2. 4.1 The Mole Concept and Atoms
• Atoms are exceedingly small
– Unit of measurement for mass of an atom is
atomic mass unit (amu) – unit of measure for
the mass of atoms
• carbon-12 assigned the mass of exactly 12 amu
• 1 amu = 1.66 x 10-24
g
• Periodic table gives atomic weights in amu
3. • What is the atomic weight of one atom of
fluorine? Answer: 19.00 amu
• What would be the mass of this one atom
in grams?
• Chemists usually work with much larger
quantities
– It is more convenient to work with grams
than amu when using larger quantities
Mass of Atoms
4.1TheMoleConceptand
Atoms
atomF
Fg10156.3
Famu1
g101.661
atomF
Famu19.00 23-24 −
×
=
×
×
4. • A practical unit for defining a collection
of atoms is the mole
1 mole of atoms = 6.022 x 1023
atoms
• This is called Avogadro’s number
– This has provided the basis for the concept
of the mole
The Mole and Avogadro’s
Number
4.1TheMoleConceptand
Atoms
5. 4.1TheMoleConceptand
Atoms
The Mole
• To make this connection we must define
the mole as a counting unit
– The mole is abbreviated mol
• A mole is simply a unit that defines an
amount of something
– Dozen defines 12
– Gross defines 144
6. Fmol1
Fatom10022.6
Famu1
Fg1066.1
Fatom1
Famu00.19 2324
×
×
×
×
−
=19.00 g F/mol F or 19.00 g/mol F
4.1TheMoleConceptand
Atoms
Atomic Mass
• The atomic mass of one atom of an element
corresponds to:
– The average mass of a single atom in amu
– The mass of a mole of atoms in grams
– 1 atom of F is 19.00 amu 19.00 amu/atom F
– 1 mole of F is 19.00 g 19.00 g/mole F
7. 4.1TheMoleConceptand
Atoms
Molar Mass
• Molar mass - The mass in grams of 1 mole of
atoms
• What is the molar mass of carbon?
12.01 g/mol C
• This means counting out a mole of Carbon
atoms (i.e., 6.022 x 1023
) they would have a mass
of 12.01 g
• One mole of any element contains the same
number of atoms, 6.022 x 1023
, Avogadro’s number
8. 4.1TheMoleConceptand
Atoms
Calculating Atoms, Moles, and Mass
• We use the following conversion factors:
• Density converts grams – milliliters
• Atomic mass unit converts amu –
grams
• Avogadro’s number converts moles –
number of atoms
• Molar mass converts grams – moles
9. 4.1TheMoleConceptand
Atoms
Strategy for Calculations
• Map out a pattern for the required
conversion
• Given a number of grams and asked for
number of atoms
• Two conversions are required
• Convert grams to moles
1 mol S/32.06 g S OR 32.06 g S/1 mol S
• Convert moles to atoms
mol S x (6.022 x 1023
atoms S) / 1 mol S
10. 4.1TheMoleConceptand
Atoms
Practice Calculations
1. Calculate the number of atoms in 1.7
moles of boron.
2. Find the mass in grams of 2.5 mol Na
(sodium).
3. Calculate the number of atoms in 5.0 g
aluminum.
4. Calculate the mass of 5,000,000 atoms
of Au (gold)
12. 4.2 The Chemical Formula,
Formula Weight, and Molar
Mass
• Chemical formula - a combination of
symbols of the various elements that make up
the compound
• Formula unit - the smallest collection of
atoms that provide two important pieces of
information
– The identity of the atoms
– The relative number of each type of atom
13. 4.2TheChemicalFormula,
FormulaWeightandMolarMass Chemical Formula
Consider the following formulas:
• H2 – 2 atoms of hydrogen are chemically bonded
forming diatomic hydrogen, subscript 2
• H2O – 2 atoms of hydrogen and 1 atom of
oxygen, lack of subscript means one atom
• NaCl – 1 atom each of sodium and chlorine
• Ca(OH)2 – 1 atom of calcium and 2 atoms each
of oxygen and hydrogen, subscript outside
parentheses applies to all atoms inside
14. 4.2TheChemicalFormula,
FormulaWeightandMolarMass Chemical Formula
Consider the following formulas:
• (NH4)3SO4 – 2 ammonium ions and 1 sulfate ion
– Ammonium ion contains 1 nitrogen and 4 hydrogen
– Sulfate ion contains 1 sulfur and 4 oxygen
– Compound contains 2 N, 8 H, 1 S, and 4 O
• CuSO4
.
5H2O
– This is an example of a hydrate - compounds containing
one or more water molecules as an integral part of their
structure
– 5 units of water with 1 CuSO4
15. Comparison of Hydrated and
Anhydrous Copper Sulfate
Hydrated copper sulfate Anhydrous copper sulfate
4.2TheChemicalFormula,
FormulaWeightandMolarMass
Marked color difference illustrates the fact
that these are different compounds
16. Formula Weight and Molar Mass
• Formula weight - the sum of the atomic weights
of all atoms in the compound as represented by its
correct formula
– expressed in amu
• What is the formula weight of H2O?
