Take-Home Exam Answer Sheet	__8 pts

Standard #	Cu(II) moles/L	Ni(II) moles/L
0	0	0
1	0.1	0.05
2	0.25	0.10
3	0.50	0.15
4	0.75	0.20

Chemical Love Story: part 1

(details at end of answers)

___/6 pts

mL unknown added	# Standard Cu additions	ppm std Cu	# Standard Ni additions	ppm std Ni
5	0	0	0	0
5	1	2.5	1	2.0
5	2	5.0	2	4.0
5	3	7.5	3	6.0
5	5	12.5	4	8.0

Chemical Love Story: part 2(detailed calculations at end of answers)

Chemical Love Story: part 3	____/7 pts

Cu conc	A(812 nm)	A(395 nm)	Ni conc	A(812 nm)	A(395 nm)
0	0	0	0	0	0
0.1	0.125	0.002	0.05	0.001	0.056
0.25	0.320	0.006	0.10	0.002	0.125
0.50	0.621	0.014	0.15	0.004	0.169
0.75	0.987	0.018	0.20	0.0055	0.220
unknown	0.2675	0.119

Fit with EXCEL trendline function

species	m(812 nm)	b(812 nm)	r2	m(395 nm)	b(395 nm)	r2
Cu(II)	1.3037	-0.0066	0.999	0.0252	-0.00006	0.993
Ni(II)	0.028	-0.0003	0.986	1.106	0.0034	0.995

Unknown [Cu(II)] = ___0.20__ M	Unknown [Ni(II)] = ____0.10___M

Chemical Love Story: part 4	_____/3 pts

AAS conditions for analysis )(these will vary spectrometer)

Copper AAS conditions : HCL lamp current__15 ma__	slit 1.8_ (nm)_216nm_

Pertinent range of linearity _____~6.5________ ppm

Nickel AAS conditions: HCL lamp current_15 ma_ slit__1.8/.6nm__ (nm)_341.48__

Pertinent range of linearity _~20_________ppm

Flame composition (fuel, oxidant and ~volume ratio): ____acetylene/air 1:4 lean_blue

scan conditions	pts averaged=___5__ time/point= 1____ delay= __0______ sec

Chemical Love Story: part 5	_____/6

Unknown Cu(II) concentration= ___200______ppm

Unknown Ni(II) concentration=_____80_____ ppm

Chemical Love Story parts 6 & 7

Identity of compound in bottle #1_____acetone____________________ ___/5 pts

Identity of impurity in bottle #1 ______water________________

Reason: 3500 broad band (acetone absorbs from bathroom atmosphere)

___/2 pts

Identity of compound in bottle #2______ethanol_______________ ___/5 pts

Infrared Spectral Assignments of Bottles #1 & 2 from Jill’s Side of the Medicine Cabinet

Bottle #1 band positions (cm-1)	Mode assignment	Bottle#2 band positions (cm-1)	Mode assignment
2950-2850 multiplet	C-C-H aliphatic stretch	3400 vs br	O-H stretch
1700 sharp	C=O stretch	2950-2850	Aliphatic C-H stretch
~1400 m, sh	CH3-C(O)-CH3 bend Or C-C(O) stretch	1470 1380	CH2 scissor CH3 scissor
1370 s, sharp	CH3 scissor	1100	C-O-H scissor

____/4	____/4

Mass Spectra Assignments of Bottles #1 and #2 from Jill’s Side of the Medicine Cabinet

Bottle #1 mass peaks (amu)	Peak assignment	Bottle #2 mass Peaks (amu)	Peak assignment
58	(CH3)2CO parent = P+1	46	C2H5OH parent =P+1
43	(CH3)CO+=P-CH3	45	P+1 - H
29	P2+ or HCOH+(chamber rxn)	31	P-CH3
27	CH3C+	29	CH2CH3+
15	CH3 +	15	CH3+

___/5	___/5

Chemical Love Story: 2.7 chapters 8 & 9	___/10 pts total

Chapter 8

1a)	guess on S identity and why (__/2)

