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THE MOLE / COUNTING IN CHEMISTRY
***A mole is 6.022 x 1023 items.***
1 mole = 6.022 x 1023 items
1 mole = 602, 200, 000, 000, 000, 000, 000, 000 items
Analogy #1
1 dozen = 12 items
18 eggs = 1.5 dz.
- to convert from eggs from dozen, we need to multiply by conversion factor
1 dz
18 eggs  18 eggs 
 15
. dz
12 eggs
Example: How many eggs in 3.2 dozen?
12 eggs
3.2 dz  3.2 dz 
 38 eggs
1 dz
Example: How many dozen is 138 gears?
1 dz
138 gears  138 gears 
 115
. dzgears
12 gears
Analogy #2
1 gross = 144 items
Conversion factors are
1 gross
144 items
and
144 items
gross
Example: How many gross is 295 pencils?
Example: How many apples is 0.473 gross?
1 mole = 6.022 x 1023 items (usually ions, atoms or molecules)
- 6.022 x 1023 is called Avogadro’s number and is abbreviated NA.
6.022 x 1023 molecules = 1 molemolecule
12.044 x 1023 molecules = 2 molemolecule
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Example: How many moles of atoms is 7.43 x 1021 atoms?
- use conversion factor
1 mol
6.022 x 1023 items
or
6.022 x 1023 items
mol
1 mole
7.43 x 1021 atoms  7.43 x 1021 atoms 
6.022 x 1023 atoms
 0.0123 mol atoms
Example: How many moles of ions is 2.5 x 1025 ions?
Example: How many molecules are in 8.333 mol of molecules?
ATOMIC MASS AND MOLAR MASS
How much mass does a hydrogen atom have?
1 p+ = m(H)  1.00 amu
What is an amu?
1 amu = 1.66 x 10-24 g BY DEFINITION
Relationship between atomic mass and molar mass
- atomic mass – mass of one atom
- molar mass – mass of a mole of atoms
How much mass does a helium atom have?
2 p+ + 2 n0  4.00 amu
166
. x 1024 g
4.00 amu 
 6.64 x 10 24 g
amu
How much mass does a mole of helium atoms have?
 6.022 x1023 atoms 6.64 x1024 g 
 4.00g 
1mol  

  1mol  
  4.00g
mol
atom
 mol 


The fact that 1 He atom has 4.00 amu of mass and 1 mole of He atoms has 4.00
g of mass is not a coincidence. Definition of amu is made to ensure this
“coincidence”.
The value of 4.00 g/mol is call the molar mass of He.
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Molar mass of the elements are found on the periodic table.
Why aren’t molar masses on periodic table integers?
1.) Molar masses are an average of isotopes.
- most important reason
2.) Nuclear energy has mass.
3.) Mass of electrons is very small, but not zero.
Converting between mass and moles
- a molar mass is a conversion factor!
Example: How many moles of atoms are in 96.3 grams of carbon?
Example: How much mass in grams does 0.0840 moles of uranium have?
FORMULA MOLAR MASS
a.k.a. molecular mass, molecular weight, formula weight, etc…
Formula Mass (Weight) – sum of atomic masses in chemical formula
Calculating Formula Mass
Example: What is the molar mass of ethylene, C2H4?
2 C:
2 x 12.0115 g/mol =
24.0230 g/mol
4 H:
4 x 1.00794 g/mol =
+ 4.03176 g/mol
28.0548 g/mol
M(C2H4) = 28.0548 g/mol
Ethylene is used to ripen fresh fruit. It is also used to make milk jugs.
1 Ba:
2 N:
6 O:
Example: What is the molar mass of Ba(NO3)2?
1 x 137.33 g/mol =
137.33 g/mol
2 x 14.0067 g/mol =
28.0134 g/mol
6 x 15.9994 g/mol =
+ 95.9964 g/mol
261.34 g/mol
M(Ba(NO3)2) = 261.34 g/mol
Barium nitrate is used to color fireworks green.
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Converting between mass and moles
Example: How many moles are in 538 g of Ba(NO3)2
First calculate molar mass.
137.33 g/mol
2 x 14.0067 g/mol
4 x 15.9994 g/mol
261.34
g/mol
538g  538g 
1mol
 2.06 mol
261.34 g
Example: How many grams is 0.147 mol of NaCl?
M(NaCl) = 22.98977 g/mol + 35.453 g/mol = 58.443 g/mol
0.147 mol  0.147 mol 
58.443g
 8.59 g NaCl
1mol
Road salt is mined under the city of Detroit.
SCHEME: Converting mass to moles to number
Mass
(g)
M
Molar
Mass
Moles
(mol)
NA
Avogadro’s
Number
Number
(atoms or
molecules)
Example: How much mass in grams does 1.38 x 1022 molecules of CO2 have?
Example: How many F atoms are in 13.19 g of CaF2?
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MASS COMPOSITION OF A COMPOUND
- Mass composition tells us percentage of mass for each element in compound.
- Mass of molecule equals the sum of the masses of the atoms.
- Given a chemical formula, one is able to find the percent mass of each element.
Strategy to find percent mass of compound
1. Assume 1 mole of substance
2. Calculate the total mass of one mole of molecules, i. e., find the molar mass of
the compound.
3. Calculate the mass of a single element by multiplying number of atoms by atomic
weight
4. Divide the mass of single element by total mass and multiply by 100% to get
percent mass.
5. Repeat for all elements.
6. Adding all percentages should equal 100%.
Example: What the mass percentages of the elements in C3H6?
1. Assume 1 mol
2. M(C3H6) = 3 x 12.0115 g/mol + 6 x 1.00794 g/mol
= 42.0821 g/mol  42.0821 g
For % C
3. 3 x M(C) = 3 x 12.0115 g/mol = 36.0345 g/mol
= 36.0345 g/mol + 6.04764 g/mol

