General Chemistry Lab Introduction PDF

Palestine Polytechnic University General Chemistry Lab: Introduction Introduction Course Syllabus Safety Rules Glassware Chemistry Lab Techniques Course Syllabus General Chemistry Lab 1, 1-credit hours, 19035,20002, 8596 and 20303, first COURSE...

Palestine Polytechnic University General Chemistry Lab: Introduction Introduction Course Syllabus Safety Rules Glassware Chemistry Lab Techniques Course Syllabus General Chemistry Lab 1, 1-credit hours, 19035,20002, 8596 and 20303, first COURSE semester 2024-2025. PREREQUISITES AND General Chemistry COREQUISITES INSTRUCTOR TEXTBOOK General Chemistry Lab Manual 1. Moodle Course (General Chemistry lab for medicine/General Chemistry lab ADDITIONAL MATERIALS for medical sciences) or EQUIPMENT NEEDED FOR THE COURSE 2. Scientific calculator. 3. Goggles, lab coat and gloves  This is a freshman practical chemistry course, student will learn basic instrumental techniques including weighing, pipeting, titration and use of COURSE DESCRIPTION different laboratory glassware, student must be aware of safety rules. Students will learn how to do laboratory calculation associated with the experiments. To introduce students to the basics of chemistry lab reactions and teach them how COURSE AIMS to deal with lab tools and instruments. 1. To use of different laboratory glassware 2. To know basic instrumental techniques including weighing, pipeting and INTENDED LEARNING titration OUTCOMES 3. To know safety rules of chemistry laboratory 4. To do laboratory calculation associated with the experiments a. Traditional classroom (95%) MODE OF INSTRUCTION b. Blended learning (5%)  1. Class attendance is mandatory 2. Three absence will make course fallen 3. Class quizzes may be given without prior notice COURSE POLICIES 4. Missing home works, tests, or exams will lost mark 5. No make-up exam 6. Copy-past reports will loss report mark Course Outline and Calendar Week # Topics Experiment # 1 Introduction to chemical laboratory - 2 Preparation of solution 1 3 Standardization of sodium hydroxide solution 2 4 Volumetric analysis of vinegar 3 5 Determination of calcium carbonate contents in limestone by back titration 4 6 Limiting reactant 5 7 Determining percent yield in a chemical reaction 6 8 Midterm exam 9 Standardization of sodium thiosulfate 7 10 Bleach analysis 8 11 Calorimetry and heats of reactions 9 12 Determining molecular weight of solid from Freezing point depression 10 13 Water of hydration 11 14 Solubility constant and common ion effect 12 15 Final exam Teaching methods Lab experiments Lab report Independent work Group work Assessment measures Quiz Homework Report Written exams All reports are due one week of experiment Grading system Evaluation Technique Percentage Reports 40% Midterm exam 20% Final exam 30% Evaluation and quizzes 10% Glassware and Equipment Beaker Flask Test Tube To carry chemical reaction To carry chemical reaction To carry chemical reaction Test Tube Rack Graduated Cylinder Volumetric Flask To hold test tubes To measure volume of liquids To prepare solutions in molar concentration (Measure accurate volume of liquids)  Burrete Pipette Pipette fillers To measure accurate volume of To measure accurate volume of To fill pipette with liquid liquids liquids (Used in titration)  Crucible Tongs Funnel To carry reactions at high To hold hot glassware To transfer substances temperature  Droppers Watch Glass Spatula To add drops from liquids To weigh solid on it To transfer solids  Brush Glass rod Thermometer For cleaning of glassware For stirring To measure temperature  Clamp Ring Burret Clamp To hold glassware To hold separatory funnels To hold burrets  Stand Washing bottle Hotplate To hold clamp, ring and burrete To carry solvents For heating clamp (Should be labeled)  Oven Electronic (Analytical Balance) For heating To measure mass of substances Chemistry Lab Techniques 1. Measure Mass 2. Measure Volume 3. Measure Temperature 4. Filtration 1. Measure Mass Unit: g, kg, mg (Interchange between them) Equipment: Balance - Set Zero (Press Esc/0 bottom) - Add substance - Record mass (Record all digits) Note: In chemistry lab mass and weight are used interchangeably for mass 2. Measure Volume Unit: ml, L, m3, cm3 (Interchange between them) Equipment: Burret, Pipett, graduated Cylinder, Volumetric Flask - Read Liquid Meniscus at eye level - Note the tool Graduate 3. Measure Temperature Unit: ℃, K, °F (Interchange between them) Equipment: Thermometer - Immerse in liquid - Be sure not to touch tool walls - Read temperature when thermometer liquid level is stable - Note the tool graduate 4. Filtration: Procedure to separate liquids from solids Types of filtration Systems: 1. Simple Filtration System (Gravity Filtration) 2. Suction Filtration System 1. Simple Filtration System (Gravity Filtration) Components: Filter paper, Funnel, Flask Folding filter paper - Fit funnel shape - Increase area of filtration 2. Suction Filtration System Components: Filter paper, Buchner funnel , Suction flask Advantages: - Faster - Dry the sample Thank You Palestine Polytechnic University General Chemistry Lab: Experiment 4: Preparation of solutions Objective : To prepare solution using mass percent (m/m%) and molarity (CM) expressions Solution: Homogenous mixture of two or more