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CHAPTER 3
STOICHIOMETRY:
CALCULATIONS OF CHEMICAL
FORMULAS AND EQUATIONS
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CONTENTS
3.1 Chemical Equations
3.2 Chemical Reactivity
3.3 Atomic and Molecular Weights
3.4 The Mole
3.5 Empirical Formulas from Analyses
3.6 Quantitative Information from Balanced
Equation
3.7 Limiting Reactants
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Learning Outcomes Able to calculate the molecular weight of any
given compound
Able to calculate the amount of reactants orproducts in mass or mole with informationgiven.
To be able to identify the limiting reactant in
a given reaction Able to determine the empirical & molecular
formula of a compound from an analysis
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Introduction
Stoichiometry study of quantitativerelationship between chemical formulas
and chemical equations. The law of conservation of mass is
observed: total mass of all substances
present before the reaction (reactants) isthe same as the total mass after the
reaction (products).
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3.1 Chemical Equations
Describe chemical reactions.
Eg: 2H2 + O22H2O
This chemical equation shows that hydrogen
react with oxygen to form water.
H2 and O2 (written to the left of the arrow) are
the reactants.
H2O (written to the right of the arrow) is the
product.
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Cont: 3.1 Chemical Equations
There are 2 types of number in a chemical
equation:
i. Stoichiometric coefficient - number infront of a chemical formula.
ii. Subscripts number of atoms of eachelement present in the chemical formula.
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Cont: 3.1 Chemical Equations
E.g:
2H2O:
2 water molecules present.
H2O - there are 2H atoms in one
molecule of water.
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3.1.1 Balancing ChemicalEquations
Formulas of the substances must be
correctly written.
The number of atoms of each type ofelement must be the same on both sides
of the arrow.
Only stoichiometric coefficients may beadjusted. Subscripts in chemical formulas
must not be changed.
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Cont: 3.1.1 Balancing Chemical
Equations
The sum of charges of ions on the left
hand side of the arrow must be the same
as on the right side.
Balancing chemical equation requires
some trial and error.
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Cont: 3.1.1 Balancing Chemical
Equations
Eg. methane reacting with oxygen.
CH4 + O2 CO2 + H2O : this is unbalanced.
o Firstly, balance the carbon and hydrogenatom without considering oxygen.
CH4 + O2 CO2 + 2H2O
o Secondly, balance the oxygen atom.CH4 + 2O2 CO2 + 2H2O
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Cont: 3.1.1 Balancing Chemical
Equations
Finally, the physical state of each chemical in
chemical equation should be written down.
Use the symbols: (g) for gas
(l) for liquid
(s) for solid (aq) for aqueous (water) solution
Eg: CH4(g) + O2(g) CO2(g) + 2H2O(l)
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Exercise 3.1
Balance the following equations:
a) HNO3(aq) + Na2CO3 (s) NaNO3 (aq) + H2O (l)+ CO2(g)
b) HCl (aq) + Al(OH)3(aq) H2O(l) + AlCl3 (aq)
c) ZnS(s) + O2(g) ZnO(s) + SO2(g)
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3.2 Chemical Reactivity
3.2.1 Using the Periodic Table
The pattern of chemical reactivity of elements
varies systematically, using the Periodic Table. Eg. All alkali metal react with water as follows:
2M(s) + 2H2O(l) 2MOH(s) + H2(g)
M= alkali metal
The reaction becomes more vigorous as wemove down the Periodic Table, i.e. from Li toCs.
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3.2.2 Combustion Reaction in Air
Combustion reactions produce a flame.
Most of the combustion reactions
involve oxygen as a reactant. Complete combustion of hydrocarbon
compounds and compounds containing C,
H and O atoms produces CO2 and H2O.Eg: C3H8(g) + 5O2 (g) 3CO2 (g) + H2O(l)
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Cont: 3.2.2 Combustion Reaction in Air
Incomplete combustion occurs when there is
insufficient amount of O2 present.
In this case, CO is produced instead of CO2.
Eg: C3H8 (g) + 7/2 O2 (g) 3CO (g) + 4H2O (l)
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3.2.3 Combination andDecomposition Reactions
i. Combination reaction
Combination reaction: reaction in which
two or more substances combine to formone product.