36.0345 g

6.04764 g
36.0345g
4. % C 
 100%  85.6290%
42.0821g
For % H
3. 6 x M(H) = 6 x 1.00794 g/mol = 6.04764 g/mol
4. % H 
6.04764 g
 100%  14.3710%
42.0821g
As a check
6. 85.6290% + 14.3710% = 100.0000%
C3H6 is propylene which is used to make polypropylene. Polypropylene is used to make wash bottles,
plastic sheet protectors, long underwear, rope and “Tupperware”.
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EMPIRICAL FORMULAS
- chemical formula with lowest possible ratio of atoms
Molecular Formula
C3H6
P4O10
SnCl2
C6H12O6
Empirical Formula
CH2
P2O5
SnCl2
CH2O
Calculating Empirical Formulas from Mass Percentages
Given: Percent Mass Composition
Find: Empirical Formula
Strategy:
1) Assume 100 g of matter.
2) Multiply 100 g by mass percent to find amount of each element.
3) Convert mass of each element to moles using molar mass.
4) Find whole number ratios by dividing each number of moles by lowest
number of moles.
5) Use these ratios to find empirical formula.
Example: Find the empirical formula for a compound with the following mass
percentages:
69.6 % O
30.4 % N
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Assume 100 g
2
m(O) = 100.0 g x 0.696 = 69.6 g
m(N) = 100.0 g x 0.304 = 30.4 g
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O:
N:
1 mol
 4.35 mol O
16.00 g
1 mol
30.4 g 
 2.17 mol N
14.01g
69.6 g 
4
molO 4.35 molO
2 molO

 2.00 
mol N 2.17 mol N
1mol N
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Empirical Formula is NO2
Nitrogen dioxide is a component of automotive exhaust.
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Example: Find the empirical formula for a compound with the following mass
percentages:
4.6 % H
40.9 % C
54.5 % O
1
2
3
4
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Empirical Formula is H4C3O3
H4C3O3 is the empirical formula for ascorbic acid, H8C6O6, the chemical name for vitamin C.
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THEORETICAL STOICHIOMETRY
- coefficients of balanced equations relate moles of reactants to moles of products
- Comparisons in chemistry must be done by comparing numbers of molecules to
each other, i. e., comparing number of moles of each substance.
Example: N2 (g) + 3 H2 (g)  2 NH3 (g)

