Phân dạng và phương pháp giải hóa học 11 Phần hữu cơ Dành cho học sinh lớp 11 ôn tập và nâng cao kĩ năng làm bài

if
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D6
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MINH
,
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62
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T?i:
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ty
TNHH
MTV in ^n
MAI THjNH
DLfC '
Dja chl: 71, Kha V^n CSn,
P.HiOp
Binh
Ch^nh,
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Dufc,
Tp.HCM.
So
DKKHXB:
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ngSy
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Quy^t djnh xuat bkn s6:

1815/QD-THTPHCM-2012
do NXB
Tong
h(?p
Thanh
ph^ H6 Chi
Minh
cS'p ng^y
28/12/2012.
In xong nOp lUU chi^u quy' I nanri 2013.
(HuMtta
4,
p DAI
CI/CJNG
VE
H6A
HpC HtfU CCJ
A. TOM TAT Lf
THUY^T
I.
MdDAUVEHbA
HOC
HirUC(]fVAH0P
CHAT H0UCCf
1.
H^p
chaft
hffu ccf la hdp chat cua cacbon (trilf CO, CO2, muoi cacbonat xianua,
cacbua, )
2.

PhSn
loaii
h^p
chSt
hffu
ccf .
^
9!^: : J^'l • ,
Hop cha't hQu
CO
Hidrocacbon
^ ^ ' Din xuS't cua hidrocacbon
Hidrocacbon
Hidrocacbon Hidrocacbon
no
khOng no
thorn
Din Ancol
Andehit
Amin
Axit
HOP
xuS't Phenol Xeton
nitro
Este
chat
halogen
Ete
tap
chile,

polime
3. Danh phap h^p
chS't
hffu
cd
a) Ten thong thiidng (nguon goc flm ra chat)
^'-''-^Msg
«*av*f«
b)
Ten
h^
thong (lUPAC) '
'':i,'1^'/^(ryi
Ten goc chiJc = Ten phan goc + T6n phan
dinh
chiJc
Ten thay the = Ten phan thay the + Ten mach C chinh + Ten phan
dinh
II.
PHAN
TICH
NGUYEN TO
1.
Ph3n tichdjnh
tinh
f''''"/''-;'
* Xdc
dinh
C : Dot chdy chat
hSu

cd CO2
CaCOjl
* Xac
dinh
H : Dot chdy chat
hihi
cd H2O ^CuS04khan^ CUSO4.5H2O
'••'^ ^ - (xanh)
* Xic dinh
nitd
: C^HyCN, (NH4)2S04 + .jV- v!* ?n.f•
*«
^ • • *
(NH,)2S04
+ 2NaOH >Na2S04 + 2NH3t + 2H2O
* Xac
dinh
halogen :
CxHyO,N,
>
CO2 + H2O + HCl 1:4 ;
?
HCl
+ AgNOj
>
AgCli
+
HNO3
vi ,
2.

PhSn
tich
djnh
Irf^ng
(A la chat
hffu
cd c6
HIA)
chffc
mc=12 i4^(g)
44
-H
= 2.i^(g)
=
i^.100%
=> %H=-2^.100%
=>%N=i^.lOO%
niA
Phan
djing
\ik
phuong ph^p
g'Ai
H6a hpc 11 HJu
co
-
B8
Xuan Hung
=> mo =
niA

-
(mc
+
mH
+
IHN)
hay
%0 =
100%
- (%C + %H +
%N) ^
; ,
III.
CONG
THLTC
PHAN
TlT HOP
CHAT
H0U
COf '^'iMj^J
1.
C6ng
thtfc
defn
gian
nhS't !
Y it
Mi
* t ,vN '
(

.
C6ng
thu-c phan tuT
(CTPT)
:
CxHyO,Ni
v,!^ j
Cong
thu-c ddn gian nhat
(CTDGN):
(CpH,OrN,)„ (n= 1,2, 3, ) ,|
*
Cach
thiet lap
:
Tim ti 1$
(C,HyO,)
so, ^^.,^
mr
mH niQ ^ %C %H %0 , ;
x:y:z=—^:—^:—^
hay
x:y:z=
: : = p : d:r
12
1 16 12 1 16
2.
Cdngthtfc
phan
tuT

(CTPT)
^ /•^^j,;' •is/Vi' — ••"^
n:!ir::.xM\i
*
Cach
thiet lap
:
-
Xdc
dinh khoi luWng mol phan tuT: MA
=
Me.dA/B
hoac
MA
=
29.dA/kk
•i^itiiH
'
- Difa
vac %
khoi lu'dng
cac
nguyen
to :
j
CxHyO,
>
xC
+
yH + zO

' ' '
I „ . 12x y 16z M
Tacdtil?:
= -^ = = => x, y, z
%C
%H %0 100% • • , J;
- Dura
vao
cong
thiJc ddn gian nha't:
i^i^Y j
j.
C,HyO,
=
(CpH,0,)„
= M =^n= ^ „,
i^'0:^
12p
+ q +
16r
,/>a F .
- Tinh tnic tiep
theo
kho'i Itfdng san pham do't
chdy
:
-H
-
'I
CxHA

+
^
y z
x
+
4
2
02 ^xCO^+^H^O '
"A
•"':i'f<(T}iD
"002 "H2O 1
pff
'
'1
'*
Tacdtiie:—=
—=
^x,y;tirMAtinhdirdcz.
^'-'^^^^^'^
'
t "A "002
2n,,20
i
IV.
CAU TRUC
PHAN
Tlf HOP
CHAT
H0U CO iii *
1.

C6ng
thitc cfiu tao
js
*
Cong
thiJc cau tao khai trien
;|
.>fiX
*
Cong
thiJc ca'u tao thu gon
2.
Ddngdang
Nhffng hdp chat
c6
thanh phan phan tuT hdn k^m nhau
mpt
hay nhieu
nhom
CH2
nhiftig
c6
tinh
chat hoa
hpc
tiidng tif nhau. "V''';^;!
' ' ,4^
3.
D6ng
phan

.,.,.:,„
,;„.,
, , . „ .,.j, .
j^w'^^
NhiJng hdp
chat
khdc
nhau nhu^g
c6
cClng
c6ng
thiJc phan til.
iti
4.
Lidn kd't
trong
phan
tuT
hflTp
chS't hflu ccf
Lien
ket
ddn, lien
ket
doi
va
hen ket ba.
'* JifJ^K hn-
A
V.

PHAN
0NG
HOU CO
1. Phan
tfng
the": CH4 + CI2CH3CI + HCl HO - : J r
2. Phan
uTng
cOng:
C2H4 + Br2 >
C2\I^BT2
'
3.
Phanurngtach:
CH3-CH2-OH-S^^ CH2=CH2 + H2O H
Dac
diem
cua
phan
iJng
h6a
hpc
trong
hoa hoc
hffu
cd : - 1
-
Phan
iJng
thifdng

xay ra
cham.
\'t
} -\{^ An
-
Phan
iJng
thudng
sinh
ra hon hdp san
pham.
; '.^^
B.
PHAN
LOAI
VA PH
JCJNG
PHAP GIAI
CAC
DANG
BAI TAP
-
(Vm
fihan
ling^hau
c^ririO'-vH^^-TH;}
,HI4 ,H;:HHD-;.ii:):-,-i-''
•
- <Jjhaii loai fditut itiig. hitii
eet ' t

-
(Vi& eMiq.
Ihiie eaii
tim,
ctmtg^ phdn^ ddtig^
dqttQ^
ede
ehdt
BAI TAP
MAU
VA BAI TAP
NANG
CAO i?
Bai 1.
Viet
phifdng
trinh
hoa hoc
ciia
cac
phan
iJng
xay ra
theo
sd do sau f.^
CH^CH
CH2=CH2
CH3-CH2-OH
CH3CH2-Br
Q j|j

Trong
cac
phan
iJng
tren,
phan
iJng
nao
thupc
loai
phan
tfng
the,
phan
iJng
cong,
phaniJngtdch?
,,,,
^^^^^^
^.^^
Giai
(1)
CH^CH
+ H2
^'^''^^^^^)
CH2=CH2:
phan
iJng
cpng
(2)

CH2=CH2 + H2O ""'^°> CH3-CH2-OH
:
phan
iJng
cpng
1,,, ?
Mrtr;
(3)
CH3-CH2-OH
+ HBr
CH3-CH2-Br
+ H2O :
phdn
tfng
thg*
' '
(4)
3CH=CH ^•^°°°^) (Q) :
phan
u-ng
cpng
Bai
2. Co ba
chat
hCu cd c6
cong
thiJc
Ian
liTdt
la : C4Hii)0, C4H9CI, C4H11N, Cho

biet
chat
nao c6
nhieu
dong
phan
hdn
?
Gidi
*
C4H|()0 : „., 'J, '(.,, . - ,M >•'•;
^ ;j;y
CH3-CH2-CH2-CH2-OH ; CH3-CH2-CH2-O-CH3
CH3-CH2-CH-CH3 ; CH3-CH2-O-CH2-CH3
Phan
djing va
phUdng
ph^p
giai
H6a hpc 11 HOu cO - D5
XuSn
HiJng
OH
^ , :Jt
CH3-CH-CH2-OH
; CH3-CH-O-CH3 ; CH3-C-CH3 i
CH3 CH3 CH3
*
C4H9CI: i ,:u ;.
- CH3-CH2-CH2-CH2-CI

; CH3-C-CH3 • „
CH3-CH2-CH-CH3
;
CH3-CH-CH2-CI
CH3-CH2-CH2-CH2-NH2
; CH3-CH2-CH2-NH-CH3
uhi^t^ya:f"••••
CH3-CH2-CH-NH2
; CH3-CH2-NH-CH2-CH3 . „ '
CH
^
,'',»:v'A,«;
tvifi^i^;
nvA
^'ihiyii,
'I'l-iM mhh,'\'vt;^5"*
•••
CH3-CH-CH2-NH2
;
CH3-CH-NH-CH3
CH3-C-CH3^"'
'.HDsHD v';^;
CH3-CH2-N-CH3
,m m
^-i^-^
fiCtelt)
I
I " 1
CH3 CH3 (iij
Vay chat

cd
nhieu dong
phan
nhat
Ik
CiH^N roi
den
C4H,„0, C4HyCl.
Bai
3.
Cho ba hdp
chat
: A (cau tao tit C, H); B (cau tao
tuf
C, H, N); C (cau tao tit
C, H, O).
Chaft
nao
thupc loai hidrocacbon, cha't
nao
thuQC loai
dan
xuat
cua
hidrocacbon.
' ' . ,
*
A la
hidrocacbon
\\

chi
cSu tao gom
hai nguyen
to C. H. 1;,)
•
jil
*
B va C la dan
xuat
cua
hidrocacbon
vi
ngoai
hai
nguyen
to C, H cdc
hcJp chat
B,
C con
chiJa
cdc
nguyen
to
khac
(N va O).
a)
Trong
cac hdp
chaft
sau. hc(p

chat
n^o la
hffu
cd,
hcJp cha't
nao la v6 cd CH4,
CHCI3. C2H7N, HCN,
CHjCOONa,
C,2H220|,,
f
C2H3Cl}-„,
AI4C3.
b)
Chat
nao sau day la
hidrocacbon
? la dan
xuat
cua
hidrocacbon
: CH2O,
C2H5Br,
CH2O2,
CfiHsBr, CfiHfi,
CH3COOH.
Gidi
a)
Hdpchathi?ucd:CH4,CHCl3,C2H7N.CH3COONa,C,2H220,i,{C2H3Cl}„.
*
Hdp chat v6

cd :
HCN,
AI4C3. '
b)
Hidrocacbon: CfiHfi.
frt')
Dan
xua't
cua hidrocacbon
: CH2O,
C2H5Br,
CH2O2,
CfiHjBr,
CH3COOH.
Bai
5. Cho 6 hdp cha't hi?u cd ddn
chiJc,
mach hcl cd cung cong thiJc phan tuT 1^
C4HXO2. Viet cdc cong thiJc ca'u tao thu gpn cua cdc chat dd. • , • ,
CH3-CH2-CH2-COOH ; CH3-COO-CH2-CH3 ; CH3-CH-COOH
ir
CH3
HCOO-CH2-CH2-CH3 ; CH3-CH2-COOCH3 ; HCOO-CH-CH3.
CH
Bai
6. Cac
phan uTng
sau day
thuOc
loai

phan iJng
nho ?
(phan iJng the, phan iJng
cong, phan iJng tach)
?
a) C2H6 + CI2 ^ C2H5CI + HCl b) C4HX + H2O C4H,oO
c)
CzHsCl
) C2H4 + HCl d)
2C2H50H^^C2H50C2H5
+ H20
e) CH2=CH2 + H2O CH3-CH2-OH f)
CfiHft
+
Brj CfiHsBr
+ HBr
g) C4HH, + 5F2 > 4C +
lOHF
h) CH3OH + HCl CH3CI + H2O
a) Phan iJng the
b)
Phan iJng cpng
c)
Phan tfng
tach
^'
d)
Phan iJng tach
e)
Phan iJng cpng

f)
Phan iJng the
g)
Phan ifng tach
h)
Phan u-ng the.
'
Bai
7.
Dtfa
vao
tinh
chat
hoa hoc cua
CH2=CH2
va
CH^CH
(da hoc d Idp
9).
Hay
Viet
phiTdng
trinh
hoa hoc
khi
cho
CH3-CH=CH-CH3
va
CH3-C^-CH3
lac dung

vdi
Br2, H2 va cho
biet
nhufng nhdm nguyen
tuT
nao
trong
phan tuf
cua hai
hdp cha't
tren
da gay nen cac
phan u'ng dd.
, ,:(
; .
*
CH3-CH=CH-CH3 + Br2 > CH3-CH-CH-CH3
-mwaot/rh
sn6:; ?iv (.
^;,H"^
' Br Br 'hi: , ' '
CH3-CH=CH-CH3 + H2 CH3-CH2-CH2-CH3 - .
t
.•ri.VH,?
„5»ni;.,.>
;
y.>t3ffi.;*):.!
=> Nhdm nguyen tuf
gay nen
phan iJng

: -C=C. 0;|
^0ktxi-nUym¥''
''
Br
Br ,:.„„ ; ,
*
CH3-CHC-CH3 +
2Br2
—> CH3-C—C-CH3
| ^.l^:,
CH3-C^C-CH3 + 2H2 CH3-CH2-CH2-CH
3
=> Nhdm nguyen
tuT
gay nen
phan uTng 1^
:
-C=C
Bai
8.
Viet phufdng
trinh
hda hoc cua cdc
phan iJng
xay ra
trong
cac
triTdng hdp
sau
va

cho
biet
cac
phan
u'ng dd
thuQC loai phdn uTng
n^o
(phan vJng
the,
phan uTng
cpng, phan u'ng tach).
Phan
dgng vji phuong
phap
giSi H6a hgc
11
HDu
co
-
D8
XuSn Hung
a) Etilen tac dung vdi hidro c6
Ni
lam xuc tdc va dun n6ng.
b) Dun n6ng axetilen
d
600"C
vdi bgt than lam xuc tac thu
diTdc
benzen.