Represented by: A + B CEg. 2 Mg(s) + O2(g) 2 MgO (s)
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Cont: 3.2.3 Combination and
Decomposition Reactions
ii. Decomposition reaction
Breakdown of a compound into two or
more components.
May be represented by: C A + B
Eg. CaCO3(s) CaO (s) + CO2(g)
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3.3 Atomic Weight and MolecularWeight
3.3.1 Relative Masses and Atomic Mass Unit
Atoms are too small to be weighed. Atomic mass unit (amu) is used in dealing with
these extremely small masses:
1 amu = 1.66054 10-24 g
1 g = 6.02214 1023 amu The atomic mass of any isotope is then
determined relative to 12C nuclide.
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3.3.2 Average Atomic Masses
Most elements occur in nature as a
mixture of isotopes.
For example, naturally occurring chlorineis 75.53 % 35Cl (atomic mass of 34.969
amu) and 24.47 % 37Cl (atomic mass
36.966 amu). The average of the masses of isotopes will
give the relative atomic masses.
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Cont: 3.3.2 Average Atomic
Masses
Average Atomic Mass ( atomic weight)
= [(isotope mass) (fractional isotope abundance)]
for all isotopes of the element. Eg:
Average atomic mass of chlorine:
= (34.969 amu)(0.7553) + (36.966amu)(0.2447)= 26.41 amu + 9.05 amu
= 35.46 amu
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3.3.3 Formula and MolecularWeight
Formula weights (FW) are the sum of atomic
weights for the atoms in the formula.
Eg Sulfuric Acid H2SO4
FW (H2SO4)
= 2AW (H) + AW (S) + 4AW (O)= [2(1.0)+ (32.0) + 4(16.0)] amu
= 98.0 amu
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Cont: 3.3.3 Formula and Molecular
Weight
Molecular weight (MW) is used in place of formula
weight if the chemical formula of a substance is its
molecular formula. E.g. MW of glucose (C6H12O6):
= 6AW (12.0 amu) + 12AW (1.0 amu) + 6AW (16.0
amu)= 72.0 amu + 12.0 amu + 96.0 amu
= 180.0 amu
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Cont: 3.3.3 Formula and Molecular
Weight
For ionic substances like NaCl which
exists as 3D array of ions, it is not a
molecule. Therefore we cannot write it asmolecular formula or molecular weight
(MW).
The formula weight of NaCl:
FW (NaCl) = 23.0 amu + 35.5 amu
= 58.5 amu
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3.3.4 Percentage Compositionfrom Formulas
The percent composition of a substance isthe percent by mass of each element in thesubstance.
The sum of all percentage of elements in asubstance must be 100 %.
% mass of an element:= mass of an element in substance 100
formula weight of substance
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Example 1
Calculate the percentage composition ofC12H22O11 (FW 342).
The percentage of a given element in a
compound:
= (atom of element) (AW)
100FW of compound
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Cont: Example 1
% C = 12 (12.0 amu) 100 = 42.1 %
342 amu
% H = 22 (1.0 amu) 100 = 6.4 %
342 amu
% O = 11 (16.0 amu) 100 = 51.5 %
342 amu
Total % = (42.1 + 6.4 + 51.5)% = 100 %
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3.4 The Mole
The unit for dealing with atoms, ions, and molecules is themole (mol).
A mole is defined as the amount of matter that contains the
same amount of particles (molecules, atoms, ions) as thenumber of atoms in exactly 12 g of12C.
In 12 g of12C, there are 6.022 1023 atoms which is calledthe Avogadros number.
1 mole = 6.022
10
23
atoms Thus a mole of water contains 6.022 1023 molecules of
water.
A mole of NaCl contains 6.022 1023 sodium ions (Na+)
and 6.022
1023
chloride ions (Cl-
).
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Example 2
How many oxygen atoms in 0.25 mol of Ca(NO3)2?
