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Chapter 3
Stoichiometry of Formulas and Equations
3-1
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mole(mol) - the amount of a substance that contains the
same number of entities as there are atoms in exactly
12 g of carbon-12
This amount is 6.0221 x 1023. The number is called
Avogadro’s number and is abbreviated as NA
One mole (1 mol) contains 6.0221 x 1023 entities
(to five significant figures)
3-2
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Figure 3.2
Oxygen
32.00 g
One mole of
common
substances.
CaCO3
100.09 g
Water
18.02 g
Copper
63.55 g
3-3
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Table: Summary of Mass Terminology (from 4th ed.)
Term
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element
weighted according to their abundance
amu
Molecular
(or formula) mass
(also called
molecular weight)
Sum of the atomic masses of the atoms
(or ions) in a molecule (or formula unit)
amu
Molar mass (M)
Mass of 1 mole of chemical entities
(atoms, ions, molecules, formula units)
g/mol
(also called
atomic weight)
3-4
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Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Table 3.1
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms/mole of
compound
Mass/molecule
of compound
Mass/mole of
compound
3-5
6(6.0221 x 1023) 12(6.0221 x 1023)
atoms
atoms
6(12.01 amu)
=72.06 amu
72.06 g
6(6.0221 x 1023)
atoms
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
12.10 g
96.00 g
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PRACTICE:
Find molar mass of caffeine, which we all need, C8H10N4O2, and
the hydrate BaCl2.2H2O. Which mole weighs more? Also do
percent composition of caffeine if you remember how.
Caffeine:
8*C = 96.08
10*H=10.079
4*N =56.026
2*O =31.9988
194.19 g/mol
49.48%
5.1903%
28.851%
16.478%
BaCl2.2H2O:
1*Ba = 137.327
2*Cl = 70.9054
2H2O = 36.0304
244.263 g/mol
NOTE: PERCENTAGES MUST OBEY SIG FIG RULES!
3-6
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MOVING BETWEEN ATOMS,
MOLES AND MASS
START OF A MOLE MAP – SEE FIG 3.3 IN
TEXT, NEXT SLIDE AND LECTURE
NOTES
NA STANDS FOR AVOGADRO’S NUMBER
M STANDS FOR MOLAR MASS
3-7
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Atoms, ions, molecules,
formula units…
/NA
X NA
MOLES
XM
/M
MASS, grams
3-8
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EXAMPLES:
How many atoms in 0.222 g of He?
0.222g *
NA = 3.34 x 1022 atoms
4.003 g/mol
How many grams in 1.0 x 109 atoms of lead?
1.0 x 109 atoms * 207.2 g/mol = 3.4 x 10-13 g
NA
3-9
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PRACTICE:
How many moles of caffeine in 0.485g? How
many molecules? How many carbon atoms?
(You previously determined molar mass to be
194.19 g/mol.)
(2.50 x 10-3 MOLES, 1.50 x 1021 molecules,
*8 = 1.20 x 1022 atoms C)
3-10
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DEFINITIONS:
Empirical Formula The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
Molecular Formula The formula of the compound as it exists, it may be a
multiple of the empirical formula.
3-11
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From molecular formula to
empirical formula
Molecular
Formula
C2H2
C6H6
C5H10
C6H12O6
N2H4
3-12
Divider
2
6
5
6
2
Empirical
Formula
CH
CH
CH2
CH2O
NH2
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Determining empirical formula from
percent composition
If we only know percent composition and want the
molecular formula, there are two possible ways:
Way 1: Assume 100 g of it, and plug in mass for
percent. Convert mass to moles, then find lowest
whole number mole ratio to get empirical formula.
Use experimental molar mass to get molecular
formula. (This is very common method because we
don’t always know the molar mass.)