'^f
'
c) Dung dich ancol etylic
d^
lau
ngoSi
khong khi chuy^n thanh dung dich axit
axetic
(giam
an).
•
ti
>
.
if iJ
;^W;>
; /
:i•0 ,^K;;»
,•
'
a)
CH2=CH2
+
H2 CH3-CH3
(phan iJng
cOng)
,li::).:.:H':;) H':)
i:X)'»
'
^

^
(CfiHfi)
[Q]
(phan iJng
cpng)
b)
3CH=CH
600°
C
•
c)
C2H5OH
+
02(kk)
^^"^
>
CH3COOH
+
H2O
^
phan ifng oxi h6a
,,
,
r); ^
<.':
, , , .
khong hoan toan.
Bai 9. Cho cac chat sau, nhffng chat nao la dong ding cua nhau
?
NhCng cha't nao

la
ddng phan cua nhau
?
,
CH3-CH3
•
ill
(1)
CH3-CH-CH3
(Ui +
jK'(2):} VJ
{.7
CH,
^IM;;;
+
QI ^
-
CH.
CH,
CH3-CH2-CH2-OH
(4)
•>1
(3
CH3-CH=CH2
(5)
CH3-CH2-O-CH3
(7)
va) CH3-CH2-CH2-CH2-OH(9)
CH3-CH2-CH2-CH3
CH2CI-CH2-CH3

(6)ii'i(t)
J
'I'd
I i.i'i I', /
411
X
\M
^m)hWfm \0- :H':)
fjv
fi'0 i
Gidi
^.rf
Anr^iiJ' ]ifh
/
Iq iHu
f£;Cic
chat
la
dong
ding
:
(1)
va
(6); (4)
va
(9).
rt'
rirt 4 t\i&
'•^'cf
f,.' II }H

* Cdc chaft
dong
phan cua nhau
:
(2)
va
(8); (4) (7), (3)
(5). ^ .
<h.
Bai 10.
'
a)
Hay viet
cong
thufc electron
va
cong
thiJc
cau tao c^c phan tuT sau
:
>~.,r)i
}f'3
,
*
CH3CI,
CH4O, CH2O,
CH5N.
b)
Hay
vie't

cong
thufc
cau
tao khai trien
va
cong
thiJc
cau
tao
thu
gon
nhat cula cac
hdp chat sau
:
C3H6, CH3CHO, CH3COOC2H5, CH3CN,
biet
ring
trong
phan
tiJf
cua chiing diu
c6
lien ket bpi. :''rf !"'v •"^'H^;-^'' ijik I'-'iy
'•^•"•.i-r.,
,
•
Gidi
a)
If/
CTPT

CT
electron
CTCT
H
M
V
,
CH3CI
H:C:C1
"
"
'
H-c-ci
••^-v:'^5'-«•
/-^'ii^gj 4;
H::) ?:HD
+
H
CH4O
H:CJO:H
H-C-O-H
' '
1
H H
b)
CH20
H:C::0:
ii
H-C=0
H

H
H
1
CH5N
H:C:N:H
H
H
1
H-(j:-N-H
H
H
CTPT
CTCT
khai
trien
CTCT
thu gon nhat
H
1
CH3-CH=CH2
C3H6
H-C-C=C-H
AAA
hoac
CH3CHO
•?
/°
H-C-C
A
hoac

CH3-CHO
H-c-c
H
0-C-C-H
1
1
H
H
CH3COOCH2-CH3
CH3COOC2H5
H-c-c
H
0-C-C-H
1
1
H
H
hoac
H
1
CH3CN
CH3CN
H-C-C=N
hoSc
=N
Bai 11. Dong phan
la gi ?
The nao
la
nhffng chat dong ding

?
Trong
so
nhuTng chat
diTdi day, nhCTng chat nao
la
dong phan cua nhau
?
Nhffng chat nao
la
dong ding
cua nhau
?
a)
CH3-CH2-CH=CH2
b)
CH3-CH=CH2
c) I ^ I ^
CH2
—CH2
i, J^'p
'f':''f(\f
.'/"iiV;'
lyi-
V/• y.i
•'
U
H
CH3
I

d)
C=CH2.
CH,
^han dang
phuong
phiip
giai
H6a hgc 11 HOu co - D8
Xufln
Ht/ng
•))
Viel
cong
thu'c
ca'u tao c6 the cd cua cac
dong
dang
cua
ancol
etylic
cd
cong
thu'c
phan
tuT
CsHxO
va C4H|„0.
I)
*
C2H6O:

CH3-CH2-OH
;
CH3-O-CH3
*
C3H6O:
CH3-CH2-CHO; CH2=CH-CH2-OH; CH2=CH-0-CH3;
OH
! CH3-C-CH3 ; CH ; ^\ P"2-CH2 -
*
C4H,,,:
CH3-CH2-CH2-CH3 ; CH3-CH-CH3 IB.'IV M
I
. , .
, CH, !
*
CH2CI2:
H-c-ci ;
! H I
O
*
C2H4O2:
H-C-C
H
^0-H
hay
CH3COOH
.&i
,
H-^-O-C-H
hay

HCOOCH3
O H
.
H-O-C-C
5fs|.;y'|a)(jl
sS wi? H H
^ , H H . .
*
C2H4CI2:
H-C-C-CI
H
CI ^
HH
•iA''VA;«(THi,v.H-C-C-H
I
I
CI CI
hay
CHg -CHO
^>|ri;
i;^ q
.i
j,
1
i.iaii!
KuHfiidq
snob iii
uVifi
}«b
jjrfMfi

iftrfj
hay
CHiCHCl:
1, ir }
hay
CH2CI-CH2CI.
.) * CjHsO :
CH3-CH2-CH2OH
; CH^-CH-CHg
in
*
C4H,oO;
CH3-CH2-CH2-CH2OH
;
CH,-CH,-CH-CH
3
—v^xa2
^~^'^'-3
OH
OH
CHg -C-CH3;
CH3
-CH-CH2OH.
•'AO. *
KHAJgG
VIET
Bai 13.
NhCng
cha't nao sau day la
dong

ding cua nhau,
dong
phan cua nhau ?
a) CH3-CH=CH-CH3 b) CH2=CH-CH2-CH3
c)
CH3-CH2-CH2-CH2-CH3
d) CH2=CH-CH3
e) CH3-CH=CH-CH2-CH3 g) CHg =CH-CH-CH3 -)
CH3
Si
I'lfU.
_n—i^n.o.
I
h)
CH3-CH2-CH2-CH2-CH2-CH3
i) CH3-CH2-CH-CH3.
Gidi
* Cac
cha't
la dong dang cua
nhau
:
d,
a va e;d, a va g;d, b va e;d, b va g;i va h;e h. .,
* Cac
chat
la dong
phan
cua
nhau

: a va b; e va g; c va i. ^- •
Bai 14.
a) Hay vie't cong
thu'c
phoi
canh
cua
CH3OH
(metanol) va
ciia
CHCI3
(ciorofom).
b) Hay viet cong
thiJc
phoi
canh
cua C2H6 (etan) va C2H5OH (etanol).
. " , .', Gidi " .' ' ,
a) *
CH3OH
H ^ V , , . *CHCl3: H V
HO^ " CI
b)
*C2H,:
H H *C2H50H: » H
\« . \
H'HilIC—C-*H r, H iii|C-C-*H
V \1
t::.ith,</-
: Hut 4 \

H
H H OH
•
,,, r
•
' < W r
Dcuia 2. . ,,
-
(Xide
(tiiih it/ ed mat lu/mjeu ta, thanh plidit ede iitfiiifjeii to' trmtq
chut hffll etf i) -f;>i vJ.(}v
.'4\
jli^f,!)
ym C.
IJ,Ofni
r;iV),(J -~ -
-
f£gfi
etniff thiie ittfii qlun iihal . ,,),,.,
„,,,,,v,y,)
.>.4,j
j.^j;,^;^
. , , ^ BAI TAP MAU VA BAI TAP
NANG
CAO
Bai 1. Oxi h6a ho^n toan 0,6g hdp chat hi?u cd A thu
duTdc
0,672
lit CO2
(dktc)

vj
0,72g
H2O. Tinh thanh phan % khoi liTdng cua cac
nguyen
to trong phan tuf cha't A.
Gidi
Khoi lifdng cac
nguyen
to : viul^h^ njo! r , -
mc=12.neo,^12.^
=
0.36(g)^
'
mH = 2.nH,o=2.^ =
0,08(g)
"
"i'^
=> mo = 0,6 - (0,36 + 0,08) = 0,16 (g)
Phan
d?ng
va
phudng
ph^p
giSi
H6a hpc 11 HQu co - D8
XuSn
Hung
Thanh
phan
%

khoi lifdng cic nguyen
16'
trong A : d,
i
jy.
-fir Ji'''J" ^rNw
%C=
•^xlOO%
=
60%
jriH./,
0.6
•••:>
;if^;r.^:jf:> ft,-:.,
%H=
100% = 13,33% ^%0 = 26,67%. £ >-
iH^*^
>^ ^^
V
0,6 ' •
Bai 2.
Tinh
khoi
lifdng
mol phSn tuf cua cdc chat sau :
jj,;.,
,,,,,
a) Chat A c5 ti
khoi
hdi so vdi khong khi bkng 2,07.

b)
The
tich
hdi cua 3,3g chat X b^ng the
tich
cua l,76g khi oxi (do cf ciang dieu
kifn
ve nhiet do, ap sua't).
Gidi
'•^''''^ '^''''^
*
a)
dA/kk
= ^ = 2,07 ^
MA
= 29.2.07 = 60
(dvC).
, ^g^,, .^.^ .
b)
Ta c6: no2 =^^ =
0,055
mol
=> nA = =
0,055
mol=>MA
== 60
(g/mol).
32
0,055
Bai 3. Dot chay hoan toan 0,282g hdp chat hiJu cd A, san pham sinh ra cho qua

binh
dtrng
CaCl2
khan va
binh
dUng
NaOH,
thay
binh
diTng
CaCla tSng them 0,194g,
binh
dufng NaOH tSng them 0,8g. Mat khac dot chay
liTdng
chat A tren thu difdc
22,4ml
khi
nitd
(dktc).
Xac
dinh
thanh phan %
khoi
liTdng
cac nguyen to trong A.
Gidi
"
Cho san pham chay qua
binh
dtfng

CaCb : mi = 0,194g =>mH20 =0'194g
=
Qua
binh
dirng
NaOH : m, = 0,8g=> mco, = 0.8g
Kho'i
lifdng
cac nguyen to :
mc=12.M
=
o,22(g);
mH =
2.
^ = 0,02 (g)
.^-Wi^
44 18
0 0224 ' . V . ^ . J' 1. t
"Nz
= = 0.001 (mol) mw = 0,001 x 28 = 0,028 (g) . | |
Thanh
phin
% kh^i
liWng
cac nguydn
ttf
f*«
"*»*'*^
" '
%C=-^^x

100% =
78,01%;
%H =-^^x 100% = 7,09%
^ ^'f^ 0,282 -*v.>M;-ii., vi^ uuu 0,282
fu,.//
iu;.'>fi
ixO ,|
ijiitt
%N=
°i^x 100% = 9,93% :^ %0 = 4,97%. ^^"f^^^^^
0,282
!ai 4. Oxi h6a hoan toan 4,92mg mot hdp chat A chiJa C, H, N O roi cho san
pham Ian
liTdt
qua
binh
chiJa
H2SO4
dam dac,
binh
chiJa KOH thi thay
khoi
liTdng
binh
chiJa
H2SO4
dSc tang them 1,8
Img,
binh
chiJa KOH tSng them

10,56mg.
d thi nghie
m
khac, khi nung 6,15mg hdp chat A do vdi CuO thi thu
difdc
0,55ml (dktc) khi
nitd.
Hay xdc
djnh
h^m
liTdng
phan trSm cua C, H, O va
N
d hdp chat A.
Gidi
*
Khoi
lifdng binh chiJaH2S04 tang them 1,8Img chinh la mH20=>mH20 =1.81mj
*
Khoi
liTdng
binh
chiJa
KOH
tang them 10,56mg chinh la mco,
=> "^C02
^0,56 mi
Khoi
lifdng
cac nguyen to :

10,56
1,81
mc=
12.^^^ = 2,88(mg); mH = 2.^ = 0,2 (mg)
44 18 • .I .
0 55 4 92
mN(4,92.g)
=
28.^x-i—=
0,55(mg)
> •
'>
mmiiy.
j:}.,H„p
: A nrv
=>
Thanh phan %
khoi
lifdng
%H
=
2,88
4,92
0.2
4.92
^xlOO%
= 58.54%
^,^^^,„
£, s^fM)ixSk^^^^^
x

100% = 4.065%
%N
=
^X
100% = 11,18% - ^^'^-'^^o^''^^' '^^^
r:>
%0 = 100% - %C -
%H
- %N = 26.215%.
Bai 5. Hai chat
hffu
cd A va B
ciing
chtJa cac nguyen to C, H, O. Khi dot chay moi
chat deu phai dung mOt
lifdng
oxi b^ng 8 Ian
lifdng
oxi c6 trong moi chat va thu
difdc
lifdng
khi
CO2
va
H2O
theo ti le
khoi
lifdng
CO2
so vdi

khoi
lifdng
nuTdc =
—.
Xac
dinh
cong thiJc ddn gian nhat cua moi chat ? ''' " " '" -
•
''' CUM '
Theo dieu
ki?n
de bai thi A, B la hai chat dong phan cua nhau. •
Dat
CTTQ
cua A, B la
CxHyO^.
. i ,, ,, ,
"Mrira?-
^' YV
.^'^'^
„
Phifdng
trinh
dot chay: , 1 . , i
CxHyO,
+
^
y
x
+

^^
4 2
O2 >
XCO2
+ -HjO ,
Theo de : 32
y
z
x
+
.
4 2)
=
8xl6z <:>4x + y=18z (1)
'IK''-
mco,
22 44x 22 44x 22 .
!
^ =— o = —
<=>
= — =>y = 2x
m
9 y 9 9y 9
The y =
2xvao(;i)
=> x = 3z=> z= - '
Tac6til$
x:y:z = 3z:6z:z = 3:6: 1
Vay
cong thiJc ddn gian nhat cua hai chat la CsHeO. ' ' ' '

1
Phan
djing
va
phuong
ph^p giai H6a hpc 11 HOu
co
- Dfi Xuan Hung
Bai
6.
Vitamin
A
(retinol)
c6 c6ng thiJc phan tuf
C2()H3()0,
vitamin
C c6 cong thiJc
phan tur la
CfiHsOfi.
a)
Vie't
cong thiJc ddn gian nhat
ciia
moi chat. 'li,, ,.
b)
Tinh
ti 1? % ve
kho'i
li/dng
va ti ie % so' nguyen tur cua cac nguyen to d

vitamin
A
va
vitamin
C.
Gidi
, "
a) Cong thiJc ddn gian nhat cua cdc cha't: ^
Vitamin
A :
C2()H3(,0;
Vitamin
C :
C3H4O3.
,0 '
•
' 7. ,
b)
* Ti
1Q
% ve
khoi
liTdng
cdc nguyen to :
+
Vitamin
A :
CajHjdO
(M = 286) ' ''''' '
%mc=-^^^x

100% =83.92%
-'*^>;':'-
,
286
%mH=—xlOO%
= 10,49% ^ ' ' ^ '
286
=>
%mo = 100% - 83,92% - 10,49% = 5,59%. ' - ,
+
Vitamin
C :
QHsOfi
(M = 176)
6x12 •
%mc = X 100% = 40,9% , , ,,
176 'I f , ^ i .1' ' i V lih
'
%mH=
—xl00%
= 4,55%=^ %mo = 54,55%. ' ^" ""' ^ '
"^^'^'
176 ^.1 .!| w 'i
*
Ti 1? so nguyen tuf cdc nguyen to: , / j
+
Vitamin
A :
C2(iH3„0
(51 nguyen tuf)

20
%nc=—X
100% =39,22%
51
' \ bt
ii' ^''.
irfih-ic^'iT
%nH=
—xlOO%
=58,82% ^ %no=
1,96%.*'
' ' ' '
+
Vitamin
C : C6Hx06 (20 nguyen
tijf)
; v ;
%nc= xl00%
=30% '
20 - ^.^^^ . .
%nH=
—
xl00%
=40% => %no = 30%. f' •
Bki
7. Chat A chiJa C, H. O c6 ti 1$
khoi
li/dng
mc : mo = 3 : 2 khi dot chdy h6't A thu
diTdc

CO2
va hdi
niTdc
theo
u'
I9
the
tich
Vco2
:
VHJO
=4:3 (cdc the
tich
do d cung
dieu
ki^n
nhi§t
dp, ip suat). Tim cong thiJc ddn gian nha't cua A. ^,
,
(TrichTSDHSuphamKythu^tTP.HCM)
Z^:
Gidi
D$tCTTQcuaA:C,HyO,.,^^,'v.
^ ^
.fr,,s^^i,r^:
':-tf
• •
r
|.i
•