+
- 1 mole of N2 is “stoichiometrically equivalent” to 2 moles of NH3.
- in other words, for every 1 mole of N2 reacted, 2 moles of NH3 are produced.
- 1 mol N 2  2 mol NH 3
- equivalence is only true for specific chemical reaction
- equivalence can be considered a conversion factor
1 mol N 2  2 mol NH 3 
1 mol N 2
2 mol NH 3
or
2 mol NH 3
1 mol N 2
- other equivalences are
1 mol N 2  3 mol H 2
3 mol H 2  2 mol NH 3
Example: a) What are all of the stoichiometric equivalences for the reaction
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g)?
2 mol C 2 H 2  5 mol O 2
1 mol C 2 H 2  1 mol H 2 O
5 mol O 2  2 mol H 2 O
1 mol C 2 H 2  2 mol CO 2
5 mol O 2  4 mol CO 2
2 mol CO 2  1 mol H 2 O
b) How many moles of carbon dioxide are formed when 5 moles of
acetylene (C2H2) is combusted?
5 mol C2 H 2 
2 mol CO2
1 mol C2 H 2
 10 mol CO2
c) How many moles of oxygen are needed to fully burn 29.8 moles of
acetylene (C2H2)?
Acetylene is a welder’s fuel.
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Example: For the reaction Pb(NO3)4 (aq) + 4 KCl (aq)  PbCl4 (s) + 4 KNO3 (aq),
How many moles of KCl are needed to form 15.6 moles of PbCl4?
PRACTICAL STOICHIOMETRY
- can’t measure moles directly in the “real” world.
- must measure amount of substance with grams.
***Cannot compare substances stoichiometrically by mass, must convert to moles.***
SCHEME:
Mass of
reactant
(g)
Mass of
product
(g)
M
Molar
Mass
Moles of
reactant
(mol)
M
Molar
Mass
Balanced
Equation
Moles of
product
(mol)
Example: For the reaction, NH3 (g) + HCl (g)  NH4Cl (s),
a) how much NH3 is needed to react with 92.3 g of HCl?
1) First convert grams of reactant to moles of reactant
92.3 g HCl 
1 mol HCl
  2.53 mol HCl
36.5 g HCl
2) Compare moles of one reactant to other reactant.
1 mol NH 3
2.53 mol HCl 
 2.53 mol NH 3
1 mol HCl
3) Convert moles of other reactant to grams.
17.0 g NH 3
2.53 mol NH 3 
  43.0 g NH 3
mol NH 3
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b) How much ammonium chloride is produced when 92.3 g of HCl is fully
reacted?
- Note with factor-label method, we can do problems all on one line.
92.3 g HCl 
. g NH 4Cl
1 mol HCl 1 mol NH 4Cl 535


