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Unformatted text preview: NAME : NOTE WELL! You Must Show Your Work And A Logical Path To The Solution Of Each Problem
To Receive Any Full Or Partial Credit. N0 Credit Will Be Given For Numerical
Answers Alone. The relative response it'actor for an individual compound is deﬁned as follows:
(signal) X (relative response factor) = (relative amount of component) Where
response factor and amount of component have consistent units (as used in lecture). I) 16 points
2) 20 points
3) 20 points
4) 10 points
5) 14 points 6) 20 points (Homework) TOTAL 100 points _ l) 16 points: The weight % nC6H14 in a complex solution of organic compounds is
measured using the method of internal standard. Gas chromatography with a thermal
conductivity detector is used. nC9H20 is used as the internal standard. Bod: compounds
give fully resolved peaks. The relative weight response factor for hexane is 1.25 and the
relative weight response factor for nonane is 1.50. 12 grams of nonane are added to 100
grams of the original sample containing hexane, and the chromatogram is run. In this
chromatogram, the AREA % hexane is 16,0 and the % nonane is 19.0 Calculate the weight % hexane in BOTH the original solution and in the solution containing
the added nonane. 2
2) 20p0in13: TheﬁpﬂowhsgisatypicalplotofﬁvasuspobminedusingtheVan
DmEqnaﬁmm==A+Bip+Cuy a) Onthisplotcluniylabelandmarktheporﬁmofthcmprﬁnmﬂycomﬂed
byﬂm“B”termANDtheportionofﬂmcurvemhnmﬂyountmﬂedbyﬂw“C”  14.
b) Onﬁleplatbelow,CLEARLYdrawﬂ1cnewchmgcdcuweﬂntwiﬂbeobsened
ifﬁmvaiueofﬁ”isdonbledmﬁ“A”and‘C”ruminumﬂmpd.Shuwcleaﬁy
whexeﬁlewrvachmgesﬁmnﬂnmigimlplotmﬂwhueitdoesm Showtime
changes (ifany)inﬂleminimlm\vahle ofHandlmwihatpointmoveswiﬂlmspeato p. onthe plot. a“.
c) Onﬂleplotbekrw,CLEARLYdrawﬂlenewdnngedcnrveﬂmeﬂlbeobsuved
ifﬂnvaimofvisqmltoﬁofﬁsodgimlvﬂueand“A”and"B”rwmin
mohanged. Showclemiywhﬂethccurvechmgesfmmiheoﬁg'mlplotandwlmeit
doesnot. Showﬂaemﬁfanyﬁnﬂnmiﬁmmnvameofﬂmdhowﬂmmim
moveswiﬂuespecttopmﬂmeplot d) Onﬂaepkytbeiow,CLEARLYdmwﬂlcnewdungdmifﬂaevalueof“A”is
doubledand“B”ami“C”remainmdmged. Slmwcleaﬂywhereﬁwmedmnges
fromﬂnoﬁgimlplotmdwhaeitdownot Showthemﬁfmy)h1ﬂw
mirﬁmmnvalw'oﬁnmﬂhowﬁmmhnmwiﬁlmmpmﬁnplot \/ WM _..—._.._ 3
3) 20 points: Calculate the relative weight response factors for CsFu and C4Hm
given the following data: A solution containing 3.15 grams of C5F12 and 1.45
grams of C4Hm was separated by gas chromatography. The AREA % of the
CsFu peak was 40.0%. The AREA% of the C4Hm peak was 60.0%. b) Normalize these two response factors so that the response factor for CdIm z 1. 4 4) 10 points: The ztbsorbance of a sample containing the absorbing compound ML:
at 0.01 molesfliter was measured and found to be 0.655 Absorbance units. The
absorbence of the blank (a solution containing everything except MLz) was found to
be 0.042 Absorbanm units. A cell with a 5.00 cm path length was used. :1) Calculate
the value of the absorptivity coefﬁcient (“a”) for ML; ﬁ'om Beer’s Law. b) Give the
units for “a” in this case. S 5) 14 points: Given that the uncertainty in T is +/— 0.005 , Calculate both AA and
AMA for the following values of A or T: a) Absorbance = 0.75 6 6) 20 points: Homework Problem 1913: Finding a 13K,l by spectrophotometry.
Consider an indicator, HJn which dissociates according to the equation KHln
Him 6—) H’*+1n‘ The molar absorbtivity, s , is 2080 M”l cm" for Hin and 14200 M“1 cm'1 for In', at a
wavelength of 440 nm. a) Use Beer’s Law to write an expression for the absorbance at 440 nm of a solution
in a 1 cm cuvet containing IIIn at a concentration [HIn] and In“ at a concentration
[10‘] b) A 1.84 X 10 4M solution ofIIin is adjusted to pH 6.23 to give a mixture ome
and In' with a total a total concentration of 1.84 X 10". The measured absorbence of
the solution at 440 nm is 0.868. Using your expression from (a) and the fact that
[Hill] + 111'] = 1.34 X 10“, calculate the pKHm for this indicator. ...
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 Spring '11
 bulman

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