Post lab 2 - post lab report covering identifying unknown using stoichiometry experiment PDF

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post lab report covering identifying unknown using stoichiometry experiment...

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.Identifying unknown using stoichiometry Experiment 2 Experiment performed on 2/22/2021 Chemistry 1215L section 12  Data analysis: f(x) = 0 R² = 0

Pressure vs molarity 0.04

Pressure aatm

0.03

f(x) = − 0.08 x + 0.07 R² = 0.91

0.03 0.02 0.02 0.01 0.01 0 0.45

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

Molarity M

Balanced equations of sodium bicarbonate and sodium carbonate with hydrochloric acid: NaHCO3(aq) + HCl(aq) = NaCl(aq)+CO2(g)+H2O(l) limiting reactant (HCl(lower molar mass than NaHCO3) Na2CO3(aq) + 2HCl(aq) = 2NaCl(aq) + CO2(g) + H2O(l) limiting reactant (HCl)

Since Na2CO3 and HCl have a 1:2 ratio then one mole of HCl will react with 0.5 moles of Na2CO3 resulting in a 0.5 moles of product. Showing that HCl is the limiting reagent.

Number of unknown 1: A46 Pressure values: 0.1, 01 0.2,0.2,0.03,0.03,0.01,0.01,0.004,0.004 Mole ratio : volume of base/volume of acid+volume of base Mole ratio values: 0.1667(for 1 ml of base and 5ml of acid) 0.333(for 2ml of bas and 4ml of acid) 0.5(for 3ml of base and 3ml of acid) 0.667(for 4ml of base and 2ml of acid) 0.883(for 5ml of base and 1ml of acid) Acid: HCl Base: unknown A46 f(x) = 0 R² = 0

Pressure vs molarity 0.25

Pressure aatm

0.2

f(x) = − 0.51 x + 0.45 R² = 0.99

0.15 0.1 0.05 0 0.45

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

Molarity M

Number of uknown 2: B35 Pressure values: 0.1,0.1,0.2,0.2,0.2,0.2,0.1,0.1,0.03,0.03 Mole ratio : volume of base/volume of acid+volume of base

0.9

Mole ratio values: 0.1667(for 1 ml of base and 5ml of acid) 0.333(for 2ml of bas and 4ml of acid) 0.5(for 3ml of base and 3ml of acid) 0.667(for 4ml of base and 2ml of acid) 0.883(for 5ml of base and 1ml of acid) Acid: HCl Base: unknown B35

Calculations:

,

Values of pressure for the first curve(unknown A46): 0.1, 01 0.2,0.2,0.03,0.03,0.01,0.01,0.004,0.004(values of pressure for the 5 values of molar fractions, 2 trials for each value) Mean(average) = (0.1,+ 01+ 0.2+,0.2,+0.03+,0.03+,0.01+,0.01+,0.004,+0.004)/10

Mean = 0.0688 Standard deviation:

0.08

Values of pressure for the second curve(unknown B35): : 0.1,0.1,0.2,0.2,0.2,0.2,0.1,0.1,0.03,0.03(values of pressure for the 5 values of molar fractions, 2 trials for each value) Mean(average) = :( 0.1,+0.1+,0.2+,0.2+,0.2,+0.2+,0.1+,0.1,+0.03+0.03)/10

Mean = 0.126 Standard deviation: 0.07

The peak point of first graph is 0.33, which indicates that 0.33 is the max mole fraction. 1-0.33 = 0.67. which indicates that the stoichiometric ratio is 0.33:0.67(1:2) The peak point of the second graph is 0.4, which indicates that 0.4 is the max mole fraction 1-0.4 = 0.6 so the stoichiometric ratio is 0.4:0.6(2:3)