substance Solute Solution Solvent Solute: Any substance dissolve in solvent Solvent: Medium in which the solutes are mixed or dissolve To prepare any solution, we need to know two basic things: 1. Amount (Volume, mass) 2. Concentration - Concentration: The amount of solute (the substance being dissolved) present in a given volume of solvent. - It can be expressed in various units such as: Mass percent (m/m%) Weight/volume percentage (w/v) Molarity (M) Volume/volume percentage (v/v) Molality (m) Parts per million (ppm) Normality (N) Parts per billion (ppb) - Mass Percent/ Weight Percent (m/m%) 𝑚 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Mass Percent 𝑚% = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Mass Percent 𝑚% = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒+𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡  (m/m%) No unit for it (g/g)  Example: 1. Prepare 12% of NaOH solution. Answer: 𝑚 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Mass Percent 𝑚% = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Assume mass of solution = 100g 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Solute 12% = × 100% (NaOH) 12g 100 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 12𝑔 Solution Solvent 88g (Water) 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 100 − 12 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 88𝑔  Example: 2. Prepare 50 gr of 8% NaOH solution. Answer: 𝑚 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Mass Percent 𝑚% = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 8% = × 100% Solute 4g 50 (NaOH) 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 4𝑔 Solution Solvent 46g 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (Water) 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 50 − 4 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 46𝑔 Homework: 1.Prepare 5% of NaOH solution. 2. Prepare 50 gr of 12% NaOH solution. - Procedure: 1. Calculated mass of solute (NaOH) and solvent (H2O) 2. Weigh NaOH in balance 3. Add water to final solution mass 4. Mix to dissolve by use glass rod 5. Calculate actual mass percent (m/m%) Beaker Water To determine density (D): (Water) NaOH D= m/v (gr/ml) Volume Mass Balance Balance - Molar concentration or Molarity (M)/(CM) Molarity (CM): The number of moles which dissolved in a certain volume of solution in liter 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑛 𝑚𝑎𝑠𝑠 𝑚 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑀 = = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑣 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 ∗ 𝑣𝑜𝑙𝑢𝑚𝑒 𝑣 Unit: mol/L (M) Note: Suitable tool to prepare a solution in Molarity is the volumetric flask  Example: Prepare 250ml of 0.2M NaCl solution Answer: 𝒎𝒂𝒔𝒔 𝒎 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 𝑴 = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂𝑪𝒍 ∗ 𝒗𝒐𝒍𝒖𝒎𝒆 𝒗 𝒎 𝟎. 𝟐 = 𝟐𝟑 + 𝟑𝟓. 𝟒𝟖 ∗ 𝟎. 𝟐𝟓 Mass of solute 𝒎 = 𝟐. 𝟗𝟐𝟐 𝒈𝒓 calculated - Procedure: Calculated mass of solute (NaCl) and solvent (H2O) Use watch glass to weigh NaCl in balance and add it to volumetric flask by funnel Add water to half volume, shake until completely dissolve Complete to final volume with distilled water (D.W) and mixed well Calculated actual concentration 1. Calculated actual concentration NaCl D.W Mixed well To determine density (D): (NaCl) D= m/v (gr/ml) Volume Balance Mass Balance To convert from m/m% to molarity , we use the following equation: 𝒎𝒐𝒍𝒂𝒓𝒊𝒕𝒚 ×𝑴𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 m/m % = 𝟏𝟎 ×𝑫𝒆𝒏𝒊𝒔𝒕𝒚 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 m/m % = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑙𝑢𝑡𝑒 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 1000 × 100% 𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛×𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑙 × 1000 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑙𝑢𝑡𝑒 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = × 100% 𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛×𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝐿 ×1000 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 10 ×𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Molarity CM = nsolute/ Vsolution(L) Density = msolution / Vsolution (ml) Thank you  Palestine Polytechnic University General Chemistry Lab: Experiment 7: Standardization of sodium thiosulfate solution (Redox Titration) Objectives: 1. To recognize oxidation reduction titration 2. To prepare and standardize thiosulfate solution Burrete (titrant) Introduction clamp Titration: Slowly addition of one reagent to the other until Stopchok sufficient amount is added stand Used to determine the unknown concentration solution Flask (Analyte) A + B  C + D Known Unknown white paper  Standard Solution Secondary Standard Primary Standard Its concentration not accurate initially or change Solution with accurate and constant with time concentration Example: NaOH, HCl, Na2S2O3 Example : KIO3, Na2CO3, KHP, K2CrO4 Titration requirements: 1. Reaction is stoichiometric 2. Reaction is rapid 3. No side reaction 4. There is end point for reaction Classification of titration (according to type of reaction): 1. Acid – base Titration 2. Oxidation – Reduction Titration (Redox Titration) 3. Gravimetric Titration (Precipitation