• •
• ;
14
PhiTdng
trinh
dot chay :
CxHyO,
+
TheodI:
""^
3
12x
•
= — <=>
mo 2 16z
y
z
X + —
.
4 2)
X
=>
z = —
2
XCO2
+
yHjO
(1)
Ta
CO :

V,
CO2
_ 4 ^ JL,^1
''H20
y
3
2
3
X
y
=
3x
(2)
,
Ta
c6 ti le x : y : z = x : ^x : ^ = 2 : 3 : 1 !
.r^:^^^^'":^
M
2 2 {
••••.i-n
Vay
cong Ihu'c ddn gian nhat cua A :
C2H3O.
^ = ' '
Bai
8. Hay
thict
lap cong thiJc ddn gian nhat
tCf
cac .so

lieu
phan
tich
sau :
a) 70,94%C; 6,4%H; 6,9%N; con lai la
oxi.
i
rion
' r ;o 5 '
b)
65,92%C; 7,75%H; t,;,con lai la oxi.
lf:>H:\tP
kvj,;tv|::,.',
'
.,!,;•!
X^:,,
,,,, Gidi .
•i^il
i orf''anf
giiW'utkj
I.'
a) Thanh phan % cua oxi •
'jiinod:?^'^
ni'^q
di'
%0 = 100% - 70,94% - 6,4% - 6,9% = 15,76%
Dat
CTTQ cua chat
hilu
cd :

C,HyO,N,.
f ; ^ ^
%C %H %0 %N
Ta
c6 ti
1^
X : y : z : t =
12 1 16 14
70,94% 6,4% 15,76%. 6,9%
12 1 16 14
=
5,91:6,4:0,985:0,49= 12:13:2:1
Vay
cong Ihu'c ddn gian nhat la
C12H13O2N.
, „ ;
b)
Thanh phan % cOa oxi: H , . , ,
X : y : z
%0 = 100% - 65,92% - 7,75% = 26,33%
Dat
CTTQ cua chat
hffu
cd :
CxHyO,.
Ta c6 ti 1?
^65^.7J5%^2M3%^ 5,49:7,75:1,64 =
10:14:3
12 1 16
Vay

cong thu-c ddn gian nhat la
C10H14O3.
Bki 9. Dot chdy ho^n loan 4,6g mpt hdp chat huTu cd A thu diTdc 3,6g
H2O.
Dan khi
CO2
sinh ra vao dung
dich
ni/dc voi trong 0,1M thi difdc 8g ket tua v^ dung
dich
Y.
nho tiep dung
dich
NaOH v^o dung
dich
Y thi difdc 3,5g ket tua
nffa.
Xac
dinh
cong Ihu'c ddn gian nhat cua X.
Gidi
Tac6:
mH=2.—=
0,4g
18
i
CO2 + Ca(0H)2 > CaCOji + H2O
0,08 mol 0,08 mol
Phan
djing

va
phudng
phip
giSi
H6a
hpc 11
HQu
co
-
D8 Xuan Hifng
2CO2
+
Ca(OH)2
>
Ca(HC03)2
^
? 0,07
mol
0,035
mol
' ' « ;
Ca(HC03)2
+
2NaOH
>
CaCOji
+
Na2C03
+
2H2O

0,035mol
' • ' '
0,035mol
yi'; i; i,^
SomolC02: nco2 =0>08
+
0,07 =0,15 (mol)
.
r=>mc=
12x0,15=
1,8
(g)
' ^ ^
^,,ffV
=>
mo
=
4,6
- 0,4 - 1,8 =
2,4 (g)
1'
Dat CTTQ
cua A la
C.HyO,.
Ta c6
ti
le : , £ , , . ,
,
v.
x:y:z

= ; : .
M M; M=
o,15 :
0,4
: 0,15
=
3
: 8 : 3
12
1 16 12 1 16 yv
Vay
cong thtfc ddn gian
nhat
cua X la
CsHgOs.
•> iah-)
~iSi
iilrfi
y
\M
J
Bai
10.
KM
cho 5,3
gam hon
hdp
gom etanol
C2H5OH ;H''i?i',fl! ;Civ^,ii'
/j;

va propan-l-ol
CH3CH2CH2OH
tac
dung vdi natri drfthu diTdc
1,12
lit khi (dktc).
a)
Vie't
phifdng
trinh
hoa hoc cua cac
phan iJng
xay ra.
b)
Tinh
thanh phan
%
khoi
lifdng
cua
moi
chat
trong hon hdp.
S'
dinMT
{a
a)
C2H50H
+ Na >
C2H50Na+-H2

;
X
mol
,
—mol
,:'S*C,d,^
,S'dt<ci_,
:ki2,
CH3CH2CH2OH
+ Na
>
CHsCHzCHzONa
+
^H,
2
S^molH2:
nH,=-^
=
0,05
=
^+y
'
Vv: viv.c. *0()( X)^'
, ,, ^,
^
22,4 2 2
^^_,,,,iii,3,jto.tfOr{iiM3Mp^
Ta
c6 h?
phudng

trlnh
:
-
+
^
= 0,05
fx
= 0,05 (mol)
.• s
;
y
•. K
2
2 => < "
<cn
CO y
= 0,05 (mol)
46x + 60y = 5,3
.t,* veV
b)
Thanh phan
%
khoi
lOdng moi
chat:
r^^-i&m
3^,*
a.!,wsFi
ibii
J(?

M«
%C2H30H=Q'"^^^^
100%
=
43.39%
:
=>%CH3CH2CH20H
=
56,61%."""
!:>tU^
Jlfln ft? ,,
^.^ .,.
.fltfc^
:
.
t.;.,,iV -:7 ,S'*jim'V;-&.rfiT
(DoMig
3.
<£Afi.
eS^ntf,
thi'te
fthdit
tilt
:
-
<T)t/a
ocLO^
cang,
thiie
fteiti

Qtatt
nhdi i
<-
'
-
<T)ita
oda khai liMtiq. hoae % khoi
lii&iig.
ede nqiujjen t»
-
^i/a i9dn the
tieh
(phiMuj
fthdfi khi nhien ke)
BAITAPMAU
-v^''
Bai
1.
Dot chay ho^n toan
5,6
lit
chat
hiJu
cd A d thi
tich
khi
thu
difdc
16,8
lit

CO2
va 13,5g,
H2O.
Xac
dinh cong thiJc phan tuT
cua A
biet
ti
khoi
so
vdi hidro
la 21
(cac khi
do a
dktc).
, , ,
Gidi
5
g y,C:VO£l
Afit!ip•^^);-
So mol
chat
hu-u
cd
A:
n = —^ =
0,25 (mol)
>, „ \
22,4
:>iA, ,.f=

,r|.v.o.H;:>
Khoi
liTdng
mol phan
tuT
A:
M =
21
x 2 = 42 =>
mA
= 0,25 x 42 = 10,5 (g)
16,8
22,4
,tyr
Khoi
lifdng
cac
nguyen to':
mc =
12. ^^^^
= 9
(g)
j
.35
06/
mH
= 2 ^ = 1,5
(g)=:> mo
=
10,5-9-1,5

= 0
nl",.i,,
^
J,
1
I'M
'
11* 1!
Vay
trong
A
khong
CO
oxi.
, • 1
Dat CTTQ
cua A : CJiy. Ta c6
ti I9
: x : y =
—:
— =
0,75:1,5
= 1:2
,
i • 12 1
^
Cong thu-c
A c6
dang
(CH2)„.

,
.l
„, ^ „ .,,
Taco: 14n
= 42 ^ n =
3. V$y CTPT
A :
C3H
'^''-^'^^^-^^-^
>
^
Bai
2.
Limonen
la mot
chat
c6 mdi
thdm diu difdc tdch
tilf
tinh
dau
chanh. Ke't
qua
phSn
tich
nguyen
to cho
thay limonen diTdc
cau tao tiJ hai
nguyen

to C va H,
trong
do C
chiem 88,235%
ve
khoi
lifdng.
Ti
khoi
hdi
cua
limonen
so
vdi khong
khi
gan
b^ng 4,69. Lap cong thiJc phan
tuT
cua
hmonen.
, '
Gidi
Thanh phan
%
kh6i
li/dng
nguyen
to
hidro
:

%H
=
100%
-
88,235%
=
11,765%
''"'^^'^
" ^
Khoi
liTdng
mol phan tur:
M =
4,69x29=
136 ,,
Dat CTTQ
cua
limonen
:
CxHy.
Ta
c(5 ti 1§
: . ,
%C
%H
88,235% 11,765%
'^'^'''^^'^^'^'''^^'^^^^
n
I 12 1
=

7,353
:
11,765
= 5 :8
=>
Cong thtfc limonen
c6
dang
(C5H8)„.
' • V,t> - '
Tacd:
(5 x 12 +
8)n
=
136=>
n = 2 . v i , "
V$y
CTPT
cua
limonen m
CjpHii.
THLT
VIEN
TIMH
BSNH THUAN
-zr::^"
' "A 9
17
Phan dgng va phuong phap
giSi

H6a hoc 11 Hau co - Pi Xuan Hung
Bai
3. Dot chay 100ml hdi cha't
hffu
cd A chtfa ba nguyen to C, H, O vdi 450ml
O2
la'y
dff trong khi nhien ke. Sau phan
ffng
the
tich
hon hdp khi va hdi thu difdc
650ml.
Sau khi cho hdi nffdc
ngffng
tu hon hdp chi con 350ml va sau khi cho Igi
tiep
qua NaOH chi con
50ml.
Xac
dinh
cong thiJc phan tuf cua A.
Gidi
' ''
The
tIch
cic chat:
uir
= 50ml
=>

Vo2(ptf)=
450-50 = 400ml
Vco2
=350-50 = 300ml
VHJO
=650-350 = 300ml
Dat
CTTQ
cua A la
CxHyO^.
' t'i
1'
I, rt ? •
C,H„0,
+
Ta
c6:
100ml
1
y
z
x
+ —
4 2
400ml
O,
XCO2
+ ^HjO
300ml
300ml

; ,.X
,<'):H
100 300
_L=_y
100
1
2.300
4~2
x
= 3
y
= 6
«
x +
i
= 4; Thex = 3,y = 6=>z= 1
4 2
100 400
Vay
CTPT cua A la
CjHfiO.
. ,• ; _ , ^
Bai
4. Hay thie't lap cong thiJc phan tuf hdp chat
hffu
cd trong moi
trifclng
hdp sau :
a)
Dot chdy ho^n toan lO.OOmg hdp chat hCu cd Y sinh ra 33,85mg CO2 6,95mg

H2O.
Ti
khoi
hdi cua hdp chat Ad doi vdi khong khi la 2,69.
b)
Dot chay hoan toan 28,2mg hdp chat
hffu
cd Z vk cho cdc san pham sinh ra Ian
Iffdt
di qua cac
binh
dtfng CaClz khan va KOH dff thi thay
binh
CaCl2 tSng them
19,4mg con
binh
KOH tSng them SOmg. Mat khac, khi dot 18,6mg chat do sinh
ra 2,24ml khi
nitd
(dktc).
Biet
rkng phan tuf chat d6 chi chffa mpt nguyen
tijf
nitd.
J\ tii t lat '
(1,1 t.i (jfi
1
a)Kho-ilffdngmolphantuf:M
= 2,69x29 = 78 ^^^^ , ,, ,
„VM,H

. U
Khoi
lifdng
cac nguyen to : ^ ; W ' ^ i .v^
33,85 ' '
mc=
12.
mH = 2.
44
6,95
18
=
9,23 (mg)
=
0,77 (mg)
y • If
=>
mo =
10,0©-
9,23 - 0,77 = 0=> khong c6 oxi.
D5t
CTTQ cua hdp chat hu»
ceJ:
C^Hy. Ta c6 d I9 :
x:y=
^j^:.2:21
= 0,77:0,77 = 1:1 =>
CTcd'dau^H)„
Ta
c6 : 13n = 78 => n = 6.

VHy
CTPT1^
CfiHfi.
.
b)
Theo de ta c6 : ,. ,
;„,,,,
;,„,
19 4
^H20
= 19,4mg => mH = 2. = 2,15 (ja0
18
80
mco2=80mg
=> mc= 12. —= 21,81
(mglr
^
44
m
2,24 , iT¥' 'M'
•)'"•'''''
:
niN
(28,2mg)
= 4,24 (mg)
=>
mo
= 28,2 - (2,15 + 21,81 + 4,24) = 0 'h ,f ;;a
.
J

Vay
khong c6 nguyen to oxi trong hdp chaft
hffi^ccfe
lili,
)i ; b 0 ',
Dat
CTTQ :
C^HyN,.
Ta c6 ti 1? : ' ' J OS! :
21,81 2,15 4,24 , „^ ^ ,^ „^ :iq
utidi
^06:)
x:y:z
=—•—:——:——
= 1,82 : 2,15 :QB .
,:;;
;
- =6:7:1 => CT cd dang
(CeBilS^/
Jri^
f)^?'
VI
trong phan tuf chi chffa mgt nguydn tuf
nitd
=>-m=^U.''' ' "P '^^ "'^^i •!
Vay
CTPT chaft
hffu
cd
m

C6H7N.
. -"5' <«^
Bai
5.
Phan
tich
chat
hffu
cd X chffa
C, H,
O ta c6 :
mc : mH : mo = 2,24 : 0,357^^'- '
'''''''
" '
a)
Lap cong
thffc
ddn gian nhat cua X. t '
b)
Xdc
dinh
cong
thffc
phSn tff
ciia
X biet 1 gam Ji^lam bay hdi c6
th^tich
1,2108
lit
a 0"Cva 0,25 atm. '

a) Dat
CTTQ
caaX:CxHA.Tac6til$:
' i : '
mc . mH . mp _ 2,24 . 0,357 .2' >, r
J
j
,
~ 12 • 1 • 16 " 12 • 1 -16.
=
0,187 :
0,375
: 0.125 = 3 :6 : Z:
,:,|*^^'
=>C6ng
thffc
ddn giSn nhat cua X :
C3H6O2.
' " •
^
^ u /
b)
S^molX : nx = — =
^'^^^^'^^"^
= 0,0135(mot)
frf^iwlspft.!
RT
^^273 v i . ,
.
, 273 , ,

Kh6i
Iffdng
mol cua X : Mx =
—^-—
= 74 , < ,
0,0135
(^ ,
Tacd:
(3 x 12 + 6 + 2 x 16)n = 74 =i> n = 1
V§y
cong
thffc
phan tff cua X la
C3H6O2.
r'ff) c ! >
19
Phan
d?ng
va
phUdng phAp
Q\i\a hgc 11
Hi?u
co - D5 XuSn
Hung
Bai
6.
TO
tinh
dau hoi, ngi^di
ta

tach difdc anetol, mpt chaft thdm du"dc diing
de
san
xuat
keo cao
su. Anetol
c6
khoi lifdng
mol
phan tuf bang 148g/mol. Phan
tich
nguyen
to
cho thay anetol
c6
%C =
81,08%; %H
=
8,1%;
con lai
la
oxi. Llipcong
thuTc ddn gian nhat
va
cong thiJc phan tuf cua anetol.
i
,,„.„:
Gidi

,.,,„,

M
=
148g/mol.
Thanh phan
%
oxi
:
%0
=
10,82%
Dat
CTTQ
:
QHyO,
"
^
. %C %H %0
81,08%
8,1%
10,82%
,
Ta CO ti 1?
: X
:
y
:
z = : : = : :
,
12 1 16 12 1 16
= 6,8

:
8,1 :0,68= 10
:
12
:
1
,
Vay cong thuTc ddn gian nhat cua anetol: Ci()H|20
=>
CT
CO
dang (CioHi20)„.
\'
Taco:
(10
x
12
+
12
+
16)n
=
148 =>n=l»,^>r
VH,
W,rrt»''i
V|iy cong thiJc phan
tiJf
cua anetol
la
CioHuO.

