Chapter 3
Formulas, Equations, and Moles
Balancing Chemical Equations
• Alphabet – elemental symbols
• Words – chemical formulas
• Sentences – chemical equations (chemical reactions)
reactants  products
limestone  quicklime + gas
Calcium carbonate  calcium oxide + carbon dioxide
CaCO3(s)  CaO(s) + CO2(g)
Balancing Chemical Equations
• Chemical reactions include
– Reactants
– Products
– Balanced – Law of Conservation of Mass
• # of atoms of an element on the reactant side must
equal the # of atoms of that element on the product
side.
– Indicate the state of matter of each chemical
in the reaction (Chapter 4)
Balancing Chemical Equations
• Write the equation without coefficients
• List the elements in each equation
– Secret: if the same polyatomic ion exists on both sides, keep it
together
• Determine the # of each kind of atom on both sides
• Balance atoms one element at a time by adjusting
coefficients
– DO NOT ALTER THE FORMULA OF THE COMPOUND!!!!!
• Only coefficients can be altered
– Secret:
• Balance atoms appearing only once on each side first.
• Save compounds comprised of only one type of element till last.
• Reduce to lowest terms if necessary
Examples
• Balance the following equations:
– Al(s) + Fe2O3(s) → Al2O3 (s) + Fe (l)
– Solid copper reacts with aqueous silver nitrate
to form aqueous copper (II) nitrate and silver
solid
– H3PO4 (l) → H2O (l) + P4O10 (s)
– C4H10(g) + O2 (g) → CO2(g) + H2O (g)
Avogadro’s Number and the Mole
•
Meaning of a chemical reaction
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O (g)
–
2 molecules of C4H10(g) reacts with 13 molecules of O2(g)
to form
8 molecule of CO2(g) and 10 molecules of H2O(g)
Avogadro’s Number and the Mole
• Molecule’s mass = the sum of the atomic
masses of the atoms making up the
molecule.
• m(C2H4O2) = 2·mC + 4·mH + 2·mO
»
= 2·(12.01) + 4·(1.01) + 2·(16.00)
»
= 60.06 amu
Avogadro’s Number of the Mole
• One mole (mol) of any substance contains 6.02
x 1023 (Avogadro’s Number) units of that
substance.
• One mole (mol) of a substance is the gram
mass value equal to the amu mass of the
substance.
– Calculated the same as amu’s for a molecule
Avogadro’s Number and the Mole
• Calculate the molar mass of the
following:
– Fe2O3 (Rust)
– C6H8O7 (Citric acid)
– C16H18N2O4 (Penicillin G)
Avogadro’s Number and the Mole
• 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)
Stoichiometry
• 4 Conversion units
– Chemical formula
– Balanced chemical equation
• Coefficients can read as;
– # of molecules
– # of moles of that molecules
• Allows conversion between compounds in an equation
– Avogadro’s # - 6.02 x 1023 of X = 1 mole of X
– Molar mass – how many grams of a substance = 1
mole of that substance
Stoichiometric Calculations
Avogadro’s Number and the Mole
• How many grams of oxygen are present in
5.961 x 1020 molecules of KClO3? How
many atoms of oxygen are present?
Avogadro’s Number and the Mole
• Calculate the number of oxygen atoms in
29.34 g of sodium sulfate, Na2SO4.
– A.
– B.
– C.
– D.
– E.
1.244 × 1023 O atoms
4.976 × 1023 O atoms
2.409 × 1024 O atoms
2.915 × 1024 O atoms
1.166 × 1025 O atoms
Problem
• Potassium dichromate, K2Cr2O7, is used in
tanning leather, decorating porcelain and water
proofing fabrics. Calculate the number of
chromium atoms in 78.82 g of K2Cr2O7.
–
–
–
–
–
A.
B.
C.
D.
E.
9.490 × 1025 Cr atoms
2.248 × 1024 Cr atoms
1.124 × 1024 Cr atoms
3.227 × 1023 Cr atoms
1.613 × 1023 Cr atoms
Stoichiometry: Chemical Arithmetic
Stoichiometry: Equation Arithmetic
• 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)
Stoichiometry: Chemical Arithmetic
•
Aqueous sodium hydroxide and chlorine
gas are combined to form aqueous sodium
hypochlorite (household bleach), aqueous
sodium chloride and liquid water.