Titration) Oxidation-Reduction titration (Redox titration) Oxidation – Reduction Loss of e- – Gain of e- Reducing agent – Oxidizing agent Example of oxidizing agents: OCl-, H2O2, Ascrobic acid, KIO3 Example of reducing agents: Na2S2O3, KMnO4 To analysis oxidizing agents we need standard reducing agents Sodium thiosulfate Na2S2O3 - Excellent, safe reducing agent used for an oxidation-reduction titrimetric analysis. - Secondary stable 1. Hydroscopic (found as Na2S2O3.5H2O) 2. Not stable (Decompose with time) 3. Has Low molar mass (158 g/mol) - Needed to standardize against primary standard oxidizing agent (KIO3) S2O32-(aq)  S4O62- + 2e- IO3- oxidize iodide ion (I-) in acidic media to iodine (I2 which react with S2O32- This Redox titration called Iodometry I2 is volatile and insoluble in water, so its concentration is not accurate It react with excess I- to produce triiodide ion I3-. I3- is nonvolatile and soluble in water, so its concentration is accurate. I-  I2  [I3]- Iodide iodine triiodide IO3-(aq) + 8I-(aq) + 6H+ 3I3-(aq) + 3H2O(l) S2O32- reduce I2 in I3- to I- 2S2O32-(aq) + I3-(aq) I-(aq) + S4O62-(aq) Red brown colorless I3- high concentration  I3- low concentration  I- Red Brown Pale yellow colorless Not clear end point Use starch indicator I3- + starch  [I3-.starch] Red brown Dark blue - [I3-.starch] is stable at high concentration, so it added near end point (when color is pale yellow) - Starch indicator is 1% solution Prepared freshly as it decompose with time - NaHCO3 or Na2CO3 is added to stabilize thiosulfate solution in acidic media Na2S2O3 + H+  SO2(g) + S(s) + H2O - H2SO4 is added to solution to make solution acidic IO3-(aq) + 8I-(aq) + 6H+ 3I3-(aq) + 3H2O(l) Red brown 3I3-(aq) + 3starch 3[I3-.starch] Red brown Dark Blue 1I- 6S2O32-(aq) + 3[I3-.starch] 3S4O62-(aq) +9I-(aq) + 3starch Dark blue color less IO3-(aq) + 6S2O32-(aq) + 6H+ I-(aq) + 3S4O62-(aq)+ 3H2O(l) Procedure: 1. Preparation of standard KIO3 solution M = 0.01 mol/L, V = 50 ml m = n. Mwt = M. V. Mwt = 0.01 * 50 *10-3 * 214 = 0.107 g 1.Weight 0.107 g KIO3 2. Transfer KIO3 to 50 ml 3. Add water to half volume 4. Complete to final Record exact mass volumetric flask and wash and dissolve well volume with distilled water with distilled water Determine the accurate concentration of KIO3 solution Mass of KIO3 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐼𝑂3 Moles of KIO3 = 𝑔 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐼𝑂3 (214𝑚𝑜𝑙) 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐾𝐼𝑂3 Molarity of KIO3 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑛 (𝐿)  2. Titration of Na2S2O3 solution against KIO3 solution. Na2S2O3 ~0.1 M Na2S2O3 ~0.1 M High conc. of I3- [I3-.Starch] Low conc. of I3- I- 10 ml of KIO3 solution(Oxidizing Agent) 20 ml of distilled water (Make sample larger) Add 2-3 ml of starch solution (1%) 0.5 g KI (Excess I-) 2 ml H2SO4 (0.5 M) (Acidic media) Titrate with Na2S2O3 Complete titration with Na2S2O3 0.1 g NaHCO3 or Na2CO3 (Stabilize Na2S2O3) untill color change to Pale yellow untill color change to Colorless Data Trial 1 2 3 Initial Burette reading (ml) Final Burette reading (ml) Volume of thiosulfate (ml) Average volume of thiosulfate(ml) - Volume of KIO3 titrated = 10 ml = 0.01 L Calculations: IO3-(aq) + 6S2O32-(aq) + 6H+ I-(aq) + 3S4O62-(aq)+ 3H2O(l) At the end point n Na2S2O3 = 6 × n KIO 3 = 6 × M KIO3 × v KIO3 n Na2S2O3 M Na2S2O3 = V Na2S2O3 n Na2S2O3 M Na2S2O3 = (vf – vi)/1000 Remember: - Before using burrete 1. Wash with distilled water 2. Wash with solution 3. Fill with solution 4. Be sure that tip is filled with solution Thank You  Palestine Polytechnic University General Chemistry Lab: Experiment 11: Standardization of Sodium hydroxide Solution Objectives To standardize a solution of NaOH Introduction Standard Solution: - It’s a solution with very known accurate concentration. - It used to determine the concentration of other solution by titration process A + B  C + D Known Unknown  Standard Solution Primary Standard Secondary Standard - Highly pure - Not match primary standard properties: - Stable Example: NaOH, HCl - Not Hygrocopic NaOH: - Has High Molar mass - React with glass - Crystaline compounds - React with CO2 - Cheap, available, not toxic - Absorb Water from surrounding Example : KIO3, Na2CO3, KHP HCl: - Accurate concentration solution - Evaporate prepared - Not stable - Approximate concentration solution prepared - Need to be standaized against primary standard Titration process is needed for standardization Titration: Slowly addition of one reagent to the other until sufficient amount is added Titration requirements: 1. Reaction is stoichiometric 2. Reaction is rapid 3. No side reaction 4. There is end point for reaction Classification of titration (according to type of reaction): 1. Acid – base Titration 2. Oxidation – Reduction Titration (Redox Titration) 3. Gravimetric Titration (Precipitation Titration) Acid – Base Titration Burrete (titrant) Acid + Base  Salt + water Equivalent point (Stoichiometric point): Point at which clamp sufficient amount of reactant is added to complete the Stopchok reaction. stand End point: Point at which indicator color change Flask (Analyte) - Equivalent point come before end point - For calculation we assume that both are the same white paper Indicator: - Dye