Download Formulas, Equations and Moles

3
Chapter
Formulas, Equations,
and Moles
Chemistry, 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
Balancing Chemical Equations
•
01
A balanced chemical equation represents the
conversion of the reactants to products such that
the number of atoms of each element is conserved.
reactants → products
limestone → quicklime + gas
Calcium carbonate → calcium oxide + carbon dioxide
CaCO3(s) → CaO(s) + CO2(g)
Chapter 03
Slide 2
Prentice Hall ©2004
Balancing Chemical Equations
•
Balancing Equations: Write the unbalanced
equation:
A2B
A2 + B2
•
Use coefficients to indicate how many formula
02
units are required to balance the equation:
2 A2 + B2
2 A2B
Chapter 03
Slide 3
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Balancing Chemical Equations
03
•
Balancing Equations - Method 01:
•
Balance those atoms which occur in only one
compound on each side
•
Balance the remaining atoms
•
Reduce coefficients to smallest whole integers
•
Check your answer
Chapter 03
Slide 4
Prentice Hall ©2004
Balancing Chemical Equations
04
•
Balancing Equations - Method 02:
•
Identify most complex compound
•
Balance this compound by placing 1 before it
•
Balance remaining compounds using fractions
•
Multiply fractions to obtain integers
Chapter 03
Slide 5
Prentice Hall ©2004
Balancing Chemical Equations
•
05
Balance the following equations:
C6H12O6
C2H6O + CO2
Fe + O2
Fe2O3
NH3 + Cl2
N2H4 + NH4Cl
KClO3 + C12H22O11
KCl + CO2 + H2O
Chapter 03
Slide 6
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Balancing Chemical Equations
•
06
Write a balanced equation for the reaction of element
A (red spheres) with element B (green spheres) as
represented below:
Chapter 03
Slide 7
Prentice Hall ©2004
Atomic and Molecular Mass
1 amu =
Mass:
01
mass of carbon −12 atom
12
proton = 1.00728 amu
neutron = 1.0086 amu
electron = 0.0005486
12C atom = 12.00000 amu
13C atom = 13.00335 amu
Chapter 03
Slide 8
Prentice Hall ©2004
Atomic and Molecular Mass
02
•
The atomic masses as tabulated in the periodic table
are the averages of the naturally occurring isotopes.
•
Mass of C = average of 12C and 13C
= 0.9889 x 12 amu + 0.0111 x 13.0034 amu
= 12.011 amu
Chapter 03
Slide 9
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Atomic and Molecular Mass
•
The mass of a molecule is just the sum of the
masses of the atoms making up the molecule.
•
m(C2H4O2) = 2·mC + 4·mH + 2·mO
03
= 2·(12.01) + 4·(1.01) + 2·(16.00)
= 60.06 amu
Chapter 03
Slide 10
Prentice Hall ©2004
Avogadro and the Mole
01
•
One mole of a substance is the gram mass value
equal to the amu mass of the substance.
•
One mole of any substance contains 6.02 x 1023
units of that substance.
•
Avogadro’s Number (NA, 6.022 x 1023) is the
numerical value assigned to the unit, 1 mole.
Chapter 03
Slide 11
Prentice Hall ©2004
Avogadro and the Mole
•
02
The Mole: Allows us to
make comparisons
between substances
that have
different
masses.
Chapter 03
Slide 12
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Avogadro and the Mole
•
03
Calculate the molar mass of the following:
Fe2O3 (Rust)
C6H8O7 (Citric acid)
C16H18N2O4 (Penicillin G)
•
Balance the following, and determine how many
moles of CO will react with 0.500 moles of Fe2O3.
Fe2O3(s) + CO(g)
Fe(s) + CO2(g)
Chapter 03
Slide 13
Prentice Hall ©2004
Avogadro and the Mole
•
04
Methionine, an amino acid used by organisms to
make proteins, is represented below. Write the
formula for methionine and calculate its molar mass.
(red = O; gray = C; blue = N; yellow = S; ivory = H)
Chapter 03
Slide 14
Prentice Hall ©2004
Stoichiometry
•
01
Stoichiometry: Relates the moles of products and
reactants to each other and to measurable
quantities.
Chapter 03
Slide 15
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Stoichiometry
•
02
Aqueous solutions of NaOCl (household bleach)
are prepared by the reaction of NaOH with Cl2:
2 NaOH(aq) + Cl2(g) → NaOCl(aq) + NaCl(aq) + H2O(l)
•
How many grams of NaOH are needed to react
with 25.0 g of Cl2?
Chapter 03
Slide 16
Prentice Hall ©2004
Stoichiometry
•
03
Aspirin is prepared by reaction of salicylic acid
(C7H6O3) with acetic anhydride (C4H6O3) to form aspirin
(C9H8O4) and acetic acid (CH3CO2H). Use this
information to determine the mass of acetic anhydride
required to react with 4.50 g of salicylic acid. How many
grams of aspirin will result? How many grams of acetic
acid will be produced as a by-product?
Chapter 03
Slide 17
Prentice Hall ©2004
Stoichiometry
•
04
Yields of Chemical Reactions: If the actual amount
of product formed in a reaction is less than the
theoretical amount, we can calculate a percentage
yield.
% yield =
Actual product yield
× 100%
Theoretica l product yield
Chapter 03
Slide 18
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Stoichiometry
•
05
Dichloromethane (CH2Cl2) is prepared by reaction
of methane (CH4) with chlorine (Cl2) giving
hydrogen chloride as a by-product. How many
grams of dichloromethane result from the reaction
of 1.85 kg of methane if the yield is 43.1%?
