#### Finding the concentration of oxalic acid in a mixture with sulphuric acid

The redox titration will be done between potassium permanganate and ethanedioic acid. This reaction requires acid catalyst because ethanedioic acid is too weak an acid to make the solution acidic enough to react at a reasonable rate. Sulphuric acid is in the mixture and provides the acid catalyst. The ethanedioic acid in the mixture will reduce the manganate(VII) ions (MnO4-) into manganese(II) ions (Mn2+). The potassium permanganate will oxidise the oxalic acid into carbon dioxide. H2C2O4(aq) + MnO4-(aq) ? CO2(g) + Mn2+(aq) Reactants Products H2C2O4: Carbon +3 CO2: Carbon +4 MnO4-: Manganese +7 Mn2+: Manganese +2 To work out the ionic equation, balanced electron-half equations for both the potassium permanganate and the ethanedioic acid have to be worked out. Adding electrons (e-), water (H2O) and hydrogen/hydroxide ions (H+/OH-) depending on the conditions, they can be created. Manganate(VII) ions are reduced to form manganese(II) ions. MnO4- + 8H+ + 5e- ? Mn2+ + 4H2O Ethanedioic acid is oxidised to form carbon dioxide. H2C2O4 ? 2CO2 + 2H+ + 2e- The two equations combine and are

• Word count: 1069
• Level: AS and A Level
• Subject: Science

#### Calculating the DHc of the combustion of alcohols.

Sam Forshaw Calculating the ?Hc of the combustion of alcohols. Aim: To work out the ?Hc of the alcohols: ethanol, methanol, pentan-1-ol, butan-1-ol and propan-1-ol by using calorimetry. The idea in this experiment is that if you burn a substance to heat water there is a link between the rise in temperature of the water and the amount of energy the substance contains. For this you need to know the amount of the substance you have burned and the volume and mass of the water you are heating. For this experimant I wil use a copper container for the water, place an aluminium lid on top (to prevent heat loss) and a spirit lamp containing the alcohol that I will be burning. There are however problems with this experiment. Not all of the heat from the reaction will go into the water, much of this energy will be lost to the surroundings and the copper container and aluminium also absorb some of the heat and rise in temperature also. Both of these factors greatly reduce the amount of heat energy produced by the combustion of the alcohol that is available to heat the water. The standard definition for enthalpy change is 'Enthalpy change is the amount of heat energy taken in or given out during any change in a system, provided the system is kept at constant pressure'. Standard enthalpy changes (?H?) are these that occur under standard conditions of 100kPa, 298K and 1 mol dm-3 and

• Word count: 426
• Level: AS and A Level
• Subject: Science

#### Application of Hess's Law to determine the enthalpy change of hydration of Magnesium sulphate

? Name: Penguin Chow Cheuk Yan ? ? Class: 6S Class Number: 14 ? ? Date: 04 - 11 - 2002 ? Chemistry Full Report Experiment 5 Title: Application of Hess's Law to determine the enthalpy change of hydration of Magnesium sulphate (VI) Aim: Using Hess's Law to find the enthalpy change of hydration of magnesium sulphate (VI) Theory: Hess's Law can be defined as the heat given off or absorbed by a reaction is independent of the route taken. In this experiment, the enthalpy change of hydration of Magnesium sulphate (VI) cannot be directly measured by calorimetry in the laboratory as hydration is a very slow process. ?H1 = enthalpy change of hydration of MgSO4(S) ?H2 = molar enthalpy change of solution of hydrous MgSO4(S) ?H3 = molar enthalpy change of solution of MgSO4• 7H2O(S) According to Hess's law, the enthalpy of the overall reaction should be equal to the sum of the enthalpies of the two sub-reactions. Thus, this value can be calculated by applying Hess's Law Procedure: A. Determine the enthalpy change of solution of MgSO4(S) . A balance was used to weigh the empty polystyrene foam cup 2. 50 cm3 of deionized water was poured from the measuring cylinder to the polystyrene cup 3. The temperature of the water in the cup was measured using a thermometer 4. 0.025 mole of anhydrous magnesium sulphate (VI) was weighed accurately by the balance and was added into

