evaporation of the liquids? HYPOTHESIS The stronger the intermolecular force, the slower the rate of evaporation Hydrochloric acid is the acid with the lowest rate of reaction compared to the other two acids, phosphoric acid and sulphuric acid. This is because HCl is a strong acid. It dissociates easily in water to form hydrogen and chloride ions. Strong acid means that HCl has strong covalent bond. So, it requires a lot of energy to break the bonds between the hydrogen and chloride ions. VARIABLES . Manipulated variable: types of solution used ; HCl [0.1 mol) , H3PO4 [0.1], H2SO4 [0.1] 2. Responding variable: time taken for the solution to evaporate 3. Controlled variable: 1) volume af the acid * Only a little acid is needed as it have a strong intermolecular forces. 2) surface area of the evaporating dish * the surface area used is the same that is in the petri dish. 3) concentration of the acid * The concentration used for the acid is the same that is 0.1M 4) temperature * the temperature for the water bath is fixed to 60°c 5) air humidity * the experiment was carried out in the same place that have the same humidity MATERIALS AND APPARATUS Materials/ apparatus Quantity/ volume Size/ concentration Hydrochloric acid, HCl 5ml 0.1 M Sulphuric acid, H2SO4 5ml 0.1 M

  • Word count: 816
  • Level: International Baccalaureate
  • Subject: Chemistry
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Determination of the purity of commercial aspirin

DETERMINATION OF THE PURITY OF COMMERCIAL ASPIRIN Student: Vu Cuong Nguyen Lab partner: Tianren Yang Due date: Monday September 10th 2007 I. Aim: To determine the purity of commercial aspirin by performing the reaction between sodium hydroxide and aspirin. II. Data collection: Table 1.Titration of Aspirin Run 1 Run 2 Run 3 Initial volume of NaOH (±0.05ml) 2.65 23.45 22.70 Final volume of NaOH (±0.05ml) 23.45 34.15 33.50 Amount of NaOH used (±0.1ml) 0.8 0.7 0.8 Mass of Aspirin (±0.0005g) 0.2060 0.2020 0.2000 Volume of ethanol using (±0.1ml) 5.0 5.0 5.0 Table 2.Observation Process State of solution Obtained ethanol in the conical flask Colorless Dropped aspirin powder into ethanol white and pink Added bromthyol into the solution yellow The solution was titrated by sodium hydroxide light blue III. Data processing: CH3COOC6H4COOH NaOH Mole ratio: 1 : 1 Run 1: % errormass of aspirin = naspirin (theoretical) = VNaOH (used) = (23.45 ± 0.05ml) - (12.65 ± 0.05ml) = 10.8 ± 0.1ml % errorvolume of NaOH used = × 100% = 0.9% nNaOH = (0.0108 L ± 0.9%) × 0.1 mol/L = 0.00108 mol ± 0.9% --> naspirin (actual) = nNaOH = 0.00108 mol ± 0.9% Percent purity = Run 2: % errormass of aspirin = naspirin (theoretical) = VNaOH (used) = (34.15 ± 0.05ml) - (23.45 ± 0.05ml) = 10.7 ± 0.1ml % errorvolume of NaOH

  • Word count: 650
  • Level: International Baccalaureate
  • Subject: Chemistry
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