Cl assical Anal ysi s
Chem 325
Preliminary examination o f phys ical and che mical
characte ris tics
De te rminatio n o f physical constants
Solubility tests
What’s in a Formula?
Classification tes ts
Preparation o f de riv atives
Elemental analys is
Co m busti o n Anal ysi s
C o mpo und: mo lecules : co llectio n o f ato ms o f v ario us types .
E le me ntal A nalys is
Q ualitative : types o f ato ms
Flame tests
So dium fus io n: co nve rs io n
N ? C N -, SC N S ? S2X ? Xfo llo we d by standard ino rg anic qualitativ e analys is
R atio o f ato m types : fo rmula, nee d quantitative info rmatio n
Similar ‘traps ’ fo r S, N ox ides
N o direct info rmatio n fo r o x yge n!
M OLE CUL AR W EIG H T DE TE RM IN ATIO N
Qual i tati ve pl us el emental anal ysis
? atom rati o s, but the si m pl est w hole num bers
empirical form ul a, e.g. ben zene C 1H 1
Need o ne m o re piece o f i nform ation:
the M OL AR M AS S o r M OLE CU L AR WE IG H T
G as behavi our (Ideal G as Law )
F reezi ng po i nt depressi on
B oili ng poi nt el evati on
Osm oti c pressure
M ass spectro m etry
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Carbo n: tetraval ent: tetrahedral geometry
molar mass
= integer n
empirical formula mass
m o lecul ar fo rm ul a = (em pirical fo rm ul a) ´ n
Attachi ng carbo ns
Al ka nes: acyclic or cycli c
H
C
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
H3C
‘strai ght chai n’ or acyclic
hy drocarbons
CnH2n+2
CH2
CH2
H 2C
H 2C
CH2
CH2
CH3
C 6H 14
C 6H 12
C nH 2n+2
C nH 2n
Degree of uns aturatio n
Al kenes: unsaturation
= E le ments o f uns aturatio n
= Index o f hydro ge n de ficiency
H 3C
CH 3
C nH 2n
Akynes: more u nsaturati o n
CH2 C
H 3C
C
CH 2
C nH 2n-2
CH 3
B ase d o n s aturate d hydro carbo n fo rmula C n H 2 n + 2
degree of uns aturatio n = ½ (2C + 2 – H )
C 6 H 14 : ½(2(6) + 2 – 14 ) = 0
C 6 H 12 : ½(2(6) + 2 – 12) = 1 ? 1 do uble bo nd OR o ne ring
C 6 H 10 : ½(2(6) + 2 – 10) = 2 ? 2 do uble bo nds
OR o ne triple bo nd
OR o ne do uble bo nd and o ne ring
OR two rings!
N o te pro ble m with cyclic alk anes vs uns aturate d co mpo unds!
?? ? ? pro perti es, chem ical tests, spectroscopy
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Heteroatoms in Formulas
Exam pl es
C 4H 7Cl D.U. = ½ (2 (4 ) + 2 – 8) = 1
O
Oxyg en
O
- could be:
C
H
C
H
O
CH 3
C
H3 C
H
-ig no re ox ygens in ato m co unt
Hali des
C
CH3
C
Cl
C 9H 14O 3 D.U. = ½ (2(9) + 2 – 14) = 3
X
- could be:
-takes place o f a hydro ge n
O
O
-co unt as o ne hydro ge n
Ni trogen
C
C
N
-triv ale nt vs tetravale nt
OH
-co unt as ½ C
H
Formulas from Masses
• Suppo se we k no w the mo lecular mass M .
C 6H 6N2O D.U . = ½ (2(7) + 2 – 6) = 5
- could be:
What fo rmulas are poss ible?
O
NH2
Vitamin B 3
N
M
r
=n +
13
13
Rul e o f Thi rteen
Base formula i s then C nH n+r
Exam pl e: M = 58,
58
13
=4 +
6
13
B ase fo rm ul a i s then C 4 H 1 0
Rule of Thirteen
M = 58 , base formula C 4 H 10
If we s us pect o ne O (1 6 u), s ubtract mass equiv ale nt C 1H 4
è C 3H 6O
If we s us pect o ne N (14 u), s ubtract mass equiv ale nt C 1 H 2
è C 3H 8 N
If we s us pect two N (2 8 u), s ubtract mass equiv ale nt C 2H 4
è C 2H 6 N 2
If we s us pect o ne S (32 u), s ubtract mass e quiv ale nt C 2H 8
è C 2H 2 S
O ur s us picio ns are g uide d by othe r s pectral and che mical
data, and che mical se nse!
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