CHAPTER THREE
STIOCHIOMETRY CALCULATIONS WITH CHEMICAL
FORMULAS AND EQUATIONS
Balanced chemical equation: is a chemical equation which has an equal
number of each element on each side of the chemical reaction
2 H2 + O2
Engin. Chemistry
Second 2013/2014
2 H2O
CO2 + 2 H2O (balanced)
CH4 + 2 O2
Indicating the states of Reactants and Products: We use the symbols (g), (l), (s), and (aq) for gas,
liquid, solid, and aqueous solution, respectively.
Some types of reactions: ‫بعض أنواع التفاعالت‬
1- Combination reactions )‫ (اإلتحاد‬: in which two or more reactants combine to form the
product(s). General Form: A + B → C
2- Decomposition reactions )‫ (التفكك‬In which the substance undergoes a reaction to produce two
or more other substances.
Sample Exercise 3.3
Write balanced equations for the following reactions:
(a) The combination reaction that occurs when Li metal and F2 gas react.
(b) The decomposition reaction that occurs when solid barium carbonate is heated (two products
form: a solid and a gas.)
(a)
Combination reaction
(b)
Decomposition
3- Combustion (burning) reaction: )‫ (اإلحتراق‬Reactions involving O2
Combustion of propane:
Combustion of methanol:
Formula and molecular weights
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Formula weight: is the sum of atomic weights of each atom in the chemical formula. For example:
formula weight of H2SO4 is 98 g/mol
Molecular weight: if the chemical formula is that of a molecule.
Molecular formula of glucose C6H12O6 is 180 g/mol
Percentage composition from formulas
% Element 
number of atoms of the element  At.Wt of the element  100%
Formula Weight of Compound
Practice Exercise:
Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
Solution:
Formula weight of Ca(NO3)2 = 164 g/mol
% N
%N 
214 100%  17.1%
164
2  14
 100% 17.1%
164
Avogadros Number and the Mole:
A mole: is the amount of matter that contains the atomic (or molecular) weight of a substance in
grams.
Avogadros Number (NA) = 6.02  1023 objects per mol.
1 mol of C = 6.02  1023 C atoms.
1 mol of H2O = 6.02  1023 molecules.
1 mol of NO3- = 6.02  1023 ions.
Practice Exercise 3.8:
How many oxygen atoms are in (a) 0.25 mol Ca(NO3)2 and (b) 1.50 mol of carbonate?
Solution:
a) No. of oxygen atoms in 0.25 mol Ca( NO3 ) 2
 0.25 mol Ca( NO3 ) 2 
6 moles of O
6.02  10 23 O atoms

 9.02  10 23 oxygen atoms
1 mol Ca( NO3 ) 2
1 mol of O
b) No. of oxygen atoms in 1.5 mol Na 2 CO 3
 1.5 mol Na 2 CO 3 
3 moles of O 6.02  10 23 O atoms

 2.71 10 24 oxygen atoms
1 mol Na 2 CO 3
1 mol of O
2
Interconverting moles and masses
The mole concept acts as a bridge between mass and the number of particles.
P.E 3.12
(a) How many nitric acid molecules in 4.20 g HNO3? (b) How many O atoms are in this sample?
Solution:
a ) No. of HNO3 molecules 
4.20 g HNO3 
1 mole HNO3 molecules
6.02  10 23 molecules

 4.01 10 22
63 gHNO3
1 mole HNO3 molecules
b) No. of O atoms 
4.20 g HNO3 
1 mole HNO3
3 mol O
6.02  10 23 O atoms


1.20  10 23.
63 gHNO3
1 mole HNO3
1 mole O
Empirical Formulas From Analyses
The general procedure for determining empirical formula:
Sample Exercise 3.13:
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H and 54.50% O by mass. What is the
empirical formula of ascorbic acid?
Solution:
Assume that we have 100 g sample.
Thus we have:
40.92 g C, 4.58 g H, and 54.50 g O
3
moles C  40.92 g C 
1 mol C
 3.407 mol C
12 g C
moles H  4.58 g H 
1 mol H
 4.54 mol H
1.008 g H
moles O  54.50 g O 
1 mol O
 3.406 mol O
16 g O
Molar ratio : C :
H
O
3.407 : 4.54 : 3.406
Divide each number by the smallest No.
C:
H:
O  CH 1.33O
Multiply each number in ( CH 1.33O ) by 3
 the empirical formula is C 3 H 4 O3
Molecular Formula from Empirical Formula
We can obtain the molecular formula of a compound from the empirical formula and molar mass of
the compound, using the relation:
Whole number multiple 
Molecular Weight
Empirical formula weight
In the sample exercise (3.13) the empirical formula of ascorbic acid = 88 amu
Experimentally, the molecular formula of ascorbic acid = 176 amu.
The molecular formula is 2 times the empirical formula.
176/88 = 2. Consequently, we multiply the subscripts in the empirical formula by 2 to obtain the
molecular formula.
Combustion analysis
When a compound containing C and H is completely combusted in plenty of O2.
The carbon is converted to CO2, and the hydrogen is converted to H2O. and the amounts of CO2,
H2O produced are determined by measuring the increase in mass of CO2 and H2O absorbers.
S.E 3.15: Isopropyl alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl
alcohol produces 0.561 g CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl
alcohol.
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Solution
The molar ratio in the compound is (C:H:O =0.0128:0.034:0.0043)
Divide by the smallest number that is: 0.0043
So, the empirical formula of isopropyl alcohol is C3 H8 O.
Quantitative information from balanced equations
2 H2 (g) +
2 molecules
2 moles
O2 (g)

