Introduction
Extraction is a technique used for separating a compound from a mixture. One of the example of extraction is liquid-liquid extraction. This process involves the distribution of compound between two solvents that are immiscible in each other. As a result of the different solubilities of the solutes in two different solvent, the compounds can be selectively transported from one phase to another.
This occurrence is quantified by the partition coefficient (K):
K = (concentration of solute in ether phase)/(concentration of solute in water phase )
During extraction, an aqueous phase and the organic phase are shaken in a flask and are the mixture are then allowed to distribute themselves between the
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Calculate the percentage recovery of each solutes.
Pre lab questions Excluding spillage and incomplete transfer of solids from the filter papers, what would be responsible for a low recovery of a component in a separation by extraction?
Answer : Other possible reason of the low recovery components in a separation by extractions is that the mixture is not shaken well or the mixture is not shaken long enough. As a result the components are not much dissolve in the each solvent. The distribution coefficient K for compound X in diethyl ether/water is 3. If 12.0g of compound X is in 400Ml water, calculate the weight of X removed into ether by : Extracting the aqueous solution once with 200Ml ether Extracting the aqueous solution twice with 100Ml ether each time
Answer : K = (concentration of compound X in diethyl ether )/(concentration of compound X in water )
3 = ((X)g/2OOmL)/((12-X)g/400mL) x = 7.2g K = (concentration of compound X in diethyl ether )/(concentration of compound X in water )
First extraction
3 = ((X1)g/ (100mL ))/((12-X1)g/400mL) X1 =
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From the first extraction, the amount of compound X left in the aqueous layer is 6.86g Second extraction 3 = ((X2)g/100 ml)/((6.86-X2)/400mL) X2 = 2.94g
Therefore the total amount of compound X extracted is X1+X2= 8.08g
Flowchart Mixture (carboxylic acid + neutral compound)
Dichloromethane
Prelab week 1 Calculations Preparation of 1.5μmol/L mixed low-level standard dilution 150μmol/L × V1=1.5μmol/L × 10ml V1=(1.5μmol/L×10ml)/(150μmol/L)=0.1ml Conversion of milliliters to microliters (0.1ml×1000)μL= 100μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=3μmol/L × 10ml V1=(3μmol/L×10ml)/(150μmol/L)=0.2ml Conversion of milliliters to microliters (0.2ml×1000)μL= 200μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=7.5μmol/L × 10ml V1=(7.5μmol/L×10ml)/(150μmol/L)=0.5ml Conversion of milliliters to microliters (0.5ml×1000)μL= 500μL Preparation of the blank samples The volumetric flask will be filled to the mark with 150μmole/L of stock solution to act as blank (reference). Additional two blanks will
Table 1 Results DDA Concentration Initial Mass(g) Time Interval Recovered Mass Cumulative Mass (g) Cumulative Recovery (%) Ln[(Rinf -R)/ Rinf] R=Rinf(1-e-kt) (M) (g) 10^(-5) 160 0 0
Using the equation m = ΔTf/Kf , the molality of the unknown solution was found. Then, moles of unknown were calculated, which was used to calculate the average molar mass of unknown. Theory: After the experiment was completed, the data
The percent recovery of the copper was calculated using the equation, percent recovery = (the mass of the copper recovered after all the chemical reactions/the initial mass of the copper) x 100. The amount of copper that was recovered was 0.32 grams and the initial mass of the copper was 0.46 grams. Using the equation, (0.32 grams/0.46 grams) x 100 equaled 69.56%. The amount of copper recovered was slightly over two-thirds of the initial amount.
Introduction 1.1 Aim: To determine the kinetic parameters, Vmax and Km, of the alkaline phosphatase enzyme through the determination of the optimum pH and temperature. 1.2 Theory and Principles (General Background): Enzymes are highly specific protein catalysts that are utilised in chemical reactions in biological systems.1 Enzymes, being catalysts, decrease the activation energy required to convert substrates to products. They do this by attaching to the substrate to form an intermediate; the substrate binds to the active site of the enzyme. Then, another or the same enzyme reacts with the intermediate to form the final product.2 The rate of enzyme-catalysed reactions is influenced by different environmental conditions, such as: concentration
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
After the reaction is finished, the percentage composition of each element in the product can be found and used to calculate the empirical formula, which is the lowest whole number ratio
(0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube
The calculated value was 1.6 x 10^-5. Conclusions The resulting Ka of the acetic acid from this experiment’s calculations was consistent with the experimental results. The experimental percent of CH3COOH was calculated at 1.6 x 10^-5, while the actual value was 1.8 x 10^-5.
Abstract — This experiment was conducted to familiarize the students with the procedures regarding distillation—to be more precise, the separation of ethanol from an alcoholic beverage—using a distillation set-up consisting of boiling chips, a Bunsen burner, a condenser, a thermometer and several other materials. In the end, it was discovered that one may actually separate a homogeneous mixture, given that the components of said mixture differ in volatility and that they utilize a complete distillation set-up and follow laboratory safety rules and regulations. Keywords — Matter, homogeneous and hetereogeneous mixtures, distillation, volatility, boiling point I. INTRODUCTION There are typically two categories of matter, these are pure substances
Introduction Drug use in sports has always been a controversial issue. With athletes pushing for the top podium position, performance enhancing drugs can be extremely enticing. One of the main types of drugs used by athletes are stimulants such as cocaine, amphetamines or ecstasy. These can create unfair advantages in sports. To keep sports even and fair, certain drugs became prohibited.
The mixture were stirred by using a glass rod until the mixture is fully dissolved. The solution were tested by using calibrated pH meter to get the pH value of the solution. Results and Discussions pH ratio between acid and base: 7.3 = 6.82 + x x = 0.48 0.48 = log ([base])/([acid]) 100.48 =base/acid salt/acid = 3.02 There, 1 acid : 3 base calculate number of mole of acid and base to find the mass : molar = mol/L 50 mM = (mol )/(0.5 L) mol = 25 mol number of mole of NaH2PO4 25/4 = acid = 6.25 mol number of mole of Na2HPO4 25/4 × 3 = salt = 18.75 mol to calculate the mass of the acid and base : Mass of NaH2PO4 (6.25 mol)/(119.98 g/mol)=0.052g
DETERMINATION OF PERCENTAGE ETHANOL IN BEVERAGES 1. Introduction to Gas Chromatography Gas chromatography is a very powerful separation technique for compounds that are reasonably volatile. The components of a sample partitions into two phases, the 1st of these phases is a immobile bed with a great surface area, and the other is a gas phase that permeates through the immobile bed. The sample is evaporated and passed by the mobile gas phase or the carrier gas through the column. Samples separates into the stationary liquid phase, based on their solubilities at the given temperature.
Rediet Legese iLab Week # 6 CRUDE OIL DISTILLATION Introduction: The aim of this week lab experiment is to experiment distill crude oil and to check how temperature determine the chemical properties of crude oil plus how the boiling point can also show physical properties. They are two major finding in this experiment. he first finding was the point at which the raw petroleum is heated to the point of boiling, at 275 0C, the gas and kerosene oil are refined, however the oil (lubricant ) stays as an unrefined feature oil.