( t
j
>l
3
K
Bai
7.
Dot chay 200ml hdi 1 chat
Mu
cd
A
chiJa
C,
H,
O
trong 900ml O2,
the
tich
h6n
hdp khi thu di/dc 1^
1,3
lit. Sau khi
cho
hdi nifdc ngiftig tu, chi
con
700ml.
Tiep
theo
cho
qua dung dich

KOH dac
chi
con
100ml
(cac
the
tich
do d
cUng
dieu
ki^n). Tim
CTPT
cua
A.
• -'Kiisn'
•
ngtfngtu
.
CO2
KOH
dac
Sd
do
phan
tich
de
bai:
m'kA
+ O2
200ml

900ml
-CO2
•
H2O
-O2 diO
-
H2O
1300ml
O2 dif
700ml
Difa
vao
sd
do
ta
tinh
difdc:
Vo^phintfng =900- 100
=
800
ml
Vco^
=
700- 100
=
600 ml
VH^O
=
1300-700
=

600
ml
Phifdng
trinh
phan iJng chdy:
O2 con dir
100ml
1'
QH,0,+
(x4~)02
1
4
2
^
4 2^
200
800
=>
X =
3;
y = 6
va
z = 1
XCO2
+ -H2O
2
J
Ml i
^
1

*h nJ^I
CTPT
cua A,la CjHfiO
''W^^^'i'^'^
on
Bai
8. Trpn
400
cm^ hon hdp hdp chat
Mu
cd
A va
nitd
vdi
900
cm' oxi duf roi dot.
The
tich
hon hdp sau phan iJng
la
1,4 lit. Sau khi
cho
hdi nufdc ngung tu thi con
800
cm^ tiep tuc cho qua dung dich KOH thi con
400
cm"\T cua
A
la:
A.C2H4

'
B.CH4
Sd
do
phan
tich
de
bai:
C.C2H6
Giai
C,Hy
(A)
N2
+ O? dU
t"
rco2
^
H20
N2
^02
dir
400cm^ 900cm^ 1400cm
Dtfa
vao
sd do
ta
tinh
difdc: s,
.
.

j
ngutig
tu
-
H2O
CO2
N2
'J
ii'ii
^ qsoflq
,i
800cm
D.
C,u,
dd
KOH
N2
02dir
400cm
3
Vco,
=
800
-
400
=
400
cm^
s!»i»>
= 1400

-
800
=
600 cm^
V(02c6ndU+
N2)= 400 cm'
"»'fH:;
^
Phi/dng
tfinh
phan iJng chay:
|i O;!!, f !
•!•
CAM ' , -
QHyO,
+ (x
+
^ )02
4
2
1
Tacd: VojphanJng
=
Vco2
+
2^H20
=> condt/ = 900 - 700 = 200 cm'
= 400
-
200

=
200 cm'
=>
Ta cd ti 1?:
Vr
» : V.
-> XCO2
+ ^
H2O
= 400+
—
=700
cm' M'u , ,„•,.•
^CxHy
= 400 - 200 = 200 cm'
C,Hy ••
V02 phan tfng
:
Vco2
:
VH^Q
=
200
:
700
:
400
:
600
= 2 : 7 : 4 : 6

Viet lai phi/dng
tnnh
phan u-ng: 2CxHy
+
7O2
-A-
4CO2
+
6H2O
Ap dung
dinh
luat bao toan nguyen to'
C va H
ta cd:
2x
= 4
2y
=
6.2'
x
=
2
y
=
6
=>
CTPTcuaAla:C2H6
Bai
9. Cho
5

cm'
C,Hy
d
the khi
vdi
30 cm' O2
(lay
diT)
vao
khi nhien ke. Sau khi
bat tia lufa di^n
va
lam lanh, trong khi nhien
ke
con 20 cm' ma
15
cm' bi hap thu
bdi
dung dich
KOH.
Phan
con lai bi
hap
thu
bdi
photpho.
Tim
CTPT
cua
hidrocacbon.

,1
Theo
de
bai:
Vco^=
15 cm'
=^
=> phan ifng = 30 - 5 = 25 cm'
Gidi
. ,
Vo
d^ =
20- 15
=
5 cm'
21
Ptiftn^ang^
phicnq
phap
giai
Hfa^pc
11
Hi?u
co
-
P8
Xuan
HiMg
•^uang trinhphan tfngchdy:
QHy

+
(x
+
^,)02
->-xC02 + ^HzO
5
25 15
=>x = 3vay = 8=>
GTPT
la
CjHs.
B^O:
Cho vao khi nhiSnske 10 cm^ ch^it hffu cd A
(chiJa
C. H, N), 25 cm^
H2
va 40
tsdl^Oz.
Bat tia
liJa
di$^nn:ho hon hcfp no. Chuyen hSn
hcJp
khi nhan
diTdc
ve dieu
aki^Ii'ban
dau, H2O
ngiteg
tu he't, thu
diTdc

20 cm^ hSn hdp khi, trong 66 c6 10
••Wff
bi
NaOH
hap thu vi'S cm^ hi photpho hap thu.
CTPT
cua A la: i^^' '
"ATCHsN
B C2H7N
C.C3H9N
D.C4H,|N
a,!
-Sd^o phan tfng:
O2
c6n dir
CO2
+ N2
40^^A(C,H,N)
+
Photpho hafp thu
O2
=>
VQ^
ji,
= 5 cm'
©oingdich
NaOH
hap
thu
CO2

=> VQO = 10 cm' v>v'
20 cm'
=> = 20-(5+ 10) = 5 cm'
25 3
Wolchayhkiro)
= —- = 12,5 Cm
^
iih
b:)
el
(dotehayea
A
vi
hidm)
= 40 - 5 = 35 Cm'
%)j
(dcfichiTy
A>
= 35
-42,5
=
22,5
cm'
-Ph«ang
trinhphin tfngxAdy: •
^gatf
naiq
ritvn^
g0biniq
ifij

fS^V;
C^yN,
+ (x
+
^)02
XCO2
+ ^HjO + -N2 •
feolriniasOTbqA
4
2 2
=>
VH,o<J *ftch4y
A)
=
^(V^,
- Vco,) =
2.(22,5
-
10)
= 25 cm' ,
Ta
cd
tH?
the tich:
VA:-Veo^:VH^o:VN^
=
10:10:25:5
=
2:2:5:1
fod

=>A:C:H:N
= 2:2 :40 : 2 = 1 : 1 : 5 : 1
CTPT
cua A la
CH5N
,rii'i\i!.> iO-;I,»vrt
'•,m ,ix^
'.},>'i
i(6o;«iT
BAI TAP NANG CAO
Bai
1. Mot hdp chat hilu cd A
chuTa
54,8%C;
4,8%H;
9,3%N
con lai la O. Cho biet
ph§n
tuf khoi cua nd la 153. Xac dinh c6ng
thiJc
phan tuf cua hdp chat. Vi sao
phSn
tuf khoi cua c&c hdp chat chtfa C, H, O la s6'
ch£n
ma phan tuf khoi
ciia
A
lai
la so' le (khong ke phan thap phan) ?
•

u
.•
*,
|,
Gidi
%0=
100%-54,8%-4,8%-9,3%
=
31,1%
irfJN
^
MA
=153
,
/••dijoo^bjri
BtV:r'rs-r?
;
Dat
CTTQ
cua A :
CH^N,.
Ta c6 ti
1?
:
.i„,:^;,fc,M,^^,g,««>d
vftrSo
m
54^,
4^,3U . 9^
X

:
y
:
z
:
t =
12 1 16 14
=
4,57:4,8:
1,94 : 0,66 = 7 : 7 : 3 : 1
=>
CT
cua A cd dang
(C7H703N)„.
Tac6
: 153n= 153 => n = 1
Vay
CTPT
cua A :
C7H7O3N.
*
Phan tuf khoi cua A la so le vl nguyen to nitd c6 h6a tri 16 => nguyen
tuT
H la so
B^i
2. Cho 400ml mOt hidro g6m
N2
C^Hy
v^o 900ml
O2

dif roi dot. The tich hon
hdp khi thu dtfdc sau khi dot la 1400ml. Sau khi cho hdi
niTdc
ngifng tu thi con
800ml hon hdp ngi/di ta cho loi qua dung dich KOH tha'y con 400ml khi. Xac
dinh
cong
thiJc
phan tuf cua hidrocacbon tren, biS't the tich cic khi deu do d ciing
dilu
ki^n nhi$t dp va ap sua't. ^^\,.
Giai
f(,5fV(
tO.O
CxHy
+
'
y^
x + -
4
O2
>
XCO2
+ ^H20;
•
f
.AS';
Si'it)
nvA A ;fe'b hy-
x + y

4
Dat
the tich cua
C^Hy
la a
=400-a
Theo
de bai ta c6 :
ax
<vjH
\'
; :Mit
iirtJ
}flA
c^,,-!
VH20
=
1400-800
= 600ml = « c#
Vco2
= 800 - 400 = 400ml = ax
^,
j,
|,
Vo2du=900-
X^I
4
11
Vo2d^+VN2
=400

<* 900-
x + y
.
4;
a
+ 400-a = 400
23.
Phan
dgng va phuong ph^p giai H6a hpc
11 HQu eo
- B5 Xufln Hung
Vaytacd:
ay = 1200 (1)
•Vivi
ax = 400 (2)
Glai(l),
(2), (3)
a = 900-a (3) ' ' •
'i«i#iffc
•
• - • ,
-itiVj
i'kih
'•'pli
ri/!;y
,fy§^
a = 200ml fSsirj
'^Iv;
Vjnv'i/i) bl u, ii lli
•

X = 2 ,:ut}
y
= 6
•=
"''r;i'
m.P
'fli
s.i'
•>]::
Vay
CTPT cua hidrocacbon la C2H6.
Bai
3. Dot chay hodn toan 3,6g chat
huTu
cd A (C, H, O) b^ng 4,48 lit O2 (dktc) thu
diTdc
hon hdp khi va hdi trong do Vco, =3Vo ; ^^^^ = 11. Tim cong thiJc
phan
tuTcua
A biet d the hdi; l,8g chat A chiem the
tich
bang the
tich
cua 0,8g oxi
cung dieu ki?n. , _ ^ , , •
Gidi
Dat
CTTQ cua A :
C^HyO,,
,

0,8
Ta
CO : nA =
\l f '1 > J, , '
lit
1'' I .', . nllt
'HW
=
0,025
mol =>
MA
=
1,8
=
72
32 ••
0,025
. , .
PhUdng tnnh dot chdy : ,
,^^,„^,
•„
O2
—^ XCO2+ ^H20 •« iTt'rf
aftd
U^;^(m.
mi&jB'jof{):,ii
~.tciq
atfrti
;ii'n>;!-> idofsi
'

y
x +
4 2
0,05 mol
y z
x +
^^
4 2
0,05 0,05x o,05.^'-'^''^«^«''^'^^«^*^^
S6
mol chat A dem dot:
nA
= ^ = 0,05 (mol)
So mol O2 da
diing
: n^^ = = 0,2 (mol)
,
22,4
1,1 1
-•i.;:t
, , H,,D
H5n
hdp khi thu difdc :
CO2
H2O
O2 dir
Theo dl
Vco2
= 3du « 0,05x = 3 0,2-
y

z
x +
4 2
0,05
"^C02 11 44.0,05x 11
^
= — <=>
m
(1)
"29 18.0,05 i
y
3
"'rrtU:
•'"•"•"-'•2
K; v': I ]• 1'X)9iaf;^j(_'^^,,V
=>
- = 1^ = - =>
y=-X.
The vao(l) => z = 8-2x
24
4
Ta
CO : 72 = 12x + y + 16z
<=>
12x + -x + 16(8 - 2x) = 72
3 ,r! ,!'• i',,,/,. :
••
•;,(]
.
=>x = 3;y = 4;z = 2 a,.

.,.,(1!
<n;'.;:q
T ;• .
Vay
CTPT cua A \k C3H4O2.
Bai
4. Dot chay hoan toan l,37g chat
hufu
cd A thu diTdc 3,08g CO2; 0,63g H2O va
0,126 lit N2 (30"C va 75 cmHg). Xac dinh cong thiJc phan
tuT A.
Biet l,37g A cho
bay hdi d 100"C va 1 atm thi the
tich
thu diTdc la 306ml.
•,.(;./
•
o«€^kir>
m4:vi>d'riif
.:^iil.K\!
n-!:'!
Sdmol
A :
HA
= ^ = = 0,01
(mol)'
^^^'^'^
RT
0,082.(100 + 273) ,A.,, Aii:'.b:
YSH

^
Khoi
lifdng
mol A
:MA
== 137
0,01 . ' >: ,A
f,Mb ^;TT
• •
Taco: mc= 12. —= 0,84(g);
mH
= 2. — = 0,07 (g) f ^n)-
44 18
S6'molN2:
n^=—^ =
0,005
,„ / ,,vi' , 1 ,
0,082.(30 + 273) .^U| , v,
*
+
:=> mN
= 28.0,005 = 0,14 (g) ; ^;
mo = 1,37 - (0,84 + 0,07 + 0,14) = 0,32 (g) '
Dat
CTTQ cua A :
C,HyO,N,
, i, /
0,84 0,07 0,32 0,14 + 7-f
Ta
cd ti I9 : X

:
y : z : t =
12 1 16 14
'
• •
' = 0,07 : 0,07 : 0,02 : 0,01 = 7 : 7 : 2 : 1
^ ^
.
Vay
cong thu-c A cd dang
(C7H702N)„.
Tacd:
(7 x 12 + 7 + 32 + 14)n = 137 n = 1 ^>
-^'''-^
"
Vay
CTPT
cua A m C7H7O2N.
AX'kD
s^HOkJ ^
Bai
5. Tri^dc kia, "pham do" dung de nhupm do cho^ng cho cac Hong y gido chu
diTdc
tdch chie't
tijf
mpt
lo^i
oc bien. Do la mpt hdp cha't cd th^nh phan nguyen to
nhiTsau :
C

: 45,7%>; H : 1,9%; O : 7,6%; N : 6,7%; Br :
38,1%.
a) Hay xac dinh cong thi^c ddn gian nhat cua "pham do".
b)
Phi/dng
phdp phd
khdi
liTdng
cho biet trong dd phan tuT "pham do" cd chiJa hai
nguyen
tur
brom. Hay xac dinh cong thufc phan tiJf cua nd.
Gidi
, „ _
a)
D^t
CTTQ
cua pham do :
CxHyO,N,Brk
45,7 1,9 7,6 6,7 38,1 =' '
Ta
cd ti 1# : x
:
y : z : t: k =
12 • 1 • 16 14 80 •
=
3,808
: 1,9 :
0,475
:

0,478
: 0,476 = 8:4:1:1:1
'hSn
dgng
vk
phtwng ph^p
giSi
H6a hpc 11
HOu
CO
-
D8
Xuan Hung
Vay
cong IhiJc ddn gian nhat cua pham do la
CKH40NBr.
j)
Vi phan tuf chiJa hai nguyen tuf Br nen A c6 dang
(CsH40NBr)2.
Vay
CTCT cua pham do la CfiHxOsNsBra.
Bai
6. Dot chdy hoan loan hon hdp X gom hai hidrocacbon A B, mach hd
cilng
day dong dang. Cho toan bp san pham chay vao 4,5 lit dung
dich
Ca(OH)2
0,02M,
thu difdc ke't tua va
khoi

liTdng
dung
dich
tang 3,78g. Cho dung
dich
Ba(OH)2
(in vao dung
dich
thu
difdc,
ket tua lai tSng them, tong
khoi
liTdng
ket
tua hai Ian la 18,85g. Ti
khoi
hdi cua hon hdp X so vdi hidro < 20. So mol cua A
b^ng
60% tong so mol cua A va B trong hon hdp X. Cdc phan iJng xay ra hoan
toan. Hay xac
dinh
cong thiJc phan
tuT
cua A va B. '
Gidi
-J
Dat
CTTQ cua A : C^Hy : 3a
(mol);
B :

+„Hy
+2n:
2a (mol)
VinA
=
60%(nA
+
nB)^
•^ = -
Phifdng
trinh
dot chay : .
CxHy
+
x
+
—
4j
O2
>XC02+^H20
, (1)
3ay
3a mol i 3axmol mol
)
1' *
M,
'',:S:
a:
2
Cx