1 molecule
1 mol
2 H2O (l)
2 molecules
2 moles
The stoichiometric equivalent quntities (2, 1, and 2) are used to convert between quantities of
reactants and products in a chemical reaction.
For example, consider the following reaction: the combustion of butane (C4H10), the fuel in
disposable general-purpose lighters H : ‫وقود أجهزة اإلشعال المنزلية‬
2 C4H10 (l) + 13 O2 (g) 
8 CO2 (g) + 10 H2O (g)
Let’s calculate the mass of CO2 produced when 1.00 g of C4H10 is burned.
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Solution
Moles of C 4 H 10 1.0 g C 4 H 10 
1 mole C 4 H 10
1.72  10  2 moles C 4 H 10
158.0 g C 4 H 10
Use the stoichiometric factor to calculate moles of CO2
Moles of CO2  1.72  10  2 moles C 4 H 10 
mass of CO2  6.88  10  2 mol CO2 
8 mol CO2
 6.88 10  2 mol CO2
2 mol C 4 H 10
44 g CO2
 3.03 g CO2
1 mol CO2
Another solution:Conversions in one step
mass of CO2  1.0 g C 4 H 10 
1 mole C 4 H 10
8 mol CO2
44 g CO2


 3.03 g CO2
158.0 g C 4 H 10 2 mol C 4 H 10 1 mol CO2
Sample Exercise 3.16
Determine how many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6:
C6H12O6(s) + 6 O2(g)
→
6 CO2(g) + 6 H2O(l)
Limiting Reactants
limiting reactant : The reactant that is completely consumed in a reaction.
It determines, or limits, the amount of product formed.
The other reactants are sometimes called excess reactants.
In the chemical reaction, the quantities of reactants consumed and products formed, are restricted by
the quantity ‫ محصورة بكمية‬of the limiting reactant.
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For example, the reaction: 2 H2(g) + O2(g)
2 H2O(g)
The limiting reactant: (H2) and 2 mol O2 (excess reactant) will remain and 10 mol H2O is the product.
Q 3.67)
Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component
elements:
2 NaN3 (s) →
2 Na (s) + 3 N2 (g)
(a) How many moles of N2 are produced by the decomposition of 1.50 mol of NaN3?
(b) How many grams of NaN3 are required to form 10.0 g of nitrogen gas?
(c) How many grams of NaN3 are required to produce 10.0 of nitrogen gas, about the size of an
automotive air bag, if the gas has a density of 1.25 g/L?
Solution:
3 moles N 2
3
a ) moles of N 2 produced  moles NaN 3 
 2.25 moles N 2
2
2 moles NaN 3
b) g NaN 3 required 10 g N 2 
2 moles NaN 3
65 g

15.5 g NaN 3
3 moles N 2
1 mol NaN 3
c ) density of gas 1.25 g / L
3
3
 12 inch   2.54 cm  
1 L  1.25 g

  
  
g N 2 gas V  density 10 ft  
 353.96 g
3 
1
ft
1
inch
1
L
1000
cm

 
 

1 mol N 2 2 mol NaN 3 65 g NaN 3
g NaN 3 required  353.96 g N 2 gas 


 547.8 g NaN 3
28 g N 2
3 molN 2
1 mol NaN 3
3
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Q 3.75)
Sodium hydroxide reacts with carbon dioxide as follows:
2NaOH (s) + CO2 (g) → Na2CO3 (s) + H2O (l)
Which is the limiting reactant when 1.85 mol NaOH and 1.00 mol CO2 are allowed to react? How
many moles of Na2CO3can be produced? How many moles of the excess reactant remain after the
completion of the reaction?
Solution:
2 NaOH (s) + CO2 (g)

Na2CO3 (s) +H2O(l)
Initially:
1.85 mol
1.0 mol
0
0
Change:
- 1.85
-0.925
0.925
0.925
Final quantities: 0
0.075
0.925
0.925
According to the chemical equation 1mol of CO 2 requires 2 moles of NaOH, but only 1.85 moles of
NaOH is present. So, NaOH is the limiting reactant.
Moles of Na2CO3 produced = 0.925 mol
Moles of CO2 remained in excess = 0.075 mol
Q 3.83)
Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method
is called the Claus process, which relies on the reaction:
8 H2S (g) + 4 O2 (g) → S8 (l) + 8 H2O (g)
Under optimal conditions ‫ وعند أفضل الظروف‬the Claus process gives 98% yield of
S8 from H2S. If you started with 30.0 grams of H2S and 50.0 grams of O2, how many grams of S8
would be produced, assuming 98% yield?
Solution:
Moles of H2S = 0.88 mol
Moles of O2 = 1.54 mol
8 H2S (g)
+
Initially
0.88 mol
Change
- 0.88
Final quantities
0
4 O2(g) 
1.54 mol
- 0.44
1.10
Mass of S8 obtained theoretic ally  0.88 mol H 2 S 
S8 (s)
0
0.11
0.11
+
8 H2O (l)
0
0.88
0.88
1 mol S8
256 g S 8

 28.16 g S 8
8 mol H 2 S 1 mol S8
actual yield
 100%  98%
Theoretical yield
Actual yield
 0.98  Actual yield  27.6 g S 8
28.16 g S 8
Selected problems: 3.11, 12, 19, 20, 23, 36-53, 63, 64, 68,75, 76, 88, 90 - 97
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