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in a Mixture. Gravimetric Determination of NaHCO 3. FINAL Final Exercise – 105 points. FINAL. CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM. Eligible Students are ONLY those who have excused absences from
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in a Mixture Gravimetric Determination of NaHCO3 FINAL Final Exercise – 105 points FINAL
CHE 133MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM Eligible Students are ONLY those who have excused absences from either a TEST (105 point) orPRELIMINARY (55 point) Exercise SHOULD HAVE SIGNED UPON SHEETS POSTED IN EACH LABORATORY ROOM Everyone does the same make-up exercise. You can download the exercise at: http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf
Purpose: Determine the percent of NaHCO3 in a mixture by Gravimetry Concepts: Thermal Decomposition Constant Weight Techniques: Weighing Thermal Decomposition Apparatus: Analytical Balance Hot Plate Crucible Crucible Tongs
This Part of the Exercise is, again ConceptuallySimple. • Weigh Sample, wSample. 2. Decompose the sample by heating 2 NaHCO3 (s) 2 NaHCO3 (s) Na2CO3 (s) + H2O (g) + CO2 (g) + NaCl(s) + NaCl(s) 3. Weigh product to get weightof CO2 & H2O lost weight loss = wCO2 + wH2O weight loss = nCO2 * 44.01 + nH2O * 18.02 Molar Massof CO2 Molar Mass of H2O weight = mol X Mol Mass
4. Get moles of CO2 , H2O and NaHCO3 lost 2 NaHCO3 (s) ½Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g) ½ ½ ½ nCO2 = nH2O= ½ nNaHCO3 (= nCO2 * 44.01 + nH2O * 18.02 ) so, weight loss = ½ nNaHCO3 * ( 44.01 + 18.02 ) or, nNaHCO3 = 2 * ( weight loss ) / 62.03 Molar Mass of NaHCO3 5.wNaHCO3 = nNaHCO3 * 84.01 g / mol 6. Compute Percent Composition of Sample PctNaHCO3 = 100 * wNaHCO3 / wSample
How is Pct NaHCO3 related to the weight loss? wNaHCO3 = nNaHCO3 * 84.01 g / mol Showed that: 2 * ( weight loss ) 62.03 nNaHCO3 = 2 * ( weight loss ) 62.03 wNaHCO3 = X 84.01 = 2.709 * ( weight loss ) wNaHCO3= ( weight loss ) / 0.3691 PctNaHCO3 = 100 * wNaHCO3 / wSample = 270.9 * ( weight loss) / wSample
wNaHCO3= ( weight loss ) / 0.3691 How much weight would 1.000 g of pure NaHCO3 lose? wNaHCO3 = 1 g 1 g = ( weight loss ) / 0.3691 weight loss = 0.3691 g PctNaHCO3 = 100 * wNaHCO3 / wSample 270.9 * ( weight loss) / wSample = 270.9 * 0.3691/ 1.000 = 100.0 %
Each mole of NaHCO3 produces one mole of CO2 and one mole of H2O when decomposed by heating. • True • False
Each mole of NaHCO3 produces one mole of CO2 and one mole of H2O when decomposed by heating. NaHCO3 (s) ½Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g) BFalse
Unlike gasometric part, this part is NOT complicated by any special adjustments or considerations. Accuracy and precision of result are solely a function of care with which weighings and decomposition are conducted!
How do the Various Substances behave at High Temperature? NaHCO3 decomposes rapidly at T > 200 oC forming Na2CO3 The melting point of NaClis 800 oC. (Molten NaClcools to produce a solid that adheres to porcelain. ) Na2CO3(product) melts at 850 oC AND begins to decompose to CO2 + Na2O at that temperature. The peak temperature on a hot plate is well below 800 oC. But make sure it is at least 200 oC Would result in additional weight loss 850 oC is bright red heat
If some sample is lost from the crucible during the heating, the resulting percentage of NaHCO3 that we calculate will be ______ the actual value. • smaller than • unchanged from • larger than
OMG ! ! ! ! ! Let’s try again. This time, with enthusiasm (and discussion)
If some sample is lost from the crucible during the heating, the resulting percentage of NaHCO3 that we calculate will be ______ the actual value. • smaller than • unchanged from • larger than
If some sample is lost from the crucible during the heating, the resulting percentage of NaHCO3 that we calculate will be ______ the actual value. PctNaHCO3 = 270.9 * ( weight loss) / wSample If sample is lost, the apparent weight loss will be too large and, therefore, the % NaHCO3 will be too large. CLarger
Last Year Right to Right 48 Right to Wrong 31 Wrong to Right 46 Wrong to Wrong 114
What 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 • weighingsdiffer by Δ mg • overheating • not likely on hotplate • weighing errors • record numbers carefully To be specified To find weight loss, you need the weight of the dry, empty crucible A word from Obviousman
This exercise depends heavily on the proper use of the analytical balance! • try to use same balance throughout a run DO NOT TINKER WITH BALANCE SETTINGS! ?? Standard Penny ?? Weigh a coin before anything else Check the weight of the “standard” coin before each subsequent weighing.
