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Summary Stoichiometry and Combustion Reactions

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The document begins with an explanation of what goes on in a combustion reaction to gain a general understanding. The document then explains the law of conservation of mass and then provides step-by-step guided examples. The document highlights the differences between unbalanced and balanced chemic...

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  • August 31, 2023
  • 5
  • 2022/2023
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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

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