 135 g NH 4Cl
36.5g HCl 1mol HCl mol NH 4Cl
The reaction of ammonia with hydrogen chloride gas is used to a “smokescreen”.
Example: For the reaction
4BaCO3 (s) + Y2(CO3)3 (s) + 6 CuCO3 (s) → 2 YBa2Cu3O7 (s) + 13 CO2 (g) + 3 O2 (g)
a) Calculate how many grams of CuCO3 is needed to fully react with 0.104 g of BaCO3,
b) Calculate how many grams of YBa2Cu3O7 is formed from 0.104 g of BaCO3 fully
reacting.
Yttrium barium copper oxide (YBCO) is a superconducting ceramic. It is superconducting below a
temperature of 95 K.
Example: For the reaction SiO2 (s) + 6 HF (aq)  H2SiF6 (aq) + 2 H2O (aq),
a) Calculate how many grams of HF is needed to fully react with 0.0140 g of
silicon dioxide,
b) How much H2SiF6 is made when 1.43 g of HF fully reacts?
Hydrofluoric acid, HF(aq), is used to “frost” glass, SiO 2.
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LIMITING REAGENTS
- Often starting materials are not available in proper stoichiometric proportions.
- Given “unbalanced” amounts of reactants, we would like to know how much product
can be produced.
Analogy: Bicycle Factory
The equation to make a bicycle is 2 wheels + 1 frame + 1 handlebar  1 bicycle
If the parts inventory is as follows: 240 wheels 150 frames
we ask ourselves
- What “reactant” limits production?
- How much product can be produced?
135 handlebars,
240 wheel + 150 frame + 135 handlebar  ?? bicycle
1 bicycle
 120 bicycles
2 wheels
1 bicycle
150 frames 
 150 bicycles
1 frame
1 bicycle
135 handlebars 
 135 bicycles
1 handlebars
240 wheels 
Limiting reactant: wheels
Production: 120 bicycles
In a limiting reactant problem, amounts of two (or more) reactant are given.
Calculate how much product is produced by each.
The reactant that yields the lowest amount of product is the limiting reactant.
***In limiting reagent problems, we need to compare moles to moles***
Example: For the reaction 2 SO2 (g) + O2 (g) + 2 H2O (l)  2 H2SO4 (aq), if 5.6 mol of
SO2, 4.8 mol of O2 and 6.0 mol of H2O are reacted together, how many moles of H2SO4
are produced?
1mol H 2SO4
For the SO2: 5.6 mol SO2 
 5.6 mol H 2SO4
1 mol SO2
For the O2:
4.8 mol O2 
2 mol H 2SO4
For the H2O: 6.0 mol H2O 
1mol O2
1mol H2SO4
1mol H2O
 9.6 mol H 2SO4
 6.0 mol H2SO4
SO2 is limiting reactant and therefore 5.6 moles of H2SO4 is produced.
Sulfur dioxide is a pollutant from burning coal that is a contributor to acid rain. Sulfur
dioxide is removed from air with calcium oxide.
SO2 (g) + CaO (s)  CaSO3 (s)
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Example: For the reaction Zn (s) + CuCl2 (aq)  ZnCl2 (aq) + Cu (s), what mass of
copper metal is produced from the reaction of 2.00 g of Zn and 2.00 g of
CuCl2?
Find limiting reactant by comparing moles to moles
Must convert mass to moles using molar mass as conversion factor.
Thus CuCl2 is the limiting reactant and the amount of copper produced is
REACTION YIELDS
- We have been calculating theoretical yields by assuming that the reactions proceed
perfectly.
- An actual chemical process is rarely perfect; therefore, the actual yield is always less
than the theoretical yield.
- We compare the actual yield to the theoretical yield by calculating percent yield.
Definition of percent yield
% yield 
actual yield
 100%
theoretical yield
Example: For the reaction Cr2O3 (s) + 2 Al (s)  2 Cr (s) + Al2O3 (s) 18.7 g of
Chromium (III) oxide reacts to form 10.8 g of chromium metal. What the
percent yield of this process?
Theoretical yield of chromium metal is
Thus the percent yield is
% yield 
10.8 g Cr
 100%  84.4%
12.8 g Cr
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BOND ENTHALPIES
When two atoms bond together, the chemical energy of the system decreases.
Consider an energy level diagram of before bonding and after bonding.
H
H
H–H
before
after
The energy of the bonded system is lower than the unbonded system.
The energy released when two unbonded atoms become bonded is called the bond
enthalpy.
Aside: Enthalpy is another word for heat.
The bond enthalpy increases as atoms are more strongly bonded together. As the
strength of the bond increases, the distance between the atoms decreases.
Bond enthalpies are an experimentally found quantity; i. e., we can’t predict bond
enthalpies from periodic table.
Bond Enthalpies and Chemical Changes
**All chemical changes involve the breaking and creation of bonds.**
If we can understand what bonds are breaking and what bonds are forming, then we can
use bond enthalpies to estimate the energy (technically, enthalpy) change of the reaction.
To break a bond, we input (add) the bond enthalpy.
When a bond is formed, the bond enthalpy is released (subtracted).
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Example: Given the table of bond enthalpies below, calculate the energy change, when
two molecules of hydrogen and one molecule of oxygen change into two
molecules of water.
H–H
+
O
H–H
–
–
O

O
O
H
H
H
TABLE OF BOND ENTHALPIES
Bond
C–H
C–C
C–O
C=C
CC
C=O
E (kJ/mol)
413
348
358
614
839
1072
Bond
H–H
N–N
N=N
NN
O–H
O=O
E (kJ/mol)
436
163
418
941
463
495
Breaking two H – H bonds means inputting 2 x 436 kJ/mol.
Breaking one O = O bond means inputting 495 kJ/mol.
Forming four O – H bonds means releasing 4 x 463 kJ/mol.
Overall the energy change is
H