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CHAPTER 3: Quantitative
Relationships in Chemical Reactions
Stoichiometry: Greek for “measure elements”
Stoichiometry involves calculations based on
chemical formulas and chemical equations
(reactions) – quantitative.
Law of conservation of mass is central to
stoichiometry.
Atoms are neither created nor destroyed in a
chemical reaction. Atoms are simply rearranged
to yield new substances.
The following is a common teaching laboratory
reaction:
Zn + HCl
ZnCl2 + H2
Reactants
Products
This reaction is not balanced:
reactant atoms ≠ product atoms
Zn + HCl
ZnCl2 + H2
Reactants
Products
1 Zn
1 Zn
1 H
2 H
1 Cl
2 Cl
Procedure for balancing a reaction:
Balance the elements that appear in only one
reactant and one product first.
Zn - Already balanced
Balance H and Cl; need 2 H atoms on
each side of the reaction
Putt a coefficient
P
ffi i nt of
f 2 in front
f nt of
f th
the
HCl molecule on the reactant side.
Also results in balancing Cl's.
Balance this reaction by the method of
inspection.
Balancing Chemical Reactions
• Formulas of reactants and products
can’t be changed (i.e., cannot
change subscripts)
• Only amounts of reactants &
products can be changed to reach
atom balance
– put coefficients (multipliers) in
front of the substances in the
reaction
Zn + 2 HCl
ZnCl2 + H2
Now, the reaction is balanced:
Reactants
Products
1 Zn
1 Zn
2 H
2 H
2 Cl
2 Cl
reactant atoms = product atoms
1
Another example:
C4H10 +
O2
CO2 + H2O
Designate physical states of each species:
Balance C's and H's first:
– only
y appear
pp
in one reactant and one product
p
each
– O's balanced last; appears in two products
Balance C's, multiply the CO2 product by 4.
C4H10 +
O2
4 CO2 + H2O
Next, balance H's:
multiply the H2O product by 5
C4H10 + O2
4 CO2 + 5 H2O
13 O’s
Balance the O's; multiply the O2 reactant
by 13/2
/ to give 13 O atoms:
C4H10 + 13/2 O2
4 CO2 + 5 H2O
Reactants Products
4 C
4 C
10 H
10 H
13 O
13 O
BALANCING CHEMICAL REACTIONS
• Typically, want the smallest whole-number
coefficients in a balanced reaction.
• Hence, fractional coefficient is removed by
multiplying the entire equation by 2.
2
2 C4H10 + 13 O2
Often, the physical state of each
substance is specified.
8 CO2 + 10 H2O
(g) – gas
(l) - liquid
(s) - solid
(aq) - aqueous (water) solution
Chemical equation with physical states specified:
2 C4H10(g) + 13 O2(g)
8 CO2(g) + 10 H2O(g)
Combustion reactions: Involve O2 as a
reactant, typically form CO2 and H2O as
products.
Examples:
CH4 + 2 O2
2 C2 H 2 + 5 O 2
2 C8H18 + 25 O2
CO2 + 2 H2O
4 CO2 + 2 H2O
16 CO2 + 18 H2O
Examples: Balancing Reactions
__NO + __Cl2
__Fe
Fe2O3 + __CO
CO
__C3H6 + __O2
__NOCl
__Fe
Fe + __CO
CO2
__CO + __H2O
Balanced chemical reactions are called
chemical equations.
2
ATOMIC AND MOLECULAR MASSES
• It is rather easy to determine the relative
masses of atoms experimentally.
• For example, C has a mass 12 times that of
H.
• O a mass that is 16 times that of H or
1.33 times that of C.
• Cl a mass that is 35.4 times that of H or
2.95 times that of C.
RELATIVE ATOMIC MASS (WEIGHT) SCALE
H
100
1
200
H2(g) + 2 Li(s)
2 LiH(s)
1.00 g
7.88g
What is the atomic mass of Li relative to H ?
6.88 to 1.00
RELATIVE ATOMIC MASS SCALE
• Developed In A Similar Manner For All
Elements
H
C
1
12 16
O
Masses relative to that for H.
Consequently, a new mass unit (atomic mass
unit, amu) is used to express the mass of
atoms/molecules/formula units.
Define the atomic mass unit (amu):
Mass of 1 C-12 atom = 12 amu
1 amu = 1
1.661
661 X 10-24 g
Some atomic masses:
Hydrogen = 1.008 amu
Helium = 4.003 amu
Fluorine = 18.998 amu
From the periodic table, the mass of C is given
as 12.011 amu, not 12.000 amu.