Way 2: If we already know molar mass and % comp,
then multiply molar mass by % to get actual mass of
each element, then convert to moles to obtain
molecular formula
3-13
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Example with caffeine using Way 1:
Make grams with percent composition from previous slide 8 about caffeine:
49.48 g of C/12.011 g/mol =4.120 mol C
5.19 g of H/1.0079 g/mol = 5.149 mol H
28.85 g of N/14.0067 g/mol =2.059 mol N
16.49 g of O/15.9994 g/mol = 1.031 mol O
Now divide all by one with lowest # of moles:
4.120 mol C/1.031 mol O = 3.996/1 C/O,
2.059 mol N/1.031 mol O = 1.998/1 N/O,
5.149 mol H/1.031 mol O = 4.99/1 H/O
4 C, 2 N, 5 H, 1 O: C4H5N2O is the empirical formula
Have to find molar mass to get molecular formula: we know caffeine has
molar mass of 194.193 g/mol
Divide molar mass by empirical mass to see how many EF units in molecule:
194.193/82.1 = ~2
Double everything in EF for molecular formula of C8H10N4O2
3-14
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Example with caffeine using Way 2:
Convert percents back to fractions and multiply by molar
mass:
194.193*.4948 = 96.087 g /12.011g/mol =8.000 mol C
194.193*.0519 = 10.079 g /1.0079g/mol= 10.0 mol H
194.193*.2885 = 56.0247g/14.0067g/mol=4.000 mol N
194.193*.1649 = 32.022g/ 15.9994g/mol=2.001 mol O
Therefore molecular formula is C8H10N4O2
3-15
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PRACTICE ON AN IONIC COMPOUND:
Magnetite is a magnetic form of an iron and oxygen compound that
is 72.4% Fe and 27.6% O. Find its empirical formula, which is
same as its chemical formula.
USE METHOD #1: 100 g of compound
72.4 g Fe/55.846 g/mol=1.30 mol Fe
27.6 g O/15.9994 g/mol = 1.73 mol O
1.73 mol O/1.30 mol Fe = 1.33 O TO 1 Fe
Have to be whole numbers, so multiply to 3: 1.33 * 3 = 4 O, 1 * 3
= 3 Fe TO MAKE Fe3O4
****Note I did not round down 1.33 to 1 to make it FeO!!! This is
important; the ratio should be within 0.05 of a whole number to
round off
3-16
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Video: nice review of method 1
• Finding Empirical Formulas - YouTube
3-17
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Figure 3.5
Combustion train for the determination of the
chemical composition of organic compounds.
m
m
CnHm + (n+ ) O2 = n CO2(g) +
H O(g)
2
2 2
A sample of compound that contains C and H (and perhaps other elements) is
burned in a stream of O2 gas. The CO2 and H2O formed are absorbed
separately, while any other element oxides are carried through by the O2 gas
stream. The increases in mass of the absorbers are used to calculate the
amounts (mol) of C and H in the sample.
3-18
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PRACTICE: a little harder than the AP
Chem examples
If 1.505 g of THIOPHENE (C,H,S) yields 3.149 g CO2, 0.645 g H2O & 1.146 g SO2
then find empirical formula.
CxHySz + O2  x CO2 +
1.505
3.149
y/2 H2O
0.645
+
z SO2
1.146
Do lots of practice problems to learn how to do this!!
3-19
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PRACTICE WITH PERCENT COMPOSITION
• Work on problems 30c, 33.
• 30c: Determine empirical formula for a
compound that is 79.9%-mass carbon and
20.1%-mass hydrogen.
• 33: Find EF & MF for cmpd that is 69.6%mass C, 8.32%-mass H, 22.1%-mass O, with
M of ~362 g/mol.
• (Practice combustion analysis: prob 34)
3-20
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Figure 3.6
The formation of HF gas on the macroscopic and molecular levels.
How to view a chemical equation: as moles, as atoms, or even as
mass.
3-21
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LEARN STEPS TO BALANCING EQ'NS IF YOU HAVE NOT
ALREADY LEARNED HOW TO DO SO!
