Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) I. UV-Vis Absrbance Spectrscpy Lab Beer s law Relates cncentratin f a chemical species in a slutin and the absrbance f that slutin A= Ebc A- absrbance Measured by spectrmeter Cmpares light entering sample t the light that made it thrugh yur sample E- mlar absrptive cnstant (L*cm /ml) Unique fr every element and wavelength b- path length (cm) (bulb) (cuvette) (detectr) pathlength: distance light travels thrugh slutin (which wuld be the length f the cuvette in this diagram) c- cncentratin (ml/l) An increase in the path length r cncentratin wuld increase the absrbance because there is a direct relatinship between thse variables and the absrptin. Path length: yu are increasing the distance where the light can be absrbed therefre the absrbance wuld be larger if yu increased it. Cncentratin: the light can t pass thrugh the substance as easily, aka it is absrbed, because there are mre mlecules, therefre the absrbance ges up if the cncentratin des. Darker clr= mre cncentrated= mre absrbance Linear relatinship between A and c Pre-Lab Questins: Why d we want t use a particular wavelength when determining the absrptin f a particular chemical species? Is it imprtant t measure absrbance fr all wavelengths frm 400 t 700 nm? (aka n bjectives list: hw taking a scan f a slutin can be used in lab & why we used a single wavelength t measure the absrptin f a particular chemical species ) We want t nly measure the absrbance f the metal with n impurities. It s imprtant t measure absrbance fr all wavelengths that way we knw what the particular wavelength is. By measuring ne wavelength, we can measure just ne chemical species, even if it s in a mixture We must first determine which wavelength t measure. 1 Made by Ashley Thmas
Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) If yu knw the mlar absrptivity f cpper sulfate at 630 nm, explain hw yu culd determine a wavelength where the mlar absrptivity is half f that simply by examining the absrptin spectrum. Yu wuld just lk at the graph and fllw it until yu get t 0.5 n the absrptivity side. Serial Dilutins Like step 4 f the lab: Frmula: Frmula derived frm: Example calculatin: see yur lab! (either step 4 r Data Questin 2) Calibratin Plt (these ntes are based n the pst-lab prblem n the back f the lab handut, als can be seen belw) Cnstructin y-intercept: Clseness t zer R^2- precisin, hw well des yur trend line fit yur data, best fit is 1 Linear relatinship between A and c Best fit line- pints may nt be perfect, it s the line that fits the pints the best, pretty self explanatry Errr analysis Ging back t what the y-intercept and R^2 shw and based n the pst-lab assessment R^2= 0.995, which means their data is precise because the trend line fits the data well (the data pints are actually n the line). Y-intercept= -0.005494, is extremely clse t zer, which means, using beer s law, that is a gd mdel fr that data. Calculatin 0.2877ml 0.05 L 159.62 g CuSO4 L 1 ml. CuSO4 = 2.296 g CuSO4 2 Made by Ashley Thmas
Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) Hw Spectrmeter wrks The light ges thrugh the substance in the cuvettte and the light that is nt absrbed by the substance ges t the detectr which shws the absrbance. Diagram n next page! (bulb) (cuvette) (detectr) II. ChemActivity 3: Culmbic PE (ptential Energy) Mdel 1: Tw Charged Particles Separated by a Distance d Ntes: The ptential energy (V) f tw statinary charged particles is given by the equatin listed belw, where q1 and q2 are the charge n the particles and d is the separatin f the particles (in pm) and k is a psitive-valued prprtinality cnstant. 1pm= 10^-12 m CTQ s 1. Assuming that q1 and q2 remain cnstant, what happens t the magnitude f V if the separatin, d, is increased? Smaller, decreases 2. If the tw particles are separated by an infinite distance (that is d= infinity), what is the value f v? Appraches/is zer 3. If d is finite, and the particles have the same charge (that is q1=q2), is V>0 r is V<0? Explain yur answer. V>0, because the number must be psitive. 4. If q r an electrn is -1 (a) What is q fr a prtn? 1+ (b) What is q fr a neutrn? 0 (c) What is q fr the nucleus f a C atm? 6+ 5. Recall that a H atm cnsists f a prtn as the nucleus and an electrn utside f the nucleus. Is the PE, V, f a H atm a psitive r negative number? Explain. Negative because a psitive times a negative must be negative. The PE f the interactin between a prtn and electrn will always be negative. Mdel 2: IE (inizatin energy) Ntes: The IE is the amunt f energy needed t remve an electrn frm an atms and mve it infinitely far way. IE is usually measure in Jules CTQ s (see table 1 pg 18, in packet fr the table references) 3 Made by Ashley Thmas
Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) 6. D yu expect the PE f the hypthetical atms in Table 1 t be psitive r negative numbers? Explain. Negative, because the ppsite charges cause negative numbers. 7. Withut using a calculatr, predict what trend (if any) yu expect fr the values f V fr these hypthetical atms. Frm A t Z the V will increase in magnitude r get mre negative. 8. Calculate the Ptential energies f the hypthetical atms t cmplete table 1. Use the value k= 2.31*10^-16 J*pm. Just use the PE equatin 9. What is the relatinship between IE and V fr these hypthetical atms? They are directly prprtinal, but ppsite in sign. 10. Which f the fllwing systems will have the large IE? Explain. (a) An electrn at a distance f 500pm frm a nucleus with a charge +2 (b) An electrn at a distance f 700pm frm a nucleus with a charge +2 (a) wuld be the crrect answer because it is clse t the nucleus, therefre harder t remve. 11. Which f the fllwing system will have the larger IE? Explain. (a) An electrn at a distance d1 frm a nucleus with a charge +2 (b) An electrn at a distance d1 frm a nucleus with a charge +1 (a) because it has a strnger attractive frce 12. Hw many times is the larger f the tw IEs frm CQT 11? Shw yur wrk s the secnd is tw times larger. 13. Cnsider a H atm and a He in, He+. Which f these d yu expect t have the larger IE? Explain. He+ because it has mre prtns fr the same number f electrns, it has a higher charge fr the same distance. III. ChemActivity 4: The Shell Mdel (I) (k s I gt lazy and am just typing up ntes yu wuld need fr the bjectives list srry!) Objective 1: Explain hw experimental inizatin energies prve the existence f electrn shells The numbers fr the inizatin energies drp nce yu start a new rw n the peridic table. Based n the first cuple values yu wuld expect the values t cntinue t increase but actually they g dwn then back up. Which tells us that the distance increases, therefre prving the existence f electrn shells. Objective 2: be able t use the CPE equatin t ratinalize/explain changes in IE data The distance, when yu add the secnd electrn shell increase, therefre the IE wuld g dwn. Hwever as yu get a larger charge with the same distance (the 2 nd shell) yu 4 Made by Ashley Thmas
Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) see an increase in the IE. This is based n the direct relatinship between charge and IE and the indirect relatinship between distance and IE. Objective 3: Definitin Catin: a psitively charges species Inizatin Energy: the minimum energy required t remve an electrn frm a gaseus atm f that element. First Inizatin Energy: The energy required t remve the utmst/ electrn frm the atm. Crrespnds t the smallest amunt f energy that a bmbarding electrn needs t be able t knck ff ne f the atm s electrns. Shell: ummm I dn t really knw hw t explain it in simplistic terms s its um a shell. 5 Made by Ashley Thmas