+nHy
+2n
+
A-
3„ rv + 2nV '
'-'''J'T^^^'ifi^^i
x
+ —+ ^ O, >(x +
n)C02+
H2O
(2)
2
4;
2a mol ! : i \ X
1
o.;);';
1 (x + n)2a
2
y
+ 2n
.2a
•Sh(V
;kv.(}:j
-{IV
nca(0H)2
=0,02.4,5 = 0,09
(mol)
, / ^ ^ , „^ „ ,
^, ^,
CO2

+
Ca(OH)2
> CaCOs +
H2O ,/!.,(:),
1-?,":)
;;l
; (3) yJ:M
b
mol b mol b
moLji
ffiourU,.
-lib
gauf; "nf>
frartq"
jnJ;
•JSuil
2
}i:H
2CO2
+
Ca(OH)2
)•
Ca(HC03)2
l
v:/M&.n.>n
f (4)
,1-^Ij^A
2c mol c mol c mol
(H'M-in
Ca(HC03)2

+
Ba(OH)2
> CaCOji +
BaCOji
+
2H2O
(5)
c mol c mol c mol
,,;sjy|i
:,:!;^}f:
'^iM,
('»
Theode: mcoj
+mH20
=3,78 +
mcaco3
(3) v^i'
'i^nf-JnM
(d
hay
mco2
+
mH20
= 3,78 + 100b "•'• i^^H (6)
Theo (3) va (4):
mcafOH),
= b + c = 0,09 (mol)
Matkhac
: 18,85 =
mcacoaO)

+
mcaC03(4)
+
mcacogrs)
„, , ,
hay 18,85 = 100b + 100c + 197c = 100b + 297c
Ta
c5 he phifdng
trinh
:
b
+ c = 0,09 fb = 0,04 (mol)
100b + 297c = 18,85 [c = 0,05 (mol)
=>
nco2
=b + 2c =
0,14(mol)=>
mcoj
(3.4)
=(b + 2c).44 = 0,14.44 = 6,16(g)
Ttf
(6)=i>
mHjO
= 3.78 +100.0,04 - 6,16 = 1,62 (g)
1,62
ni
-
= 0,09 (mol)
Theo phifdng
trtnh

(1), (2) ta c6: ncoj = 3ax + 2a(x + n) = 0,14 1 ri
1
3ay
-i-2a.
y
+ 2n
2
hay
nco2
= 5ax + 2an = 0,14
Ma
ta c6 :
dx/Ha
< 20 Mx < 40 ,
(12x
+ y).3a + [12(x + n) + y + 2n].2a
hay 60x + 5y + 28n < 200
5ax-i-2an
= 0,14
Ta
c6 he phifdng
trlnh
: •
•
F
;/;,,;,)
r;/!':'
I'i'ji'i,
'jLnJ.
>•

(7)
(9)
Say = 4an = 0,18
Giai
ra ta c6 :
<40
^'
60x
+ 5y + 28n<200 , „ , , ,
a = 0,01
n
= 2
x
= 2
V$y
CTPT
cua A m
C2H2,
B la
C4H6.
, ^ , , g, W.
Bai 7. Dot chay ho^n toan m gam chat hiJu cd X chtfa ba nguyen to
C, H,
O thu
difdc
a gam
CO2
va b gam
H2O,
biet a = va b = Hay xic

dinh
cong
thtfc
phan tvl cua
X.
Biet
3,6g hdi X cd the
tich
b^ng the
tich
cua l,76g
CO2
ciJng
diluki?n.
^ ^
. j •'J:\
Gidi
Li
Dat
CTTQ
cua X
: CxHyO,.
22m
^ 3m m , .
Tac6:mc=
12.^^ = 0,4m(g); mH = 2. —= —(g)
15.44 J-io
=>
mo = m
—

0,4m
+ —
15j
8m
I5
(g)
27
Phan
dgng
phuong
ph&p
g\k\a hgc 11 HOu cO- D8 XuSn Hang
Tacdtil? : X :y : z = : — := 0,033 : 0,066 : 0,033 = 1:2:1
12
15.1 15.16
=> X c6
dang
(CH20)„. ,
176 't.
Matkhac:
nx =
nco2
== 0.04 (mol) ,
=> Mx = = 90 30n = 90 n = 3
- " ' 0,04 ''
V$y
CTPT
cua X m CjHfiOs.
v*/rv,-,i.
v * -' f . ,.t •

Bai
8. Oxi hoa
hoan
loan
18,6g
chat
hffu
cd A
thu
diTdc
52,8g CO2 va 12,6g H2O.
Mat
khac
khi
phan
tich
liTdng
cha't
hffu
cd A
thu
difdc
khi NH3.
Dan
toan
bp khi
nay
vao 125ml
dung
dich

H2SO4
2M
thi
phan
axit
duT
difdc
trung
hoa
vilfa
he't bdi
100ml
dung
djch
NaOH 3M.
a) Xac
dinh
thanh
phan
% N cd
trong
A. u> ' , ^,
b) L$p
cong
thiJc
phan
tuT
A biet
dA/kk
< 3,25.

, rr, Gidi • " :
Tacd:
nH2S04
= 0,125.2 = 0,25 (mol)
'''";"'']•"/;'';://'''
^,
,f
"NaOH
-0,1.3 = 0,3 (mol) ^ ^ ^
2NH3 +
H2SO4
> (NH4)2S04 ' ' ' (1)
0,2 mol 0,1 mol ;
H2SO4
+
2NaOH
> Na2S04 +
2H2O
V'*(2)
0,15 mol 0,3 mol
: , -
nH2SO4(^pu(i)=0,25-0,15
= 0,l(mol) '''\
=>
mN = 0,2 X 14 = 2,8 (g) " ' , - M n iM)
2,8
18,6
Thanh
phan %N trong A : %N = x 100% = 15,05% \
T

A 52,8 ^ 12,6 ,,,,
.sfolTOxJl^-
Tacd: mc=12. =
14,4(g);
mn = 2.—^ =
l,4(g)
^
=> mo = 18,6-(14,4+1,4+ 2,8) = 0 . .„
VSy trong A khong cd oxi. :
^l-
Dat
CTTQ
cua A :
C^HyN,
rf:^='''^'M
^-^^
''"^f"^'*
- f*-^^'
{!•&;)•&•
n.^ij^
ptjrfl
^
•
Ta cd ti le : X : y : z =
:
M: M
=
1,2 :1,4: 0,2 = 6 :7
:1
^ ^

.
12 1 14
m^y'.'-^'^
^ A cd dang (C6H7N)„. X
fiio
pTTO
Theo
de :
dA/kk
< 3,25 => MA < 3,25 x 29 = 94,25
Ta cd : (6 x 12 + 7 + 14)n < 94,25
V§y A cd
CTPT
:
C6H7N.
7Sl
93n
< 94,25 => n < 1,01 => n = 1
Bai
9. Dot chdy 560 cm^ hon hdp khi (dktc) gom 2 hidrocacbon cd cilng so nguyen
tiJ
cacbon ta thu duTdc 4,4g
CO2
va 1,9125g hdi niTdc.
•
'
a) Xdc
dinh
CTPT
cac chat

hiJu
cd ,.,, ,
,,
j ^y
j:
•
b) Tinh %
khoi
lifdng cac cha't
c) Neu cho lifdng CO2
tren
vao 100 ml dd KOH 1,3M; Tinh
CM
muoi tao
thanh
Giai
6 bki nay, ta
dilng
phifdng phap so nguyen tuf H
trung
bmh ke't hdp vdi phiTdng
phap bi^n
luan
de giai. ij(|,;,,,|v) d d
a) Xdc
dinh
CTPT
cdc hidrocacbon : _ , ||, ^ ,j ., ,,1
Dat
CTPT

2 hydrocacbon
tren
:
B:C,Hy.
CTPT
trung
binh
2 hydrocacbon
tren
: C-H-
Giasur
y<y'^y<y <y'
So mol hon hdp khi nhh = = 0,025 mol
nco2
=4,4/44 = 0.1 (mol) _
nH20
= 1,9125/18 = 0,10625 (mol)
x+y O,
-!—).
XCO2
+ ^H20
0,025 / • ' ^
"002
=0,025 = 0,1
nH2o =0,025^ = 0,10625
0,025x-^ 0,025 y/2
x
= 4 '
y
= 8,5 '

CTPT
A, B cd dang : A :
C4H,
va B :
C4H,.
-^^^ ^^^'^
^^^^^^''•''''^
Tacdy < y
<y'hayy<8,5<y'(1)
^
8,5 <y' chan^v^v ;^
'''^l'
_y'^2x + 2 = 2.4 + 2 = 10' '
t<««-''
• ' '
=^y'=10r=>CTPTB:C4H,o
tentO,0
TiWng tir bi?n
luan
tim
CTPT
A : y < 8,5, y chSn
>|.ii^||''ai
'
Bi^n
luHn
tim
CTPT
B :
y

2 4
6 8
A
C4H2
C4H4
C4H6
C4HS
Vay cd 4 cap nghiem :
A:C4H2
B:
C4H,()
va
A:
C4H4
B: C4H
va
10
A:C4H6
B:C4H
va
10
A:C4HH
B:
C4HI()
29
PhSn
d^ing
phUOng
phAp
gi^i H6a hgc 11 HOu co

-
iJu Xuan
c) Tinh
CM
cdc muol tao thanh r
'
i,
,
nKOH
=
V.CM = 0,1.1.3= 0,13 (mol)
:
Tac6:
^^ = ^ = 1,3 ^Taothanh2mu^.
^
X C*
,
, "CO2 ^'-^ '''^ •
CO2
+
2KOH->
K2CO3
+
H2O 'ol\ir,';(yr
o: .
.'ryj'^fiir's.
• (••
a
2a a (mol)
CO2

+
K0H->KHC03
^y '
:r;^:A:q
yuni
a hyi()
b b b (mol)
••f'*;;
a + b =
nco2
= 0,1
Ta
c6 :
2a
+ b =
nKOH
=0-13
0,03
^ 0,07
a = 0,03
0.;
(mol)
, : ,
b
=
0,07
^^•''•^•-••••-•••-'••.•H;:T-ir^VK;(r'
CM(K2C03)
= -5J
=

0.3(M);
CM(KHCO3)
= -^
=
0,7(M)
, ,.
„^.j
Bai 10. Mot hidrocacbon
A
d
the khi c6 the tich gap
4
Ian the tich cda lUU huynh
dioxit
CO
khoi li/dng tifdng difdng trong cung dieu ki?n. San pham chay cua
A
dan qua binh diTng nifdc voi trong
diT
thi c6 Ig ket tua dong thdi khoi li/dng binh
tang 0,8g. Tim CTPT
A.
*
Tim MA : '
^\
^:^ «f«'n-i"fOt^::r:r
Ji^
1VA =
4.VSO2
(dcungdieukien)=>nA =

4.nso2
,
("y
1 .
=> = 4"^^"^ =>
-i-
=
(A
SO2
c6
khoi liTdng ti^dng diTdng nhau)
MA
MSO2
MA
MSO2
:^MA=^ =
^
= 16
•
^.'^-<^V^'^>^^-^.0:>=^j
D3tA:QHy
v
',
,-e&ri(.
J
,u
•^:^::rG ^-
,;v:,,n|
Binh dtfng
Ca(OH)2

hap
thu
CO2
va
H2O
^, ^,
.
, ,^ •
.
Ca(0H)2
+
CO2
->
CaCOai
+
H2O
r '
m-i^=
mcaco3
=
Ig
nco2
=
"CaCOs
=
1/100=
0.01mol
" "' h
a/TT! ^m)) J^ri
ndiH

=>nc
=
nco2
= 0,01mol mc = 12.0,01 = 0,12g
^
j , ,^
p;
r ^ ,.
mcoj = 0,01.44 = 0,44g
,,
,
,^.
^ ,
^'[''i
rr.^'
n|:.)]
flfic!
;j«\arr'
Ambi„ii=
mco2
+
mH20
=>
mH20
= 0.8
-
0,44 = 0,36g
it
mH =
2^

=
2^
= 0'04g
"^MS"
nio
f
5.
vifV
DLBT khoi lifdng
(A):
mA
= mc +
mH
= 0,12 + 0,04 = 0,16
' {
30
W
12,
y MA
MA-mc
16.0,12
,
Xa c6
—=
=
—— => X
=
—-—— = = I
m^
m^ mA 12.mA 12.0»16

MA-mH
_
16.0,04
^
^
/
mA
" 0,16 " : wV''
VayCTPTA:CH4
| ,
<T)cum
4.
j£4Lfi
emtQ^
thi'f£
pliikn
lit:
•'•'A
fid
i
' j;
ri'fj;
-
^i/a
ixda
phiiiAig.
phdfL bifii
liiati
' '
-

<T)ifii
oa»
dftih.
liiai
bao^
ioAti
khoi
''•'''y'':
'
BAr TAP
MAU VA
BAI TAP
NANG
CAO
Bai 1. Dot chdy hoan toan ba hidrocacbon A, B, C c6 ciing so nguyen
tuT
cacbon ta thu
diTdc ti le
CO2
va
H2O
c6 gid tri tiTdng iJng la
0,8
: 1 : 2. Xdc dinh cong thiJc phan
tufcua A, B, C.
, , , ,
^''^^ Gidi
S,0
* ChS't A CO
:

—£2L
=
0,8 => nco, < "HIO =^ A1^ ankan :
CnH2n+2-
nH20
CnH2n+2
+
r3n+n
' '
.
2
O2
>
nC02 + (n
+
1)H20
•,^^,]^p
y:
,,5., ^ fff;
Tac6:
-
= 0,8 =>n = 4
.
M/
,f'>
"
in,,
ii\
f ,nr
n +

1
'
Vay CTPT
A
la
C4Hi„.
* Chat B c6
:
=
1
Ucoj
=
nH20
=>
B la
anken hoSc xicloanken:
CnH2n.
/
Vi A, B ciing so nguyen
tuT
cacbon => B la
C4H&.
'
f f
*
I:
* Chat C c6
:
= 2
=>

nco2
=
2nH20
hay
nH20
< "coz
"H2O
,T
DSt CTTQ cua C : C„H2^^2-2k(k Ih so lien ket 71)
r3n + l-k\
CnH2n+2-2k
+
2
O2 >
nC02
+ (n + 1 -
k)H20
n
.••„'';|,,,„t
IS '
I
,;U,> /
:/.,'
,.4: •• >••
Taco:
= 2
.
Vi
A,
B, C ciing so cacbcn => n = 4 => k = 3 ^ -;;

if
-; •
'
V§y
CTPT cua CmC4H4. v t
J-
Phan dgng
vji
phuong
pUip
giSi
H6a
hi?c
11 HOu ca - P5
Xuan Hang
Bai 2. Dot chay 3,7g chat hQ^a cd X (C, H, O) dung vUa du 6,72 h't oxi (dktc) va thu
diTdc
0,25 mol
H2O.
Xac dinh
cong
thiJc phan
tuT
X bie't 70 < Mx < 83.
Gidi -i
Tacd:
= —=
0,3(mol)
''''^^
'v^;-

X + O2 >• CO2 + H2O
Ap dung dinh luat bao toan khoi liWng : • ^idvM •'iimb» '^/.i^,.
Ta CO : mx + =
mco2
^^lO '
^.i^TavAicit
'fw^io
n#J' -
=>
mco2
=3J+0,3x32-0,25x18 = 8,8(g),t^uVAttUvWi'.*
Kv^^^^^
-
8 8
"'^ = 7^:=2.4(g),^.,k\'i wn^(
miiitm'ti^
mH = 2 X 0,25 = 0,5 (g) .A^
i^li.ii
id 1
i.i«'n
? wH
'/a*
==> mo =
3,7-(2,4
+ 0,5) =
0,8(g)
* t'f ' j J)M «. M ?'1 . j'i ,
Dat CTTQ cua X : C^HyO, i ^ ii'^ <• '
Ta c6 ti le : X : y : z = —: —: — - 0.2 :0,5:0.05 = 4 :10:1
12 116

Xc6dang(C4H,oO)„. 'kU
„ ,'•
O^H"
Theode
70<74n<83 => 0,95 < n < 1,12 => n = 1
;jif,f'^
Vay CTPTcua X la C4H,oO. }/0'}.J$i
t^r'^sf^'
'••r I *
Bai 3. Cho bon hdp cha't
hiJu
cd A, B, C, D deu ben, mach
cacbon
lien tuc, kho'i
lifdng
phan
tijf
cua chiing lap thanh mot cap so cpng. Khi do't chdy mot lifdng bat
ki
moi chat deu chi thu difdc CO2 va
H2O
ma khoi liTdng CO2 bang
1,8333
Ian
khoi
liTdng
H2O.
Xdc dinh
cong
thiJc phan tuf cua A, B, C, D.