– 16.00 amu + 2(1.008 amu) = 18.02 amu
• Molar mass – mass of a mole of compound in
grams / mole
– Numerically equal to the formula weight in amu
• What is the molar mass of H2O?
– 18.02 g/mol H2O
4.2TheChemicalFormula,
FormulaWeightandMolarMass
17. Formula Unit
• Formula unit – smallest
collection of atoms from which
the formula of a compound can
be established
• When calculating the formula
weight (or molar mass) of an
ionic compound, the smallest
unit of the crystal is used
4.2TheChemicalFormula,Formula
WeightandMolarMass
What is the molar mass of (NH4)3PO4?
3(N amu) + 12(H amu) + P amu + 4(O amu)=
3(14.01) + 12(1.008) + 30.97 + 4(16.00)=
149.10 g/mol (NH4)3PO4
18. 4.3 The Chemical Equation and the
Information It Conveys
A Recipe For Chemical Change
• Chemical equation - shorthand notation of a
chemical reaction
– Describes all of the substances that react and all
the products that form, physical states, and
experimental conditions
– Reactants – (starting materials) – the substances
that undergo change in the reaction
– Products – substances produced by the reaction
19. 4.3TheChemicalEquation
andtheInformationItConveys
Features of a Chemical Equation
1. Identity of products and reactants must
be specified using chemical symbols
2. Reactants are written to the left of the
reaction arrow and products are written
to the right
3. Physical states of reactants and products
may be shown in parentheses
4. Symbol ∆ over the reaction arrow
means that energy is necessary for the
reaction to occur
5. Equation must be balanced
20. )(O)2Hg()2HgO( 2 gls +→∆
ProductsProducts – written on the right
Reactants – written on the left of arrow
Products and reactants must be
specified using chemical symbols
Physical states are shown in parentheses
∆ – energy is needed
4.3TheChemicalEquation
andtheInformationItConveys
Features of a Chemical Equation
22. 4.3TheChemicalEquation
andtheInformationItConveysEvidence of a Reaction Occurring
The following can be visual evidence of a reaction:
•Release of a gas
– CO2 is released when acid is placed in a solution
containing CO3
2-
ions
•Formation of a solid (precipitate)
– A solution containing Ag+
ions mixed with a solution
containing Cl-
ions
•Heat is produced or absorbed
– Acid and base are mixed together
•Color changes
24. Writing Chemical Reactions
• We will learn to identify the following
patterns of chemical reactions:
– combination
– decomposition
– single-replacement
– double-replacement
• Recognizing the pattern will help you
write and understand reactions
4.3TheChemicalEquation
andtheInformationItConveys
25. A + B → AB
• Examples:
2Na(s) + Cl2(g) → 2NaCl(s)
MgO(s) + CO2(g) → MgCO3(s)
4.3TheChemicalEquation
andtheInformationItConveys
Combination Reactions
• The joining of two or more elements or
compounds, producing a product of
different composition
26. 4.3TheChemicalEquation
andtheInformationItConveys
Types of Combination
Reactions
1. Combination of a metal and a nonmetal
to form a salt
2. Combination of hydrogen and chlorine
molecules to produce hydrogen chloride
3. Formation of water from hydrogen and
oxygen molecules
4. Reaction of magnesium oxide and
carbon dioxide to produce magnesium
carbonate
27. AB → A + B
• Examples:
2HgO(s) → 2Hg(l) + O2(g)
CaCO3(s) → CaO(s) + CO2(g)
4.3TheChemicalEquation
andtheInformationItConveys
Decomposition Reactions
• Produce two or more products from a
single reactant
• Reverse of a combination reaction
29. 1. Single-replacement
• One atom replaces another in the
compound producing a new compound
• Examples:
Cu(s)+2AgNO3(aq) → 2Ag(s)+Cu(NO3)2(aq)
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
A + BC → B + AC
4.3TheChemicalEquation
andtheInformationItConveys Replacement Reactions
30. 1. Replacement of copper by zinc in
copper sulfate
2. Replacement of aluminum by
sodium in aluminum nitrate
4.3TheChemicalEquation
andtheInformationItConveys
Types of Replacement
Reactions
31. 2. Double-replacement
• Two compounds undergo a “change
of partners”
• Two compounds react by
exchanging atoms to produce two
new compounds
4.3TheChemicalEquation
andtheInformationItConveys Replacement Reactions
AB + CD → AD + CB
32. AB + CD → AD + CB
4.3TheChemicalEquation
andtheInformationItConveys
Types of Double-Replacement
• Reaction of an acid with a base to
produce water and salt
HCl(aq)+NaOH(aq) →NaCl(aq)+H2O(l)
• Formation of solid lead chloride from
lead nitrate and sodium chloride