Why ???
Carbon consists of two stable isotopes (C-12 &
C-13))
Mg
24.3
These very small numbers are awkward to use.
use
The mass of one C-12 atom is assigned a value of
12 atomic mass units (amu), exactly !
Derived from early experiments involving known
substances; e.g.
6.88g
Atoms have very, very small masses. The O-16
atom has a mass of only 2.6560 X 10-23 g.
35
Isotope
C-12
Mass (amu)
12.00000
% abundance
98.890
C-13
13.00335
1.110
3
Average Atomic Mass of C:
Avg. Mass C = (12.00000)(0.98890) + (13.00335)(0.01110)
Avg. Mass C = 12.01 amu (Sig. Figures)
This number, 6.022 X 1023, is given the special
name mole.
6.02214199 X 1023
Average Atomic Mass of chlorine (2 isotopes):
Isotope
I
t
Cl-35
Mass (amu)
M
(
)
34.968
% abundance
b d
75.53
Cl-37
36.956
24.47
Mole SI base unit for the amount of a
Mole:
substance (mol).
1 mole of anything is 6.022 X 1023 of whatever
we a considering.
Avg. Mass Cl = (34.968)(0.7553) + (36.956)(0.2447)
Avg. Mass Cl = 35.45 amu
Average atomic mass is also called atomic
weight for an element.
Example Problem
Copper is
C
i a metal
t l used
d in
i the
th production
d ti of
f
cables, pennies, etc. The atomic masses of its
two stable isotopes, Cu-63 (69.09 %) and Cu65 (30.91 %), are 62.93 amu and 64.93 amu,
respectively. Calculate the average atomic
mass of Cu (63.55 amu).
It is difficult to work with individual atoms-too
small.
Consider a very large number of C-12 atoms.
One C-12 atom has a mass of 12 amu exactly.
H
How
many C-12
C 12 atoms are iin exactly
l 12 g ??
Mole:
The amount of matter that contains as many
elementary units (atoms, molecules, etc.) as
there are atoms in exactly 12 g of C-12.
1 mole = 6.022 X 1023 (Avogadro’s number)
1 mole C-12 = 6.022 X 1023 atoms
= 12.0000 g
KNOW !!!!!
Molar Mass
Molar mass (g) Î mass (g) of one mole of a
substance
Molar mass in grams is numerically equivalent
to atomic mass in atomic mass units (amu).
One C-12 atom has a mass of 1.9927 X 10-23 g.
1 C-12 atom = 12.0000 amu;
1 mol C-12 = 12.0000 g
⎛ 1 C - 12 atom ⎞
⎜⎜
⎟ (12 g ) = 6.022 X 10 23 C atoms
- 23 ⎟
⎝ 1.9927 X 10 g ⎠
1 H-1 atom = 1.0078 amu;
1 mol H-1 = 1.0078 g
4
We can use the result (5.81 X 10-23
g/Cl-35 atom) to calculate the mass of
the amu is terms of grams:
gram ⎛⎜ 5.81 X 10 − 23 g ⎞⎟⎛ 1 Cl - 35 atom ⎞ 1.66 X 10 - 24 g
=
⎜
⎟=
amu ⎜⎝ 1 Cl − 35 atom ⎠⎟⎝ 34.968 amu ⎠
amu
1 amu = 1.660538728 X 10-24 g
1 g = 6.02 X 1023 amu
Avogadro’s number
Figure 3.1
CONVERSION FACTORS: Molar Mass
1 mol C = 12.011 g
1 avg. C atom = 12.011 amu
amu/atom is numerically equivalent to grams/mole
1 mol = 6.02 X 1023
ANOTHER CALCULATION
How many moles O are in 37.00 g O?
First look at given and desired units:
gO
These concepts apply to all elements!!!
Example: What is the mass in grams of one Cl35 atom?
Given: 1 mol Cl-35 = 34.968 g
conversion
factor
moles O
1 mol O ⎞
(37.00 g O ) ⎛⎜
⎟ = 2.312 mol O
⎝ 16.00 g O ⎠
Desired: mass in g/Cl-35 atom
Solution to Cl-35 example
ANOTHER CALCULATION
How many O atoms are in 37.00 g O?
