Using the information in chemical equations
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Expressing mass in chemistry
◦ Formula weight
◦ Molecular weight
◦ Molar mass
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Concept of the mole
Atomic mass %
Empirical formula (simplest formula)
Molecular formula (actual formula)
Combustion analysis
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Formula weight (FW)
◦ sum of masses of all atoms in the chemical formula
e.g. NaCl FW = 23.0 amu + 35.5 amu = 58.5 amu
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Molecular weight (MW)
◦ sum of masses of atoms in the molecular formula
e.g. H2O MW = 2×1.0 amu + 16.0 amu = 18.0 amu
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Molecular compounds have FW and MW
Non-molecular compounds have only FW
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Number of atoms in 12 g of
12C
◦ 1 mole = 6.022×1023
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Avogadro’s number (NA)
◦ 6.022x1023 mol-1 (note the units)
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Molar mass (M)
◦ mass of 1 mol of molecular formula
◦ e.g. MHCl = 36.5 g mol-1
NOTE: Molar mass (in g/mol) is numerically
equal to the FW (in amu)
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Mass Moles: use molar mass (n = m/M)
e.g. How many moles in 168 g of water?
1 mol
m
168 g
 9.32 mol
 168 g 
n 
-1
18.02 g
M 18.02 g mol
Moles Particles: use Avogadro’s number
e.g. How many molecules in 168 g of water?
1 mol
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23
-1
168 g 
 6.022  10 mol  5.61 10
18.02 g
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How many moles of nitrate present when 5.00 g
of aluminum nitrate dissolved in water?
Aluminum Nitrate = Al(NO3)3 (or AlN3O9)
Molar Mass = MAl + 3MN + 9MO
= (26.982) + 3×(14.0067) + 9×(15.9994)
= 212.997 g/mol
Moles of Al(NO3)3 = (5.00 g)/(212.997 g/mol)
= 0.02347 mol
Moles NO3- = 3 × Moles Al(NO3)3 = 0.0704 mol
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Empirical formula: Relative numbers of
atoms in a substance
◦ can be obtained from elemental analysis data
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Elemental analysis data
◦ relative masses of elements in compound
◦ can convert to relative numbers of atoms using the
atomic masses
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A compound contains 75.9% carbon, 17.7%
nitrogen, and 6.4% hydrogen by mass.
What is the empirical formula?
Step 1: Assume 100 g of sample.
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100 g contains
◦ 75.9 g of C
◦ 17.7 g of N
◦ 6.4 g of H.
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Step 2: Convert the masses of atoms in the
sample to moles, using the molar masses of
the atoms
1 mol C
75.9 g C 
 6.32 mol C
12.011g C
1 mol N
17.7 g N 
 1.26 mol N
14.0067 g N
1 mol H
6.4 g H 
 6.3 mol H
1.0079 g H
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Step 3: Calculate the mole ratio
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Divide moles from step 2 by the smallest
value
6.32 6.3 1.26
C:H :N 
:
:
 5.01: 5.0 :1
1.26 1.26 1.26
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The empirical formula is C5H5N
Note:
◦ 5.01 is not exactly 5
◦ Remember:
 It is a measurement
 Measurements always have errors
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A compound contains C (63.8%), H (6.4%) and
N (29.8%). What is the empirical formula?
In 100 g of sample there are:
 63.8 g C(1 mol/12.011 g) = 5.31 moles C
 6.4 g H(1 mol/1.0079 g) = 6.3 moles H
 29.8 g N(1 mol/14.007 g) = 2.13 moles N
The mole ratio is therefore:
C : H : N = 2.49 : 3.0 : 1
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2.49, is not a whole number ratio of N
◦ In between 2 & 3
◦ Not likely experimental error
◦ Close to 2.5, or 5/2
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Multiply the ratio by 2 to get whole numbers
2x(C:H:N)=5:6:2
The empirical formula is C5H6N2
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Mass %
Analysis
Data
Report result
(empirical
formula)
Divide by lowest
number of moles
to get mole ratio
Work out moles
of atoms in
100g of sample
Yes
Multiply by smallest
factor that will result
in whole-number ratio
Is
mole ratio
simple, whole
numbers
?
No
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Need one more thing: Molar mass of cmpd
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e.g. MW of compound is 136.15 g/mol.
Empirical formula is C4H4O. What is
molecular formula?
FW = (4x12.01)+(4x1.01)+16.00
= 68.08 g mol-1
About half the molar mass
Therefore, the molecular formula is C8H8O2
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Used to find empirical formulae of molecules
containing only C, H & O
◦ C converted to CO2
◦ H is converted to H2O
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CO2 and H2O trapped and weighed
◦ mass % of C and H are found from stoichiometry
◦ 1 C atom per CO2 trapped; 2 H atoms per H2O
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Remaining mass attributed to O
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Stoichiometric equation for combustion of a
compound of C, H and O is:
CXHYOZ + {X+Y/4-Z/2} O2  X CO2 +Y/2 H2O
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moles CO2 (RHS) = moles C atom (LHS)
moles H2O (RHS) = 1/2 moles H atom (LHS)
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1.663 g of a C-H-O compound is combusted,
producing 3.48 g of CO2 and 0.712 g of H2O.
What is the empirical formula of the compound?