–
How many grams of NaOH are needed to react
with 25.0 g of Cl2?
Problem
•
Sulfur dioxide reacts with chlorine to produce thionyl chloride (used
as a drying agent for inorganic halides) and dichlorine monoxide
(used as a bleach for wood, pulp and textiles).
SO2(g) + 2Cl2(g) → SOCl2(g) + Cl2O(g)
If 0.400 mol of Cl2 reacts with excess SO2, how many moles of Cl2O
are formed?
–
–
–
–
–
A.
B.
C.
D.
E.
0.800 mol
0.400 mol
0.200 mol
0.100 mol
0.0500 mol
Problem
• Nitrogen gas and hydrogen gas are combined to form
ammonia (NH3), an important source of fixed nitrogen
that can be metabolized by plants, using the Haber
process.
How many grams of nitrogen are needed to produce 325
grams of ammonia?
–
–
–
–
–
A.
B.
C.
D.
E.
1070 g
535 g
267 g
178 g
108 g
Lab Homework
• MISC 486 Problem Set 2 – Due
Yields of Chemical Reactions
• 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.
Actual product yield
% yield 
 100%
Theoretica l product yield
Yield of Chemical Reactions
• 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%?
Problem
• What is the percent yield for the reaction
PCl3(g) + Cl2(g) → PCl5(g)
if 119.3 g of PCl5 ( MM = 208.2 g/mol) are formed when
61.3 g of Cl2 ( MM = 70.91 g/mol) react with excess
PCl3?
–
–
–
–
–
A.
B.
C.
D.
E.
195%
85.0%
66.3%
51.4%
43.7%
Reactions with Limiting Amounts of
Reactants
• Limiting Reagents: The extent to which a
reaction takes place depends on the reactant
that is present in limiting amounts—the limiting
reagent.
• Process
– Convert each reactant into a single product
– The one that forms the least is the limiting reactant
– Complete all other calculations using the limiting reactant
Reactions with Limiting Amounts of
Reactants
• 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)
Reactions with Limiting Amounts of
Reactants
• Limiting Reagent Calculation: Cisplatin is an anti-cancer
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 [Pt(NH3)2Cl2] = 299.9 g/mol
MM(NH3) = 18.04 g/mol
Problem
•
What is the percent yield of H2O if 97.2 g CH4S reacts
with 183 g of O2 to produce 58.5 g H2O according to
the following chemical reaction? (Reaction may or may
not be balanced.)
CH4S + O2 → CO2 + H2O + SO3
–
–
–
–
–
–
–
Write the balanced chemical reaction?
Calculate the molar mass of CH4S?
Calculate the molar mass of H2O?
Calculate the molar mass of O2?
Calculate the theoretical yield of H2O?
What is the limiting reagent?
Calculate the percent yield of H2O?
Lab Homework
• MISC 486 Problem Set 4 - Due
Concentrations of Reactants in
Solution: Molarity
• Molarity: The most common way of
expressing the amount of a substance
dissolved in a solution
– Conversion factor between moles and volume
Moles of solute
Molarity (M) 
Liters of solution
•
It is important to note that the final volume of solution must be used,
not volume of solvent.
Solution Stoichiometry
• Known: molarity of solution, volume of solution,
and balanced chemical equation
Concentrations of Reactants in
Solution: Molarity
• 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?
Problem
• A 0.150 M sodium chloride solution is referred to
as a physiological saline solution because it has
the same concentration of salts as normal
human blood. Calculate the mass of solute
needed to prepare 275.0 mL of a physiological
saline solution.
–
–
–
–
–
A.
B.
C.
D.
E.
41.3 g
31.9 g
16.1 g
8.77 g
2.41 g
Concentrations of Reactants in
Solution: Molarity
• Solution Preparation
– Determine the mass of solid needed to obtain the
desired # of moles
– Mass the solid
– Add it to a volumetric flask
– Add water to the volumetric flask until it is about halffull
– Cap and shake to dissolve the solid
– Add water to the line
– Cap and shake
Diluting Concentrated Solutions
• Dilution: process of reducing a
solution’s concentration by adding
more solvent.