• Word count: 754
• Level: AS and A Level
• Subject: Science

#### F336- aspirin individual Investigation

Does Temperature Affect The Yield And Purity Of Aspirin? Contents Objectives ) Research using various sources on aspirin with reference to; history ,use in medicine and methods of synthesis 2) produce a sample of aspirin using a method found during research 3) Verify the presence of aspirin and use various method to measure the purity of the manufactured aspirin 4) Carry out investigation to see if temperature effects the yield and purity of aspirin Research History of aspirin 400 BC, Greece - Hippocrates recommended a brew made from willow leaves to ease the pain of child birth. 763, England - A reverend called Edward Stone was walking through a meadow in Oxfordshire while suffering from an acute fever. Stone removed a small piece of bark from a willow tree and nibbled on it. He was struck by its extremely bitter taste, as an educated man he knew that he bark from Peruvian cinchona tree (which has a similar taste) is used in the treatment of malarial fevers. He surmised that the willow might also have therapeutic properties. He gathered and dried a pound of willow bark and created a powder which he gave to about fifty people. It was consistently found to be a "powerful astringent and very efficacious in curing agues and intermitting disorders". He had discovered salicylic acid, the active ingredient in aspirin. 1828, Germany- Johann Büchner isolates pure

• Word count: 10517
• Level: AS and A Level
• Subject: Science

#### Experiment to determine the ethanol content of wine

Experiment to Determine Ethanol Content of Wine The purpose of this experiment is to determine the ethanol content of each of the wines and compare the value determined to the value quoted on the label. These results can then be used to conclude which region is more accurate in quoting the value of the ethanol content of the wine. This experiment takes advantage of the fact that ethanol is less dense than water in solution. The density of ethanol at 20°C is 0.789 g/cm3 while the density of water at the same temperature is 0.998 g/cm3. It then follows that different solutions of ethanol and water will have different densities also, because the relative volume of ethanol increases and water decreases so the density of higher percentage ethanol solutions will be less than the density of lower percentage ethanol solutions. This occurs because water molecules are much smaller than ethanol molecule, meaning more water molecules can "pack" into a smaller volume than ethanol molecules, meaning there is more mass per unit volume of water compared to ethanol, meaning it has a higher density. In this experiment, solutions of ethanol in water were made up, going from 0% to 20%. These were then weighed, and the density of the ethanol was calculated. From this, a graph of percentage ethanol solution against density was made. This graphs later compared to the density of the wine, so

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• Word count: 2599
• Level: AS and A Level
• Subject: Science

#### Equilibrium Constant for esterification

Chemistry Experiment 15 (a) Title: Equilibrium Constant for esterification (b) Aim: To find out the equilibrium constant for esterification of ethanoic acid and propan-1-ol. (c) Theory: In the presence of conc. H2SO4 and under reflux, ethanoic acid undergoes a reversible reaction with propan-1-ol to form ester and water. As equilibrium is reached, CH3COOH(l)+CH3CH2CH2OH(l) CH3COOCH2CH2CH3(l)+H2O(l) Kc = By titrating the reaction mixture with standard alkali solution before and after refluxing, the quantity of the acid used, and thus the quantities of the alcohol used as well as of the ester and water formed can be determined. Knowing the quantities or molar concentrations of the species present in the equilibrium mixture, the equilibrium constant, Kc can be calculated. Set-up: (d) Result: Titration of 8 drops of H2SO4 against NaOH Titration (trial) 2 Final burette reading (cm3) 5.1 30.2 Initial burette reading (cm3) 0 5.1 Volume of NaOH used (cm3) 5.1 5.1 Titration of final reaction mixture against NaOH Titration (trial) 2 Final burette reading (cm3) 8.2 6.0 Initial burette reading (cm3) 0.0 8.2 Volume of NaOH used (cm3) 8.2 7.8 No. of moles of CH3COOH in 1 cm3 of reaction mixture before heating = [(density x volume) / molar mass] / volume = [(1.05 gcm-3 x 15 cm3) / (12x2+1x4+16x2)] = 8.75 x 10-3 mol 2 NaOH + H2SO4 Na2SO4 + 2

• Word count: 831
• Level: AS and A Level
• Subject: Science

#### Preparation of propanone from propan-2-ol

Date: 11/10/2011 Exp. No.: 23 Title: Preparation of propanone from propan-2-ol Aim: The aim of this experiment is to prepare propanone by oxidizing propan-2-ol with acidified potassium dichromate solution. Introduction: Ketones and aldehydes are important series in preparation of other compounds and they are commonly prepared by oxidizing alcohol which is done in this experiment. The experiment is an oxidation reaction where a secondary alcohol (propan-2-ol) is oxidized by acidified potassium dichromate. The reaction does not need to be heated but should be placed in an iced water bath as the reaction is highly exothermic. The product is propanone and no catalyst is needed for the reaction. The propanone is serparated from the reaction mixture by simple distillation and is purified using anhydrous cacium chloride. The equation of this reaction is as follow: Chromic acid is produced in situ by adding potassium dichromate (VI) with sulphuric acid and water. K2Cr2O7 + H2O + 2H2SO4 ? 2 H2CrO4 + 2 NaHSO4 The term chromic acid is usually used for a mixture made by adding concentrated sulfuric acid to a dichromate, which may contain a variety of compounds, including solid chromium trioxide. Chromic acid features chromium in an oxidation state of +6. It is a strong and corrosive oxidising agent. Apparatus and chemicals: Apparatus: Quick-fit distillation setup,

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• Word count: 2193
• Level: AS and A Level
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#### Experiment to Determine Acidities of Wine. The purpose of this experiment is to determine the total and volatile acidities of each of the wines and compare them.