1 mole of C atom produces 1 mole of CO2,
therefore,
1 mol C
1 mol CO 2

Amt. C = 3.48 g CO 2 
44.01 g CO 2 1 mol CO 2
= 0.0791 mol C
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1 mole of H2O produced implies 2 moles of H
in the compound
Amt H = 0.712 g H 2O 
2 mol H
1 mol H 2 O

18.02 g H 2 O 1 mol H 2 O
= 0.0790 mol H
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H and C are in a 1:1 ratio in the original cmpd
Combined mass of H and C is:
◦ (0.0791 mol)(12.01 g/mol) C +
(0.0790 mol)(1.008 g/mol) H = 1.03 g
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The remainder of the compound is O
Mass of O = Mass of Cmpd - Mass of C+H
= 1.663 g - 1.030 g
= 0.633 g
Moles of O = (0.633 g)/(16.00 g mol-1)
= 0.0396 mol
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The mole ratio, C:H:O is
0.0791 : 0.0790 : 0.0396
Or, dividing by 0.0396,
2.00 : 1.99 : 1
Therefore, the empirical formula is C2H2O
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Review Questions: 16-22.
Problems by Topic, Cumulative Problems,
Challenge Problems: 87, 89, 91, 95, 99, 107,
109, 113, 115, 117, 121, 125, 131, 139,
149.
Note: answers to all odd-numbered problems
are found in Appendix III.
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“Molar” interpretation of stoichiometric
coefficients
Stoichiometric ratios
Limiting reagents
Theoretical Yield
Percentage yield
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Please review § 4.2 (self-study) to remind
yourself how to write balanced chemical
equations.
The balanced chemical equation is key to
stoichiometry.
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Quantitative aspects of reactions
Relative moles of species in the reaction
Key: balanced chemical equation
From the equation can derive mole ratios
◦ Like “conversion factors” in dimensional analysis
◦ Convert reagent consumed into product formed
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Found from stoichiometric equation
Convert between equivalent molar amounts
of reactants and products in a particular
reaction
Treated like unit conversion factors
◦ e.g., in the previous example, we used
2 mol H
1 mol H 2 O
Top and bottom are equivalent
factor equivalent to 1
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Before constructing stoichiometric ratios:
◦ Ensure equation is balanced
Interpret values as moles of
e.g. for H2(g) + 1/2O2(g)  H2O(l)
1 mol H2  0.5 mol O2  1 mol H2O
Ratios:

0.5 mol O 2
1 mol H 2
1 mol H 2 O
0.5 mol O 2
1 mol H 2
1 mol H 2 O
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Students sometimes use stoichiometric ratios
with mass data, without converting to moles
e.g. Find mass O2 req’d to form 16.7 g Fe2O3
from Fe3O4?
2 Fe3O4(s) + 1/2 O2(g)  3 Fe2O3(s)
Balance First
Amateur error: Do not do this in a test!
0 .5 O 2
mass O2 = 16.7 g Fe2O3 
3 Fe 2 O 3
= 2.78 g O2
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Stoichiometric ratios are MOLAR ratios
Convert moles of one substance into moles of
another substance
0.5 mol O 2
1 mol Fe2 O3
moles O2 = 16.7g Fe2O3 

