Chapter 3
Mass Relations in Chemistry;
Stoichiometry
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3.1
Atomic and Formula Masses
Meaning of Atomic Masses
Relative masses of different atoms and molecules
Atomic Masses from Isotopic Composition
Atoms occur in nature as a mixture of two or more isotopes
Masses of Individual Atoms
Avogadro’s number represents the number of atoms of an
element which is equal to the atomic mass of the element.
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3.2
Meaning of Atomic Masses
Relative masses of atoms of different elements are expressed in
terms of their atomic masses
Standard value is based on C-12 scale
Most common isotope of carbon is assigned an atomic mass of
12 amu.
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3.3
The mass spectrometer
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3.4
Meaning of Atomic Masses
A nickel atom is 58.69 / 40.08 = 1.464 times as heavy as a
calcium ion
It is 58.69 / 10.81 = 5.429 times as heavy as a boron ion
element
B
atomic mass 10.81 amu
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Ca
40.08 amu
Ni
58.69 amu
3.5
Mass spectrum of chlorine
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3.6
Atomic Masses from Isotopic Composition
%
A.M. = (A.M. isotope 1)(
)
100
%
+ (A.M. isotope 2)(
)+…
100
Isotope
Ne-20
Ne-21
Ne-22
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Atomic Mass
20.00 amu
21.00 amu
22.00 amu
Percent
90.92
0.26
8.82
3.7
Atomic Masses from Isotopic Composition
A.M. Ne =
20.00 (0.9092) +
21.00 (0.0026) +
22.00 (0.0882)
= 20.18 amu
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3.8
Masses of Individual Atoms
The atomic masses of H, Cl, and Ni are
H = 1.008 amu
Cl = 35.45 amu
Ni = 58.69 amu
Therefore 1.008 g H, 35.45 g Cl, and 58.69 g Ni all have the
same number of atoms: NA
NA = Avogadro’s number = 6.022 × 1023
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3.9
Masses of Individual Atoms (cont.)
Mass of H atom:
1.008 g H
–24 g
1 H atom ×
=
1.674
×
10
6.022 × 1023 atoms
Number of atoms in one gram of nickel:
6.022 × 1023 atoms Ni
1.00 g Ni ×
= 1.026 × 1022 atoms
58.69 g Ni
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3.10
The Mole
Meaning : a collection of 6.0122 x 1023 items
Molar Mass : the mass of one mole of a substance
Mole-Mass Conversions
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3.11
Meaning
1 mol = 6.022 × 1023 items
1 mol H = 6.022 × 1023 atoms; mass = 1.008 g
1 mol Cl = 6.022 × 1023 atoms; mass = 35.45 g
1 mol Cl2 = 6.022 × 1023 molecules; mass = 70.90 g
1 mol HCl = 6.022 × 1023 molecules; mass = 36.46 g
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3.12
Molar Mass
Generalizing from the previous examples, the molar mass, M, is
numerically equal to the sum of the atomic masses
CaCl2
C6H 12 O 6
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sum of atomic
masses
110.98 amu
180.18 amu
molar mass ( M)
110.98 g/mol
180.18 g/mol
3.13
Mole-Mass Conversions
Calculate mass in grams of 13.2 mol CaCl2
110.98 g CaCl2
mass = 13.2 mol CaCl2 ×
= 1.47 × 103 g
1 mol CaCl2
Calculate number of moles in 16.4 g C6H12O6
1 mol C6 H 12O 6
moles = 16.4 g C6H12O6 ×
180.18 g C6 H 12O 6
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= 0.0919 mol
3.14
Formulas
Mass % from Formula
Simplest Formula from % Composition
Simplest Formula from Analytical Data
Molecular Formula from Simplest Formula
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3.15
Mass % from Formula
Percent composition of K2CrO4?
molar mass = (78.20 + 52.00 + 64.00) g / mol
= 194.20 g / mol
78.20
%K =
× 100% = 40.27%
194.20
52.00
%Cr =
× 100% = 26.78%
194.20
64.00
%O =
× 100% = 32.96%
194.20
Note that percents must add to 100
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3.16
Simplest Formula from % Composition
Simplest formula : the simplest whole-number ratio of the atoms
present
Find mass of each element in sample of compound.
Find numbers of moles of each element.
Find mole ratio.
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3.17
Simplest Formula from % Composition
Simplest formula of compound containing
26.6% K, 35.4% Cr, 38.0% O
Work with 100 g sample:
26.6 g K, 35.4 g Cr, 38.0 g O.
1 mol
moles K = 26.6 g ×
= 0.680 mol K
39.10 g
1 mol
moles Cr = 35.4 g ×
= 0.681 mol Cr
52.00 g
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3.18
Simplest Formula from % Composition (cont.)
