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SBU CHE 133 - Gravimetric Determination of NaHCO3

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1in a MixtureGravimetric Determination of NaHCO31Purpose:Determine the percent of NaHCO3in a mixture by GravimetryConcepts:Thermal Decomposition Constant WeightTechniques:WeighingThermal DecompositionApparatus:Analytical Balance Hot PlateEvaporating Dish Crucible Tongs2This Part of the Exercise is, again Conceptually Simple.1. Weigh Sample, 2. Decompose the sample by heating3. Weigh product to get weight of CO2 & H2O lostweight loss = wCO2+ wH2O= nCO2* 44.01Molar Mass of CO2Molar Mass of H2O+ nH2O* 18.022 NaHCO3(s) Na2CO3 (s) + H2O (g) + CO2 (g)wSample.weight = mol X Mol Mass+ NaCl (s)+ NaCl (s)2 NaHCO3(s)3weight loss4. Get moles of CO2, H2O and NaHCO3 lostNaHCO3(s)  ½ Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g)nCO2= nH2O= ½ nNaHCO3= ½ nNaHCO3* ( 44.01 + 18.02 )or, nNaHCO3=5. wNaHCO3=6. Compute Percent Composition of SamplePctNaHCO3= 100 * wNaHCO3/ wSampleMolar Mass of NaHCO3(= nCO2* 44.01+ nH2O* 18.02 )so, weight loss2 * ( weight loss ) / 62.03nNaHCO3* 84.01 g / mol½½½4wNaHCO3=PctNaHCO3= 100 * wNaHCO3/ wSamplenNaHCO3* 84.01 g / mol2 * ( weight loss )62.03X 84.01= 2.709 * ( weight loss )= 270.9 * ( weight loss ) / wSampleHow is Pct NaHCO3related to the weight loss?5wNaHCO3= ( weight loss ) / 0.3691nNaHCO3 =wNaHCO3=2 * ( weight loss )62.03 Showed that:Accuracy and precision of results are solely a function of care with which weighings and thermal decomposition are conducted!627The melting point of NaCl is 800 oC so don’t overheat But, the peak temperature on a hot plate is well below 800 oC.Molten NaCl cools to produce a solid that adheres to porcelain. Hard to clean! Also, Na2CO3(product) melts at 850 oC ANDbegins to decompose to CO2+ Na2O at that temperature.NaHCO3decomposes rapidly at T > 200 oC forming Na2CO3How do the Various Substances behave at High Temperature?But make sure it is at least 200 oC Would result in additional unwanted weight loss.8What precision & accuracy should you expect?The only sources of error are:• loss of sample during run• don’t spill any sample• incomplete heating• reheat until two successive weighings differ by < 5.0 mg• decomposition due to overheating• not likely on hotplate•weighing errors!!!!!• record numbers carefullyTo find weight loss, you need the weight of the dry, empty crucible5.0 mg difference in two successive weighings is our criterion for establishing that decomposition is complete• try to use same balance throughout a runDO NOT TINKER WITH BALANCE SETTINGS!This exercise depends heavily on the proper use of the analytical balance!?? Standard Penny ??Weigh a coin before anything elseCheck the weight of the “standard” coin before each subsequent weighing.9 10How do we know when the reaction is done?When all the H2O and CO2have been lost, the sample will cease to lose weight.I.e., two successive weighings before and after heating should be the sameHow should we define “the same” (Δ)?Our operational definition is that they differ by less than ± 5.0 mg. I.e., |w2–w1| < 0.0050 gw116.5486w216.5294- 0.0192 w2 16.5294w3 16.5242- 0.0052 w316.5242w4 16.5214- 0.0028 w316.5242w4 16.5276+ 0.0034- 9.2 mg - 5.2 mg - 2.8 mg + 3.4 mge.g.XXNOT 0.5 mg !!Reproducibility wsampleanalytical balance (0.0004 / 1.0000) < 0.1%wresidueanalytical balance (0.0004 / 0.7000) < 0.1%Our criterion for final weight: two successive weighingsdiffer by less than 5.0 mg, so our weight loss should differ from the “true” weight loss by less than 5.0 mg .E.g., for a 2.0 g sample with 50% NaHCO3the minimum weight loss is 369.1 mg. So, 5.0 mg constitutes less than a 100 X 5.0 / 369.1 = 1.3 % error in the weight loss and same error in %NaHCO311Uncertainty in Weighing:But reproducibility is determined by:Do the analysis twice,and12Handle the evaporating dish ONLY with crucible tongs!3Wt of crucible + sample 16.0755 gWt of crucible 14.9842 gWt of sampleWt of crucible + residue –aft heat1 15.9013 gWt of crucible + residue –aft heat4 15.8689 gWt of crucible + residue –aft heat3 15.8723 gWt of crucible + residue –aft heat2 15.8816 g19.7 mg9.3 mg3.4 mgWt of residue [ 15.8689 – 14.9842 ]Wt loss [ 1.0913 – 0.8847 ]Wt NaHCO3[ 0.2066 / 0.3691 ]% NaHCO3[100 X 0.5597 / 1.0913 ]DATA SHEET0.8847 g0.2066 g0.5597 g51.29 %1.0913 g174.2 mgWt loss per gram of NaHCO313Suppose our second result is 52.41% 51.29 + 52.41 Average = ------------------ = 51.85 %20.56 + 0.56Avg Dev = ----------------- = 0.56 %2100 X 0.56Pct Dev = ------------------- = 1.1 %51.85Review - Avg, Avg Deviation, Pct Deviation14RFS


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