Pb(NO3)2(aq) + 2NaCl(aq) →
PbCl2(s) + 2NaNO3(aq)
33. Types of Chemical Reactions
Precipitation Reactions
• Chemical change in a solution that
results in one or more insoluble products
• To predict if a precipitation reaction can
occur it is helpful to know the
solubilities of ionic compounds
4.3TheChemicalEquation
andtheInformationItConveys
34. Predicting Whether Precipitation
Will Occur
• Recombine the ionic compounds to
have them exchange partners
• Examine the new compounds formed
and determine if any are insoluble
according to the rules in Table 4.1
• Any insoluble salt will be the precipitate
Pb(NO3)2(aq) + NaCl(aq) →
PbCl2 (?) + NaNO3 ( ?)(s) (aq)
4.3TheChemicalEquation
andtheInformationItConveys
37. The H+
on HCl was
transferred to the oxygen
in OH-
, giving H2O
4.3TheChemicalEquation
andtheInformationItConveys
Acid-Base Reactions
• These reactions involve the transfer of a
hydrogen ion (H+
) from one reactant (acid)
to another (base)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
38. Two electrons are
transferred from Zn to
Cu2+
4.3TheChemicalEquation
andtheInformationItConveys
Oxidation-Reduction Reactions
• Reaction involves the transfer of one or
more electrons from one reactant to
another
Zn(s) + Cu2+
(aq)→ Cu(s) + Zn2+
(aq)
39. Writing Chemical Reactions
Consider the following reaction:
hydrogen reacts with oxygen to produce water
• Write the above reaction as a chemical
equation
H2 + O2 → H2O
• Don’t forget the diatomic elements
4.3TheChemicalEquation
andtheInformationItConveys
40. 4.3TheChemicalEquation
andtheInformationItConveys Law of Conservation of Mass
• Law of conservation of mass - matter
cannot be either gained or lost in the
process of a chemical reaction
– The total mass of the products must equal
the total mass of the reactants
41. A Visual Example of the Law of
Conservation of Mass
4.4BalancingChemical
Equations
42. 4.4 Balancing Chemical Equations
• A chemical equation shows the molar
quantity of reactants needed to produce a
particular molar quantity of products
• The relative number of moles of each
product and reactant is indicated by
placing a whole-number coefficient
before the formula of each substance in
the chemical equation
43. )(O)2Hg()2HgO( 2 gls +→∆
Coefficient - how many of that substance
are in the reaction
4.4BalancingChemical
Equations
Balancing
• The equation must be balanced
– All the atoms of every reactant must also
appear in the products
• Number of Hg on left? 2
– on right 2
• Number of O on left? 2
– on right 2
44. 4.4BalancingChemical
Equations
Examine the Equation
H2 + O2 → H2O
• Is the law of conservation of mass obeyed
as written? NO
• Balancing chemical equations uses coefficients
to ensure that the law of conservation of mass is
obeyed
• You may never change subscripts!
• WRONG: H2 + O2 → H2O2
45. Step 1. Count the number of moles of
atoms of each element on both
product and reactant sides
Reactants Products
2 mol H 2 mol H
2 mol O 1 mol O
4.4BalancingChemical
Equations
Steps in Equation Balancing
The steps to balancing:
H2 + O2 → H2O
46. H2 + O2 → H2O
H2 + O2 → 2H2O
This balances oxygen, but is hydrogen
still balanced?
4.4BalancingChemical
Equations
Step 2. Determine which elements are not
balanced – do not have same number on
both sides of the equation
– Oxygen is not balanced
Step 3. Balance one element at a time by
changing the coefficients
Steps in Equation Balancing
47. Reactants Products
4 mol H 4 mol H
2 mol O 2 mol O
4.4BalancingChemical
Equations
Steps in Equation Balancing
H2 + O2 → 2H2O
How will we balance hydrogen?
2H2 + O2 → 2H2O
Step 4. Check! Make sure the law of
conservation of mass is obeyed
50. 4.5 Calculations Using the
Chemical Equation
• Calculation quantities of reactants and
products in a chemical reaction has many
applications
• Need a balanced chemical equation for the
reaction of interest
• The coefficients represent the number of
moles of each substance in the equation
51. 4.5CalculationsUsingthe
ChemicalEquation
General Principles
1. Chemical formulas of all reactants and
products must be known
2. Equation must be balanced to obey the
law of conservation of mass
• Calculations of an unbalanced equation
are meaningless
1. Calculations are performed in terms of
moles
• Coefficients in the balanced equation
represent the relative number of moles of
products and reactants
52. Using the Chemical Equation
• Examine the reaction:
2H2 + O2 → 2H2O
• Coefficients tell us?