⎞
grams
⎛ 34 .968 g Cl - 35 ⎞⎛ 1 mol Cl - 35
⎟ = 5.81 X 10 - 23
⎟⎜
⎜
atom Cl
⎝ 1 mol Cl - 35 ⎠⎝ 6.02 X 10 23 atoms ⎠
(given) X (conversion factor) = desired
Use the correct conversion factor to go
from given to desired units.
g O Î moles O Î atoms O
Avogadro’s #
molar mass
conversion
factors
1 mol O ⎞ ⎛ 6.022 X 10 23 O atoms ⎞
⎟
(37.00 g O ) ⎛⎜
⎟ ⎜⎜
⎟
⎝ 16.00 g O ⎠ ⎝
1 mol O
⎠
= 1.393 X 10 24 O atoms
5
Molar Mass:
– Mass, in g's, of one mole of a substance.
– Numerically equal to the formula mass (in
amu) or molecular mass (in amu).
– Examples:
Molar mass [[Fe(NO
(
3)3] = 241.88 g
Molar mass (NH3) = 17.03 g
– Another example:
– What is the molar mass of C2H5OH ?
(46.07 g)
Figure 3.2
Formula and Molecular Masses
Mass, Moles, Number of Particles
molar mass
Avogadro' s #
Grams ←⎯ ⎯ ⎯⎯→ Moles ←⎯ ⎯ ⎯ ⎯
⎯→ Molecules
Formula Mass (FM):
sum of the atomic masses of all atoms in the
chemical formula (formula unit)
Example:
Fe(NO3)3
FM = 1(AM of Fe) + 3(AM of N) + 9(AM of O)
FM = 1(55.85 amu) + 3(14.01 amu) + 9(16.00
amu)
FM = 241.88 amu
Molecular Mass (MM):
–The sum of the atomic masses of all atoms
of the molecular formula of a molecular
substance.
–Example: NH3
–MM = 1(AM of N) + 3(AM of H)
= 1(14.01
1(14 01 amu) + 3(1
3(1.008
008 amu)
= 17.03 amu
–Another example:
What is the molecular mass of aspartame
(nutrasweet), (C14H18N2O5), the artificial
sweetener ? (294.32 amu)
1. What is the average mass in grams of one avg.
chlorine atom ? (5.89 X 10-23 g)
2. What is the avg. mass in grams of one ethanol
(C2H5OH) molecule ? (7.65 X 10-23 g)
3. How many moles of PbCrO4 (Lead Chromate) are
in 45.6 grams ? (0.141 mol)
4. How many HCl (hydrogen chloride) molecules are
in 46.0 grams ? (7.60 X 1023 molecules)
Calculations:
Percent Composition of elements by mass:
The percent by mass of each element in a
compound.
Example: ammonia,
ammonia NH3
Divide the mass of each element in one
mole of the compound by the molar mass
of the compound, then multiply by 100 to
get a percent.
6
General Approach:
%=
(# moles of element)(MM element)
X 100
MM of compound
%H =
3(1.008 g)
X 100 = 17.76 %
17.03 g
First, calculate mass of each element:
• Consider a 100 g sample (any sample
size will work).
• For a 100 g sample:
p
= mass of the element:
• % composition
85.63 g C
1(14.01 g)
X 100 = 82.27 %
%N=
17.03 g
Percent composition by mass
• sum of % composition must equal 100 %
• 17.76 % + 82.27 % = 100.03 %
• This small deviation from 100 % is due
to round-off error.
• Another example:
What is the percent composition of
hydrogen in pentylacetate, CH3CO2C5H11
(responsible for the odor in bananas) ?
(10.84 %)
Determine Empirical Formulas from a
given Percent Composition by Mass (from
experiment)
• What is the empirical formula of a compound
that is 85.63 % C and 14.37 % H by mass ?
• How
H
tto solve
l ??
% mass Î mass Î moles Î empirical formula
14.37 g H
Second, convert grams of each
element into moles of each element:
Conversion:
molar mass
Mass
Moles
1 mol C ⎞
(85.63 g C ) ⎛⎜
⎟ = 7.130 mol C
⎝ 12.01 g C ⎠
1 mol H ⎞
(14.37 g H ) ⎛⎜
⎟ = 14.256 mol H
⎝ 1.008 g H ⎠
Third, determine empirical formula:
Use the mole amounts as a subscript for
each chemical symbol:
C7.130
7 130 H14.256
14 256
Convert subscripts to smallest whole-number
ratio; divide each by the smallest subscript.
