Preparation of a Haloalkane

Experiment 2 Preparation of a Haloalkane Date: 11-9-2006 Objective To produce 2-chloro-2-methylpropane from 2-methylpropan-2-ol. To find the yield of 2-chloro-2-methylpropane. Introduction 2-chloro-2-methylpropane is formed when 2-methylpropan-2-ol and concentrated hydrochloric acid are added together. Because alcohols undergo substitution rapidly, the reaction takes place at room temperature. Afterwards, the haloalkane would be separated with a separating funnel, dried with anhydrous sodium sulphate and extracted by distillation. Procedure . About 9 ml of 2-methylpropan-2-ol was poured into a measuring cylinder and the measuring cylinder was weighed. 2. The 2-methylpropan-2-ol was poured into a 50ml separating funnel. The mass of the empty measuring cylinder was weighed again. 3. About 20ml of concentrated hydrochloric acid was added into the separating funnel, 3ml at a time. This procedure was carried out by the window side. 4. After each addition, the funnel was sealed and was shaken. The tap was opened at intervals to allow the gas produced to be released. 5. The separating funnel was allowed to stand near the window for 20 minutes. 6. A distillation apparatus was set up, as in the diagram below. 7. The small flask in the above set-up was weighed. 8. The lower aqueous layer in the separating funnel was discarded into a beaker. 9. Excess

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Chemistry Module 1 revision notes - salts and redox reactions

Chemistry Module 1 Salts - A salt is an ionic compound with the following features: The positive ion or cation in a salt in a salt is usually a metal ion or an ammonium ion NH4 The negative ion or anion in a salt is derived from an acid Formation of Salts: Salts can be produced by neutralising acids with: - Carbonates - Bases - Alkalis Salts from bases: Acids react with bases to form a salt and water Salts from Carbonates: Acids react with carbonates to form a salt, CO2 and water Salts from alkalis: Acids react with alkalis to form a salt and water Salts from metals: Salts can be formed from the reaction of reactive metals with acids. There are known as redox reactions. Ammonia salts and fertilises: - Ammonia salts are used as artificial fertilisers - Ammonia salts are formed when acids are neutralised by aqueous ammonia Water of crystallisation -Water of crystallisation refers to water molecules that form an essential part of the crystalline structure of a compound. Often the compound cannot be crystallised if water molecules are not present. -The empirical formula of a hydrated compound is written in a unique way: -The empirical formula of the compound is separated from the water of crystallisation by a dot. - The relative number of water molecules of crystallisation is shown after a dot. Oxidation number: in a chemical formula each atom has an

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Investigation to determine the Relative atomic mass of Li

Investigation to determine the Relative atomic mass of Li Results and calculations for the first method. Mass of Li used : 0.12g Start volume of Water : 55.5ml Final volume of water : 250.0ml Volume of hydrogen produced : 194.5ml Now based on these results I t is possible to calculate the concentration of the limewater. Calculate the number of moles of hydrogen produced: We have 195ml of hydrogen that has been produced. - This is equal to 0.195 dm3 Since 24dm3 is one mole of hydrogen at room temperature. - 0.195 dm3 = 0.008125 mol Calculate the number of moles of Li reacted: Using the equation: 2Li (s) + 2H2O (l) --> H2 (g) + 2LiOH (aq) 2 : 1 One can see the reacting ratio is 2:1, the no. moles of Li will be twice that of H2. - Moles of Li = 2 x 0.008125 mol - Moles of Li = 0.01625 mol Calculate the relative atomic mass of Li: As the mass of Li and No of moles used are know, it is possible to calculate the relative atomic mass of Li. - Mass of Li = 0.12g - No Moles of Li = 0.01625 mol - Relative atomic mass = mass/no moles - 0.12/0.01625 = 7.3846 - Relative atomic mass of Li = 7.38 Results and calculations for the second titration method. 2 3 End Burette vol (ml) 46.70 43.95 43.70 Start Burette vol (ml) 06.50 01.65 01.30 Amount used (ml) 40.20 42.30 42.40 Using the two best titres (within 0.1 ml),

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Redox titration of copper evaluation