Gidi
„;>:;.,
;e 3Mt) *
Dat CTTQ cua A, B, C, D la CxHyO,.
f
y
J
,
I, 4 2.
Theodl:^ =
1,8333
= ^ =
1,8333
"-''^-"^
'
n^HzO
18.1
x_3 • • '
C6ng thu-c cua A, B, C, D c6 dang
(C3H«)„0,
m^taluoncd: 8n < 2 x 3n + 2 n < 1
••$^1^:^,
•^f
Vay
cong
thiJc cua bon chat c6 dang CSHRO,,.
Vi
khoi liTdng phan tuf cua chiing lap thanh mpt cap so
cgng
nen ta c6 z = 0; 1; 2; 3.

Vay
CTPT
cua A, B, C, D Ian
iMdl
Ik :
CaHg,
CjHgO,
CjHsOz.
CJHHOJ.
32
B^i
4. Dot chay hoan to^n 3,24 gam hon hdp X gom hai chat
hiJu
cd A va B khac
day dong ding, trong dd A hdn B mot nguyen tuT cacbon, ngu'di ta chi thu
diTdc
H2O
va
9,24g
CO2.
Bie't ti khoi hdi cua X doi vdi H2 b^ng 13.5. Tim
cong
thiJc
ciia A, B va
tinh
% khoi lifdng cua m6i chat trong X. 1 i '
„/sCt:,
W '! li
^
h M f (Trich TS DHQG TP HCM, dat 2)

'•'•'ts,,.,,
Gidi (
',lin^
nj' M> r in
4
nco,=
—=
0,21(mol)
' ' ^
o>,^r,
,Mf . , */t.i *U ,
Khoi lifdng mol phan
tuT
cua X1 (Jxytiji fiX^
>
r;> Mx = 13,5 x 2 = 27
nx=^ =
0.12(mol)
^^^^.^
_ nco, 0 2r^'''''-^^^''M'W^^^^^^^
=> So nguyen tif C trung binh : n = ^ = —— = 1,75 '
"X
0,12 -V
n,
= 1 < n = 1,75 < n2 = 2 ma n =
-
vol a la so mol cua A, b la so mol cua B. , ; ;
^.U5^|a=0.09.oi;l,i^^
a
+ b = 0.12 ll' =

».03™ol'
-V.
Ta c6 h? phufdng
trinh
:
mx = 0,09MA + 0,03MB = 3,24
'^Mk^^A^^
p 'i
'^^m'^:>i^i
itik Wl
=>3MA
+ MB=108 MB= 108-3MA>0 ''.•'iy-^^t
=>
MA < 36 => A khong c6 oxi va trong A c6 2 nguyen
tuT
cacbon. •
Nen A c6 the la C2H2, C2H4
hoac
C2H6.
+ A la C2H2 (MA = 26) =^ MB = 30 (CH20)^*'"*"' '
'^'^'^"
'''' '
+ Ne\ A la
C2H4
(MA = 28) ^ MB = 24
(loai)
'^''"""^
+ A m
C2H6
(MA = 30) =^ MB = 18

(loai)
^'^''^
« ' •
V$y
CTPT
cua A la C2H2, B la
HCHO
(CH2O).
Thanh phan % khoi liTdng moi chat trong X: %C2H2 =
-—^.100%
= 72,22%
, 3,24.
=>
%HCHO
= 27,78%.
Bai 5. MQt hidrocacbon X cl the khi c6 ti khoi hdi so vdi hidro la 15. Lap
cong
thiJc
phSn tuf ci5a hidrocacbon X.
itMra
Gidi ") .ft J, AUM'*
.ri.
1 ' '
D$t
CTTQ
cua
X
1^
C^Hy.
\m\\x4\J ^

\m I irA mh\ ' ^
Mx=15x2
= 30 ).Mfyj,ifj,.f?^(lJft/.j *KS )l ' '
Tac6:
12x + y = 30 => y = 30 - 12x ;
Di^u
ki?n y > 0 => 30 - 12x > 0 => X < 2,5 .it-sdatooifc-,? sA
:uwz
'^XnO m •.
Phan
d jng
vS
phuong
pMp
giSi
H6a hpc 11 HOu
cd
- B5 Xuan Himg
;/i;,;'Neu
x=l => y = 18
(loai)
(KAqno
ipi^ynsn^
x
= 2.=>y = 6 (nhan)
ivs,t5|vvw;(fa
.H
n>^.n
.A , ^ 'i'M:
V$y

CTPT cua X :
C2H6.
Bai
6. Mpt hon hdp khi A gom hai hidrocacbon thupc
ciing
day dong dSng. Dot
chay 2
lit
khi A can
diing
7,2
lit khi
oxi thu diTdc 4.8
lit CO2
mgt
lUdng
hdi nifdc
(cdc khi do 3
Cling
dieu
ki#n).
sb©
a)
Tinh
the
tich
hdi nUdc thu dUdc. , , ,, , ,^ ^.^
b)
Xdc
dinh

day dong
ding
cua hai hidrocacbon trong A.
c)
Lap cong thiJc phan tuf
tinh
thanh phan % theo thi
tich
cac chat trong A.
Biet
hai
chat
dtfng
cdch nhau mOt chat trong day dong
ding.
Gmi
, 'rr
a)
DatCTTQtrungbinhciia
Ala C-H «
,„
••• n , daUi mmi Din.tZ
x+y
O2
>
XCO2+
^HzO
2
lit 7,2 lit ^ 4,8 lit
^ ^

iaif:>
Sswt^ia
41
d ,A afc
|r»i:rro*
1
'^•'7 x - - '
,
^,."
Ta c6 ti 1? : - = —f- = — => x = 2,4; y = 4,8 .
j,.^a/rfq
ho tsT,
•
ThI Uch hdi
node
:
VH20
= f " "
^'"^ ^^'^^
'
"•'
_ 24 r
• •'<AM£ *]i=;:flM:
, ^
b)
Ta c6
til?
:-£• = — = -
=>
HH

= 2nc thuOc day dong dang anken. ^
HH
4,8 2
c)
VI nc = 2,4 =
X
la anken nen n, = 2 n: = 4 (hai chat c^ch nhau rngt
chlft).
Vay
CTPT
ciia
hai hidrocacbon la
C2H4
C4Hg. ,,. ,,
"^r
, ^
j^.^
^
Dat
a la nc2H4 trong
1
mol hon hdp A (1 - a) la sd'mol
C4H8.
^
^ ^.j^
^
.^^^
Ta c6 : nc =
"""'^"^
=2,4 ^ a = 0.8 (mol)

'^•^^^•^•^
^1A .v.
TITO
v£V
Thanh
phan % theo thi
tich
cdc
khi:
^.
=> %Vc,H8=20%.
:: ,>,,
Bai
7. Cho ba hidrocacbon A. B. C 5 thg khi trong
dilu
ki^n
thiTdng.
A. B, C c6
ph^i
.
; la dong
ding
cua nhau hay khong ? Bi^t rJlng khi phan hOy deu tao ra cacbon va
j
!
hidro.
the
tich
hidro gap ba Ian \hi
tich

hidrocacbon ban diu (5
cilng
dilu
ki§n).
D$t
CTTQ cua ba hidrocacbon la
CjHy.
xC
+
ax
^
2
Theo de :
ay
=
3a
y =
6
vay
A, B, C Ian
liTdt
c6 CTPT
la :
C2H6,
CjHfi,
C4H6.
BSi
8. Khi phan
tich
chat

hilu
cd A chi chtfa C,
H,
O thi c6 mc + mw = 3,5.mo
a) Tim cong thrfc ddn
giin
cua A. /
b)
Lay hai
nfdu
ddn chi?c X, Y dem dun nong vdi
H2SO4
dam dac d nhipt dO
thich
hdp thl thu difdc A. Xac
dinh
cong thiJc cau tao mach hcl cua A, X, Y.
Biet
ring
Aiaete.
m^'^'^^^i.:'m
.uuici-a^;-*i
,m<n: (j^i,,,
DHQG,
dot
1)
'm<Mi' i:
K»
j'bi rofl.'
itjU^'dKm^m

vi'
jf.iofi
•< 's.i
Giai
D|t
CTTQ cua A : C.HA- ' ' '' ' '
„
, , ^ ¥
&\r>
Jlidsj Bhbj,
i}rn^
'rfj;!!
ui?
Ta
CO
: mc +
mH
= 3,5.mo
Hay
12x + y = 3,5.16z => 12x + y = 56z
+
N^u z = 1=> 12x + y = 56 y = 56 - 12x
I
wfif,,
,'.f:ft
ijOHb
X
1
2
3

4
5
y
44
(loai)
32
(loai)
20
(loai)
8
nh$n
-4
(loai)
=>
A cd CTPT :
C4H8O.
'
•^•^^w^,/-
liw.^'^*^;*^^^^ ^^^
iCr ;
+
N^u z = n=> CT cua A c6 dang
(C4HgO)„.
t •
Vay
cong thtJc ddn gian cua A
la
C4H8O.
cn^ii
'

Theo de : X la
CH3OH,
Y la
CH2=CH-CH20H.
' '
A
cd CTCT :
CH30-CH2-CH=CH2.
PhUdng trinhphan iJng : T'" r''"'':^'
CH3OH
+
CH2=CH-CH20H
^l^^o*c>
CH30-CH2-CH=CH2 +
H2O.
Bai
9. Dot
chiy hoan toan
1 hidrocacbon A
cin
dilng
28,8g oxi thu dUdc 13,44 lit
CO2
(dktc).
Bi^t ti
khd^i
hdi cua A doi vdi khong khi la d vdi 2 < d < 2,5. Tim
CTPTcia
A.
;fi

•;,i-ffM"
•-'r-3
Gidi
VIA
la
hidrocacbon.
mi
khi
66t
chdy
1
hidrocacbon
ta
lu6n
c6:
1
—1
2
"02
phJntfng
-
"coj + 7"H20
"H2O
= 2( noj
phiniTng
-
nco2
) - 2
'28,8 13,44'
,

32 22,4 ,
=
0,6 mol
35
Phan
djng
va
phiiOng
ph^p
g\i\a hpc 11 HOu CO - D5
Xuan
Hung
13 44
Tacd:
mc=12.nco2= 12.^=
7.2(g)
niH
= 2.nH20=
2.0,6=
1,2 (g)
Dat
CTPT
cua A la C,Hy ta cd: x : y = — : — =1: 2
1 12 1
=>
cong
thiJc nguyfin cua A la
(CH2)n.
Ma2<d<2,5
=^

2<—^<2,5
^ 58 < MA < 72,5.
29
Hay58<
14n<72,5 =^ 4,l<n<5,2 ' ' ' ^'^
.Ma n nguyen =^ n = 5 =^
CTPT
cua A la
CsH.o.
.
/'''^^^
' "^
Bai 10. Oxi hoa hoan loan mpt chat hu^ cd Y can l,28g oxi, chi tao thanh
0,66g
^ ^H20 va l,76g CO2. Cho loan bp san pham sau phdn iJng sue vao 250ml dung
dich
Ba(0H)2
nong dp a mol/lit thu
diTdc
b gam ke't tua, Ipc ke't tua roi dun nong
dung dich loc dufdc c gam ke't tua nffa. Cho biet b + c = 5,91. Tinh a, b, c va tim
cong
thiJc ddn gian nha't cua Y ? ' *''
•>
;M Giai . , ^
Ap dung dinh luat bao loan khoi liTdng : my +
TIIQ^
=
nicQ^
+ m^^Q ,^

my = 1,76 + 0,66- 1,28 =
1,14(g)
\ t i
CO2 +
Ba(OH)2
> BaCOji + H2O * ' ' ^.
^ ^ X i ,^4^ j
2CO2
+
Ba(OH)2
>
Ba(HC03)2
'
^^H,,',-^
^
2y. y y ,. , ^
Ba(HC03)2
—^ BaCOjl + CO2 + H2O
1,76 0 66 0 22
^'I'^rOcHa:
A.#;; >
"^^=12. —=
0,48(g);
mH = 2 i^ =
-:^(g)
>
^^I.^^^^^^^^
=^
mo= 1,14- (oAS^^yhll^g)
,,ri^6mol k^ tua hai lln : .^^^^

,,,^5^,^'
^3 ,
5 91 1 76
mBaC03 =-j^ = x + y = 0,03; nco2 =-^ = x+ 2y = 0,04
TacohephiTdngtrinh:
= 1^ = °'^^
.nfc«f
AJV
[x + 2y = 0,04 [x = 0,02mol
v»y'^,
r pi
0 03
'-""'H
"Ba(0H)2
= X + y = 0,03 (mol) a = —— = 0,12
(mol/lit)
,
' b = 0,02 X 197 = 3,94 (g)
t
« c =
5,91
- 3,94= 1,97 (g) '
'•'•V^''^'}:^'.:
KiiAigfiiniirr
DatCTTQcua
Y :
C^HyO,,
0,48 0,22 1,76 0,22 0,11 ,^ , ,
Ta rd ti le : x : Y : z = : : = 0,04: : = 12 :22 :11
Tacouie

.X y ^ ^ ^
Cong
thiJc ddn gian nhat cua Y :
Ci2H220n
'i/^v^;
Bai 11. Bot chdy hoan toan 0,1 mol hdp chat hCu cd A chi chiJa C, H, O vdi oxi
theo
ti 1? mol 1 : 2.
Toan
bp san pham chay
diTdc
cho qua binh 1 difng dung dich
pdCl2
dii roi qua binh 2 di/ng dung dich
Ca(OH)2
dir. Sau thi nghiem, binh 1 tang
0,4g va xua't hi^n 21,2g ket tua, c6n binh 2 cd 30g ket tua. Tim
cong
thiJc phan
tuTcua A.
Binh
1 du-ng
PdCb
hut CO va H2O, giai phdng CO2.
j,,
,
;
,^ „ ,^,,
CO +
PdCl2

+ H2O ^ Pd4 + COjt + 2HC1 (1)
C02d
phufdng
trinh
(1) va CO2 d phan iJng chay vao dung dich
Ca(OH)2.
CO2 +
Ca(OH)2
->
CaCOal
+ H2O
30
100
Snc02
=
ncaC03
= — = 0,3 mol i
.(I
- m 0< "
212
Theo
ptpi? (1), ta cd: ncojsinhra = "co = nipd = —^ = 0,2 mol •
=> So mol CO2 do phan tfng chay la:
0,3-0,2
= 0,1 mol. , ,
Khoi lifdng binh
PdCh
tSng la: mco + mH20 " i"C02 = ^,4 j
=> mH20 = 0,4 + mco2 -
mco=

0,4 + 0,2.44 -
0,2.28
= 3,6 (g) i >t
3,6 , ^< "
=>
nH,o=
— = 0,2 mol ' '
Ap dung dinh luat bao toan khoi lifdng ta cd: St xi Y
niA+mo2
= mco2 +
'"co+mH20
' • ^' S
=>
mA=mco2
+ mco + mH20-mo2 = 0,1.44 +
0,2.28
+ 3,6 -
0.2.32
=
7,2(g)
Gpi
cong
thuTc phan
til
cilaA
1^
C^HyO^.
,„,,.„,,»ri,
AJSj.•
Tacd:mc=