– 2 mol H2 reacts with 1 mol O2 to produce 2
mol H2O
• What if 4 moles of H2 reacts with 2 moles of
O2?
– It yields 4 moles of H2O
4.5CalculationsUsingthe
ChemicalEquation
53. 2H2 + O2 → 2H2O
4.5CalculationsUsingthe
ChemicalEquation
Using the Chemical Equation
• The coefficients of the balanced equation
are used to convert between moles of
substances
• How many moles of O2 are needed to
react with 4.26 moles of H2?
• Use the factor-label method to perform
this calculation
54. 2H2 + O2 → 2H2O
=×
2
2
2
Hmol__
O__mol
Hmol26.4 1
2
2.13 mol O2
4.5CalculationsUsingthe
ChemicalEquation
Use of Conversion Factors
• Digits in the conversion factor come
from the balanced equation
55. 4.5CalculationsUsingthe
ChemicalEquation
Conversion Between Moles
and Grams
• Requires only the formula weight
• Convert 1.00 mol O2 to grams
– Plan the path
– Find the molar mass of oxygen
• 32.0 g O2 = 1 mol O2
– Set up the equation
– Cancel units 1.00 mol O2 x 32.0 g O2
1 mol O2
– Solve equation 1.00 x 32.0 g O2 = 32.0 g O2
moles of
Oxygen
grams of
Oxygen
56. 4.5CalculationsUsingthe
ChemicalEquation
Conversion of Mole Reactants to
Mole Products
• Use a balanced equation
• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
• 1 mol C3H8results in:
– 5 mol O2 consumed 1 mol C3H8 /5 mol O2
– 3 mol CO2formed 1 mol C3H8/3 mol CO2
– 4 mol H2O formed 1 mol C3H8/4 mol H2O
• This can be rewritten as conversion
factors
57. 4.5CalculationsUsingthe
ChemicalEquationCalculating Reacting Quantities
• Calculate grams O2 reacting with 1.00 mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles O2
– Moles of O2 to grams O2
– Set up the equation and cancel units
– 1.00 mol C3H8 x 5 mol O2x 32.0 g O2 =
1 mol C3H8 1 mol O2
– 1.00 x 5 x 32.0 g O = 1.60 x 102
g O
moles
Oxygen
grams
Oxygen
moles
C3H8
58. 4.5CalculationsUsingthe
ChemicalEquation
Calculating Grams of Product
from Moles of Reactant
• Calculate grams CO2 from combustion of 1.00
mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles CO2
– Moles of CO2 to grams CO2
– Set up the equation and cancel units
– 1.00 mol C3H8 x 3 mol CO2x 44.0 g CO2 =
1 mol C3H8 1 mol CO2
– 1.00 x 3 x 44.0 g CO = 1.32 x 102
g CO
moles
CO2
grams
CO2
moles
C3H8
59. 4.5CalculationsUsingthe
ChemicalEquation
Relating Masses of Reactants
and Products
• Calculate grams C3H8 required to produce
36.0 grams of H2O
• Use 3 conversion factors
– Grams H2O to moles H2O
– Moles H2O to moles C3H8
– Moles of C3H8 to grams C3H8
– Set up the equation and cancel units
36.0 g H2O x 1 mol H2O x 1 mol C3H8x 44.0 g C3H8
18.0 g H2O 4 mol H2O 1 mol C3H8
– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8
moles
H2O
grams
C3H8
moles
C3H8
grams
H2O
60. 4.5CalculationsUsingthe
ChemicalEquation
Calculating a Quantity of Reactant
• Ca(OH)2 neutralizes HCl
• Calculate grams HCl neutralized by 0.500 mol
Ca(OH)2
– Write chemical equation and balance
• Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l)
– Plan the path
– Set up the equation and cancel units
0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl
1 mol Ca(OH)2 1 mol HCl
Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl
moles
Ca(OH)2
grams
HCl
moles
HCl
62. Na + Cl2 → NaCl
4.5CalculationsUsingthe
ChemicalEquation
Sample Calculation
1. Balance the equation
2. Calculate the moles Cl2 reacting with 5.00
mol Na
3. Calculate the grams NaCl produced when
5.00 mol Na reacts with an excess of Cl2
4. Calculate the grams Na reacting with
5.00 g Cl2
2Na + Cl2 → 2NaCl
64. 4.5CalculationsUsingthe
ChemicalEquation
Sample Calculation
If the theoretical yield of iron was 30.0 g
and actual yield was 25.0 g, calculate the
percent yield:
2 Al(s) + Fe2O3(s) → Al2O3(aq) + 2Fe(aq)
• [25.0 g / 30.0 g] x 100% = 83.3%
• Calculate the % yield if 26.8 grams iron
was collected in the same reaction