Ethyl benzene…HPLC peak near that seen in previous experiments (~1.0 here vs 1.0 min retention in dry lab HPLC experiment with ethyl benzene)

1b)	Likely concentration of S in unknown in v/v % (__/3 pts)

4670=485715Vs +250=> Vs~0.0091 uL out of 20 uL injected=> (0.0091/20)*100~0.0455%

Chemical Love Story: part 9	___/5

Five reasons why S is, or isn’t what you thought from HPLC

1) parent mass of 106 = mass of ethyl benzene parent P, not benzene

2)m/e= 91 suggest P-methyl

3) multiple overtone bands in IR near 1800=> ring with substituent (benzene has only 2)

4) C-H stretch for alkyls near 2950 => not just C=C-H

5) both CH2 and CH3 scissor bands at 1460 and 1370 respectively

.

Provide an outline of a GC protocol (flow rate, column type, temperature(s) of column, detector) he should use to best see and separate the `S’ component if he injects a 1 uL: sample from bottle #2 into the GC. Sketch what you think the GC chromatogram will look like, including ~ retention times and magnitudes. Hint: you will need to look up boiling points of each component in bottle #2

Chemical Love story: 2.8 chapter 10 ___/10 pts

He Flow rate: ________0.9________ ml/min (typical of runs done this year)

Initial Column T__801___ Final Column T__1501__Ramp rate (if needed) 15-25oC/min3

GC detector type_______PID2______________

Stationary phase choice ____capillary column, OV-17 best or general OV-1 or 5_____

Sketch of expected GC chromatogram (include times and qualitative magnitudes)

Bottle #2 contains ethanol (bp=78 C) and trace ethyl benzene (bp=136 C).

1ramp temperature from ~80145 C (65 C span) with a suggested 1 minute hold to allow evaporation and removal of ethanol . The bp of ethyl benzene is 136 C so at 150 C we should have entirely evaporated our target.

2To avoid a massive ethanol peak, use PID detector (sensitive mostly to aromatics) 3For temperature range covered (65 C) these ramp ranges produce total ramp times of 2.6-4.3 minutes. If we add an the initial 1 minute hold, total runs span no more than 3.6-5.3 minutes

pA*s	ethyl benzene

tr=	(136o -80o) 136o -80o = 2.2-3.7 minutes

25C/min	15 C/min

\Assuming column temperature must reach bp of

Ethyl benzene to detect

Time

0	2.2-3.7 minutes

Chemical Instrumentation Jeopardy ( ______/9 pts)

Best GC detector for permanent gases (e.g. CO2, CO, O2…)_TCD___

Most common detector for MS__dynode multiplier___________

The three key features defining a good chromatogram____sharp, symmetric and separatedf (SSS)____________________________________________

Basic relationship defining how many possible vibrations an N atom, non-planar molecule can have: _______3N-6_______________________

What is the HETP implied for component X given a 48.55 ft column whose retention Time is 50 seconds and whose width (w0.1)=2.5 seconds) __0.2-0.23 cm_ N=18.5(tr/w0.1)2 =18.5(50/2.5)2=7400. HETP=L/N= 48.55*12*2.54 cm/7400=0.2 cm (can use N=16(tr/w0.1)2 per text also=>N=6400 and HETP=48.55*12*2.54/6400=0.23 cm

A magnetic sector mass spectrometer has a resolution of 10,000. What is the m resolvable at 100 amu ? 0.01________ 10,000=m/m =100/m

m=100/10,000=0.01 amu

A S/N (signal to noise) ratio of 5 is achieved using 20 FTIR scans. How many scans are needed to achieve a S/N of 100 ?_20*(202)=8000____

Best plumbing to use for high vacuum work__Con-flat flanges______

What kind of HPLC separation is implied when the solvent is non-polar and the column is polar ? ____normal phase__________________

Alternative way to make mass fragments that generally provides a better chance of a large parent mass in mass spectroscopy.___CI (chemical ionization)____________

Main problem with doing multiple cations using uv-vis…the __N2______ problem.