There are 6 O atoms in 1 molecule of Ca(NO3)2. 0.25 mol Ca(NO3)2 6.022 1023 molecule Ca(NO3)2
1 molCa(NO3)26 Oxygen atom
1 molecule Ca(NO3)2
= 9.0 1023 Oxygen atoms in 0.25 mole of Ca(NO3)2
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3.4.1 Molar Mass
Molar mass (MM) of a substances is themass (in grams) of 1 mole of that substance.
Unit g/mol (g mol-1
). The mass of a single atom of an element (in
amu) is numerically equal to the mass (ingrams) of 1 mol of atoms of that element.
E.g: One 12C atom weighs 12.0 amu
One mol 12C weighs 12.0 g
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Cont: 3.4.1 Molar Mass
E.g:
One 24Mg atom weighs 24.0 amu
One mol 24Mg weighs 24.0 g
The molar mass (in grams) of any substance
is always numerically equal to its formulaweight, FW (in amu).
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Cont: 3.4.1 Molar Mass
One H2O molecule weighs 18.0 amu
One H2O mol weighs 18.0 g
One NO3-
ion weighs 62.0 amu One mol NO3
- ion weighs 62.0 g
One Cl2 molecule weighs 70.90 amu (35.45 amu 2)
One mol Cl2 weighs 70.90 g
One BaCl2 formula weighs 208 amu
One mol BaCl2 weighs 208 g
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Formula Formula
weight
(amu)
Mass of 1
mol of
formulaunits (g)
Number and kind of
particles in 1 mol
N 14.0 14.0 6.022 1023
atoms
N2 28.0 28.0 6.022 1023
molecules
2 (6.022 1023
)atoms
BaCl2 208.2 208.2 6.022 1023
BaCl2 unit
6.022
10
23
Ba
+
ions2 (6.022 10
23) Cl
-ions
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3.4.2 Interconverting Masses,Moles, and Numbers of Particles
Use dimensional analysis
Units involve are:
mass : g
moles : mol
molar mass : g/mol
number of particles: 6.022 1023 mol-1
(Avogadros number)
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Cont: 3.4.2 Interconverting Masses,
Moles, and Numbers of Particles
a) To convert between gramsand moles,use molar mass.
mol g: mol x g / mol = gg mol: g g/mol = mol
b) To convert between molesandmolecules, use Avogadros number.
moles Avogadros number = molecules
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Cont: 3.4.2 Interconverting Masses,
Moles, and Numbers of Particles
use use Ions/Grams Moles Molecules/
Molar Avogadros AtomsMass Number
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Example 3
1. How many atoms in 1 mole N2 and O3
2. How many molecules in 1 mole N2 and O3
Answers
1. Atoms : 2(6.022 1023 atoms) N.
: 3(6.022 1023 atoms) O.2. Molecules: 6.022 1023 molecules N2.
: 6.022 1023 molecules O3
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Example 4
Ethanol C2H6O (C:12.01, H:1.01, O:16.00)
a) What is its molecular weight?
b) What is the mass of 1 mol of ethanol molecules?
c) Calculate the number of moles of ethanol in 1.00 g.
d) Calculate the number of molecules in 1.00 g of
ethanol.e) Calculate the percentage of carbon in one molecule
of ethanol.
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Example 4 (Answer)
a) Molecular Weight (MW) of ethanol, C2H6O:
b) Therefore 1 mol of C2H6O = 46.08 g
Hatom
amuHatomsCatom
amuCatoms
1
01.1
61
01.12
2
Oatom
amuOatom
1
00.161
amu08.46
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Example 4 (Answer)
c) You need a conversion factor that will change
1.00g of ethanol to moles.
Conversion factor is 1 mol ethanol
46.08 g ethanol
Moles ethanol = 1.00 g ethanol 1 mol ethanol
46.08 g ethanol
= 0.0217 mol ethanol
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Example 4 (Answer)
or use simple direct proportionality
46.08 g ethanol equals to 1 mol ethanol1.00 g ethanol equals to 1.00 g 1 mol ethanol
46.08 g
= 0.0217 mol ethanol
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Example 4 (Answer)
d) The conversion from:
grams moles molecules
46.08 g ethanol = 1 mol ethanol= 6.022 1023 molecules of ethanol
Molecule of ethanol in 1.00 g= 1.00 g ethanol 6.022 1023 molecules ethanol
46.08 g ethanol
= 1.31 1022 molecules ethanol
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Example 4 (Answer)
e) The percentage of element carbon, % C :
= (no of C atoms per molecule) (AW of C) 100MW of one C2H6O molecule
= (2 atom C/1 molecule C2
H6
O)(12.01 amu/1 atom C) 100
46.08 amu/1 molecule
= 52.13 %
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3.5 Empirical Formulas FromAnalyses
Empirical formula is the relative numberof atom in the molecule. Empirical formula
gives the ratioof atoms of each type ofcompound.