1. Predict the products if not given (will learn how to
do this)
2. Formulas: determine that each has the correct
chemical formula
3. Conservation of matter: balance the # of atoms of
each kind on each side of arrow, usually by
inspection. (Never change subscripts on formulas to
make it balance.)
(If an element occurs in only one compound on
each side of equation, balance it first; if an element
occurs as free element on either side, balance it last.)
3-22
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Sample Problem 3.7
PROBLEM:
Balancing Chemical Equations
Within the cylinders of a car’s engine, the hydrocarbon octane
(C8H18), one of many components of gasoline, mixes with oxygen
from the air and burns to form carbon dioxide and water vapor.
Write a balanced equation for this reaction.
C8H18 +
O2
H2O
C8H18 + 25/2 O2
8 CO2 + 9 H2O
2C8H18 + 25O2
16CO2 + 18H2O
2C8H18 + 25O2
16CO2 + 18H2O
2C8H18(l) + 25O2 (g)
3-23
CO2 +
16CO2 (g) + 18H2O (g)
Hint: memorize products for combustion of hydrocarbons!
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TYPES OF CHEMICAL EQUATIONS:
(NOT IN CHP 3 BUT YOU NEED TO
MEMORIZE ALL FIVE OF THESE)
1. COMBINATION OR SYNTHESIS: A + B  C
A. formation of compounds from element
METALS + HALOGENS : Zn + I2  ZnI2
METALS + OXYGEN (oxidation - rusting of iron, etc)
Fe + O2 (CAN GO TO FeO OR Fe3O4)
NONMETALS + OXYGEN : C OR N OR S + O2 -->
gases that make acid rain
2. DECOMPOSITION: compound decomposes to simpler
compounds or elements CA+B
CaCO3  CaO + CO2
3-24
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TYPES OF CHEMICAL EQUATIONS:
3. COMBUSTION: A rapid reaction with oxygen.
Organic compounds:
C3H8+ O2  CO2 + H2O
Practice balancing it: balance C, then H, leave O til last
4. SINGLE REPLACEMENT: A + BC  AC + B (Use of
Activity Series)
__Cu + __AgNO3  __Ag + __Cu(NO3)2
5. DOUBLE REPLACEMENT: AB + CD  AD + CB. Notice
how written: C is positive ion and gets written first, goes with
B, the negative ion. (ppt, neut, or gas-forming reactions)
3-25
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PRACTICE WITH CHEMICAL EQUATIONS
• Work on problems 37bd and 38a. Then ID
each type of reaction they are.
• 37b: ___P4O10(s) + ___H2O(l)  ___H3PO4(l)
• 37d: ___CH3NH2(l) + ___O2(g)  ___CO2(g) + ___H2O(l) + ___N2(g)
• 38a: ___Cu(NO3)2(aq) + ___KOH(aq)  ___Cu(OH)2(s) + ___KNO3(aq)
3-26
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Stoichiometry Steps – A Review
1. Start by writing the balanced chemical
equation.
2. Convert mass (or whatever) to moles (divided
by molar mass).
3. Use stoichiometric conversion factors:
mole/mole ratio from balanced equation.
4. Convert moles back to mass (or whatever).
3-27
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Practice Stoichiometry:
Suppose we want to make 10,000. kg of CaO to use in the
chemical industry. Theoretically, how much CaCO3 is needed
as starting material?
1. Start by writing the equation, predict products, balance:
CaCO3  CaO + CO2
2. Convert mass to moles (dividing by molar mass) – we want
CaO:
10,000. kg * 1000 g/kg * mol CaO/56.08 g = 1.783 x 105
mol CaO
3. Use stoichimetric conversion factors:
1.783 x 105 mol CaO * 1 CaCO3/1 CaO = 1.783 X 105 mol
CaCO3 starting
4. Convert moles back to mass (multiply by molar mass)
1.783 x 105 mol CaCO3 * 100.09 g/mol = 1.784 x 107 g
3-28
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Practice Stoichiometry:
Find the mass of CO2 produced by the combustion of
one 10.0 gallon tank of octane (C8H18), DENSITY =
0.75 g/mL.