that has different color in acidic and basic media - It Change color at or extremely close to end point - It is a weak organic acid or base that color different when it protonated or ionized. (just drops of it added) - It chosen according to pH at end point Indicator examples: Strong acid + Strong base  pH = 7 Bromothymol Blue Strong Acid + Weak base  pH < 7 Methyl Orange Weak Acid + Strong base  pH > 7 Phenolphthalene – change color at pH = 8-8.2 - In acidic media colorless -In basic media pink Notes: - Titration is volumetric analysis use volumetric glass ware: (Burrete, Pipette, volumetric flask) - All reagent should be in solution - Burrete reading graduation is from up to down - Accurate reading is the bottom edge of liquid meniscus when it at the eye level  In this experiment we want to standardize NaOH solution (secondary standard) against KHP (primary standard). KHC8H4O4 + NaOH → KNaC8H4O4 + H2O Weak Acid Strong Base Known ?? Primary standard Secondary standard Pure ,Stable Not Stable Not hygroscopic Hygroscopic High molar mass Low Molar mass (204.24 g/mol) (40 g/mol) Procedure - Before using burrete 1. Wash with distilled water 2. Wash with ≅ 0.1 M NaOH solution 3. Fill with ≅ 0.1 M NaOH solution 4. Be sure that tip is filled with solution  NaOH around 0.1 M Add 0.2-0.3 g KHP (Record acurate mass) 30 ml of water (Heat to dissolve) 1-2 drops ph.ph Titrate with NaOH untill color change to faint pink (perminant color for 30s)  NaOH + KHC8H4O4 → KNaC8H4O4 + H2O From the balanced equation: MNaOH = nNaOH/VNaOH n NaOH = n KHP 𝑚𝑎𝑠𝑠 𝐾𝐻𝑃 MNaOH * V NaOH (L) = 𝑀𝑚 𝐾𝐻𝑃 𝑚 𝐾𝐻𝑃 M NaOH = 𝑀𝑚 𝐾𝐻𝑃 × 𝑉 𝑁𝑎𝑂𝐻 𝐿 - The mass of KHP is known (weighted) - The volume of NaOH will be known from the burette experimentally = vf – vi / 1000 Example: To standardize NaOH solution, a student weight 0.274 g of KHP, dissolve in water and add 2 drops of phenolphthalein. Then he titrate it with NaOH. He start titration with burrete reading of 0.2 ml and color change to faint pink when burrete volume is 13.8 ml. Calculate NaOH molar concentration. m KHP = 0.274 g Vi = 0.2 ml Vf = 13.8 ml VNaOH = (Vf – Vi)/1000 = (13.8-0.2)/1000 = 0.0136 L 𝑚 𝐾𝐻𝑃 0.274 M NaOH = = = 0.0986 mol/L 𝑀𝑚 𝐾𝐻𝑃 × 𝑉 𝑁𝑎𝑂𝐻 204.24 ×0.0136 𝐿 Thank You Palestine Polytechnic University General Chemistry Lab: Experiment 12: Volumetric Analysis of Vinegar Objective : To determine the concentration of acetic acid in vinegar as a m/m% and molarity. Introduction Vinegar: is an aqueous solution of acetic acid (CH3COOH) in water. (normally around 5%) Solvent (Water) 95 % Solution (Vinegar) Solute 5% (Acetic Acid) Concentration of acetic acid in vinegar: 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 m/m% = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑣𝑖𝑛𝑒𝑔𝑎𝑟 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 (𝑚𝑜𝑙) Molarity = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑣𝑖𝑛𝑒𝑔𝑎𝑟 (𝐿) We will use NaOH solution of known molarity (standardized) to determine the mass % of vinegar. CH3COOH (aq) + NaOH (aq) → CH3COONa(aq) + H2O(l) ?? Known As we see CH3COOH is a weak acid and NaOH is a strong base So, pH at the end point > 7 and we can use phenolphthalein as indicator Note: ph.ph in acidic media  colorless basic media  pink Step 1 : Standarization of NaOH At end point : Titrant (NaOH) n KHP = n NaOH nKHP = masskHP / molar massKHP vNaOH(L) = (vf -vi )NaOH * 10-3 0.2- 0.3 g KHP (analyte) CNaOH= nNaOH/vNaOH (L) 30 ml water 1-2 drops phenolphthalein Step 2: Determination concentration of acetic in vinegar Titrant (NaOH) 5ml vinegar (accurate volume by pipette)/ weighted 20ml Water 1-2 drops phenolphthalein Calculation : From the equation CH3COOH (aq)+ NaOH (aq) →CH3COONa(aq)+ H2O(l) 1 mol NaOH → 1 mol CH3COOH mol NaOH = CNaOH. VNaOH - CNaOH: from standardization - VNaOH (L) = (vf-vi)*10-3 mol NaOH = moleCH3COOH mass CH3COOH = molesCH3COOH * molar massCH3COOH mass vinegar = D vinegar * V vinegar (ml) (Or weighted at balance) m/mCH3COOH% =( massCH3COOH / mass vinegar) * 100% Molarity CM = n CH3COOH/ Vvinegar(L) Density = mvinegar / Vvingar (ml) To convert from m/m% to molarity , we use the following equation: 𝒎𝒐𝒍𝒂𝒓𝒊𝒕𝒚 ×𝑴𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 m/m % = 𝟏𝟎 ×𝑫𝒆𝒏𝒊𝒔𝒕𝒚 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 Finally, compare your experimental mass percent of acetic acid with the manufacturers value. Error % = (experimental value - True value )/ True value *100% To convert from m/m% to molarity , we use the following equation: 𝒎𝒐𝒍𝒂𝒓𝒊𝒕𝒚 ×𝑴𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 m/m % = 𝟏𝟎 ×𝑫𝒆𝒏𝒊𝒔𝒕𝒚 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 m/m % = × 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑙𝑢𝑡𝑒 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 1000 × 100% 𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛×𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑙 ×1000 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑙𝑢𝑡𝑒 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = × 100% 𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛×𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝐿 ×1000 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 10 ×𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Molarity CM = nsolute/ Vsolution(L) Density = msolution / Vsolution (ml) Astudent carried