7
C:
H:
Molecular Formula:
7.130
=1
7.130
14.256
= 1.9994 ≈ 2
7.130
The Empirical Formula is:
28.05 g
= 1.9993 empirical formula units ≈ 2
14.03 g
The molecular formula is:
(CH2)2 Î C2H4
CH2
(smallest whole-number ratio of atoms in
the Chemical Formula)
Determine Empirical Formulas from a given
Percent Composition by Mass (from experiment)
Example: What is the empirical formula of a
compound that is 48.6% C, 8.2% H, and 43.2%
O by mass?
Calculate moles of each element in 100 g sample:
(48.6 g C)⎛⎜
1 mol C ⎞
⎟ = 4.047 mol C
12.01
g C⎠
⎝
(8.2 g H )⎛⎜
1 mol H ⎞
⎟ = 8.135 mol H
⎝ 1.008 g H ⎠
(43.2 g O )⎛⎜
1 mol O ⎞
⎟ = 2.700 mol O
⎝ 16.00 g O ⎠
Desire a Molecular Formula ?
To determine it we need to know the molar
mass.
# of empirical formula
units in the molecular
formula
=
Molar Mass
Empirical Molar Mass
Molar mass = 28.05 g (given)
Empirical molar mass (CH2) = 14.03 g
(determined from the
above example)
Convert to small mole ratios by dividing each by
2.700:
C1.50H3.01O1
Multiply mole ratios (subscripts) by an
appropriate multiplier to get smallest whole
number ratios.
Multiply mole ratios by a multiply of 2 to yield
the empirical formula:
C3H6O2
8
Required Multipliers for Fractional
Subscripts
Subscripts Multipliers Whole #’s
1.50
2
3
1.33
.33
3
3.99 Î 4
1.67
3
5.01Î 5
1.25
4
5
1.75
4
7
Calculations Involving Chemical
Equations
Examples:
P1O2.50 X 2 Î P2O5
N1O1.50 X 2 Î N2O3
C1H2.67 X 3 Î C3H8
C1H1.33 X 3 Î C3H4
Mass %
elements
100 g
sample
Grams each
element
Molar
mass
Molecular
formula
Molar
atomic
mass
Empirical
formula
Calculate
mole ratio
Moles each
element
Stoichiometric relationships in chemical equations:
CH4 + 2 O2
CO2 + 2 H2O
ff
in the equation
q
give the
g
The coefficients
ratios by molecules or moles of reactants and
products. For stoichiometric calculations, the
coefficients represent mole quantities.
“1 mole methane reacts with 2 moles of
diatomic oxygen to form 1 mole carbon
dioxide and 2 moles water”
Stoichiometric Relationships
• Also called “stoichiometric equivalents”
• Show how reactants and products are
related quantitatively (number of molecules
or moles).
moles)
• How ?
• Coefficients in the equation gives ratios
by molecules or moles of reactants and
products.
9
CH4 + 2 O2
CO2 + 2 H2O
One “Stoichiometric equivalent” is:
1 mole CH 4 ⇔ 2 moles O 2
“One mole methane reacts with (is
stoichiometrically equivalent to) two moles
oxygen”
CH4 + 2 O2
CO2 + 2 H2O
Stoichiometric Equivalents
CH4 + 2 O2
conversion
factor
1 mole CH 4 ⇔ 2 mole O 2
CO2 + 2 H2O
conversion
factor
conversion
factor
1 mole CH 4 ⇔ 1 mole CO 2
1 mole CH 4 ⇔ 2 mole H 2 O
2 mole O 2 ⇔ 1 mole CO 2
2 mole O 2 ⇔ 2 mole H 2 O
1 mole CO 2 ⇔ 2 mole H 2 O
Use these relationships to construct conversion
factors for stoichiometric (quantitative)
calculations.
STOICHIOMETRIC CALCULATIONS
• Construct conversion factors for calculations
from stoichiometric equivalents
CH4 + 2 O2
CO2 + 2 H2O
• E
Example:
l How
H
many grams of
f CO2 is
i
produced from 10.0 g of O2 as a reactant ?
given
grams O2
X
conversion ratios
& stoichiometric equivalents
given
i
g O2
desired
d
i d
g CO2
⎛ 1 mol O 2 ⎞⎛ 1 mol CO 2 ⎞⎛ 44.01 g CO 2 ⎞
⎟⎜
⎟⎜
⎟
⎝ 32.0 g O 2 ⎠⎝ 2 mol O 2 ⎠⎝ 1 mol CO 2 ⎠
(10.0 g O 2 )⎜
= 6.88 g CO 2
Another example:
The smelting (refining) of chalcopyrite
ore to produce copper metal.