Experiment to Determine Acidities of Wine The purpose of this experiment is to determine the total and volatile acidities of each of the wines and compare them. Acidity is a major contributor to the taste of wines. This is especially important in white wines, because there are very little tannins found in it, so acidity can affect the taste of the wine much more than in red wines. In this experiment, 0.1M Sodium Hydroxide solution is needed for titrations. Because this is not a standard solution, it is first standardised using oxalic acid. In order to calculate the total acidity of the wine, a titration with 0.1M sodium hydroxide is carried out with a pH meter. The pH of the wine and sodium hydroxide solution is measured when a certain volume of NaOH is added each time, and a titration curve of volume against pH is plotted. The volume for the solution to reach a pH of 8.2 is recorded. This is because NaOH is a strong alkali and wine is a weak acid, so the pH lies more to the side of the alkali. A pH of 8.2 as the equivalence point is a value agreed on by winemakers. In order to calculate the total acidity of the wine, a representative acid must be used. This must be chosen because wine contains multiple different acids, which require different moles of NaOH to neutralise them. Tartaric acid was chosen as the representative acid for the wine, because it is thought to

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• Word count: 3117
• Level: AS and A Level
• Subject: Science

#### Finding the concentration of an acid sample

Finding the concentration of an acid sample Concentration is the amount of a substance in a given volume1. There is different concentration in all solids, liquids and gases. And here lies the aim of my coursework. "to find the accurate concentration of a given sample of acid solution" The solution of acid I have been given is Sulphuric Acid (H2SO4 herein). We have been told that the concentration lies in between 0.05 and 0.15 mol dm-3. To find the concentration accurately I will titrate the solution using a known indicator with a known endpoint and Sodium Carbonate (Na2CO3 herein) using the following equation: H2SO4(aq) + Na2CO3(aq) › Na2SO4(aq) + H2O(l)+ CO2(g) Background Theory A titration is when and acid is run into a base or an alkali with an indicator added. This indicator will have a known end point. So once you have measured how much of the acid or base required to make the solution reach the endpoint. From there we can measure the concentration. So what I need to decide is which indicator I will use. There can be strong and weak acids, and the same for alkalis and bases. The Bronsted-Lowry theory2 states that acids are H+ donors and that bases are a H+ receiver. A strong acid is defined as having a strong tendency to donate a H+. H2SO4 is a well known acid with a strong base. While the sodium Carbonate is the opposite as it is relatively weak. So now that I

• Word count: 2728
• Level: AS and A Level
• Subject: Science

#### Composition of Hydrates

Composition of Hydrates Purpose: The purpose of this lab was to determine the percentage of water in a hydrate and to calculate the coefficient of the water of hydration of a hydrate. Data Table: Mass of Crucible & Cover 22.03 g Mass of Crucible & Cover & Hydrate 24.45 g Mass of Crucible & Cover & Anhydrous Salt 23.71 g Analysis: Hydrate Mass (Before Heating): 24.45-22.03 = 2.42 g Anhydrous Salt Left (After Heating): 23.71-22.03 = 1.68 g Water Lost: 24.45-23.71 = .74 g Molar Mass of Hydrate used: CuSO4?5H20 = 63.55+32.07+64+10.10+80 = 249.72 g/mole Molar Mass of Anhydrous Salt: CuSO4 = 63.55+32.07+64 = 159.62 g/mole Molar Mass of all the Waters in Compound: 5H2O = 80+10.10 = 90.10 g/mole Theoretical Percentage of Water in Hydrate: 90.10 * 100 = 36.98% 249.72 Experimental Percentage of Water in Hydrate: .74 * 100 = 30.58% 2.42 Percent Error in the Lab: 36.08 - 30.58 = 15.24% 36.08 Questions: ) Regardless of the mass of hydrate, the percentage of water lost stays the same. But, the value is not always the theoretical value, often close; it may vary, depending on the time you have kept the hydrate over the fire. 2) If I used 5 grams of hydrate, I would expect to lose around 36.98% of water. That's about 1.849 grams. Conclusion: The purpose of this lab was to determine the percentage of water in a hydrate and to calculate the coefficient of the water

• Word count: 459
• Level: AS and A Level
• Subject: Science