2 C8H18 + 25 O2  16 CO2 + 18 H2O
(Answer 8.63 x 104 g)
Then do problem 43 in chp 3. (See note below)
3-29
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STEPS TO DETERMINING THE LIMITING
REAGENT
1. Write the balanced chemical equation
2. Go to moles of each of the reactants of interest
3. Use stoichiometric coefficients to determine moles of product
each reactant would form, assuming other reactants present in
excess
4. Choose the least moles of product case. That reactant limits
the reaction, and others are present in excess
5. May have to determine excess by subtracting what was used
with the limiting reagent
3-30
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LIMITING REAGENT:
EXAMPLE: given exactly 1 mol iron (III)sulfide, 2 mol water, 3
mol oxygen gas, find moles of iron (III) hydroxide produced and
moles of excess reagents.
Fe2S3(s) + H2O(l) + O2(g)  Fe(OH)3(s) + S(s)
1. Balance: 2, 6, 3, 4, 6
2. Already in moles!
3. 1.0 mol Fe2S3 * 4 Fe(OH)3/2 Fe2S3 = 2.0 mol Fe(OH)3
2.0 mol H2O * 4 Fe(OH)3/6 H2O = 1.33 mol Fe(OH)3 **limiting
3.0 mol O2 * 4 Fe(OH)3/3 O2 = 4.0 mol Fe(OH)3
4. see ** marked above. (Could convert to mass of product:
1.33 moles Fe(OH)3 * 106.069 g/mol = 142 g)
5. 2.0 mol H2O * 2 Fe2S3/6 H2O = 0.67 mol Fe2S3 consumed.
1.00 mol - 0.67 mol = 0.33 mol remaining
2.0 mol H2O * 3 O2/6 H2O = 1.0 mol O2 consumed.
3.00 mol - 1.0 mol = 2.0 mol remaining
3-31
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LIMITING REAGENT (assuming you
know M and molar mean mol/L)
If 1.50 g Al reacts with 342 mL of 0.350 Molar HCl, what mass
of AlCl3 could be recovered? Which reactant is present in
excess and how much is left over at the end of the reaction?
(make 5.32 g AlCl3, 0.4236g of Al leftover)
If 10.00 g of Cu is put in 1.00 L of 0.100 M AgNO3, what mass
of Ag will be produced? (Cu(NO3)2 is the other product.)
How many moles of excess reagent is left? (Is it Cu?)
(10.79 g Ag, 0.1074 moles Cu leftover)
When a reactant runs out, the reaction stops!
3-32
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THEORETICAL AND ACTUAL YIELDS,
PERCENT YIELD:
THEORETICAL YIELD: amount possible based on
stoichiometric calculations
ACTUAL YIELD: mass of product obtained (always <
theoretical)
PERCENT YIELD = ACTUAL/THEOR * 100 (ALWAYS <
100%)
WHY DON'T WE ALWAYS GET 100% YIELD:
1. Reaction may not be driven to completion (equilibrium,
heat evolved, whatever)
2. Side reactions use up reactants to make unwanted products
3. Purification/recovery process causes loss.
3-33
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PERCENT YIELD EXAMPLE:
Given a reaction with 0.473 grams of phosphorous with excess
chlorine gas that makes only 2.12 g of phosphorous
pentachloride. What is percent yield?
1. Balance reaction: 2 P + 5 Cl2  2 PCl5
2. Convert mass to moles: 0.473 g/30.97g/mol = 1.527x10-2
mol P
3. STOICH RATIO: 1.527x 10-2 mol P * 2 PCl5/2 P = SAME
4. Moles to mass:
1.527x 10-2 mol PCl5 * 208.2 g/mol = 3.18 g
5. % YIELD = 2.12 g/3.18 g * 100 = 66.7% (very typical)
3-34
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PRACTICE WITH PERCENT YIELD
Work on problem 59.