– Key concept: # of moles remains
constant
Diluting Concentrated Solutions
Concentrated solution + Solvent  Dilute solution
Moles of solute (mol) = molarity (M) x volume (V)
Mconcentrated x Vconcentrated = Mdilute x Vdilute
only use if the initial solution and the final solution are the
same substance
Diluting Concentrated Solutions
• 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?
Problem
• Calcium chloride is used to melt ice and snow on
roads and sidewalks and to remove water from
organic liquids. Calculate the molarity of a
solution prepared by diluting 165 mL of 0.688 M
calcium chloride to 925.0 mL.
–
–
–
–
–
A.
B.
C.
D.
E.
3.86 M
0.743 M
0.222 M
0.123 M
0.114 M
Titration
• Titration: A technique for determining
the concentration of a solution.
– Process –
• a carefully measured volume of an unknown
solution is allowed to react with of a standard
solution (concentration is known).
• The volume of the known is measured.
• Stoichiometry calculations are performed
Titration
• 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) + KOH(aq)  K2SO4(aq) + H2O(l)
Problem
• How many milliliters of 1.58 M HCl are needed
to react completely with 23.2 g of NaHCO3 (
MM = 84.02 g/mol)?
HCl(aq) + NaHCO3(s) → NaCl(s) + H2O(l) + CO2(g)
–
–
–
–
–
A.
B.
C.
D.
E.
638 mL
572 mL
536 mL
276 mL
175 mL
Lab Homework
• MISC 486 Problem Set 5 - Due
Percent Composition and Empirical
Formulas
• 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.
Percent Composition and Empirical
Formulas
• Sugar is 42.1% C, 6.4% H, and 51.5% O
• Meaning – out of 100 g of sugar 42.1 g of
it is due to C
Percent Composition and Empirical
Formulas
• From the percent composition empirical
formulas are developed
– empirical formula gives the smallest whole
number ratio of the atoms of each element in
a compound.
• Same as ionic formula for ionic compounds
Percent Composition and Empirical
Formulas
• Compound Formula Empirical Formula
• Hydrogen
H 2O 2
OH
peroxide
• Benzene
C 6H 6
CH
• Ethylene
C 2H 4
CH2
• Propane
C 3H 8
C 3H 8
Percent Composition and Empirical
Formulas
• 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.
Percent Composition and Empirical
Formulas
• 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?
Problem
• Gadolinium oxide, a colorless powder which
absorbs carbon dioxide from the air, contains
86.76 mass % Gd. Determine its empirical
formula.
–
–
–
–
–
A.
B.
C.
D.
E.
Gd2O3
Gd3O2
Gd3O4
Gd4O3
GdO
Problem
• Hydroxylamine nitrate contains 29.17 mass % N,
4.20 mass % H, and 66.63 mass O. If its molar
mass is between 94 and 98 g/mol, what is its
molecular formula?
–
–
–
–
–
A.
B.
C.
D.
E.
NH2O5
N2H4O4
N3H3O3
N4H8O2
N2H2O4
Determining Empirical Formulas:
Elemental Analysis
• 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.
Determining Empirical Formulas:
Elemental Analysis
• Terephthalic acid, used in the production of
polyester fibers and films, is composed of
carbon, hydrogen, and oxygen. When 0.6943 g
of terephthalic acid was subjected to combustion
analysis it produced 1.471 g CO2 and 0.226 g
H2O. What is its empirical formula?
–
–
–
–
–
A.
B.
C.
D.
E.
C2H3O4
C3H4O2
C4H3O2
C5H12O4
C2H2O
Optional Homework
• Text –3.30, 3.32, 3.34, 3.36, 3.40, 3.42,
3.54, 3.58, 3.62, 3.64, 3.70, 3.72, 3.74,
3.76, 3.80, 3.82, 3.86, 3.88, 3.90, 3.92,
3.94, 3.96, 3.100, 3.106, 3.108, 3.112,
3.116
• Chapter 3 Homework – from website
Required Homework
• MISC 486 Problem Set 3 – Due
• Assignment #3