How do we know when the reaction is done? When all the H2O and CO2 have been lost, the sample will cease to lose weight. I.e., two successive weighings before and after heating should be the same How should we define “the same” (Δ)? Our operational definition is that they differ by less than ± 5 mg. I.e., |w2 – w1| < 0.0050 g NOT 0.5 mg! e.g. w1 16.5386 w216.5294 - 0.0092 w3 16.5242 w4 16.5214 - 0.0028 w2 16.5294 w3 16.5242 - 0.0052 w3 16.5242 w4 16.5276 + 0.0034 X X - 9.2 mg - 5.2 mg - 2.8 mg + 3.4 mg
Bonus Question (5 extra points) What are the last 3 digits of your student ID?
Reproducibility Uncertainty in Weighing: wsample analytical balance (0.0004 / 1.0000) < 0.1% wresidue analytical balance (0.0004 / 0.7000) < 0.1% But Our criterion for final weight is one that differs from the previous weight by less than 5mg, so our weight loss should differ from the “true” weight loss by less than 5mg . For a 1 g sample, the minimumweight loss is ~369.1/2 mg, and 5mg constitutes less than a 100 X 5 X 2 / 369.1 = 2.7 % error in the weight loss 50% NaHCO3
Do the analysis twice, and Handle the crucible ONLY with crucible tongs!
DATA SHEET Wt of crucible + sample 16.0755 g Wt of crucible 14.9842 g 174.2 mg Wt of sample 1.0913 g Wt of crucible + residue –aft heat1 15.9013 g 19.7 mg Wt of crucible + residue –aft heat2 15.8816 g 9.3 mg Wt of crucible + residue –aft heat3 15.8723 g 3.4 mg Wt of crucible + residue –aft heat4 15.8689 g Wt of residue [ 15.8689 – 14.9842 ] 0.8847 g Wt loss per gram of NaHCO3 Wt loss [ 1.0913 – 0.8847 ] 0.2066 g Wt NaHCO3 [ 0.2066 / 0.3691 ] 0.5597 g % NaHCO3 [100 X 0.5597 / 1.0913 ] 51.29 %
If the NaHCO3 is not completely decomposed when we record our final weight, the % NaHCO3 we compute will be ________ the actual value. • Smaller than • Equal to • Larger than
If the NaHCO3 is not completely decomposed when we record our final weight, the % NaHCO3 we compute will be ________ the actual value. PctNaHCO3 = 270.9 * ( weight loss) / wSample If NaHCO3 is not completely decomposed, the apparent weight loss will be too small and, therefore, the % NaHCO3 will be (too) small. ASmaller than
Review - Avg, Avg Deviation, Pct Deviation Suppose our second result is 52.41% 51.29 + 52.41 Average = ------------------ = 51.85 % 2 0.56 + 0.56 Avg Dev = ----------------- = 0.56 % 2 100 X 0.56 Pct Dev = ------------------- = 1.1 % 51.85
At the end of the exercise, the crucible should contain only NaCl. • True • False
At the end of the exercise, the crucible should contain only NaCl. NaHCO3 (s) ½Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g) The crucible contains NaCl and Na2CO3 BFalse
Next Week – No Lecture Nov 19 & 20 - Make-up Labs Fri, Nov 30, Mon, Dec 3 – REVIEW Attendance Optional
When you prepare, and eat your holiday turkey, tofu, tuna, tomatoes, turnips, pita, pasta, whatever peas, potatoes, WEAR YOUR
It has been fun learning with you Have a good rest of the semester