Which method would you want to use to measure multiple elements simultaneously at low concentration in water samples ? ____ICP_________

Aromatic compounds have several diagnostic bands: C=C-H stretches above 3000 cm-1; sharp ring breathing modes at 1600 and 1500 cm-1, and: overtones near 2000 cm

14.	Device used to smooth out the reciprocating HPLC pump’s natural pressure waves:

___pulse damper_____________________

15. X’s Parent mass: 92, strong MS peaks also at 91, 77, 65, 39,26	amu__toluene_________

1500 cm-1 X is (Name or structure):

_

Part 1

Copper (Cu)

Standard #1:	initial stock mol Cu/ L= (5 g/170.44 g mol-1 )/0.05 L =0.5867 M

C1V1 =Cstock Vstock

C1 = Cstock Vstock/V1	= 0.5871*8.521/50=0.100 M

Standards #2-4	molarity of initial stock = (15.885 g Cu/63.54 g Cu/mol)/0.1 L=2.50 M Cu

Dilutions follow form; C2-4V2-4 =Cstock Vstock

For C2= CstockVstock/V2= 2.5*5/50= 0.25 M, C3 =0.50 M, C4=0.75 M

Nickel (Ni)

Nickel standard = 100 ppt = 100 mg/mL = 0.1 g/mL=0.1 g/0.001 L => (0.1 g Ni/58.96 g mol-1)/0.001 L=1.696 M

Nickel stock created by dilution of standard => CstandardVstandard =Cstock Vstock	1.696*29.48= Cstock(100)=> Cstock=0.50 M

Final standards #1-4	C1-4V1-4 =Cstock Vstock

C1 = Cstock Vstock/V1 = 0.5*5/50= 0.05 M; similarly, C2=0.10, C3=0.15 M and C4= 0.20 M

Part 2

Copper (std add #1-3)

Initial dilution of standard#1 = (0.1 M)*(3.934/100 )=0.003934

Second dilution of above = 20/100* (0.003934)=7.868*10-4 M Cu

Cs =standard addition stock for Cu = 7.868*10-4 mol Cu *(63.54 g Cu/mol Cu )/L =0.050 g/L= 50 mg/L=50 ppm

Standard add concentrations, Cstd 1-3 computed with usual CstdVstd = CstockVstock => Cstd =CstockVstock/Vstd = 50*Vstock/100

For C1 , Vstock = 5=> C1= 50*5/100 =2.5 ppm; C2= 5.0, C3= 7.5 ppm

Copper (std add #4)

Initial dilution of standard#4 = (0.75 M*3.277/100 )

Second dilution of above = 10/50* (0.75 M*3.277/100 )=4.915*10-3 M Cu

=standard addition stock for Cu = 4.915*10-3 mol Cu *(63.54 g Cu/mol Cu )/L =0.31233 g/L= 312.3 ppm

For C4 , Vstock = 4=> C1= 312.3*4/100 = 12.5 ppm

Nickel std add #1-4

Cs= Cstd add stock= 0.15* 5/1105.4=6.7848*10-4 M Ni = 6.7848*10-4 M Ni *58.97 g Ni/mol Ni=0.040 g/L= 40 ppm

Standard add concentrations, Cstd 1-4 computed with usual CstdVstd = CstockVstock => Cstd =CstockVstock/Vstd = 40*Vstock/100

For C1 , Vstock = => C1= 40*5/100 =2.0 ppm; C2= 4.0, C3= 6.0 ppm, C4= 8.0 ppm

Chemical Love Story: part 5	_____/6

Ni std add	A
0	0.074
1	0.113
2	0.151
3	0.188
4	0.224

Nu = -b/m = 0.075/0.0375=2 => Nu*Cs(Ni)* =2*40 = 80 ppm= Unknown Ni conc *Cs(Ni) = 40 ppm from part 2

Cu std adds	A
0	0.2
1	0.25
2	0.301
3	0.352
5	0.45

Nu(Cu) = -b/m =0.2004/0.0501=4

Nu*Cs(Cu)=4*50*

*Cs(Cu) = 50 (ppm) from part 2 =200 ppm= unknown Cu conc