E.g: the empirical formula of C2H4 is CH2.
An empirical formula is determined fromexperimental percent composition data.
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Cont: 3.5 Empirical Formulas
From Analyses
a) If we have the mass % of the elements, it iseasier if we assume that we start with 100 g
of sample.
b) Translate the mass % as the no. of grams of
each element in 100 g of sample.
c) From these masses, the number of moles
can be calculated.
The subscripts in an empirical formula are
calculated as follows:
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Cont: 3.5 Empirical Formulas
From Analyses
d) Determine the simplest whole-number ratio of
atoms by dividing the number of moles of each
element by the smallest number of moles.
e) If the ratios are not whole numbers, multiply the
ratios by an integer that clears the denominators
of the fractions
e.g: 0.5 and 1.75 are expressed as fractions: 1/2and 7/4
Multiply by 4 1/2 4 and 7/4 4, the ratios are converted to whole
numbers 2 and 7, respectively.
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Example 5
Ascorbic acid contains 40.92 % C, 4.58 % H
and 54.50% O by mass. What is the empirical
formula of the ascorbic acid?
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Example 5 (Answer)
i. Assume we have 100 g of ascorbic acid. Wehave 40.92 g of C, 4.58 g H and 54.50 g of O.
ii. Calculate the number of moles:Moles C= (40.92 g C) 1 mol C = 3.407 mol
12.01 g C
Moles H= (4.58 g H) 1 mol H = 4.544 mol
1.008 g H
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Example 5 (Answer)
Moles O
= (54.50 g O) 1 mol O = 3.406 mol
16.0 g Oiii. Determine the simplest whole-number.
The smallest number of moles = 3.406
C = 3.407 =1
H = 4.544 =1.33
3.406 3.406
O = 3.406 = 1
3.406
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Example 5 (Answer)
The ratio of H is 1.33 which is equal to 4/3.
Multiply the simplest whole number by 3.
C : H : O = 3(1 : 1.33 : 1)
= 3 : 4 : 3
The whole-number ratio gives us the subscripts
for the empirical formula.Therefore, the empirical of ascorbic acid
is C3H
4O
3.
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3.5.1 Molecular Formula fromEmpirical Formula
Molecular formula is the actual ratio ofelements in the molecule.
The empirical formula may not be the molecular
formula. Eg The empirical formula of ascorbic acid is
C3H4O3 and its molecular formula is C6H8O6. To obtain the molecular formula from the
empirical formula, we need to know themolecular weight, MW.
The ratio of the molecular weight (MW) to
formula weight (FW) must be a whole number.
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Example 6
A compound contains only aluminium
and oxygen elements. Its elemental
composition is determined to be 53.0 %
aluminium and 47.0 % oxygen. The mass
of one mole of the compound is 102 g.
What is the molecular formula?
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Example 6 (Answer)
First determine the empirical formula.
Assume 100 g of compound:
Al = 53.0 g ; O = 47.0 g
Moles Al = (53.0 g Al) 1 mol Al = 1.96 mol Al
27.0 g Al
Moles O = (47.0 g O) 1 mol O = 2.94 mol O
16.0 g O
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Example 6 (Answer)
Divide by 1.96:
Al = 1.96 = 1.00O = 2.94 = 1.50
1.96 1.96
Al 1.00 O 1.50
Multiply by 2 to convert 1.50 into an integer.
The empirical formula is Al2O3
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Example 6 (Answer)
Calculate the mass of Al2O3 to be 102 g.
Thus the empirical and molecular formulaare identical, Al2O3.
MW = 1
FW
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Example 7
Given the empirical formula is C3H4. Fromexperiment, the molecular weight is 121 amu.