Redox Titration of Sodium Thiosulphate against Copper (II) Sulphate My results from this titration can be seen in the table below: Rough st 2nd 3rd Final Reading/cm3 29.30 25.25 30.40 28.80 Initial reading/ cm3 05.05 0.80 6.20 4.55 Titre/ cm3 24.25 24.45 24.20 24.25 The concordant results I will be using in my calculation are the titres from my 2nd and 3rd titration as there are within 0.01 cm3 of one another. To calculate the average titre I will use the following method: Average Titre = (24.20 + 25.25) 2 = 24.225 cm3 In this titration there are two half equations that are involved, the first is when the copper (II) sulphate is added to the potassium iodide: 2Cu2+ + 4I- --> 2CuI + I2 The second half equation takes place when the sodium thiosulphate is titrated into the solution containing the Cu2+ and iodine: I2 + 2S2O32- --> S4O62- + 2I- As I need to find the concentration of the Cu2+ I have to look at both half equations to find the ratio of Cu2+ to S2O32-, which is 1:1. This is because in each step of the reaction there are 2 moles of each. This will mean that the number of moles of Cu2+ will be the same as the number of moles of S2O32-: n = c x v n of S2O32- = 0.102827763 x 24.225 1000 = 2.491002571 x 10-3 mol n of Cu2+ = 2.491002571 x 10-3 mol Now that I have the number of moles of Cu2+ I can

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Investigation of Enzyme Activity

An Investigation of Enzyme Activity with different Substrate Concentrations. Aim To investigate how different concentrations of substrate affects the rate of enzyme activity. Objective My objective is to determine how different concentrations of substrate, affects the rate of enzyme activity. I will do this by using different concentrations of Hydrogen Peroxide, and mixing it with the enzyme catalyse. Then I will measure how much gas is produced. Theory What is an Enzyme? Enzymes are proteins which are biological catalysts. A catalyst in chemical terms substantially reduces the energy barrier which exists between atoms and which prevents the atoms from getting close enough to react and form a bond with one another. An enzyme lowers the energy of activation of a reaction but the catalyst is not changed in any way in the process .Therefore, when the atoms of molecules are acted upon by enzymes, an identical reaction occurs as would have occurred without the enzyme but, the energy hill required to overcome the getting-close barrier, is much, much smaller than would have been true without the enzyme's help. The structure of the enzyme is such that atoms of molecules can get close enough to interact, but the energy required to allow this closeness is relatively small. Its like going into an empty closet with someone relative to going into an empty auditorium with someone.

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Assessed Practical (Skill P)

AS - Assessed Practical (Skill P) The aim of this experiment is to find out which reaction is correct for the thermal decomposition of copper carbonate out of the following two equations. 2CuCO3(s) ==> Cu2O(s) + 2CO2(g) + 1/2O2(g) CuCO3(s) ==> CuO(s) + CO2(g) Avagadro's constant states that one mole of gas under standard conditions will fill 24dm3 under standard conditions so it is possible to find the amount of gas evolved by measuring the volume. From this it is also possible to find which version of the reaction has taken place. How much copper carbonate should be used? The first equation will produce more gas so that is the maximum amount to be taken into account when deciding how much copper carbonate needs to be used. If one mole of copper carbonate is used then one mole of CO2 and 0.25 moles of O2 will be evolved. This will occupy 30dm3 in total. One mole of copper carbonate has a mass of 123.664g. Because most gas syringes are 100cm3 the amount of gas evolved needs to be less than this. To make the experiment such that 60cm3 will be evolved the mass of copper carbonate needs to be divided by 500. Having 60cm3 gas evolved will leave room in the gas syringe in case the conditions are not perfect for the experiment. If the volume was calculated so that 90cm3 gas would be evolved there is a chance that the gas syringe would not be able to take it. 30000/500 = 60

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Determining the concentration of acid in a given solution

Determining the concentration of acid in a given solution Planning 4 I have been given a sample of sulfuric (VI) acid solution with a concentration between 0.05 and 0.15 mol dm-³. I am going to find out the accurate concentration of the sulfuric acid. To find out the concentration of the acid I will react it with a known volume and concentration of a base and see how much base was needed to neutralise the acid. The acid is a strong acid which means that I know all the H+ ions have been disassociated and are in the solution. The H+ ions will react with the OH- ions in the alkali which will neutralise the solution. I am provided with solid, hydrated sodium carbonate with the formula Na2CO3·10H2O.1 This is a readily available base and I can dilute it down to achieve the concentration I want to react with the acid. The formula of the reaction that will take place is H2SO4 (aq) + Na2CO3 (aq) --> Na2SO4 (aq) + CO2 (g) + H2O (l) So 1 mole of H2SO4 reacts with 1 mole of Na2CO3. A titration will give me the most reliable and accurate results with the available equipment. To do my titration I will need: A Burette 7 I will need a burette to add the sodium carbonate to the sulfuric acid solution. The burettes provide me with very accurate results of volume of solution added. The class set of burettes measure 50cm3. I want to do a titration

