Investigation to Determine the Percentage of Citric Acid in Lime Juice
Investigation to Determine the Percentage of Citric Acid in Lime Juice A – The Standardisation of Sodium Hydroxide Discussion and Method A primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and has a high molecular weight. This enables a concentration to be calculated from a direct weighing of its pure form. NaOH is cannot be classed as a primary standard as a concentration cannot be calculated directly from a weighing. This is due to the fact that over time, in its pure form, NaOH will absorb atmospheric water and react with atmospheric CO2. With NaOH solutions, again, over time, they will absorb atmospheric CO2 and will also slowly react with glass bottles in which it is stored. Therefore, before using NaOH as a standard in a titration, the solution must be standardised to determine the concentration which is being used. Using good laboratory techniques, potassium hydrogen phthalate was weighed by difference into a conical flask. 25cm3 distilled water was added and was then mixed until the potassium hydrogen phthalate had dissolved. Phenolphtalein indicator was added and the solution was carefully titrated with an unknown concentration of NaOH solution until a pale-pink end point was reached. The titration was repeated for accuracy. Results Potassium hydrogen phthalate: . Weight by difference = 0.5525g 2. Weight by
Comparison between Cyclohexane and Cyclohexene
Alkenes also known as olefin is an unsaturated hydrocarbon which contains at least one carbon-carbon double bond. The general formula for alkene is CnH2n. Alkenes are insoluble in water but are soluble in orgainic solvent. The presence of double bond between carbon-carbon atoms is a strong link. However, the second pi bond which is not as strong as the first sigma bond is more vulnerable to attack by suitable reagent, causing alkene to become relatively unstable. Thus, alkenes generally are more reactive than alkanes which contain only carbon-carbon single bond. Majority of the reactions including the addition reactions involve the rupture of the pi bond to form new single bond in which the unsaturated carbon atoms become saturated. Alkenes and alkanes in fact all hydrocarbons are non-polar molecules. This is because the difference in electronegativity between carbon and hydrogen is very small, meaning they share electrons fairly evenly. Alkenes are highly flammable and burn readily in air. With sufficient amount of oxygen gas, combustion of alkenes forms carbon dioxide and water. In our experiment, the combustion of cyclohexene burns in an orange flame and produces relatively high amount of black soot. The black soot produced is actually carbon monoxide. Cyclohexene undergoes incomplete combustion to form carbon, carbon monoxide and water. The combustion of cyclohexene
Calorimeter lab. What is the change in enthalpy for magnesium after the reaction with vinegar?
Introduction: The first law of thermodynamics states the total energy of an isolated system cannot be created or destroyed. Since chemical energy is the sum of heat plus work many chemical reactions involve the evolution of heat or production of work. An example is a car engine that converts the chemical energy stored in gasoline into heat and work. This heat that is exchanged can be measured with an apparatus called calorimeter. A calorimeter is an insulated vessel preventing the heat from escaping into the surrounding. The thermometer is used to measure the temperature change and the stirrer is used to ensure the heat is uniformly distributed. Question: What is the change in enthalpy for magnesium after the reaction with vinegar? Hypothesis: If the magnesium reacts with vinegar then the change in enthalpy will be a very high value. Procedure: All materials were gathered and cleaned. The vinegar of 20 ml was measured using two 10 ml graduated cylinder to get a precise amount of vinegar. The vinegar was put into a Styrofoam cup. Another Styrofoam cup was put over the first cup for insulation. The piece of the magnesium was measured and recorded. The apparatus was moved to the front of the class where the teacher set up the thermometer attached to the laptop showing the temperature. A Styrofoam cup with two holes for the thermometer and another for the
Comparing the enthalpy change of combustion
Comparing the enthalpy changes of combustion of different alcohols The aim of this experiment is to investigate how the length of an alcohol molecule and the position of its OH group affect its enthalpy change of combustion. This will be done using the comparison method, using a known value for the enthalpy change of combustion for one fuel (here, propan-1-ol) and replicating the experiment for each alcohol, then accurately comparing the results. The 4 alcohols that will be compared to propan-1-ol (CH3CH2CH2OH) are: o methanol (CH3OH) o ethanol (CH3CH2OH) o propan-2-ol (CH3CH(OH)CH3) o butan-1-ol (CH3CH2CH2CH2OH) The propan-2-ol differs from propan-1-ol because its OH group is positioned on the 2nd carbon atom in the chain. Combustion reactions involve breaking the intra-molecular bonds (bonds between atoms, within the molecule) of a fuel, then making new bonds with oxygen. The formula for the combustion reaction is: Alcohol + Oxygen ==> Carbon Dioxide + Water e.g. CH3OH + 11/2O2 ==> CO2 + 2H2O Breaking bonds takes energy in from the surroundings, and making bonds gives energy back out. The enthalpy change is the net energy transfer of the reaction; it is negative if the reaction is exothermic, like combustion, and positive if the reaction is endothermic. Variables The independent variable in this experiment is the alcohol used. The dependant variable is the
Working with Hess's Law.
