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# 10Post - CHE 133 MAKEMAKE-UP LABORATORY EXERCISE(12/5 or 7...

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Unformatted text preview: 11/16/2011 11/16/2011 CHE 133 MAKEMAKE-UP LABORATORY EXERCISE (12/5 or 7) Eligible Students are ONLY those who have excused absences from Gasometric Determination of NaHCO3 in a Mixture either a TEST (105 point) or or PRELIMINARY (55 point) Exercise MUST SIGN UP ON SHEET OUTSIDE OF DR. DR. AKHTAR’S OFFICE (213 CHEM) BY November 24 Last Update: 11/16/2011 12:02 PM (details will follow) 1 ? QUESTIONS ? 2 Objective: Determine the percent of NaHCO3 percent in a mixture by Gasometry How can the composition of a mixture be determined by measuring the volume of gas evolved when that mixture undergoes a chemical reaction? Concepts: Ideal Gas Law Vapor Pressure Henry’s Law Stoichiometry Techniques: Capture Gaseous Product Corrections to volume What principles must be considered in using the gas volume as a measure of the amount of gas liberated? Apparatus: Gas Syringe Thermometer Barometer 3 The Exercise is Conceptually Simple. Simple. 4. Use Ideal Gas Law to get number of moles of CO2, nCO2 The unknowns consist of a uniform Pv=nRT mixture of NaHCO3 and NaCl n CO2 = Pv CO2 CO2 / RT 5. From Stoichiometry of reaction, get moles of NaHCO3 nNaHCO3 = nCO2 1. Weigh Sample, wSample. 2. Do Chemistry – Reaction with HCl 6. From number of moles, get weight From get wNaHCO3 = nNaHCO3 * 84.0 g / mol NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g) + Cl- (aq) + Cl- (aq) NaCl(s) 4 7. Compute Percent Composition of Sample Na+ (aq) + Cl-(aq) PctNaHCO3 = 100 * wNaHCO3 / wSample (NaCl will dissolve in, but not react with, HCl) 3. Capture Liberated Gas and Measure its volume, vCO2 5 6 1 11/16/2011 11/16/2011 The Basic Experimental Arrangement But - it Involves Some Important Concepts Limiting Reagents Gas Mixtures – Partial Pressure Gas Solubility – Henry’s Law Syringe Does CO2 behave ideally? Accuracy & Reproducibility NaHCO3 / NaCl HCl 7 How Much HCl Insures that Unknown is the Limiting Reagent? NaHCO3 (s) + H+ (aq) Na+ 1 mol NaHCO3 (aq) + H2O (l) + CO2 (g) 8 How Much CO2 is produced? You weigh ~ 0.2 g of unknown. What is maximum volume of CO2 we can expect? ( with P = 1.0 atm and T = 25oC = 298oK ) 1 mol HCl Stoichiometery is 1 : 1. Assume unknown is pure NaHCO3 200 mg NaHCO3/(84 mg/mmol ) = 2.4 mmol max of unknown need 2.4 mmol of HCl max to consume it On the NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g) analytical balance balance 1 mol NaHCO3 1 mol CO2 0.200 g = 200 mg = 2.4 mmol HCl is 1.0 M = 1.0 mmol /mL Therefore, need at most 2.4 mmol/ 1.0 mmol/L = 2.4 mL v = n R T/ P = 2.4 X 0.0821 X 298/1.0 = 59 mL (Syringe capacity = 60 mL but unknowns < 100% NaHCO3) Are using 10 mL of 1.0 M HCl ( 10 X 1.0 = 10 mmol ) 10 – a significant excess Unknown will be the limiting reagent 9 How How much H2O(l) is produced by the reaction? At STP: 1 mole occupies 22.4 L 1 mmole occupies 22.4 mL 10 What Percent of Gas Volume is the H2O we Produce in the Reaction? NaHCO3 (s) + H+ (aq) Na+ (aq) + H2O (l) + CO2 (g) Reaction produces water. Is volume of water produced enough to large enough to affect volume of gas measured (max 60 mL)? 