What is the molecular formula?Solution:
The formula weight of C3H4 = 40.0 amu
Thus MW = 121 = 3
FW 40.0
The molecular formula is C9H
12.
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3.5.2 Combustion Analysis
i. Empirical formulas are routinelydetermined by combustion analysis.
ii. Hydrocarbon compounds and compoundscontaining C, H and O are combusted inexcess oxygen to produce CO2 and H2O.
All the carbon is converted to CO2All the hydrogen is converted to H2O
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Cont: 3.5.2 Combustion Analysis
O2 sample CuO H2O CO2
furnace absorber absorber
(Mg(ClO4
)2
) NaOH
iii. The amount of CO2 produced can bemeasured by determining the mass
increase in the CO2 absorber. This givesthe amount of C originally present in thesample.(1 mol CO2 contains 1 mol C).
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Cont: 3.5.2 Combustion Analysis
iv. The amount of H2O produced can bemeasured by determining the mass
increase in the H2O absorber. This givesthe amount of H originally present in thesample. (1mol H2O contains 2 mol H).
v. The amount of O originally present in the
sample is given by the difference in theamount of sampleand the amount of Cand Haccounted for.
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Example 8
Isopropyl alcohol contains C, H and O.Combustion of 0.255 g of isopropyl alcohol
produces 0.561 g CO2 and 0.306 g H2O.
1. Calculate the quantities of C, H and O in the
sample.
2. The empirical formula of isopropyl alcohol.
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Example 8 (Answer)
a) No of grams of C and H in 0.561 g of CO2 and0.306 g of H2O.
CgCmol
Cg
COmol
Cmol
COg
COmolCOg 153.0
1
0.12
1
1
0.44
1561.0
22
2
2
HgHmol
Hg
OHmol
Hmol
OHg
OHmolOHg 034.0
1
0.1
1
2
0.18
1306.0
22
2
2
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Example 8 (Answer)
oxygen amount = 0.255 g - (0.153 g + 0.034 g
= 0.068 g O
Empirical Formula:
Calculate the moles of each element first:
moles C = 0.153 g C 1 mol C = 0.01275 moles C
12.0 g C
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Example 8 (Answer)
moles H = 0.034 g H 1 mol H = 0.034 moles H
1.0 g H
moles O = 0.068 g O 1 mol O = 0.00425 moles O
16.0 g O
Divide each no by the smallest no. ie 0.00425
The ratio of H : O : C
= 8 : 1 : 3
Therefore the empirical formula is C3H
8O
3 6 Q tit ti I f ti f
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3.6 Quantitative Information fromBalanced Equations
Objective: To calculate the mass of aparticular substance produced or used in a
chemical reaction.
A balance chemical equation gives the
number of molecules which react to formproduct(s).
C t 3 6 Q tit ti I f ti
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Cont: 3.6 Quantitative Information
from Balanced Equations
This can be interpreted as the ratio of number
of moles of reactant required to give the ratio
of number of moles of product.
e.g. consider the following balanced equation:
CH4(g) + 2O2(g) CO2(g) + 2H2O (l)
C t 3 6 Q tit ti I f ti
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Cont: 3.6 Quantitative Information
from Balanced Equations
On the atomic-molecular level, the equation
states:
1 molecule CH4 + 2 molecules O2 give 1molecule CO2 + 2 molecules H2O.
The numerical coefficients in a balanced
chemical equation can be interpreted both as
the relative numbers of molecules and as therelative numbers of moles.
C t 3 6 Q tit ti I f ti
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Cont: 3.6 Quantitative Information
from Balanced Equations
Thus:
1 mol CH4 + 2 mol O2 1 mol CO2 + 2 mol H2O
From the equation, we know that 2 mol of O2 is
required to react with 1 mol of CH4.
This can be represented as
1 mol CH4 2 mol O2
means stoichiometrically equivalent to.