 Summary of experimental procedure and results) :

results(summary

of

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   

 Gass ens ormus tbeatchannel 1( connectgaspr ess ur es ens ort o l abquest )  Connec tt hel abques tt oy ourl apt op  Connec tt heUSBt ot hel apt op  Opent heLogger Pr os of t war eony ourl apt op  Cl i ckonex per i ment‘ c hangeuni t ’  Cl i ckonl abquest2ands el ectat m  Pi cksomebas eus i ngabeak er  Fi l l1mLofNaHCO3 i nt oy ours yr i nge  Scr ewi tatt hecor kandi nser tt he5mLofbas ei nt ot he Er l enmey erfl as k  Tur nt hest opcor kt os t opt heflowi nt ot heEr l enmey erflask  Rec or dt hev al ueofpr es sur e  Dot hes ames t eps( 11, 12, 13and14)butwi t h2mLofbas e  Keepr epeat i ngt hes ames t epswi t h3mL4and5mLofbas e  Rec or dt hev al ueofpr es sur eaf t erpour i ngeac hamounti nt ot he fl ask  Pi cksomeac i dusi ngabeak er  Fi l l5mLofHCli nt oy ours yr i nge  Scr ewi tatt hecor kandi nser tt he5mLofaci di nt ot heEr l enmey er fl askt hathast hebas ei ni t .  Tur nt hest opcor kt os t opt heflowi nt ot heEr l enmey erflask  Rec or dt hev al ueofpr es sur e  Dot hes ames t eps( 19, 20, 21and22)butwi t h4mLofaci d  Keepr epeat i ngt hes ames t epswi t h3mL2and1mLofaci d  Rec or dt hev al ueofpr es sur eaf t erpour i ngeac hamounti nt ot he fl ask  Swi r l t hefl askal i t t l ebi tsot heaci dandbas es ol ut i onmi xwel l t oget her  Wai tf ort hel oggerpr oc ur v et obecomeas t r ai ghtl i ne  Cl i cks t op,t henr ecor dt hedat a  Subt r actt hei ni t i al v al ueofpr es sur ef r om t hefi nal onet oobt ai n t hec hangei npr ess ur e  Cl eanev er y t hi ngont het abl eusi ngawetnapki n  Spi l lt heaci dandbas ei nt ot hechemi c al sc ont ai ner( onl ywat er goesi nt ot hesi nk )

 Results and conclusion: Thel owv ar i at i onofmys t andar ddev i at i on( 0. 08and0. 07)and t he mean( 0. 0688)meanst hatmyv al ueshav eal owv ar i at i onofdi ffer enc eand ar ecl os et ot hemean,mak esmeal i t t l econfi dentaboutmyr es ul t st oi dent i f y t heunknowncompoundbas edont hepr ess ur eofCO2r el eas edand s t oi chi omet r i cr at i o.Thest oi c hi omet r i cr at i of ort hefi r s tgr aph( unknownA46) i s( 1: 2)meani ngt hatunk nown1i sNa2CO3( s odi um car bonat e)andt he s t oi chi omet r i cr at i ooft hes ec ondgr aph( unknownB35)i s( 2: 3) ,al l udi ngt oa s l i ghter r ori nmydat a.Thepeakpoi nti s0. 4,whi chgi v esar at i oof( 10. 4) 0. 4: 0. 6,howev ert hepeakpoi nts houl d’ v ebeen0, 5l eadi ngt oar at i oofof ( 10. 5)0. 5: 0. 5( 1: 1)l eadi ngt ot hei dent i fic at i onoft hes econdunknownwhi c hi s s odi um bi car bonat e NaHCO3.

Post Lab discussion questions: Hypothesis: The only difference between sodium carbonate and sodium bicarbonate is the amount of CO2 produced during the reaction. Since the CO2 production is the only component of the reaction that would cause a significant change in pressure, the pressure measurements of the reactions of both unknown chemicals with HCl can provide a direct link to the stoichiometry of CO2. Question B: The jobs method lead to the identification of the unknown by locating the peak point on the curve first and then subtracting it from 1 to find the stoichiometric ratio (stoichiometry of CO2)which is 1:2 for the first graph referring to sodium bicarbonate and a ratio of 1:1 for the second graph which refers to sodium carbonate, but I had a slight error in my data so I got a ratio of 2:3. And the other way of identifying the unknown from the

known, we can use the graph created during the design lab of either sodium carbonate or sodium bicarbonate. The graph of the unknown matches that of the known....