Combustion
C' ✗ Hy +
¥2 →
CÉz +
H2O
I *
or
Cx Hy Oz E. z + →
CO2 H2O +
•
In a combustion analysis all the
original C gets trapped in coz and all of the
H in H2O
An of Oz in used compound needs found
•
excess
gas so any 0 from the to be
by difference ( ie : cannot be found directly from the CO2 or Hzo) or with moissperent
data
Any compound need to be found difference
•
other elements found in the
by Law of
using
conservation of Matter or with mass percent data
•
Combustion analysis can only determine mass percent values and empirical formula 's
Law Of Conservation Of Mass : Combustion of 247.2
ex .
analysis a
mg
Cxlty :
sample of an unknown compound
100% t.ctt.tl
yielded 440.6 of and
180.4mg
=
mg CO2
gcxlty =
gc +
gH of -120 This compound
. was known to contain
only C, H and 0 Determine the empirical formula
. .
Step 1 .
1g -01000mg
1g 1m01 CO2 1m01 C
12.019C
( 440.6mg CO2 Step 4-
0.12029C
: ✗ × × ✗
grams
=
1000mg 1m01 coz lmolc
44.01g coz
MOI C
1.498 Moto
,ooYmg É
0×-2=12
=
Ym°%t¥ 0.02019g
'
Hgrams 180.4mg H2O H
: ✗ ✗
✗ =
✗
Moto -2=2*6--3 MOI
18.016g H2O
MOIH 2-99
( ✗ Hy Oz
grams
:
247.2mg ( ✗ Hy Oz ✗ 19 =
0.24729 C. ✗ Hyoz =
moto ✗ 2--6 =D
1000mg Empirical
Formula :
step 2 .
Step -3
1m01 C C , Hoo,
g. C×HyOz= gc
+
gH +
go 0.1202g C ✗ =
0.01000 mac
12.01g ,
90 =
0.2472g Cxltyoz
-
(0.1202g Ct 0.02019g H)
1m01 It
90
=
0.1068g 0 0.02019g Hx = 0.02 002m01 H
1.
oogg µ
0 ✗ 1m01 0
0.1068g = 0-006675 moi O
16 .
00g 0
, BALANCING
matter is neither created nor destroyed
CH 4+02 →
CO2 + H2O
Bhai
⑧•*•o¥o%⑥• + → →
§
Bea
+
•OB•
unbalanced chemical equation : CGH , 206+02 → coz H2O +
balanced chemical equation :
CGH , 206 +602 → GCOE 6h20
-
adding coefficients for equal atoms
on both sides
AIKO-7.01-5
* distinct form of element in the same physical State
Diamonds and graphite are alltropes of carbon and
differ in bonding S Structure
MASS PERCEN
-
percent by mass of element in compound
( element)
Mass of element ✗ in 1m01 of compound
mass percent = ✗ 100%
Mass of 1m01 of compound
( compound)
Mass of the compound
mass percent = ✗ 100
mass of the mixture
C' ✗ Hy +
¥2 →
CÉz +
H2O
I *
or
Cx Hy Oz E. z + →
CO2 H2O +
•
In a combustion analysis all the
original C gets trapped in coz and all of the
H in H2O
An of Oz in used compound needs found
•
excess
gas so any 0 from the to be
by difference ( ie : cannot be found directly from the CO2 or Hzo) or with moissperent
data
Any compound need to be found difference
•
other elements found in the
by Law of
using
conservation of Matter or with mass percent data
•
Combustion analysis can only determine mass percent values and empirical formula 's
Law Of Conservation Of Mass : Combustion of 247.2
ex .
analysis a
mg
Cxlty :
sample of an unknown compound
100% t.ctt.tl
yielded 440.6 of and
180.4mg
=
mg CO2
gcxlty =
gc +
gH of -120 This compound
. was known to contain
only C, H and 0 Determine the empirical formula
. .
Step 1 .
1g -01000mg
1g 1m01 CO2 1m01 C
12.019C
( 440.6mg CO2 Step 4-
0.12029C
: ✗ × × ✗
grams
=
1000mg 1m01 coz lmolc
44.01g coz
MOI C
1.498 Moto
,ooYmg É
0×-2=12
=
Ym°%t¥ 0.02019g
'
Hgrams 180.4mg H2O H
: ✗ ✗
✗ =
✗
Moto -2=2*6--3 MOI
18.016g H2O
MOIH 2-99
( ✗ Hy Oz
grams
:
247.2mg ( ✗ Hy Oz ✗ 19 =
0.24729 C. ✗ Hyoz =
moto ✗ 2--6 =D
1000mg Empirical
Formula :
step 2 .
Step -3
1m01 C C , Hoo,
g. C×HyOz= gc
+
gH +
go 0.1202g C ✗ =
0.01000 mac
12.01g ,
90 =
0.2472g Cxltyoz
-
(0.1202g Ct 0.02019g H)
1m01 It
90
=
0.1068g 0 0.02019g Hx = 0.02 002m01 H
1.
oogg µ
0 ✗ 1m01 0
0.1068g = 0-006675 moi O
16 .
00g 0
, BALANCING
matter is neither created nor destroyed
CH 4+02 →
CO2 + H2O
Bhai
⑧•*•o¥o%⑥• + → →
§
Bea
+
•OB•
unbalanced chemical equation : CGH , 206+02 → coz H2O +
balanced chemical equation :
CGH , 206 +602 → GCOE 6h20
-
adding coefficients for equal atoms
on both sides
AIKO-7.01-5
* distinct form of element in the same physical State
Diamonds and graphite are alltropes of carbon and
differ in bonding S Structure
MASS PERCEN
-
percent by mass of element in compound
( element)
Mass of element ✗ in 1m01 of compound
mass percent = ✗ 100%
Mass of 1m01 of compound
( compound)
Mass of the compound
mass percent = ✗ 100
mass of the mixture