Given: 200. grams of PCl3 react with excess
H2O to form 128 g of HCl and some aqueous
phosphorous acid. Find the percent yield.
PCl3 + 3 H2O  3 HCl + H3PO3
200.g(1 mol/137.33g)(3HCl/1PCl3)(36.46g/mol)
= 159.3 g HCl theoretical yield
% yield = (128/159.3) * 100 = 80.4% yield
3-35
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AQUEOUS REACTIONS/COMPOSITION
Solutions have concentrations – we will use Molarity
Molarity = moles solute/liter solution
M = n/V
Go between moles, volume and molarity: n = M*V; V = n/M
Example: How many moles of NaOH are in 25.0 mL of 0.555 M
NaOH?
25.0 mL * 1L/103 mL * 0.555 mol/L = 0.0138 mol NaOH
3-36
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Sample Problem 3.13 Calculating the Molarity of a Solution
PROBLEM:
PLAN:
Glycine (H2NCH2COOH) is the simplest amino acid. What is the
molarity of an aqueous solution that contains 0.715 mol of
glycine in 495 mL of solution ?
Molarity is the number of moles of solute per liter of solution.
mol of glycine
divide by volume
concentration(mol/mL) glycine
103mL = 1L
molarity(mol/L) glycine
3-37
SOLUTION:
0.715 mol glycine
1000mL
495 mL soln
1L
= 1.44 M glycine
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IONS IN SOLUTIONS:
SOLUTION INVENTORY: WHAT IONS AND WHAT IS
CONCENTRATION?
Example 1: if NaOH dissolved and is 0.100 M, then the
concentration of each ion is 0.100 M
Example 2: if Na2CO3 is dissolved and is 0.100 M,
then [Na+] = 2 x 0.100 M = 0.200 M,
and [CO32-] is 0.100 M
Practice 1: Do the solution inventory for 0.022 M iron (III)
chloride.
Practice 2: Do the solution inventory for 0.0015 M nitric acid.
3-38
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MAKING SOLUTIONS BY TWO METHODS
Method 1: determine concentration and volume needed
1) convert to moles needed, then to grams
2) put some water in a volumetric flask, add solid and
let it completely dissolve, then dilute with water to
mark
Example: Need 0.500 L of 0.500 M NaOH
(Add about 10 g to a 500-mL volumetric flask)
3-39
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Figure 3.12 from 4th ed.
Laboratory preparation of molar solutions.
Note there is
already some water
in the flask!
A
•Weigh the solid needed.
•Transfer the solid to a
volumetric flask that contains
about half the final volume of
solvent.
3-40
C Add solvent until the solution
reaches its final volume, then mix
thoroughly again.
B Dissolve the solid
thoroughly by swirling.
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MAKING SOLUTIONS BY TWO METHODS
Method 2: C1V1 = C2V2 Dilution Equation
C is for concentration in any units
V is for volume in any units
Example: Make 1.0 L of 0.100 M NaOH from 6.0M
NaOH
6.00 M x ? L = 0.100 M x 1.0 L
Use 0.0167 L or 16.7 mL
3-41
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Figure 3.11
Converting a concentrated solution to a dilute solution.
When a solution is
diluted, only solvent
is added. The
solution volume
increases while the
total number of
moles of solute
remains the same.
Therefore, as shown
in the blow-up views,
a unit volume of
concentrated
solution contains
more solute particles
than the same unit
volume of dilute
solution.
(
3-42
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MAKING SOLUTIONS BY TWO METHODS
How would you make 250.0 ml of a 0.555 M
standard NaOH solution? You have some 6.00
M NaOH available and some solid NaOH
available. Show both methods. (Be prepared
to show your work to class on document
camera.)
(Method 1: 5.52 g solid dry NaOH; Method 2:
23.1 mL of 6.0 M NaOH)
3-43