Chapter 03
Slide 19
Prentice Hall ©2004
Stoichiometry
•
06
Limiting Reagents:
The extent to which
a reaction takes
place depends on
the reactant that is
present in limiting
amounts—the
limiting reagent.
Chapter 03
Slide 20
Prentice Hall ©2004
Stoichiometry
•
07
In this figure, ethylene oxide is the limiting reagent.
Chapter 03
Slide 21
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Stoichiometry
•
08
Limiting Reagent Calculation: Lithium oxide is a
drying agent used on the space shuttle. If 80.0 kg
of water is to be removed and 65 kg of lithium
oxide is available, which reactant is limiting?
Li2O(s) + H2O(l) →
•
MM(Li2O) = 29.88 g/mol
•
MM(H2O) = 18.02 g/mol
2 LiOH(s)
Chapter 03
Slide 22
Prentice Hall ©2004
Stoichiometry
•
10
Limiting Reagent Calculation: Cisplatin is an anticancer agent prepared as follows:
K2PtCl4 + 2 NH3
→
Pt(NH3)2Cl2 + 2 KCl
•
If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed
to react: (a) which is the limiting reagent? (b) How
many grams of the excess reagent are consumed?
(c) How many grams of cisplatin are formed?
•
MM(K2PtCl4) = 415.08 g/mol
MM(NH3) = 18.04 g/mol
Chapter 03
Slide 23
Prentice Hall ©2004
Solution Concentrations
•
01
Concentration
allow us to measure
out a specific
number of moles of
a compound by
measuring the mass
or volume of a
solution.
Chapter 03
Slide 24
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Solution Concentrations
•
Molarity: The most useful way of expressing the
amount of a substance dissolved in a solution is
with molarity.
Molarity (M) =
•
02
Moles of solute
Liters of solvent
It is important to note that the final volume of
solution must be used, not volume of solvent.
Chapter 03
Slide 25
Prentice Hall ©2004
Solution Concentrations
•
03
How many moles of solute are present in 125 mL of
0.20 M NaHCO3?
•
How many grams of solute would you use to
prepare 500.0 mL of 1.25 M NaOH?
Chapter 03
Slide 26
Prentice Hall ©2004
Solution Concentrations
•
Dilution:
•
Is the process of
04
reducing a solution’s
concentration by
adding more solvent.
Chapter 03
Slide 27
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Solution Concentrations
04
Concentrated solution + Solvent ⇒ Dilute solution
Moles of solute (mol) = molarity (M) x volume (V)
Mconcentrated x Vconcentrated = Mdilute x Vdilute
Chapter 03
Slide 28
Prentice Hall ©2004
Solution Concentrations
•
05
What volume of 18.0 M H2SO4 is required to
prepare 250.0 mL of 0.500 M aqueous H2SO4?
•
What is the final concentration if 750 mL of 3.50 M
glucose is diluted to a volume of 400.0 mL?
Chapter 03
Slide 29
Prentice Hall ©2004
Solution Stoichiometry
•
01
Solution Stoichiometry uses molarity as a
conversion factor between volume and moles of a
substance in a solution.
Chapter 03
Slide 30
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Solution Stoichiometry
•
02
Stomach acid, a dilute solution of HCl
in water, can be neutralized by
reaction with sodium hydrogen
carbonate, NaHCO3. How many
milliliters of 0.125 M NaHCO3 solution
are needed to neutralize 18.0 mL of
0.100 M HCl?
Chapter 03
Slide 31
Prentice Hall ©2004
Solution Stoichiometry
•
Titration:
•
A technique for
determining the
concentration of
a solution.
03
Chapter 03
Slide 32
Prentice Hall ©2004
Solution Stoichiometry
•
04
What is the molarity of a sulfuric acid solution if a
25.0 mL sample is titrated to equivalence with 50.0
mL of 0.150 M potassium hydroxide solution?
H2SO4(aq) + NaOH(aq) → KNO3(aq) + H2O(l)
Chapter 03
Slide 33
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Percentage Composition
•
01
Percent Composition: Identifies the elements present
in a compound as a mass percent of the total
compound mass.
•
The mass percent is obtained by dividing the mass of
each element by the total mass of a compound and
converting to percentage.
Chapter 03
Slide 34
Prentice Hall ©2004
Percentage Composition
•
02
Glucose has the molecular formula C6H12O6. What is its
empirical formula, and what is the percentage composition
of glucose?
•
Saccharin has the molecular formula C7H5NO3S. What is its
empirical formula, and what is the percentage composition
of saccharin?
Chapter 03
Slide 35
Prentice Hall ©2004
Empirical Formula
•
01
The empirical formula gives the ratio of the number
of atoms of each element in a compound.
Compound
Formula
Hydrogen peroxide H2O2
Benzene
C6H6
Ethylene
C2H4
Propane
C3H8
Chapter 03
Empirical Formula
OH
CH
CH2
C3H8
Slide 36
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles
Empirical Formula
•
A compound’s empirical formula
can be determined from its
percent composition.
•
A compound’s molecular formula
is determined from the molar
mass and empirical formula.
02
Chapter 03
Slide 37
Prentice Hall ©2004
Empirical Formula
03
•
Combustion analysis is one of the
most common methods for
determining empirical formulas.
•
A weighed compound is burned in
oxygen and its products analyzed
by a gas chromatogram.
•
It is particularly useful for analysis
of hydrocarbons.
Chapter 03
Slide 38
Prentice Hall ©2004
Empirical Formula
•
04
A compound was analyzed to be 82.67% carbon and 17.33%
hydrogen by mass. An osmotic pressure experiment
determined that its molar mass is 58.11 g/mol.
•
What is the empirical formula and molecular formula for the
compound?
Chapter 03
Slide 39
Prentice Hall ©2004
Chapter 03: Formulas , Equations, and Moles