C t 3 6 Q tit ti I f ti
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Cont: 3.6 Quantitative Information
from Balanced Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O (l)
Therefore, the stoichiometrically equivalentratio for the products are as follows:
1 mol CH4 2 mol H2O
1 mol CH4 1 mol CO2
2 mol O2 1 mol CO2
1 mol CH4 2 mol H2O
2 mol O2 2 mol H2O
C t 3 6 Q tit ti I f ti
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Cont: 3.6 Quantitative Information
from Balanced Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O (l)
These stoichiometric relations can be
converted to mass equivalences by
converting the mole of a substance to its
mass using molar mass.
HOW?
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Example 9
Combustion of butane C4H10 is given bythe equation:
2C4H10(l) + 13O2(g) 8CO2(g) + 10H2O (l)
Calculate the mass of CO2 that is producedin burning 1.00 g of C4H10.
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Example 9 (Answer)
Ensure that the equation is balanced!!!
2C4H10(l) + 13O2(g) 8CO2(g) + 10H2O (l)
Write the stoichiometrically equivalent ratio
between C4H10 and CO2.
2 mol C4H10 8 mol CO2
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Example 9 (Answer)
104
104
104104
0.58
100.1
HgC
HCmolHCgHCmoles
104
21072.1 HCmol
104
2
104
2
2 2
81072.1
HCmol
COmol
HCmolCOmoles
2
21088.6 COmol
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Example 9 (Answer)
Mass of CO2 in grams:
Thus the conversion sequence is:gram moles moles gramreactant reactant product product
producedCOg
COmol
COgCOmol
2
2
2
2
2
03.3
1
0.44
1088.6
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Exercise 3.2
Determine how many grams of HI are requiredto form 1.20 moles of H2 when HI reacts
according to the equation:HI(g) H2(g) + I2(g)
(1 mol HI = 127.91 g HI)
g984.306HImol1
HIg91.127x
Hmol1
HImol2x)Hmol20.1(
22
Answer: Balance the equation first
= 307 g
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3.7 Limiting Reactants
Limiting Reactant - The reactant in a chemicalreaction that limits the amount of product(s)that can be formed. The reaction will stop
when all of the limiting reactant is consumed. Excess Reactant - The reactant in a chemical
reaction that remains when a reaction stopswhen the limiting reactant is completelyconsumed. The excess reactant remainsbecause there is nothing with which it canreact.
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Cont: 3.7 Limiting Reactants
No matter how many tires there are, if there areonly 8 car bodies, then only 8 cars can be made.
Likewise with chemistry, if there is only a certainamount of one reactant available for a reaction,the reaction must stop when that reactant isconsumed whether or not the other reactant hasbeen used up.
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Example 10
A 2.00 g sample of ammonia is mixed with
4.00 g of oxygen. Which is the limitingreactant and how much excess reactant
remains after the reaction has stopped?
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Example 10 (Answer)
First, we need to create a balanced equation for
the reaction:
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Next we can use stoichiometry to calculate how
much product is produced by each reactant.
NOTE: It does not matter which product is
chosen, but the same product must be used forboth reactants so that the amounts can be
compared.
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Example 10 (Answer)
The reactant that produces the lesseramount of product: in this case theoxygen.
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Example 10 (Answer)
Next, to find the amount of excessreactant, we must calculate how much of
the non-limiting reactant (ammonia)actually did react with the limitingreactant (oxygen).
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Example 10 (Answer)
We're not finished yet though. 1.70 g is theamount of ammonia that reacted, not what isleft over. To find the amount of excess
reactant remaining, subtract the amount thatreacted from the amount in the originalsample.
2.00g NH3 (original sample) - 1.70g (reacted)
= 0.30g NH3 remaining
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3.7.1 Theoretical Yield
The quantity of product that is calculated to
form when all the limiting reactant reacts is
called the theoretical yield. The amount of product actually obtained in a
reaction is called the actual yield.
The percent yield of a reaction:Percent yield = actual yield 100
theoretical yield
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Exercise 3.3
Calculate the % yield of Al2(SO4) from the reaction:
2Al(OH)3(s) + 3H2SO4(aq) Al2(SO4)3(s) + 6H2O(l)
given that 205 g of Al(OH)3 react with 751 g of
H2SO4 to yield 252 g Al2(SO4)3.
Answer: % yield Al2(SO4) =56.0%
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END of CHAPTER 3
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END of CHAPTER 3