HMWK 08 Answers
CHEM101, B2
2008 03 03
3.) Do a similar analysis for NClO (go by the connectivity rules given in class).
HT
More Resonance Practice
"all octets"
Cl
N
1.) For the molecule HSCN ( connectivity as indicated) derive a Lewis structure
using the rules given in class.
Include reasonable resonance structures.
Make sure to show formal charges if any.
Cl
N
O
Skeleton: H-S-C-N; Nt = 1 + 6 + 4 + 5 = 16
Cl
N
O
Cl
N
"all octets" :
H
S
C
No = 20 ; remove 2e- from central atom
O
has FC's; "naturally" positioned;
can be removed by double bond formation
No = 20, must remove 4 e-,
N
several options; most reasonable : both C and N have neg. FC; so take 2e- away from each
overall
O
Cl
S
C
N
H
S
C
N
O
minor
major
H
N
both C and N have incomplete octets; they can be made complete by making double
bonds; use 2 e- from C and 2 e- from N
this looks very reasonable and is the major contributing resonance structure;
a resonance structure that might make a minor contribution is one
where we have a neg. FC on N :
4.) Consider a substance with molecular formula C2H4O2.
a. What is its empirical formula? (CH2O)
b. What is the mass percentage of C in this substance?
to show it as resonance hybrid:
H
S
C
N
H
major
S
C
N
minor
2.) Do a similar analysis for POCl3 ( where P is the only central atom)
Cl
P
Cl
Nt = 5 + 6 +(3 x 7) = 32
34.00% C in C2H4O2.
Cl
O
"all octets"
Cl
P
Cl
Cl
overall
No = 32; Nt = No ; but FC on P and O;
FC's can be removed by forming double bond;
now P has 10 e- in valence shell; OK since row 3
O
Cl
P
Cl
doesn’t matter; empirical is simpler
imaginary reaction:
O
Cl
Cl
P
Cl
CH2O + ...→ C + ...
1mol CH2O
1mol C
12.01g C
100g CH2O x 30.03 g CH O x 1mol CH O x 1mol C = 34.00 g.
2
2
O
Skeleton:
can be based on empirical or molecular formula;
Cl
Nt = (2 x 4) + 4 + (2 x 6) = 24
c. What is the number of C-13 atoms in 5 g (exact) of this substance (estimate).
H
To get the number of molecules, convert mass to moles then to molecules
H
O
C
C
same
H
O
H
H
H
1mol C2H4O2
6.022 x 1023 molecules
5g C2H4O2 x 60.05 g C H O x
1mol C2H4O2
2 4 2
Since there are 2 C atoms per molecule,
H
O
H
O
C
C
H
O
H
H
C
generate alterntive resonance structure by
forming double bond between C and O
H
H
d. Two (2) different constitutional isomers can have this molecular formula
H
O
C
C
H
H
H
H
O
H
A
H
C
H
O
C
B
e. For each constitutional isomer, derive the dominant resonance structure
as well as a minor one.
Justify your answer using the Precedence Rules I gave in class.
H
O
C
O
H
H
H
O
C
O
C
H
H
C
O
C
H
O
H
H
O
H
overall
each has a CH3 grouping). Show the connectivities for each.
two reasonable possibilities
C
same
H
( I assume you have some remnant knowledge of Org. Chem.; try your best;
H
C
O
O
H
the total number of C atoms is 1.003 x 10 .
Approx. 1% of these are C-13, so their total number is 1.003 x 1021.
H
same
H
23
C
H
No = 26, remove 2e-, from C since it has
"unnatural" FC
= 5.014 x 1022 molecules
O
C
major:
no FC; all-octet
H
C
H
H
O
C
O
H
H
C
O
O
C
H
H
H
major:
no FC; all-octet
H
C
H
O
O
C
H