1 mol
moles O = 38.0 g ×
= 2.38 mol O
16.00 g
Note that 2.38 / 0.680 = 3.50 = 7 / 2
Simplest formula: K2Cr2O7
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3.19
Simplest Formula from Analytical Data
A sample of acetic acid (C, H, O atoms) weighing 1.000 g
burns to give 1.446 g CO2 and 0.6001 g H2O. Simplest
formula?
Solution:
Find mass of C in sample (from CO2)
Find mass of H in sample (from H2O)
Obtain mass of O by difference
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3.20
Simplest Formula from Analytical Data
12.01 g C
mass C = 1.466 g CO2 × 44.01 g CO = 0.4001 g C
2
mass H = 0.6001 g H2O ×
2.02 g H
= 0.0673 g H
18.02 g H 2O
mass O = 1.00 g – 0.400 g – 0.067 g = 0.533 g
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3.21
Simplest Formula from Analytical Data
1 mol C
moles C = 0.4001 g C ×
= 0.0333 mol C
12.01 g C
1 mol H
moles H = 0.0673 g H ×
= 0.0666 mol H
1.008 g H
1 mol O
moles O = 0.533 g O ×
= 0.0333 mol O
16.00 g O
Simplest formula is CH2O
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3.22
Molecular Formula from Simplest Formula
Must know molar mass. Simplest formula is CH2O
For acetic acid:
M = 60 g/mol , M(CH2O) = 30 g/mol
60 / 30 = 2
Molecular formula = C2H4O2
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3.23
Chemical Equations
Balancing : any calculation involving a reaction must be based
on the balanced equation for that reaction
Mass Relations in Reactions : the coefficients of a balanced
equation represent numbers of moles of reactants
and products
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3.24
Balancing
Must have same number of atoms of each type on both sides.
Achieve this by adjusting coefficients in front of formulas.
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3.25
Balancing
Example:
Combustion of propane in air to give carbon dioxide and water:
C3H8(g) + O2(g) → CO2(g) + H2O(l)
Balance C:
Balance H:
Balance O:
C3H8(g) + O2(g) → 3CO2(g) + H2O(l)
C3H8(g) + O2(g) → 3CO2(g) + 4H2O(l)
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Meaning:
1 mol C3H8 reacts with 5 mol O2 to form 3 mol CO2 and 4 mol
H2O.
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3.26
Mass Relations in Reactions
Example:
How many moles of CO2 are produced when 1.65 mol C3H8
burns?
Use coefficients of balanced equation to obtain conversion factor:
3 mol CO2
1.65 mol C3H8 ×
= 4.95 mol CO2
1 mol C3 H 8
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3.27
Mass relations in Reactions (cont.)
Example:
What mass of O2 is required to react with 12.0 g of C3H8?
12.0 g C3H8 ×
1 mol C3 H 8
32.00 g O2
5 mol O2
×
×
44.09 g C3 H 8
1 mol O 2
1 mol C3 H 8
= 43.6 g O2
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3.28
Yield of Product in a Reaction
Limiting Reactant : the least abundant reactant based on the
equation for a reaction
Theoretical Yield, Actual Yield, Percent Yield
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3.29
Limiting Reactant, Theoretical Yield
Ordinarily, reactants are not present in the exact ratio required
for reaction.
Instead, one reactant is in excess; some of it is left when the
reaction is over.
The other, limiting reactant, is completely consumed to give the
theoretical yield product.
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3.30
Limiting Reactant, Theoretical Yield (cont.)
To calculate the theoretical yield and identify the limiting
reactant:
Calculate the yield expected if the first reactant is limiting
Repeat this calculation for the second reactant
The theoretical yield is the smaller of these two quantities
The reactant that gives the smaller theoretical yield is the limiting
reactant
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3.31
Limiting Reactant, Theoretical Yield (cont.)
2Ag(s) + I2(s) → 2ΑgI(s)
Calculate the theoretical yield of AgI and determine the
limiting reactant starting with 1.00 g Ag and 1.00 g I2.
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3.32
Limiting Reactant, Theoretical Yield (cont.)
Theoretical yield if Ag is limiting:
469.54 g AgI
1.00 g Ag ×
= 2.18 g AgI
215.74 g Ag
Theoretical yield if I2 is limiting:
469.54 g AgI
1.00 g I2 ×
253.80 g I 2
= 1.85 g AgI
Theoretical yield = 1.85 g AgI; I2 is limiting reactant
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3.33
Actual Yield, Percent Yield
actual yield
% yield =
× 100
theoretical yield
Suppose actual yield of AgI were 1.50 g:
1.50
% yield =
× 100 = 81.1
1.85
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3.34