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Analysis of sulphur dioxide content in wine.doc

Analysis of sulphur dioxide content in wine Objective To determine the amount of sulphur dioxide, an antioxidant, present in wine by using volumetric analysis. Principle All free SO2 molecules is first convert into SO32- by NaOH solution: SO2 + 2OH- › SO32- + H2O Acidification of the solution liberates all SO2: 2SO32- + 2H+ › 2SO2 + 4H2O which is then titrated with 0.0057M iodine solution in which starch is used as end-point indicator: SO2 + I2 + 2H2O › 2HI + H2SO4 Chemicals white wine (carbonated), 1M NaOH, 2M H2SO4, 0.0057M I2, starch solution Apparatus volumetric apparatus, pipette, measuring cylinder, dropper, white tile Procedure 1.> Find out the volume of wine from the label on the bottle. 2.> Pipette 25cm3 of white wine into a conical flask. 3.> Add about 12cm3 of 1M NaOH and stand for about 15 minutes. 4.> Add about 10cm3 of 2M H2SO4 to the mixture and then few drops of starch solution as indicator. Quickly, titrate the mixture with 0.0057M iodine solution. 5.> Record the titre required to produce pale blue colour. 6.> Repeat steps 2-5 for 2-3 times. Data Analysis Trial st 2nd Final reading /cm3 3.05 2.80 26.30 Initial reading /cm3 0.25 9.70 23.25 Volume of I2 added /cm3 2.80 3.10 3.05 Average volume of I2 added /cm3 3.075 Concentration of I2 solution: 0.0057 M

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Deducing the quantity of acid in a solution

Sofia Gaggiotti Chemistry coursework: Deducing the quantity of acid in a solution 20/03/2008 Index Aim and Background information 3 Hazards 3 Protection 4 Method 5 Previous calculations 5 Making the Solution 6 Equipment needed 6 Quantities of materials needed 8 Procedure 8 Making the Titration 9 Equipment needed 10 Procedure 13 References 15 Results and calculations 16 Evaluation 19 Chemistry coursework: Deducing the quantity of acid in a solution Aim and background information The aim of this experiment is to find how to develop and determine an accurate, precise and reliable concentration of an acid rain solution. 1 To do this, we are going to make first a solution of sodium carbonate with distilled water and then a titration in order to calculate the concentration of sulphuric acid in a solution. Solution: a solution is a homogeneous mixture composed of two or more substances. In this mixture, a solute is dissolved in a solvent. Solutions are characterized by interactions between the solvent phase and solute molecules or ions that result in a net decrease in free energy. 2 Titration: a titration is a laboratory technique by which we can determine the concentration of an unknown reagent using another reagent that chemically reacts with the unknown. At the equivalence point (or endpoint) the unknown reagent has been reacted with the known reagent.

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Finding the number of Moles of Magnesium and Oxygen in Magnesium Oxide

Finding the number of Moles of Magnesium and Oxygen in Magnesium Oxide Table of Results: SUBSTANCE MASS/g Before After Crucible + lid 39.48 39.48 Crucible + lid + Magnesium 39.61 N/A Magnesium 0.13 N/A Crucible + lid + Magnesium Oxide N/A 39.63 Magnesium Oxide N/A 0.15 Oxygen N/A 0.02 Calculations for Empirical Formula: . Number of moles of Magnesium in Magnesium oxide: Moles/mol = Mass/g ____ Relative atomic mass/ g/mol Moles/mol = __ 0.13g__ = 0.00541 moles ( 3sf) 24g/mol 2. Number of moles of Oxygen in Magnesium oxide: Moles/mol = ____Mass/g______ Relative atomic mass/ g/mol Moles/mol = ___0.02g__ = 0.00125 moles ( 3sf) 16 g/mol 3. Put into ratio: Mg : O 0.00541 : 0.00125 0.00541 = 4.33 0.00125 4.33 : 1 12.99 : 3 13 : 3 Empirical Formula of Magnesium Oxide = Mg O Comparison between calculated empirical formula and literature empirical formula and Sources of Error: The literature empirical formula for Magnesium oxide is MgO meaning the ratio between Magnesium and Oxygen is 1:1. However the results from my experiment differed greatly. Our results ended up in a ratio of 13:3. This could be a result of numerous sources of error and the limitations of the method.

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