out of a titration of 5.015 ml of a vinegar solution having a density of 1.00 g/ml. The acetic acid in vinegar required 10.5 ml of 0.403 M NaOH solution to reach the first lasting pink color of acid-base indicator , phenolphthalein. CH3COOH (aq)+ NaOH (aq) →CH3COONa(aq)+ H2O(l) a. Calculate the mass % acetic acid present in the vinegar sample. b. Calculate the % error based upon the manufactures 5% by mass. c. Calculate the molarity of acetic acid solution ( vinegar ). vvinegar = 5.015 ml Dvinegar = 1g/ml VNaOH = 10.5ml CNaOH=0.403 mol/l a. m/mCH3COOH% =( massCH3COOH / mass vinegar) * 100% mol NaOH = CNaOH. VNaOH = 0.403 * 10.5 * 10-3 = 4.231 *10-3 mol moleCH3COOH = 4.231 *10-3 mol mass CH3COOH = molesCH3COOH * molar massCH3COOH = 4.231 *10-3 * 60 = 0.254 g mass vinegar = D vinegar * V vinegar = 1 * 5.015 = 5.015 g m/mCH3COOH% =( massCH3COOH / mass vinegar) * 100% = ( 0.254 / 5.015 )*100% = 5.065 % b. Error % = (experimental value - True value )/ True value *100% = ( 5.065 - 5 ) / 5 * 100% = 1.3 % 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 ×𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 C. m/m % = 10 ×𝐷𝑒𝑛𝑖𝑠𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 × 60 5.065 = 10 × 1 Molarity (CM )= 0.844 mol / L M = n/v = 4.231 * 10-3/ 5.015 * 10-3 Thank You Palestine Polytechnic University General Chemistry Lab: Experiment 4: Water of hydration Introduction Salts : ionic compounds that are formed as a result of the neutralization reaction of acids and bases composed of : metal ion and non metal: NaCl, AgBr, KCl metal ion and negatively charge polyatomic ion: K2SO4, FePO4, NaNO3 positive charge polyatomic ion and nonmetal: NH4Cl positive and negative polyatomic ions: NH4NO3 Hydrous salts (hydrates): the salts that combine with water to form crystalline solid compounds. (Anhydrous salt + Water ⇄ Hydrate Salt ) Hydrate salt express as Salt. nH2O n: The amount of water connected to one mole of the salt Differs from one salt to another. Specific for each salt Integer number Naming : Salt + prefix(n)+hydrate Dehydration: Process of losing water from hydrate salt 1. Spontaneously (Efflorescent salt) 2. Thermally by heating: Through heating pressure of evaporated water increase until exceed vapor pressure of water in atmosphere. At this point dehydration occur Hydrate dehydration Anhydrous Salt. nH2O Hydration Salt The number of H2O moles in the hydrate affect the physical properties and not affect the chemical propertiess. For examples: Density : inversely proportional Melting point : inversely proportional Color: Change CoCl2.6H2O CoCl2.2H2O  CoCl2 Violet Red Blue Some anhydrous ionic compounds shows a great affinity to water and may absorb it from the surrounding medium. So they can be used as drying agents of liquids (organic solvents) or gases. Example: MgSO4 , Na2SO4 , CaCl2 , NaOH. Experimental part You have a hydrate salts 1. CrCl3.nH2O 2. CoCl2.nH2O 3. MgSO4.nH2O 4. CuCl2.nH2O 5. CaCl2.nH2O And you need to determine n, Water% in hydrate and Molar mass of hydrate Tool and material: Oven Crucible Tongs Mass of crucible mass of crucible heating 110°C for Cool to room mass of crucible + Hydrate 15min Temperature anhydrous m1 m2 m3 Mass of Hydrate = m2 – m1 Mass of anhydrous = m3 – m1 Mass of water = m2 – m3 Moles of water = mass of water/molar mass of water Moles of anhydrous = mass of anhydrous/molar mass of anhydrous Moles of water per one mole of anhydrous = nwater/nanhydrus = unknown (n) Mass % of water in the hydrate = (mass of water/mass of hydrate)*100% Molar mass of hydrate = molar mass of anhydrous + n(molar mass of water) Thank You  Palestine Polytechnic University General Chemistry Lab: Experiment 4: Determination of CaCO3 in Lime stones (Back titration) - Lime stones: type of carbonate sedimentary rocks - Dolomite is type of limestones compose of CaCO3 and MgCO3 - We deal with it as all CaCO3 - In this experiment we want to determine CaCO3% in lime stone - CaCO3 is weak base - Problems when titrated against acid: 1. Reaction is slow 2. Not easy to notice endpoint because base is very weak - To Solve this problems we use Back titration Back titration (Indirect titration): A method used to determine amount of analyte through the addition of known amount of excess reagent. Then titrate excess amount of reagent with other one. When we use back titration? (In Acid base titration) 1. When acid or base are insoluble solid (e.g: CaCO3) 2. When reaction is slow 3. When end point is not easy to observe (pH change is small) Steps of Back titration: 1. Analyte is allowed to react with excess reagent 2. Excess reagent remain unreacted is titrated, to determine quantity react with analyte. CaCO3 Analysis by back titration: 1. CaCO3 reacted with excess HCl CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) 2. Excess HCl determine by titration with standard NaOH HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) (Suitable indicator is: Bromophenol Blue or Phenolphthalein) Reaction of CaCO3 with excess HCl CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) Heating (Not Boiling) Cool to room temperature Add 0.5 g of CaCO3 untill no CO2 bubbles appear Add 50 ml of 0.25 M HCl ( To fast dissolution process) Determination of excess HCl by titration with standard NaOH HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) N