100 X 0.042 mL --------------------- = 0.07 % 60 mL 1 mol NaHCO3 1 mol H2O Still, considering pure NaHCO3, (200 mg NaHCO3 = 2.4 mmol) Volume of water produced by the reaction is < 0.1% of volume of gas collected We produce at most 2.4 mmol of liquid H2O 2.4 Is this volume significant compared to the ~ 60 mL of CO2 produced? 11 12 2 11/16/2011 11/16/2011 Gas Mixtures – Partial Pressure What’s in the System? Initial (mmol) Final H2O (l) Species 556 mmol 556 + x mmol HCl (aq) 10 mmol 10 – x mmol x mmol Reaction is conducted in a closed system • at constant external pressure - atmospheric (P ~ 1 atm) • at constant temperature - room temperature (T ~ 25oC) 0 mmol NaHCO3 (s) NaCl (s) y mmol 0 mmol Air (g) w mmol w mmol NaCl (aq) 0 mmol mmol x + y mmol mmol CO2 (g) 0 mmol x - z mmol CO2 (aq) 0 mmol z mmol Initially: System contains air & water (HCl) water Pressure in system is potentially due to: • water (from HCl) PH2O • the air in the system Pair And after reaction, • the liberated CO2 Patm PH2O + Pair = Patm PH2O, Pair and PCO2 PH2O + Pair + PCO2 = Patm 13 14 What is the Magnitude of PH2O? Table showing vapor pressure of water as a vapor Vapor function function of temperaturePressure of Water lab. is posted in P P H2O PIH2O ,T, vI, nair PH2O is a function of temperature temperature Vapor Pressure (mm Hg) 60 PFH2O,T, vF, nair, H2O P nCO2 nCO2s H2O(l) H2O(l) 50 Initially, we have P = PIair + PIH2O Finally, we have P = PFair + PFH2O + PFCO2 P,T & nair don’t change, so PIH2O = PFH2O = PH2O and 40 30 20 10 P – PH2O = PIair 0 0 5 10 15 20 25 30 35 40 Temperature (oC) Over the range 20oC – 30oC, PH2O increases from: 17.5 to 31.8 mm Hg (0.023 to 0.042 atm) 2.3% to 4.2% for P ~1 atm P = n RT/v P – PH2O = PFair + PFCO2 PIair = PFair + PFCO2 PFCO2 = PIair - PFair 15 nCO2 RT/vF = nair RT/vI - nair RT/vF 16 What are the various volumes in the exercise? nCO2 RT/vF = nair RT/vI - nair RT/vF nCO2 /vF = nair /vI - nair /vF nCO2 vSYS: The volume of just the large test tube and the right angle elbow = nair vF (1/vI - 1/vF) But, nair = (P - PH2O) vI / RT P nCO2 = [(P - PH2O) vI / RT] vF (1/vI - 1/vF) P nCO2 = (P - PH2O) (vF - vI) / RT P You will measure this! Equation 4 vI: The gas phase* volume of the entire closed system before reaction What is partial pressure of CO2 at end of the reaction? PCO2 = nCO2 RT/vF PCO2 = [(P - PH2O) (vF - vI) /RT ] RT/vF P PCO2 = (P - PH2O) (vF – vI)/vF vSYS initial vSYS – vHCl + vsyringe Equation 6 17 *We must exclude the volume of the liquid HCl but can ignore the solid NaHCO3 18 3 11/16/2011 11/16/2011 Gas Solubility – Henry’s Law vF: The gas phase* volume of the entire closed system after reaction nCO2s vSYS – vHCl + vfinal syringe *We must exclude the volume of the HCl again. Equation 4 final initial vF – vI = vSYS – vHCl + vsyringe – vSYS + vHCl - vsyringe final initial vF – vI = vSYS – vHCl + vsyringe – vSYS + vHCl - vsyringe 1- initial vsyringe final vsyringe 19 PCO2 = vI = ( P - PH2O ) ( 1 – vI / vF) SCO2 = kH * PCO2 ( kH = 3.2 X 10-2 mol / L-atm ) Suppose: • vSYS - volume of large test tube & right angle elbow is SYS 95.0 mL • Initial syringe reading is 5.0 mL and the final reading is 53.0 mL • We use 10.0 mL of HCl NOTEthat: : Note