159.7 g Fe2 O3 3 mol Fe2O3
= 1.7410-2 mol O2
Convert to
Convert to
32.00 g O
mass O2 = 1.7410-2 mol moles
O2  Fe2O3 2 moles O2
1 mol O 2
= 0.557 g O2
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In chemistry, the fundamental unit of
“amount of a substance” is the mole
Stoichiometry shows quantitative
relationship between reactants and
products in a reaction
Stoichiometric info is summarised in the
balanced chemical equation
Data in mass units usually has to be
converted to moles before it is of ANY use
The conversion factor between mass and
moles is the molar mass (units: g mol-1)
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Limiting reagent (reactant) is completely
consumed in a reaction
◦ Other reagents are said to be in excess
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Limiting reagent limits amount of product
that can form
◦ Basis for “theoretical yield”
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What mass of H2SO4 is produced from the following
reaction if 21.0 g of SO2, 5.00 g of O2 and excess
H2O are available, and the reaction goes to
completion?
(MSO2 = 64.06; MH2SO4 = 98.08; MO2 = 32.00)
Rxn: 2 SO2 (g) + O2 (g) + 2 H2O (l)  2 H2SO4 (aq)
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Convert masses to moles:
◦ amt. SO2 = (21.0 g)/(64.06 g mol-1) = 0.328 mol
◦ amt. O2 = (5.00 g)/(32.00 g mol-1) = 0.156 mol
◦ amt. H2O = excess (given)
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Determine whether O2 or SO2 is the limiting
reagent from the stoichiometry.
(0.328 mol SO2)(1mol O2)/(2mol SO2)
= 0.164 mol O2
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Need 0.164 mol O2 to react with available SO2
◦ Have only 0.156 mol O2
◦ Therefore, O2 is limiting reagent
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How much H2SO4 if all the O2 is consumed?
(0.156 mol O2)×(2 mol H2SO4)/(1 mol O2)
= 0.312 mol H2SO4
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Convert to mass:
(0.312 mol H2SO4)×(98.07 g/mol) = 30.6 g H2SO4
This is the Theoretical Yield of H2SO4
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Convert masses to moles
Divide moles of each reagent by its
stoichiometric constant
Lowest ratio is limiting reagent.
e.g. In the preceding example,
◦ SO2: (0.328 mol)/(2 mol) = 0.164
◦ O2:
(0.156 mol)/(1 mol) = 0.156
(limiting reagent)
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Theoretical yield: Mass of product formed if
a reaction goes to completion (limiting
reagent completely consumed)
Percent yield: Ratio of actual mass of product
formed, divided by the theoretical yield,
expressed as a percentage.
N.B. To find the theoretical yield, you must
first determine the limiting reagent.
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4.21g of cyclohexene (C6H10) reacts in
presence of 10.0 g Br2 to form 12.1 g of
1,2-dibromocyclohexane (C6H10Br2) as:
C6H10 (l) + Br2 (l)  C6H10Br2 (l)
Find the theoretical and percent yields:
McHx=
82.14 g/mol
MBr2=
159.8 g/mol
McHxBr2 = 241.9 g/mol
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Step 1: Convert masses to moles
C6H10: (4.21 g)(1 mol/82.14 g) = 0.0512 mol
Br2: (10.0 g)(1 mol/159.8 g) = 0.0625 mol
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Step 2: Find the limiting reagent
C6H10 (both stoichiometric constants are 1)
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Step 3: Find theoretical yield of C6H10Br2
amount: (0.0512 mol C6H10)(1mol C6H10Br2/1mol
C6H10)
= 0.0512 mol C6H10Br2

mass: (0.0512 mol)(241.9 g/mol) = 12.4 g
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Step 4: Find % Yield
= [(12.1 g)/(12.4 g)]x100% = 97.6%
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Stoichiometry is the quantitative relationship
between the reactants and products
Think in MOLES!
◦ molar mass converts mass to moles and vice versa
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Limiting reagent determines theoretical yield
◦ Know how to determine the limiting reagent quickly
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Review Questions: 1-4, 16.
Problems by Topic, Cumulative Problems,
Challenge Problems: 25, 29, 33, 35, 37, 61,
65, 67, 71, 73, 75, 77, 79, 83, 85, 107, 109,
125.
Note: answers to all odd-numbered problems
are found in Appendix III.
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