12.(Ucoj + nco) = 12.(0,1 + 0.2) = 3,6 (g)
7{ioffl?
b / ,
mH = 2. nH20 = 2.0,2 = 0.4 (g) ; "
mo =
7,2-(3.6+
0,4) = 3,2 (g)
12x _ y _ 16z _ M _ 72
mc niH mo m 7,2
12x y 16z ,^ A
o = ^ = = 10 =>x = 3 ;y = 4 ; z = 2. •: ;
3,6 0,4 3,2 ^ V,,,. , R.'-^
• ^
;
Vaiy:
CTPT
cua A la
C3H4O2
n*!''-^^/'. • '
11
Phan dgnp va phudng phap
giai
H6a hgc
11
Hgu
co
-
B8 Xufln Hung
Bai
12. Mpt hon hdp gom niQt

so
hidrocacbon hen tie'p trong day d6ng ding c6 khoi
Itfdng phan tit trung binh
(M)
=
64.
6
100"C thi hon hdp nay
6
the khi, lam lanh
den nhift
dp
phong thi mpt
so
chat
bi
ngiftig tu. C&c chat khi c6 khoi Itfdng phan
tuf
trung binh bkng 54.
Cic
chat long c6 bKng 74. Tong khoi lifdng c^c chat trong
hon
hdp dau la
252. Diet khoi liTdng phan
ti3f
chat nang nhat
g^p
d6i chat nhe
nh^t.
Tim CTPT

cic
chat
%
the
tich
cdc chat trong hon hdp
'
K'^-^
Giai
Goi
aj, 32,an 1^ kh^i lifdng phan
tijf
cua c^c hidrocacbon trdn.
*
Ap
dung
tilth
chott
toan
hgc
:
Cdc
hidrocacbon lien tiep thupc ciing mpt day
dong
dSng
se
tao nen mpt cap
so
cOng
c6

cong sai
d
=
14
an
=
at
+
(n
-
l)d
S=
—"
^
" ' "
"'^
*
"^''^
*
Vdian
=
2ai
=>2ai
=
ai
+(n-1).14
!
,f.
M) . , i
=>ai

=
14(n-
1)
•
=>
S
=
l,5na,
=
252
' ' , ,u
'n, „• 11 . ^, (U,
n
i!
( '
Hay 1.5.14n(n-
1)
=
252 =>21n,^-21ni-252
= 0
=> n
=
4
(nhan)
va n =-3
(loai)
'
Vay:a,
=
14(4-0

=
42
" "
dat hidrocacbon dau
la
A|
:
QHy
('
In ,
in "
M,
=
12x
+
y
=
42
ychin
,, , , ,, " " ' H
y<2x
+
2
1
v,.')K'l
»A(
I
w/,'oi
i\i
in;/), ynyb

qA
m^
:
,
My . -
X
1
2
3
>4
y
30
18
6
<0
V§y
A
la
C3H6,1^
hidrocacbon dau tidn trong cap cpng tren.
Cdc
dong
dSng
ke
Uep cua n6 \k
C4H8, C5H,,,,
C^Hu (M
=
84)
^

,<'
* T-inA
%
rfcA
cdc
chitttroneh3n
h<n>:
'^'''^
^"^^
'
Gpi
a, b, c,
d
(mol) Ian li/dt 1^
so
mol cdc hidrocacbon tifdng tfng:
C3H6,
C4HS,
— 42a
+
56b
+
70c
+
84d
'
^M-VrOj-
.
M
= =

64
(1)
a
+
ii
+
c
+
d
' ^ .
M«,.=^^^i^
=
54=*b
=
6a
(2)
• •/:
.
"^CJ^U/
K€
•
,
Ml
=
70c
+
84d^^4
^c
=
2,5d

(3)
c
+
d
Thay(2).(3)vao(l):
42a
+
56.6a+ 70.2,5d
+
84d 378a
+
259d
M
= =
04
=>
— 54
a
+
6a
+
2,5d
+
d,,, 7a
+
3,5d
=>d
=
2a
(4) . ,

c
=
2,5.2a
=
5a
(3')
i'^^^
Hhh
= a
+
b
+
c
+
d
=
a
+
6a
+
5a
+
2a
=
14a
*'
d Cling dieu ki#n, ti
I9 ve so
mol b^ng ti 1§
vl

the
tich
%VC3H.=
^xI00%
=
7.14%
M
•
•
%VC4HK=
xl00%
=
42,85%
«A,
'#,1,1-11,,
.
%vc3H,o=-^xioo%=35,7i%•
•
%VC6H,2=—xl00%
=
14,28%
,
14a
Bai
13: Dot chdy hoan toan m(g) hdp cha't
hiJu
cd
A
chi Ihu diTdc
a(g)

CO2
va b(g)
H2O.
Biet
3a
=
1
lb va
7m
=
3(a
+
b).
Tim CTPT cua
A.
(biet
li
khoi hdi cua
A
d5i
vdi khong khi nho hdn 3).
'
•
'X', rr:,,.
J,,,,
•<
•
Gi&i
::;/V'!^,.V;-;,YA''"1^
Ta c6:

dA/khongkhi
<
3
=>
MA
<
3.29
=
87
, , „ ,
,^
, ,^
ta
c6:
mc=
12.nroT
=
12.—
=
—(g)
li; i
VI
3a
=
1
lb =>
mf.
=
=
b

(g)
: i 'J
'
>
'
mH
=
2.nH20=2.—=
-(g)
V17m
=
3(a
+
b)
=
3(—+ b)=
14b m
=
2b
=>
mo
=
mA
-
(mc
+
mH)
=
2b
-

Dat CTPT cua
A
Id CxHyO,;.
b
b 8b
T2'9'9.16
12'9
18
8b
, ,
m^H^%''^''''
_
Ai ^
=
—-i—=
3:4:2
=> X
:
y : z
= . .
1
o •
o ' 18
*
i
'
=>
Cong
thuTc
nguyen

ciia
A
la
(C3H402)n
, , j
Matkhdc:MA<87 72n<87
=>
n<1.21
Ma
n
nguyen
=5.
n
=
1
=^
CTPT cua
A
la:
C3H4O2
Phan
dgng
va
phuong
phap
giSi
H6a hpc 11 HOu co - B5 Xuan
Htfng
Bai 14. Dot
chay hdp chat

hiJU
cd
A
(chuTa
C, H, O)
phai dung
1
li/dng oxi bang
8 Ian
lUOng
oxi
c6
trong
A va thu
dUdc lu-dng
CO2 va H2O
theo
d le
khoi lUOng
mco2
: mHjo = 22
:
9
.
Biet
ti khoi hdi cua X so vdi
H2
la
29.
Tun

CTPT
cua X.
Gia
suf
chat
hffu
cd
A
chiy
sinh ra 22g
CO2
va
9g
HzO:'-''-*
*
'
,
=>mc=12.nco2
= 12.—=
6(g)
44
mH
=
2.nH20=2 j^=l(g)
Theo djnh luat hio tohn nguyen to la cO:
modrongA)
+ mophintfng =
mO(trongC02)
"'OltrongH2O)
,

:„ „,
^,,,,
,. v;,: = (22 - 6) + (9 - 1) = 24 (g) J^'i' ::-j-f,->V;!S5
Ma
mopha„.„g
= 8.mo, „gA, hay = i
mo(.,„gA,
= ^ = ^ (g)
•"O
phaniJng' ' 'Ov -
•S'.'C.i.,

'4^&.:J'9
D$t
CTPT cua Ala QHyO,tac6: "r
12 • 1 "16 12 T 9.16 2'r6 \
.
^ Cong thiJc nguyen c,5a A la
(C3H.O)„
'
;,,eUV.r4v iM
«/.(Mf
Ma
M, = 29.2 = 58 => 58n = 58 :=> n = 1 ' , , . .
^^^^,^„
. ^
VayiCTPTcuaAlaCjHfiO
,
.^^^ ^
BAI

TAP TRAC NGHI|;M
cau
1. Mpt chat
heru
cd X khi d trang thai hdi c6 ti
khoi
hdi so vdi chat
hd^i
cd Y la
2.
Biet 2,2 gam chat Y c6 the
tich
b^ng the
tich
ciia
1,6 gam oxi trong ciang dieu
ki?n
nhi?t dp, ap
suat.
Vay ti
khoi
cua X doi vdi
CH4
la:
A.
11 B. 10 . C.6 D.5,5
Cau
2. Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chdt khi a
d'lhu
kien

thuong) r6i dem toan bp san pham chay hip thy h^t vao binh dung dung djch
Ba(0H)2.
Sau cac phan
irng
thu dugc 39,4 gam k^t
tiia
va
khoi
lugng phin dung
djch
giam bat 19,912 gam. Cong thirc phan
tCr
cua X la
A.C3H4.
B. CH4.
'
^:
^
'
C. C^,:f'^mrt
Q^^^.
'nm'i
i,,
' "
Trich
di thi tuyin sinh Dai hgc khoi A nam 2012"
cau
3.
Nicotin
cd trong thuoc la la mot hdp chat rat doc, c6 the gay ung

thiT
phoi.
Dot
chdy 16,2 gam
nicotin
b^ng oxi v^a dii thu difdc 44 gam
CO2,
12,6 gam
nifdc
va 2,24 lit
N2
(dktc).
Cho 85 <
M„i,.,ii„
< 230. Cong thiJc phan
tuT
dung cua
nicotin
1^:
A.C5H7NO
B.C5H7NO2
C.C,OHMN2 D.CH,H,4N3
'
Cfiu
4. Do't chdy hoan toan 5,2 gam hdp chat huTu cd X roi cho san pham chdy Ian
lifdt
qua binh (1) di/ng
H2SO4
dam dac, binh (2) chu"a nifdc voi trong diT, thay
blnh

(1) tang ISn 1,8 gam, binh (2) thu diTdc 15 gam ket tua. Khi hoa hdi 10,4
gam A thu di/dc mot the
tich
dung bang the
tich
cua 3,2 gam oxi trong cung dieu
kien
nhiet dp dp
suat.
V$yc6ng thu'c phan tuTcua A
1^:
'i
'i/i
• i
•*
A. C5H12O2 i B.
C3H4O4
C.
C7H4O6
D.
C;HS03
cau
5.
Khi d^t chdy hoan toan 100ml hdi cha't B can 250ml oxi, tao ra 200ml
CO2
va 200ml hdi niTdc (cac the
tich
do d cung dieu ki?n). Vay cong thu'c phan lij
cua B la:
A.

C2H4O2
B.
C3H4O4
' • C.
C2H4O
D.
C3H6O
cau
6. Do't chay hoan toSn 3,5 gam mpt hidrocacbon thu difdc 10,68 gam khi
cacbonic va 5,25 gam
niTdc.
Khoi
liTdng
oxi can
diing
de dot chdy la:
A.
6,21gam,; M
B.
11,04 gam C. 12,43 gam D. 12,73 gam
cau
7. Dot chdy hoan toan
0,366
gam chat hiJu cd X, thu du-pc
0,792
gam
CO2
v^
0,234 gam
H2O.

Mat khdc phan buy
0,549
gam X do thu difdc 37,42cm^
nitd
(do
d
27°C
va 750 mmHg), biet trong phan tuf cua X chi chiJa
1
nguyen
tijf
nitd.
Vay
cong thu'c phan tuf cua X la:
A.C9H13O3N
B.C9H15O2N
C.CyHi303N2
D.
CKH12O3N
cau
8.
Dot chay 400 ml mpt hon hdp gom
nitd
va mpt hidrocacbon Y 3 the khi
bing
900 ml khi oxi
(dif).
The
tich
hon hdp thu dUdc sau khi dot la 1,4

lit.
Sau
khi
ngu-ng tu hdi niTdc thi the
tich
hon hdp con 800 ml, tiep tuc dan qua dung
dich
KOH thay con 400 ml
khi.
Biet rkng cac khi do d
cilng
dieu ki?n nhi?t dp,
dp
suat.
Vay cong thu'c cua Y Id:
A.
C2H6
B.
C3H4
riici
C.
C2H4
D.
C4HX
Cau9.Sod6ngphancuaC5H,2la:
, . •
A.
3 B.4 C.5 D.7
cau
10. Do't chdy hodn toan mpt

liTdng
hidrocacbon X. Hap thu toan bp san
pham chay vao dung dich Ba(0H)2 (dU) tao ra 29,55 gam ket tua, dung djch
sau phan iJng c6
khoi
liTdng
giam 19,35 gam so vdi dung dich Ba(0H)2 ban
dau. Cong thiJc phan
tur
cua X Id '
A.C3H4.
B.C2H6.
'^^-'''^^^
C.C3H6:'^'-^^'^^^^
^
'
-
''^iOslli
* „ y^^^ ^^^^ ^.^^
f.^.^^
2070"
cau
11. Dot chdy hodn todn 0,9 gam hdp cha't hiJu cd A chuTa C,H,0 thu diTdc 1,32
gam
CO2
vd 0,54 gam
H2O.
Ti
khoi
hdi cua A so vdi hidro la 90. Vay A cd cong

thu'c phan tuf
Id:
A.CeHnOft
B.
C,oH,203
C.
C»H2o04 jin
• , D.
CTHifiO,
Phan
dgng va phuong ph^p
giai
H6a hpc
11
HQu cd
-
D8
Xufln
Hung
Cfiu
12. Khi dot cMy 1 lit khi X can 5 lit khi
oxi,
sau phan
tfng
thu dUdc 3 lit
CO2
va
4 lit hdi
nifdc.
Biet

cdc khi diTdc do d cdng dieu
ki#n
ve
nhi^t
dp va dp
suS't.
V§y
cong thiJc phan
tuT
cua X la:
A.
C3H6
B.
CjHx
'
C.
CJHKO
D. C3H6O2
CSu 13. Trong mot
binh
kin chiJa hdi chat
hOu
cd X (c6 dang
C„H2n02)
mach hcl
O2 (s6
mol
O2
gap doi so mol can cho phan uTng chdy) 6 139,9"C, dp
suat

trong
binh
la 0,8 atm. Dot chdy hoan toan X sau do diTa ve nhi?t dp ban dau, ap
suat
trong
binh
luc nay la 0,95 atm. X cd cong thuTc phan
tijf
la:
A.C2H4O2. B.CH2O2. C.C4H8O2. D.C3H6O2.
(Trich
de thi tuyen sink dai hoc khoi B)
CSu 14. Dot chay hoan toan 100 ml hdi chat A, can
diing
250 ml oxi, chi tao ra 200
ml CO2
va 200 ml hdi nifdc (cac the
tich
khi do d cOng dieu
ki?n).
Xic
dinh
cong thtfc phan tur
ciia
A. ,^ , , ^
A.C2H4 B.C2H6O C.C2H4O
D.CjHfiO
CSu 15. Dot chay hoan toan 1
lit
khi X can 5 h't khi

oxi,
sau phan
uTng
thu diTdc 3
lit CO2
v^
4 lit hdi nufdc.
Biet
cac khi dUOc do d
ciing
dieu ki?n ve
nhi^t
dp va ap
suat.
Cong
thifc
phan
tiy
cua X la: f .
A.C3H6
B.C3H8
,
C.C3HSO
D.CjHfiOz^"'"'
Cau 16. Co 3 chat
hOU
cd
A, B,
C ma phan
t&

ciia
chijng
I§p
th^
1
cap so cpng. Bat
ci?
chat
n^o khi chay cung chi tao
CO2
v^
H2O,
trong do n^o^:
n^^^o
=2:3.
CTPT
ciia
A, B,
ClanlUdtla:
, , .
A. C2H4, C2H4O, C2H4O2.
I i/
:>
B.
C2H4.
CzHfiO.
CjHfiOj.
\„J'
c.
C3HS, C3HS0, C3HXO2.