2 CuFeS2 + 5 O2
2 Cu + 2 FeO + 4 SO2
Chalcopyrite ore
= desired
grams CO2
10
An analogous part to this problem is to
calculate how much of the pollutant, sulfur
dioxide (SO2), is produced when 1.00 g of
raw ore is refined ?
0.699 g SO2
Chalcopyrite ore (CuFeS2)
How many g Copper (Cu) can be produced from
1.00 g of CuFeS2 ore ?
Set up the conversion scheme:
g CuFeS2
mol CuFeS2
2 CuFeS2 + 5 O2
mol Cu
g Cu
2 Cu + 2 FeO + 4 SO2
⎛ 1 mol CuFeS 2 ⎞⎛ 2 mol Cu ⎞⎛ 63.5 g Cu ⎞
⎟⎜
⎟⎜
⎟
⎝ 183 g CuFeS 2 ⎠⎝ 2 mol CuFeS 2 ⎠⎝ 1 mol Cu ⎠
(1.00 g CuFeS 2 )⎜
= 0.347 g Cu
How many g oxygen (O2) is required to react
with 1.00 g of CuFeS2 ore ?
mol CuFeS2
2 CuFeS2 + 5 O2
mol O2
g O2
2 Cu + 2 FeO + 4 SO2
⎛ 1 mol CuFeS 2 ⎞⎛ 5 mol O 2 ⎞⎛ 32.00 g O 2 ⎞
⎟⎟⎜⎜
⎟⎟⎜⎜
⎟⎟
⎝ 183 g CuFeS 2 ⎠⎝ 2 mol CuFeS 2 ⎠⎝ 1 mol O 2 ⎠
(1.00 g CuFeS 2 )⎜⎜
Dictate how much product can be produced.
Consider the following reaction:
Set up the conversion scheme:
g CuFeS2
Limiting Reactants
N2 + 3 H2
2 NH3
1 mole N2 reacts with 3 moles H2 to yield 2 moles NH3
Stoichiometric amounts of reactants
N2:H2 mole ratio of 1:3
Both reactants are completely consumed in the reaction.
= 0.437 g O 2
11
N2 + 3 H2
2 NH3
What would happen if 3 moles N2 and 3 moles
H2 are reacted?
H2 is completely consumed, however, 2 mol
of N2 remain unreacted.
H2 Î Limiting reactant
N2 Î Excess reactant
Limiting reactant dictates (limits) how much
product can be produced.
N2 + 3 H2
2 NH3
Consider that 3 moles N2 and 3 moles H2 are
reacted together. How many moles of N2 react
with the H2?
moles H2 Î moles N2
⎛ 1 mol N 2 ⎞
⎟⎟ = 1 mol N 2
⎝ 3 mol H 2 ⎠
(3 mol H 2 )⎜⎜
How many moles N2 remain unreacted?
3 mol N 2 − 1 mol N 2 = 2 mol N 2 unreacted
Limiting Reactants in Chemical Reactions
• How is the limiting reactant
determined?
• By calculating the amount of product that
each reactant would produce if it reacted
completely.
p
y
• The reactant yielding the least amount of
product is the limiting reactant.
Consider our previous reaction:
N2 + 3 H2
2 NH3
Consider 3 mol N2 reacting with 3 mol H2
Calculate the amount of product produced from
each reactant:
⎛ 2 mol NH 3 ⎞
⎟ = 6 mol NH 3
⎝ 1 mol N 2 ⎠
N2
(3 mol N 2 )⎜
H2
(3 mol H 2 )⎜
⎛ 2 mol NH 3 ⎞
⎟ = 2 mol NH 3
⎝ 3 mol H 2 ⎠
H2 is the limiting reactant, N2 remains in
excess, and 2 moles of NH3 is produced.
Reaction Yield
Products
TWO TYPES
THEORETICAL YIELD
PERCENT YIELD
Theoretical Yield:
product that can be
Maximum amount of p
obtained from a given amount of reactants based on the limiting reactant.
For the N2 + H2 reaction, the theoretical yield
of NH3 is 2 mol.
Typically, the actual yield of a reaction is
somewhat less than the theoretical yield Î
Percent Yield
% yield =
actual yield
X 100
theoretical yield
Consider that 1.5 mol of NH3 was obtained.
% yield =
1.5 mol
X 100 = 75 %
2.0 mol
% yield < 100 %
12