a O H 0.2 M A d d 1 - 2 d r o p s o f p h.p h T itra te w ith S ta n d a rd N a O H Vi = 1.00 ml u n till c o lo r c h a n g e to fa in t p in k Vf = 20.2 Calculations: CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) - nHCl total = n HCl react + nHCl excess Known amount ?? By titration added with NaOH MHCl × VHCl added = n HCl react + MNaOH × VNaOH 0.25 × 50/1000 = n HCl react + 0.2 × (vf – vi) /1000 1 nCaCO3 = × n HCl react 2 mCaCO3 = nCaCO3 × MwtCaCO3 𝑚 𝐶𝑎𝐶𝑂3 CaCO3% = × 100% 𝑚 𝑠𝑎𝑚𝑝𝑙𝑒 Calculations: CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) nHCl total = n HCl react + nHCl excess Known amount ?? By titration added with NaOH - nHCl total = MHCl × VHCl added = 0.25 × 50/1000 = 0.0125 mol - nHCl Excess = n NaOH (HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)) = MNaOH × VNaOH = 0.2 × (vf – vi) /1000 = 0.2 × (20.2-1.0) /1000 = 0.00384 mol - n HCl react = nHCl total - nHCl excess = 0.0125 – 0.00384 = 0.00866 mol From equation: 1 - nCaCO3 = × n HCl react (CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)) 2 1 = × 0.00866 2 = 0.00433 - mCaCO3 = nCaCO3 × MwtCaCO3 = 0.00433 × 100 = 0.433 𝑚 𝐶𝑎𝐶𝑂3 - CaCO3% = × 100% 𝑚 𝑠𝑎𝑚𝑝𝑙𝑒 0.433 = ×100% 0.5 = 86.6% Remember: - Before using burrete 1. Wash with distilled water 2. Wash with solution 3. Fill with solution 4. Be sure that tip is filled with solution Remember: - NaOH Should be Standardize against KHP NaOHnear 0.2 M 𝑚 𝐾𝐻𝑃 MNaOH = 𝑀𝑤𝑡 𝐾𝐻𝑃 × 𝑣𝑓 −𝑉𝑖 /1000 0.3-0.4 g KHP 30 ml of water 1-2 drops ph.ph Titrate with NaOH untill color change to faint pink - HCl should be standardize before using by NaOH standard NaOH nHCl = nNaOH 0.2 M MHCl × vHCl = MNaOH × vNaOH MNaOH × vNaOH MHCl = vHCl 20 ml HCl 1-2 drops of ph.ph Titrate with NaOH untill color change to faint pink Procedure: Video Link: https://www.youtube.com/watch?v=DjBXw7toD_0 Thank You Palestine Polytechnic University General Chemistry Lab: Experiment 8: Bleach Analysis (Redox Titration) Introduction Bleach: is a solution of oxidizing agent (OCl-) that: 1. Oxidize color compounds (whitening) 2. Kill microorganism (Disinfiction) Types of bleach: 1. Liquid bleach: 5.25% NaOCl solution. 2. Powder Bleach: mixture of Ca(OCl)2 and Ca(OCl)Cl. Whitening mechanism: - Energy from the visible light excites electrons to a higher energy level in the molecule. - These electron return to its level and transmit radiation. - The unabsorbed visible radiation transmitted and detected with the eye - The hypochlorite ion removes these “excitable” electrons from the substance, and the clothes appear whiter. - ClO-(aq) + H2O(l) + 2 e- → Cl-(aq) + 2OH-(aq) To determine concentration of NaOCl in liquid bleach (Strength of bleach) we use Redox titration/ Iodometry Oxidizing agent: NaOCl (Unknown ??) Reducing agent: Na2S2O3 (standardize against KIO3) ClO-(aq) + 3I-(aq) + 2H+ I3-(aq) + Cl-(aq) + H2O(l) 2S2O32-(aq) + I3-(aq) 3I-(aq) + S4O62-(aq) ClO-(aq) + 2S2O32-(aq) + 2H+ Cl-(aq) + S4O62-(aq) + H2O(l) Bleach oxidizing strength measured relative to Chlorine (Cl2) as if Cl2 is the oxidizing agent Available Chlorine: oxidizing strength of the bleach that is equivalent to the oxidizing strength of the same mass of Cl2 per unit volume or mass Cl2(aq) + 3 I-(aq) 2 Cl-(aq) + I3-(aq) I3-(aq) + 2 S2O32-(aq) 3 I-(aq) + S4O62-(aq) Cl2(aq) + 2 S2O32-(aq) 2 Cl-(aq) + S4O62-(aq) Calculation according to this equation Procedure: 1. Bleach dilution (10 fold): Why? – Reduce active ingredient concentration which is: A. more safe B. Less chemical is used 1. Withdraw 10 ml of 2. Add 10 ml of bleach 3. Complete to final original Bleach by pipette to 100 ml volumetric volume with distilled water Flask  Total Volume Dilution factor = Volume of concentrated solution 100 = 10 = 10 Other dilution factors can be used depending on bleach concentration  2. Titration against Na2S2O3/ 3-trials Na2S2O3 0.1 M Na2S2O3 0.1 M High conc. of I3- [I3-.Starch] Low conc. of I3- I- 25 ml of diluted bleach (Oxidizing Agent) 20 ml of distilled water (Make sample larger) Add 2-3 ml of starch solution (1%) 2 g KI (Excess I-) 10 ml H2SO4 (3 M) (Acidic media) Titrate with Standard Na2S2O3 Complete titration with Standard Na2S2O3 0.1 g NaHCO3 or Na2CO3 (Stabilize Na2S2O3) untill color change to Pale yellow untill color change to Colorless Data Trial 1 2 3 Initial Burette reading (ml) Final Burette reading (ml) Volume of thiosulfate (ml) Average volume of thiosulfate(ml) - Volume of bleach titrated = - Mass of bleach titrated = - Molar concentration of Na2S2O3 (mol/L) = Calculations: Cl2(aq) + 2 S2O32-(aq) 2 Cl-(aq) + S4O62-(aq) Cl2(aq) + 2 Na2S2O3(aq) 2 NaCl(aq) + Na2S4O6(aq) At the end point n Na2S2O3 = M Na2S2O3 × v Na2S2O3 From standardization From titration n Na2S2O3 = M Na2S2O3 × (vf – vi)/1000 1 nCl2 = × n Na2S2O3 2 mCl2 = nCl2 × MwtCl2 𝑚 𝐶𝑙2 Cl2% = × 100% 𝑚 𝑑𝑖𝑙𝑢𝑡𝑒𝑑 𝑏𝑙𝑒𝑎𝑐ℎ 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚 𝐶𝑙2 Available Cl2% = ( × 100%)×Dilution Factor 𝑚 𝑑𝑖𝑙𝑢𝑡𝑒𝑑 𝑏𝑙𝑒𝑎𝑐ℎ 𝑠𝑎𝑚𝑝𝑙𝑒  𝑛 𝐶𝑙2 M Cl2 in diluted bleach = 𝑉 𝑑𝑖𝑙𝑢𝑡𝑒𝑑 𝑏𝑙𝑒𝑎𝑐ℎ (𝐿) 𝑛 𝐶𝑙2 M Cl2 n original bleach = × Dilution factor 𝑉 𝑑𝑖𝑙𝑢𝑡𝑒𝑑 𝑏𝑙𝑒𝑎𝑐ℎ (𝐿) - Compare your value with label one - Note: You compare Cl2 not NaOCl Video link https://www.youtube.com/watch?v=txWCzKHEZwo Remember: - Before using burrete 1. Wash with distilled water 2. Wash with solution 3. Fill with solution 4. Be sure that tip is filled with solution Remember: - Na2S2O3 solution should be Standardize against KIO3 solution using iodometery 6 𝑀𝐾𝐼𝑂3𝑉𝐾𝐼𝑂3 M Na2S2O3 = 𝑣𝑓 −𝑣𝑖 Na2S2O3 0.1 M Na2S2O3 0.1 M 25 ml of 0.01 M KIO3 20 ml of distilled water Add 2-3 ml of 1 g KI starch solution (1%) 5 ml H2SO4 (0.5 M) 0.1 g NaHCO3 or Na2CO3 Titrate with Standard Na2S2O3 Complete titration with Standard Na2S2O3 untill color change to Pale yellow untill color change to Colorless Notes - For powder bleach, grind sample well and dissolve in distilled water. Then analyze sample using the same procedure. - Sodium hypochlorite decompose with time to produce Cl2 gas sample, so be aware when open bleach bottle. 