Working with Hess's Law It is often possible to calculate ?H for a reaction from listed ?H values of other reactions (i.e. you can avoid having to do an experiment) Enthalpy is a state function It depends only upon the initial and final state of the reactants/products and not on the specific pathway taken to get from the reactants to the products Whether one can arrive at the products via either a single step or multi-step mechanism is unimportant as far as the enthalpy of reaction is concerned - they should be equal Consider the combustion reaction of methane to form CO2 and liquid H2O CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) This reaction can be thought of as occurring in two steps: In the first step methane is combusted to produce water vapor: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) In the second step water vapor condenses from the gas phase to the liquid phase: 2H2O(g) -> 2H2O(l) Each of these reactions is associated with a specific enthalpy change: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) ?H = -802 kJ 2H2O(g) -> 2H2O(l) ?H = -88 Kj (Note: under conditions of standard temperature and pressure the liquid state of water is the normal state. Thus, the gas would be expected to condense. This is an exothermic process under these conditions. In a related process, it should get warmer when it rains) Combining these equations yields the following: CH4(g)+2O2(g)+2H2O(g) ->
Comparing the enthalpy change of combustion of alcohols down a homologous series.
Comparing the enthalpy change of combustion of alcohols down a homologous series Aim The aim of this investigation is to discover what effect of increasing the length of the carbon chain in an alcohol has on the energy produced. To investigate this, the enthalpy of combustion will be calculated, by burning the fuels and heating a measured volume of water. Preliminary Investigation A preliminary investigation was carried out in order to show any weaknesses in the method that is used for this investigation. Diagram of Apparatus Method The apparatus was assembled as shown in the diagram above. 100cm3 of cold water was measured out into a calorimeter. The temperature of the water was recorded. The spirit burner was weighed and recorded, then placed under the calorimeter. The wick was then lit and left until the water had reached a temperature 15ºC higher than the original recorded temperature. When this temperature was reached the flame was put out and the spirit burner reweighed and the final weight recorded. Results Methanol -274.67Kjmoldm-3 Ethanol -475.38 Kjmoldm-3 Propanol -515.00 Kjmoldm-3 Butanol -1172.6 Kjmoldm-3 Pentanol -1437.2 Kjmoldm-3 Prediction Combustion occurs due to oxidation; in this case the oxidation is of alcohols. When combustion of alcohols occurs an exothermic reaction takes place. An exothermic reaction releases energy in the form of
To Determine the Enthalpy of combustion of different alcohols.
To Determine the Enthalpy of combustion of different alcohols. Skill Area P: Planning Prediction: I predict that as the number of carbon atoms in the alcohols increases, the enthalpy (or heat content) increases at a fairly equal rate. Detailed Scientific Knowledge: For any reaction to take place bonds must be broken and made, bond breaking requires energy while bond making releases energy. Bonds between different atoms require or release different amounts of energy when broken or made because the bonds are different in strength. The energy that is stored in chemical bonds is called enthalpy and given the symbol H. By looking at the equation for the reaction and more importantly looking at the bonds that are being broken and made, it is possible to work out an estimate for the amount of energy that will be released in the reaction. This is called the change in enthalpy. The enthalpy of a reaction is the change in energy going from reactants to products. For exothermic reactions, the enthalpy is negative while for endothermic reactions, the enthalpy is positive. A negative value for the enthalpy, or ?H denotes that energy has been released into its surroundings, bonds have been made and therefore there is an increase in temperature. A positive value of ?H shows that energy has been taken in from its surroundings and bonds have been broken, as there is now more energy
COMPARING THE ENTHALPY CHANGE OF COMBUSTION OF DIFFERENT FUELS.