?! r,
D. CZH^.
CZHSO,
C2H6O2.
'I'^f .,
Hi'SUMi
HUdNGDlNGlArxa
cau 1. Ta c6: Mx = 2My ^,.
,„
.
•
A,!
:Jii:jifiuv,
rMq -^nhb o^ uk')
MY=—,32
= 44
=>Mx=
2.44 = 88 u
' ^'''''v'
1.6 , ' ' -
=>
ChpnD. • :<
Cfiu 2 : So mol BaCOs = 0,2 mol so mol COj = 0,2 mol ->
khdi
liTdng
= 8,8 gam
Khoi
liA;Jng
dung dich giam =
khoi

liMng
kel tua -
(CO2
+
H2O)
, 0 ; •
Tdngkhoi
ItfdngCO2
+
H2O
= 19,488
=>44.nco2
+
18.nH20=
19,488 (1)
Dot
X thu diidc
CO2
va
H2O. sfjri ^irf;,,
rp
^
.jsiLiJ*^
.OO'nik^ ' '
Ap
dung
DLBTKL
ta c6 : mx + moj = mcoj +
^^20
;4i

j!
rna^q
=> mo2
=
nico2
+
^»20
~'"x= 14,848 g =>
IXQ^
=0,464 mol
Vflij')
A
Bao toSn nguyen to
oxi:
2.
nco2
+
"H2O
~ 2.
no2
= 0,464 'f'
TO
(1).
(2)
nco2
= 0,348mol;
nH20
= 0'232 mol. ' "
=>nc:
nH

= 0,348:0,464 = 3
:4=>XmankinC3H4
r
-<
, .
Cflu3,ChonC.
,„,„„.••.
„: ,1,. ;T •
Nicotin
c6 cong thtfc cau tao
nhiT
sau:
Cfiu 4, Ta c6
phiMng
trinh
phan
iJng:
' ' : = =
.
c >'
CO2
+
Ca(OH)2
-> CaC03i + H2O ,
(,;00f
_mCaC03^
15 Oud
100 100
mH=7^-2 =
0,2(g)

mo =5.2-(0,2+ 1,8) =
3.2(g)
M,=iM:^
=
104
A
32
12x y 16z 104
1,8 0,2 3,2 5,2
Vay CTPT
la
C3H4O4
=>
Chon
B.
Cfiu
5.
Goi
CTPT
cua B
Ik
CxHyO,
x
= 3;y = 4;z = 4,
X
CxHyO,+
1
y z
y z
Oj-J-xCOs+lHjO

2
200
200
100 250
x
= 2; y = 4
Taco: 2,5 = x+| |=>z = l =>ChonC.
CSu
6. Ap dung
dinh
luat bao toSn
khdi
lifdng:
TOHC
+mo2 =mco2
+"1820
=^'"02
= (10,68 + 5,25)-3,5 = 12,43
=>
ChonC • - ' • "
Cfiu 7. Goi CT cua X
Ih :
CxHyO,N
•
"
''i
' ' '
750 37.42 ,,„v
,„,„*,vfc /;l
X;/.;Mfi^^^;:HV't•^.^vV^J.o;.;«

<
(
nx
= — =760 1000
=oooi5inol
7,fHr;'^i; s'ro fcf:?r'i!f-,.;. ,.•>
^
RT 0,082.300 ' '
=> Vi
khi phan
tich
0,366g X thi
nN,
=
"'""^^"'^^^
=
O.OOlmol
^
0,549
(2)
Phan
djing
phuang
ph^p
giSi
H6a
hpc
11 HOu
co
- D5

Xuan
Hiflig
^
m„ = 0,028(g) ; mc = 0,216(g) ;
ITIH
=
0,026(g); = 0,096(g)
0,216
0,026 0,096
0,028
,^ ^ ,
x:y:z:t=-^
:— :— :— =9 13 : 3 • 1
12 1 16 1
=>
Cong thiJc nguyen
(C9Hi303N)n.
*
Do
X CO mot nguyen
tu"
N n6n
CTPT
la C9H13O3N
=>
Chon A. -
Cau
8. Ta c6: V^^ +
Vco2
+ ^H^O +

VQ^UU
= 1-4
•
Khingufngtu:
VH2o=600ml
/
•
Din qua
KOH:
Vcoj
= 800 - 400 = 4()0ml . , , .
•
,
+Vo2du=400ml
. U
^
^ /« !.(, ,1 ' , A<,
VO,P/,=1VH,O+VCO,~
+400 = 700ml j^, ,
=>Vo2du=
200ml „
.H^•).!,^.^
=
200ml
'
The
tich
cua Y = 400 - 200 = 200ml
y
2 V 4,

O2
->xC02+-H2O
200 400 600 ' a' ^i( m
/WF^')^,.)
??.<»ir
=>x
= 2;y = 6=> Chon A.
CHu
9.
CH3-CH2-CH2-CH2-CH3
CH3-CH-CH2-CH3
.
CH3-CH-CH3
CH3
, V . CH3
CH3
I
4
<
=>Chpn
A.
i,:;,,:v/;-!ri;,::;;;:y:
,^ '>
>
29 55 * ' i I'J I f
Cau
10. Taco: ng^coj == 0.15mol =>
nco2
= 0.15mol
1-, ,,

. , , /b um ' ill' 4, '1 \'At I i'* , u'l''
Matkhac:
mdjgii„=
mBaco3 -
(mco2
+
"IH20)
=>
mco2
+
mH20=
mBaC03 ~
"^Jg""^
=
29,55
- 19,35 = 10,2 g
^
^^^^^
^
Ma
mco2
=
0,15.44
= 6,6
g=i>
mH20=
3.6 g=>
nH20
= 0.2 mol - „
Ta thay: n^jo > "002 ^ hidrocacbon X la ankan va:

nx = n^jo -
nc02
= 0,2 - 0,15 = 0,05 mol
til
I
ii
so nguyen tuf C trong X = = — = 3 X la
C3H8
^ Dap an
D
v.
12x
CSu 11. Ta
CO,
dp dung cong thiJc:
niyf
mQ
=i>
x = 6; y = 12; z = 6 ^ CTPT
:
CftHijOg
Chon A. > .'1. ' ••"1 . J- .
Cau
12. Goi CTPT cua X la CxHyO,.
16z _ M
a
Ta c6:
C.HyO,
+
1

x
+
4 2.
5
x
= 3;y = 8.
O2
^xC02 +IH2O
X
y
2
3
4 •
,,^', ,••1'
"
Matkhac: x +
4
= 5=>3 + 2 = 5=>z = 0
•'('5
Vly
X
Id
C3H8
=>
Chon B.
/wlfe
1
i'»/iifi
r|v' '
if J

fit/HQ
Caul3.
C„H2„02 + ^^02
1
mol
2
3n-2
nC02 + nH20
mol n
mol n mol
Gid
suf c6
Imol
X tham gia phan iJng. Theo phifdng
trinh
phan
tfng
ta c6:
no2
bandlu
= ^^Y~ -2 = ;(3n - 2) mol
3n-2 ,
n02
phan
iJng
- 1 mOl
=>
Tong so mol cua X va
O2
luc dau la: 3n - 2 +

1
= (3n - 1) mol
Sau phan iJng chdy va d
139,9°C
thi
H2O
dang
d the hdi nen tong so mol khi va
hdi
sau phan iJng la:
Skhi
=
nco2
+
nH20
+"02
J"
= n + n + ^"^ ^ = (3,5n - 1) mol
Nhiet
dO
bmh tnfdc vd sau khong
doi,
gia suf thi
tich
binh khong
doi,
ta c6:
'sau
0,8
3n-l

0,95 3,5n-l
=>
n = 3
V|iy
CTPT cua X Id
C3H6O2
=> Dap an D.
Caul4.
Tac6sdd6sau:
C,HyO,
+
2,502
->
2CO2
+ 2H2O
Cdn ciJ vdo h0 so phan iJng vd dp dung dinh luat bao todn nguyen to, de
dang
c6
AldC2H40
Dap an
C.
45
Phan
djng \A
phudng
phip
giil
H6a
hpc
11

HiTu
co
- Pg
Xuan
Hung
CSu 15: Goi cong
thiJc
phan
ttif
cua
X
\k
:
C,H„0, (z cd the bkng
0)
QHyO,
+ (x +
^-^)02
->
XCO2
+ ^HzO
K-f)
^ f
=>
x =
3vay
= 8
y
4
2

Matkhac:
x+ = 5
=:>3
+ 2- - =
5=>z
= 0
V$y:XmC3H8=>
DapanB,
n
• ' ' v: ! •
Caul(».Tac6:-^
= -
nH20
>
"CO,
Y ;^
^.
=>
A,
B, C
deu no, deu
c6
cilng
so
nguyen
tuT
C
va H.
NhiT
vay

de
phan
tuT
khoi
cua chiing l|ip
th^nh
1 cap so
cgng thl chiing khdc nhau
\i so
nguyen
tiJ
oxi
trong
phan tuf.
.' ^'"'^
'^1
-''''f
''''
J*.'^
'''
DStcongthiJctongqudtcua
A,B,CiaC„H2„
+
20x
(x>0)
C„H2„
+ 20x
—'^^2_>
nCOz
+

(n+1)H20
,
Ujoi
"coi
n 2
n
T;:T = T^" = 2=> ^^"^ '^^"g
C2H6OX.;,;;
m fn, m>
"H20 n +
1
j
Vi
C2H6OX
la
hdp chaft
no
nen di^u ki^n
de
t6n tai
Ik so
nguydn
tiJ
oxi
< so
nguyen
turC
=> x £ 2 => x = 0;
1;
2. , , ,r

V|y:
CTPT
cua
3
cMt A,
B. C
lln IU*?t
^
C2H6.
C2H6O.
CjHfiOa. „ f,,,,,
da
HIDUOCACBON
NO
A.T6MTATLfTHUY§'T
I.
ANKAN
(parafin):
C„H2„+2
(n ^ 1)
*
CH4.
C2H6,
CjHg.C„H2„+21$P thanh day dong d^ng ciia ankan.
* TCf
C4
tr3 di cd dong phan cau tao.
j ^,
* Danhphdp:
^^^j.,.,,,

,.1^:).::.^}^:).
<
igii
+
Ankan khdng phan nhdnh
: ,
Ten ankan
=
ten mach chinh
+
an
. . ,,
+
Ankan phan nhdnh
:
T6n
ankan
= s6
chi vi tri nhdnh
+
ten nhdnh
+
ten mach chinh
+
an
*
Tinh
cha't hda hoc
:
phan

tfng
the', tdch,
0x1
hda.
' .
,1
'
+
Phan
itng
the':
CH4
+
CI2 CH3CI
+
HCi
->
phan
tfng
halogen hda.
j.^.j (
+
PhaniJngtich:
, \, d
1'ih
i^l'^r
CaH2a+2
+
ChH2b
. ,

H
f(
iS'.J)".'!-!:,.)
:
M.:i
sy?.
ankan thap anken
phan
iJng crackinh. ^
.r^, jy,^.^
,q^^.m
&MiH^\V
I.AOJ
i-'
r3n+n
CnH2n+2
rt
t"
CnH2n+2
xt
t"
(n>3)
+ Phan i?ng oxi hda:
C„H2n+2
+
O2
>
nC02
+
(n

+
1)H20
2
Khi
cd xdc tdc thich hdp:
CH4
+
O2
HCHO
+
H2O
* Ph5niJngdiluche ankan:
; i
jAfi
R-COONa
+
NaOH R-H
+
NajCOj
(R Ik
C„H2„+,-)
hO^C
CnH2„
+
H2
CnHznrf.
^ '
a
XICLOANKAN:
CoHto

(n ^ 3)
* Xicloankan
la
nhOhg hidrocacbon
no
mach v6ng (g6m mono xicloankan (ddn
v6ng)
poll
xicloankan (da v6ng)).
; . 1
* Danhphdp:
T6n
xicloankan
= s6
chi vi
tri
nhdnh
+
ten nhdnh
+
xiclo
+
ten mach chinh
+
an:
Chiiy:
- Mach chinh
la
mach v6ng.
-

Ddnh
s^
sao cho
tdng
c&c
s6
chi vi
tri
cdc mach nhdnh
Ik
nh6 nhSt.
*
Tinh
ch^t hda hoc
: ^
41
Phan
dgng
vS
phuong ph^p
gl5i
H6a
hgc
11
HOu co-
B6
XuSn
Hung
+
Phan

u-ng the : , , » , ^
Xiclopentan
brom xiclopentan
+
Phan
u-ng
cpng
md vong :
THIYUHF
ii
'(M f4C: .fi
/\
Br2(dd)
>
CH2-CH2-CH2
i;y>:f«»Bt«tJ
1,3-dibrompropan
A
+ HBr '-^
CH3-CH2-CH2-Br
'
•MJ^^q.::«fla
^
,:
1-brompropan
I
I ' '
into
a
ninfjiit n&T

+
H2 ^
CH3-CH2-CH2-CH3
Xiclobutan
rfix-fU
1 f butan i {
iin
Au
>
Mi^ii)/i.
-0 i^n
fl'Vi
+
Phan
u-ng tdch :
/
r!
ur
,#4*-fi
»'{.'ii*
i^fi
i
Sihi\
^^
,
JIMI
T
I
t
Metylxiclohexan

metylbenzen (toluen) ri
<^
>
i^ir
"i ri\ +
+
Phan
u-ng oxi h6a :
C„H2„
+ —Og
nCO^
+
nH20
* V'
*
Dieuche:
CH3-(CH2)4-CH3
fj +H2. H*^. : ;
B. PHAN LOAI
VA
PHOONG
PHAP
GIAI CAC DANG
BAI TAP ' '
<ViA
eAng.
thi'te
emi t^a im gnl llii
atihati^
adehmnkati , ,«

BAI
TAP
MAU VA BAI
TAP
NANG CAO "^^^^
IS''*^
Bai
1.
Viet
cong thiJc phan tuf cua ankan va goc hidrocacbon
tifcfng
i?ng •
.
^)Ch^^lOH
b) Ch,fa8C
,.,H ,&:>':«^^
•
c)
Chifa
n nguyen tur C ' d) ChiJa (x + 1) nguydn
t)!^C.
"'"2^
C4HH,
=>
C4H9-
b)
QH,8
=>
QHn-*'
B^infioiiia.

*
'
:
C)
C„H2„.2
^
C„H2„.,-
: . d)
Q„H2x^=^ Q„H2x.3 '
^^^1''
'
a)
Viet
cong tMc phan
tijf
cua hidrocacbon
tifdng
iJng vdi cic goc
ankyl
sauf '^^^^
-CH3,
-C3H7, -C6H13.
b)
Viet
c6ng
thuTc
cau tao cua c^c ankan sau : pentan, 2-metylbutan, isobutan. Cic chat
tren
c6 ten goi nao khdc khong ?
KlfAMfiVlET

a)
-CH3:
hidrocacbon :
CFL,
•
5^;^,;
.^i:/:?^^
y,^;.;
;^eiVi:l:i
v
/Ths,
-CsHv
: hidrocacbon :
C3HS vjj^;;;)/
-C6H13
: hidrocacbon :
C6H14. v,/;,, •,,gjs#,:Sil'k:
v^ukn
itds
.• a, .,5
b)
:.il;:>
,fiHf.:;>,.,AH;:>
,.a
^Ten goi khdc
.^^.J;,.
CH3-CH2-CH2-CH2-CH3
pentan n-pentan
HJ.M
!

CH3-CH-CH2-CH3
2-metylbutan isopentan , : ;
I
2-metylpropan.
CH3
CH3-CH-CH3
, isobutan
CH3
Bai
3.
a) 0ng vdi propan c6 hai nh6m
ankyl
la
propyl
va isopropyl. Hay
viet
cong thiJc cau
tao cua
chiing
va cho biet b|c cua nguyen tuf cacbon mang hoa tri tif do.
b)
Hay
vie't
c6ng thiJc cau tao thu gon va thu gon nhat cua cac chat sau : ,
*
isopentan, neopentan, hexan.
*
2,3-dimetylbutan, 3-etyl-2-metylheptan, 3,3-dietylpentan.
Gidi
Propan

:C3Hs
a)
Propyl:
CH3-CH2-CH2-
(bac 1) '
Isopropyl:
CH3-CH-
(bac 2)
,
011,11, .K rfl 'j( ' \ii \
b)
Isopentan :
Neopentan :
CH3
Cong
thtfc cau tao
CH3-CH-CH2-CH3
CTCT
thu gon nh^t
I
CH3
CH3-C-CH3
I
CH,
Hexan:
2,3-dimetylbutan
:
CH3-CH2-CH2-CH2-CH2-CHsi ,
;
CH3-CH-CH-CH3

,'
I
CH,
CH3
3-etyl-2-metylheptan:
CH3-CH-CH-CH2-CH2-CH2-CH3
CH3
C2H5
• •
•j;v'-
3,3-dietylpentan :
C2H5
CH3-CH2-C-CH2-CH3
C2H5
49