2ClO- + 4H+  Cl2(g) + 2H2O - Its important to use distilled water in this experiment, as hypochlorite ion in tap water could be an interferences. Thank You  Palestine Polytechnic University General Chemistry Lab: Experiment 10: Determination of Molecular mass of solid from Freezing-point deperssion Phase diagram of pure solvent Critical point: The set of condition under which a liquid and its vapor become identical Triple point: The temperature and pressure at which the three phases (gas, liquid and solid) of substance co-exist Normal freezing point, of the pure solvent: Is the temperature at which the vapor pressures of the solid and liquid phases of the solvent are equal under conditions. Where the total pressure of the system is 1 atm. When a nonvolatile solute is added to a solvent, the vapor pressure of the resulting solution is lower than that of the pure solvent. Phase Diagram for pure solvent and solution Colligative properties: Properties of solution that depend on the number of solute particles and not on the identity of them. Which are: 1. Lowering of vapor pressure 2. Elevation of boiling point 3. Depression of freezing point 4. Osmotic pressure The addition of a solute lowers the vapor pressure and reduces the freezing point. The magnitude of the vapor-pressure lowering of a solution depends only on the number of solute particles in the solution, not upon their identity. Example: C6H12O6  C6H12O6 (1 particle) NaCl  Na+ + Cl- (2 particle) CaCl2  Ca2+ +2Cl- (3 particle) CaCl2 lower freezing point of water more than NaCl more than glucose ∆Tf α moles of solute ∆Tf α molality ∆Tf = Kf × molality  For diluted solution: ∆Tf = Kf × m ΔTf : freezing-point depression (℃) Kf: molal freezing-point depression constant (Kg.℃/mol) m: Molality (mol/Kg solvent) Kf : - The depression of freezing point by the addition of one mole of solute to 1000 grams of solvent. - The value of Kf depends on the particular solvent - If one mole of nonvolatile solute particles is dissolved in 1000 grams of water, the freezing point is lowered 1.86°C. (Kf water = 1.86 Kg.℃/mol) - If one mole of nonvolatile solute particles is dissolved in 1000 grams of cyclohexane, the freezing point is lowered 20.0°C. (Kf cyclohexane = 20.0 Kg.℃/mol) Molality (m): Total number of moles of solute particles dissolved in 1000 grams or 1 kg of solvent.  𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 m= 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝐾𝑔) 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 m = 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝐾𝑔) ΔTf = Kf × m 𝐾𝑓 × 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 ΔTf = 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝐾𝑔) 𝐾𝑓 × 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝟏𝟎𝟎𝟎 Molar Mass solute = ∆𝑇𝑓 ×𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝒈) Liquid Freezing Solid Super Cooling Solvent cooling curve  Procedure Prepare Icebath In large test tube weight Put test tube in ice bath (add ice and water) 20 ml of Cyclohexan Meaure temperature by thermometer Determine freezing point Note: slow cooling by stirring with thermometer to prevent freezing on the interior walls of the test tube. - Repeat freezing point determination 3 times - Added 0.1 g of unknown solid, dissolve it in cyclohexane and measure freezing point of the solution - determine the molecular mass using the equation 𝐾𝑓 × 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 ×1000 - Molar Mass solute = ∆𝑇𝑓 ×𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑔) Thank You Palestine Polytechnic University General Chemistry Lab: Experiment 6: Limiting Reactant Objectives: 1. To determine the limiting reactant L.R of precipitation reaction 2. To Analysis unknown mixture using limiting reactant concept Introduction A + B  C + D Stoichiometry1:1 1 mol A + 1 mol B  1 mol C stoichiometric amount 2 mol A + 2 mol B  2 mol C stoichiometric amount 1 mol A + 2 mol B  1 mol C A is L.R B is E.R (1 mol B react and 1 mol B unreacted)  A + 2B  C + D Stoichiometry1:2 1 mol A + 2 mol B  1 mol C stoichiometric amount 2 mol A + 2 mol B  1 mol C A is E.R B is L.R (1 mol A react and 1 mol A unreacted) Limiting Reactant (L.R): Reactant that restrict the amount of product can be form from chemical reaction ( when it finish, Reaction Stop) Excess reactant (E.R): Some amount of it remain unreacted Advantages of using excess reactant in chemical reactions: 1. Increase the rate of reaction 2. Increase the percentage yield Example: 2Na3PO4(aq) + 3BaCl2(aq)  Ba3(PO4)2(s) + 6NaCl(aq) 2 mol Na3PO4 3 mol BaCl2 1 mol Ba3(PO4)2 6 mol of NaCl (stoichiometric