COMPARING THE ENTHALPY CHANGE OF COMBUSTION OF DIFFERENT ALCHOHOLS. Aim The aim of my experiment will be to find out which alcohols have a higher enthalpy change of combustion. The comparison of the enthalpy changes of these fuels will then determine the alcohol efficiency and effectiveness. I will experiment on the first 5 consecutive primary alcohols. These are; methanol, ethanol, propan-1-ol, butan-1-ol, pentan-1-ol and hexan-1-ol. The reason why I chose these fuels is because they are the most reliable and accurate fuels to compare within the group; which also have the smallest variable, 'add one carbon' each time, to the aliphatic chain. All combustion reactions are exothermic which is why I am expecting all the values for the enthalpy change of combustion to always be negative. CH3OH + 1.5O2 CO2 + 2H2O METHANOL CH3CH 2OH+ 3O2 2CO2 + 3H2O ETHANOL CH3CH 2 CH 2OH+ 4.5O2 3CO2 + 4H2O PROPAN-1-OL CH3CH 2 CH 2 CH 2OH+ 6O2 4CO2 + 5H2O BUTAN-1-OL CH3CH 2 CH 2 CH 2 CH 2OH+ 7.5O2 5CO2 + 6H2O PENTAN-1-OL CH3CH 2 CH 2 CH 2 CH 2 CH 2OH+ 9O2 6CO2 + 7H2O HEXAN-1-OL Prediction I believe that hexan-1-ol will have the highest enthalpy of
The Synthesis Of Phenacetin From p-Acetamidophenol
CH1751B - EXPERIMENT 15 The Synthesis Of Phenacetin From p-Acetamidophenol Introduction: "Phenacetin is an analgetic and an antipyretic reagent which acts in the body by temporaririly inhibiting prostaglandin synthesis in the central nervous system". Aim: * To synthesise Phenacetin from p-Acetamidophenol and purify it by recrystallisation. * Analyse purity of Phenacetin produced, by comparing the melting point with the literature values, the appearance and the Infrared Spectrum. Method: Sodium metal (0.6g, cut into small pieces) was cautiously placed into a dry 100cm3 round flask by taking the flask to the "sodium" balance and placing the sodium directly into the flask. A dry reflux condenser was attached and industrial methylated spirits (IMS, 15cm3) was added. A vigorous reaction occurred for 5-10 minutes under reflux until most of the sodium had dissolved. Once the solution had cooled 3.8g of p-acetamidophenol was added. Through the top of the condenser ethyl iodide (3.0cm3) was introduced to the pale green mixture and the mixture boiled at reflux temperature (20-25°). Ethyl iodide was obtained from a burette in the fume hood few minutes before its use, due to its flammable nature and harmful vapour. A darkening green colouration occurred while p-acetanilide dissolved changing through to brown and then yellow. Once the reflux was completed 40cm3 of water was
HYDROLYSIS OF ORGANIC HALOGEN COMPOUNDS
HYDROLYSIS OF ORGANIC HALOGEN COMPOUNDS Name: Ho Ka Wing (9) Group:3 Grades: Date: 07-09-10 Objective The purpose of this experiment is to find out how the rate of hydrolysis of an organic halogen compound depends on: a the identity of the halogen atom. b the nature of the carbon-hydrogen 'skeleton'. Theory Experiment: In this experiment, the rate of hydrolysis of 1-chlorobutanne, 1-bomobutane, 1-iodobutane, Chlorobenzene is compared. A general equation for the hydrolysis is: R-X + H2O ???? R-OH + H+ + X- (Where R = alkyl or aryl group; X= halogen atom) Hydrolysis of organic halogen compounds is a nucleophilic substitution reaction. In a nucleophilic substitution a lone pair of electron on a nucleophile, H2Ois attracted to a carbon atom, with a partial positive charge. The nucleophile is then substituted for the atom or group attached to the carbon atom. Since halide ions are being substituted out. By following the rate of the reaction by carrying it out in the presence of silver ions, so that any halide ions produced form a silver halide precipitate. Ag+(aq) + X-(aq) ???? AgX(s) By comparing the time for appearance of precipitate, we can compare the rate of hydrolysis. The smaller the time needed, the faster the rate of hydrolysis. Hydrolysis is a chemical process in which a certain molecule is split into two parts by the addition