THAT mol of liberated CO2 dissolves in water. Using Henry’s Law, can calculate concentration of CO2 dissolved in water. • P = 1.000 atm, = We can estimate the deviation from ideality by examining the van der Waals constants. vF = 95.0 mL – 10.0 mL + 53.0 mL = 138.0 mL ( 1.000 – 0.026) ( 1 – 90.0 / 138.0) PCO2 = 20 Can CO2 be treated as an Ideal Gas? 90.0 mL PCO2 = 0.339 atm ( P + a / v2 ) ( v – b ) = RT v We use 10.0 mL of HCl CO2 nCO2s = 10.0 mL * 0.011 mmol/mL = 0.11 mmol 0.6% 0.18% Small corrections compared to others 21 Review variable 1 mol For CO2 at Room Temperature, the corrections to P and v are: a / v2 b / v SCO2 = 3.2 X 10-2 mol/L-atm * 0.339 atm = 0.011 M Issue 0.026 0.026 Hg 20 mmatm Departure of gases from Ideality? Equation 6 95.0 mL – 10.0 mL + 5.0 mL PH2O = • T = 22oC 22 Reproducibility (Factors affecting Precision) Effect Wsample analytical balance (0.0004 /0.3000) ~0.1% volume of H2O Produced vCO2 0.1% Non Ideality of CO2 PCO2 0.6% vapor Pressure of Water PCO2 2.3 – 4.2% Solubility of CO2 nCO2 3.0% PCO2 barometer (1 / 760) ~0.1% PH2O table of values (1 / 760) ~0.1% T thermometer (1/ 300)* ~0.3% (0.5 / 50) ~1% vCO2 If we seek accuracy to within 1% syringe Precision should be about 1% Accuracy should be less than 3% 23 * Assuming no ambient temperature changes – see Prob 2 24 4 11/16/2011 11/16/2011 Calculations Weight of Sample Volume of gaseous CO2 v Pressure, P = 752 mm Hg = Temperature, T = 23oC = [email protected] 23oC (from Table) 21 mm Hg 0.2147 g 45.7 mL 0.989 atm 296 K 0.028 atm mmol CO2 (gas) = (P - PH2O)v / RT mmol CO2 (liquid) (Henry’s Law) 1.81 mmol 0.10 mmol Tot CO2 1.91 mmol To calculate the Henry’s Law correction, we need the Volume of the System. Suppose it is 100.0 mL That makes the initial volume, VI = 100.0 -10.0 + 5.0 = 95.0 mL and the final volume, VF = 100.0 -10.0 + 50.7 = 140.7 mL Vsys = 100.0 mL Syringe: Initial = 5.0 mL Final = 50.7 mL PCO2 = ( P - PH2O )( 1 – vI / vF) = ( 0.989 – 0.028)( 1 – 95.0 / 140.7) = 0.312 atm SCO2 = 3.2 X 10-2 * 0.312 atm = 0.010 M 25 Calculations Calculated Weight of Sample v Volume of gaseous CO2 Pressure, P = 752 mm Hg = Temperature, T = 23oC = [email protected] 23oC (from Table) 21 mm Hg 0.2147 g 45.7 mL 0.989 atm 296 K 0.028 atm mmol CO2 (gas) = (P - PH2O)v / RT mmol CO2 (liquid) (Henry’s Law) 1.81 mmol 0.10 mmol Tot CO2 1.91 mmol mmol NaHCO3 (from stoichiometry) 1.91 mmol 0.160 g Weight of NaHCO3 1.91 X 84.0 % NaHCO3 = 100 X 0.160 / 0.2147 = 74.5 %27 nCO2s = 10.0 mL * 0.010 mmol/mL = 0.10 mmol 26 Procedure - Notes Everyone should weigh ~ 200 mg = 0.2 g – ACCURATELY – for their initial run Depending on your sample, you may need to adjust the weight in subsequent runs to insure that you get between 30 and 50 mL of CO2, but not more than 50 mL. Test that system is air-tight before using Set syringe at 5.0 mL initially – read to 1 decimal – remember to subtract from final volume Do test run - then 4 which you report. 28 Last Exercise Determination of NaHCO3 in a Mixture Part 2 - Gravimetric Gravimetric Final Exercise – 105 points Do SUSB-054 - Pre-lab Assignment 2 Final Quiz will be given at the beginning of the check-out laboratory meeting 29 5 ...
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