amount) 3mol of Na3PO4 3 mol of BaCl2 1 mol of Ba3(PO4)2 6mol of NaCl 2 mol 1 mol 3 mol react React Unreact Excess reactant Limiting reactant At the moment when the limiting reactant BaCl2 completely reacts, the reaction stops and in the final reaction mixture only the products and 1mol of Na3PO4 will exist in excess. Molecular equation: 2Na3PO4(aq) + 3BaCl2(aq)  Ba3(PO4)2(s) +6NaCl(aq) Ionic equation: 6Na+(aq) +2PO43-(aq) + 3Ba2+(aq) + 6Cl-(aq)  Ba3(PO4)2(s) + 6 Na+(aq) + 6Cl- (aq) Net ionic equation: 3Ba2+(aq) + 2PO43-(aq)  Ba3(PO4)2(s) In this experiment, the following reaction will be carried: BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq) Ba2+(aq) + 2Cl-(aq) + 2Na+(aq) +SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq) Ba2+(aq) +SO42-(aq)  BaSO4(s) In this experiment, you have an unknown Na2SO4 - BaCl2 mixture You need to determine: 1. The composition of the mixture 2. The limiting reactant when they react 3. Amount of excess unreact You will: 1. Carry a precipitation reaction between these compounds 2. Collect the precipitate result from reaction 3. Determine the composition according to limiting reactant concept The limiting reactant determine experimentally by testing the clear solution after reaction complete with precipitation agent of reactants (Na2SO4 - BaCl2 )  Precepitate Supernatant Precepitate (BaSO4) In 250 ml Beaker add Stirr well 1 g of mixture (BaCl2-Na2SO4) Wait forprecepitate to appear Filtrate 100 ml of distilled water Simple Filtration Add drops of Ba2+ precepitation agent (BaCl2 solution) If precepitate appear then Ba2+ ions is the limiting ractant SO42- ions is the excess ractant 1 1 2 Add drops of SO42- precepitation agent (Na2SO4 solution) Two portions of filtrate for limiting reactant test If precepitate appear then SO42- ions is the limiting ractant 2 Ba2+ ions is the excess ractant Calculation: The equation: BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq) Part A: Mass of mixture = Mass of dried Filter paper (m1) = After the filter paper is dry Mass of Filter paper + BaSO4 (m2) = Mass of BaSO4 = m2 – m1 Moles of BaSO4 = mass of BaSO4/ molar mass of BaSO4 Moles of BaSO4 = moles of Na2SO4 react (according to reaction equation) Mass of Na2SO4 react =moles of Na2SO4 react* molar mass of Na2SO4 Moles of BaSO4 = moles of BaCl2 react (according to reaction equation) Mass of BaCl2 react=moles of BaCl2 react* molar mass of BaCl2 Part B Assume a precipitate is appear when adding Na2SO4 precipitating agent, then L.R. is Na2SO4 Part C. Mass percent of limiting reactant = (mass of Na2SO4/mass of mixture) *100% Mass of excess reactant unreact =mass Mix – (mass L.R + mass reacted from the other ) Mass of excess reactant unreact =mass Mix – (mass Na2SO4 react + mass BaCl2 react ) Thank You  Palestine Polytechnic University General Chemistry Lab: Experiment 7: Determining Percent Yield in a Chemical Reaction Objectives: To determine the percentages yield of chemical reaction Introduction:  Theoretical yield: - the maximum amount of product that can be formed in a chemical reaction based on the amount of limiting reactant - Calculate theoretically based on balance chemical equation - from known mass reactant Experimental yield: (Actual yield) : Actual mass of product that can experimentally obtained measured Percentage yield: the relationship between these two yields Experimental yield Percentage Yield (PY)= * 100% Theoretical yield Percentage Yield usually is less than 100 % PY < 100 % Experimental yield < Theoretical yield for these reasons  Incompletion of reaction  Side reaction  Slight solubility of product in water  Loss of substance (Reactants and products) Mechanical: material stick to glass road Evaporation of volatile compound Example: Calculating percent yield A students mixes 25.0 ml of 0.1 M BaCl2 with 25.0 ml of 0.2 M of Na2SO4 and collect 0.54 grams of BaSO4 as precipitate. Calculating percent yield. The balanced equation for the reaction is shown below. Step 1: Find moles of the reactants Mole of BaCl2 = C*V = 0.1 * 0.025 = 0.0025 Mole Mole of Na2SO4= C*V = 0.2 * 0.025= 0.005 mole Step 2: Find the limiting reactant The limiting reactant is BaCl2 Step 3: Calculate the theoretical yield (for product) BaSO4 - Form the balanced equation -1 mole of BaCl2 produce 1 mole of BaSO4 -Mole of BaSO4 =mole of BaCl2= 0.0025 mole -Mass of BaSO4 = mole * molar mass = 0.0025* 233.35 = 0.583 g (theoretical yield ) Step 4: Calculate the percent yield 0.54 = *100% = 92.6% 0.583 In this experiment In this experiment, we will prepare two solutions, as we learned in experiment 2  The first solution is Na2CO3  The second one is CaCl2 These solution will react according to this equation Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq) Example: Prepare 50 ml of 0.2 M CaCl2 solution Moles of CaCl2 = M * V = 0.2 * 50/1000 = 0.01 mol Mass of CaCl2 = n * Mwt = 0.01 * 111 = 1.11 g 1.11 g of CaCl2 weighted and dissolve in water to final volume 50 ml How to prepare solution Procedure Precepitate CaCO3 CaCl2 Na2SO4 Solution Solution Filtrate Prepare CaCl2 and Na2SO4 Solutions Add given volume of each solution Simple Filtration using clean pipetes Stirre Well Data: Mass of Filter paper (m1) Mass of Filter paper + CaCO3 (m2) Mass of CaCO3 (Actual Yield) = m2 – m1 Dry Filter Paper Weigh it Thank You